Easy mathematics

BASES AND DIMENSIONS

2022-07-03T03:45:00.002-07:00

BASES AND DIMENSIONS

(a) Bases.

A non empty subset of S a vector space V(F) is said to be a (linear) basis or a base ( or coordinate system) of V iff

(i) S is L.I

(ii) V is generated or spanned by S

i.e. V = [S] i.e. each element of V is a linear combination of elements of S.

(b). Dimension of a vector space.

If a basis of vector space V(F) is a finite set containing n elements, then V is called an n-dimensional vector space or a vector space of dimension n and we write dim V = n

A vector space V(F) is said to be finite dimensional or finitely generated if there exists a finite subset S of V such that V = L (S).

A vector space V(F) which is not finitely generated is known as a finitely dimensional vector space.

Example 1:

The set S = {(1, 2, 1), (2, 1, 0), (1, -1, 2)} forms a basis for R³(R).

Solution:

For L.I let a₁, a₂, a₃ ∈ R

then a₁(1, 2, 1) + a₂(2, 1, 0) + a₃(1, -1, 2) = (0,0,0)

⟹ (a₁ + 2a₂ + a₃, 2a₁ + a₂ + -a₃, a₁ + 2a₃) = (0,0,0)

⟹ a₁ + 2a₂ + a₃ = 0,

a₁ + 2a₃ = 0

⟹ a₁ = a₂ = a₃

therefore S in L.I

for any vector (a₁, a₂, a₃) ∈ R³(R)

(a₁, a₂, a₃) = a₁(1, 0, 0) + a₂(0, 1, 0) + a₃(0, 0, 1)

where

(1,0,0) = x(1, 2, 1) + y(2, 1, 0) + z(1, -1, 2)

= ( x + 2y + z, 2x + y -z, x + 2z)

i.e. x + 2y + z = 1, 2x + y -z = 0, x + 2z = 0

Solving these we get

x = -2/9,

y = 5/9,

z = 1/9

therefore (1,0,0) = -2/9(1, 2, 1) + 5/9(2, 1, 0) + 1/9(1, -1, 2)

Similarly (0,1,0) = 4/9(1, 2, 1) - 1/9(2, 1, 0) - 2/9(1, -1, 2)

(0,0,1) = 1/3(1, 2, 1) -1/3(2, 1, 0) + 1/3(1, -1, 2)

Using these unit vectors in (1), we get that (a₁, a₂, a₃) can be expressed as linear combination of elements of S.

Hence S is a basis of R³.

Example 2:

Show that the set of vectors

e₁= (1, 0, 0, ....0), e₂= (0, 1, 0, ....0), ....., eₙ =(0, 0, 0, ....1)

is a basis of Vₙ (F). This is known as standard basis of Vₙ (F).

Solution:

Let S = { e ₁, e ₂, ...... e ₙ}

Let ( a ₁, a ₂, ......,a ₙ) ∈ V ₙ, then for L.I

a ₁e ₁, a ₂e ₂, ......,a ₙe ₙ = 0 = (0, 0, 0, ....0)

or a ₁(1, 0, 0, ....0) + a ₂(0, 1, 0, ....0) + .....+ a ₙ(0, 0, 0, ....1) = (0, 0, 0, ....0)

⟹ ( a ₁, a ₂, ......,a ₙ) = (0, 0, ....0)

⟹ a ₁ = 0, a ₂ = 0, ......,a ₙ = 0

Hence the set S is L. I

Also we have

( a ₁, a ₂, ......,a ₙ) = a ₁e ₁ + a ₂e ₂ + ......+ a ₙe ₙ

i.e. x = ( a ₁, a ₂, ......,a ₙ) ∈ V ₙ is expressed as a linear combination of elements of S.

Hence S is a basis of Vₙ (F).

Note: (i) Also dim Vₙ (F) = n

(ii) We can prove that the set of unit vectors

{ (1, 0, 0), (0, 1, 0), ( 0, 0, 1)} is basis of Vₙ (F).

and { (1, 0), (0, 1)} is a basis of Vₙ (F).

Example 3:

If { x, y, z} is basis of R³, where R is the set of real numbers, then { x + y, y + z, z + x} is also a basis of R³.

Solution:

For L.I , let

a(x + y) + b(y + z) + c(z + x) = 0 = (0, 0, 0)

⟹ (a + c)x + (a + b)y + (b + c)z = (0, 0, 0)

⟹ a + c = 0, a + b = 0 , b + c = 0, Since { x, y, z} is L.I

⟹ a = 0, b = 0, c = 0

Therefore the set of vectors {x + y, y + z, z + x} is L.I

Since { x, y, z} is a basis of R³

and so there exists a ₁, a ₂, a ₃ ∈ R s.t.

any vector p ∈ R³ can be expressed as

p = a₁x + a₂y + a₃z ..........(1)

Again suppose that

p = b ₁ (x + y) + b ₂ (y + z) + b ₃ (z + x) ......(2)

where b ₁, b ₂, b ₃ ∈ R³

From (1) and (2)

b ₁ (x + y) + b ₂ (y + z) + b ₃ (z + x) = a₁x + a₂y + a₃z

or (b ₁+ b ₃ - a ₁)x + ( b ₁+ b ₂ - a ₂)y + (b ₂+ b ₃ - a₃)z = 0 = (0, 0, 0)

Therefore

b ₁ + b ₃ = a ₁

b ₁ + b ₂ = a ₂

b ₂ + b ₃ = a ₃

solving these equations, we get

b ₁ = (a ₁ + a ₂ - a ₃)/2

b ₂ = (a ₂ + a ₃ - a ₁)/2

b ₃ = (a ₃ + a ₁ - a ₂)/2

Thus any p ∈ R³ can be expressed as a linear combination of vectors of the set {x + y, y + z, z + x}

Hence the set {x + y, y + z, z + x} is a basis of R³.

Example 4:

Determine whether or not the following vectors from a basis of R³.

(1, 1, 2) , (1, 2, 5), (5, 3, 4)

Solution:

We know that dim R³ = 3

For L.I , we have

a ₁ (1, 1, 2) + a ₂(1, 2, 5) + a ₃ (5, 3, 4) = (0, 0, 0)

a ₁, a ₂, a ₃ ∈ R s.t.

⟹ ( a ₁ + a ₂ + 5a ₃, a ₁ + 2a ₂ + 3a ₃, 2a ₁ + 5a₂ + 4a ₃) = (0, 0, 0)

therefore

a ₁ + a ₂ + 5a ₃ = 0 ...........(1)

a ₁ + 2a ₂ + 3a ₃ = 0 .........(2)

2a ₁ + 5a₂ + 4a ₃ = 0 ........(3)

From (1) and (2) on subtracting, we get

- a ₂ + 2a ₃ = 0 .........(4)

Multiplying (1) and (2) and subtracting from (3) we get

3a ₂ - 6a ₃ = 0 or a ₂ - 2a ₃ = 0 .........(5)

which is same as (4)

putting a ₂ = 2a ₃ in (1) we get

a ₁ = -7a ₃

If we put a ₃ = 1, we get a ₂ = 2 and a ₁ = -7

Thus a ₁ = -7 , a ₂ = 2, a ₃ = 1is not a zero solution of equations (1), (2) and (3).

Hence the given set is linearly dependent and so it does not form a basis of R³.

Example 5:

Show that the vectors (2, -1, 0) , (3, 5, 1), (1, 1, 2) from a basis of R³(R).

Solution:

We have dim R³ = 3

For L.I , let a ₁, a ₂, a ₃ ∈ R then

a ₁ (2, -1, 0) + a ₂(3, 5, 1) + a ₃ (1, 1, 2) = (0, 0, 0)

⟹ ( 2a₁+ 3a₂+ a₃, -a ₁ + 5a₂ + a₃, 0 + a₂ +2a₃) = (0, 0, 0)

therefore

2a ₁ + 3a ₂ + a ₃ = 0 ...........(1)

-a ₁ + 5a ₂ + a ₃ = 0 .........(2)

a₂ + a ₃ = 0 ........(3)

From (1) and (2) , we have

13a ₂ + 3a ₃ = 0 .........(4)

Multiplying (3) by 3 and subtracting from (4), we get a₂ = 0 and so a ₃ = 0

therefore from (1), a ₁ = 0

Thus the three given vectors are L.I and so they form a basis for R³(R).

Example 6:

Prove that

S = {(1, 0, 0), (1, 1, 0), (1, 1, 1), (0, 1, 0)}

spans the vector space R³(R) but is not basis of R³.

Solution:

let (a ₁, a ₂, a ₃) ∈ R and ( a₁, a₂, a₃) = x₁ (1, 0, 0) + x₂ ( 1, 1, 0) + x₃ (1, 1, 1) + x₄ (0, 1, 0) where x₁

, x₂, x ₃, x₄ ∈ R.

Then (a₁, a ₂, a ₃) = (x₁ + x₂ + x ₃, x₂ + x ₃ + x₄, x ₃)

Therefore

x₁ + x₂ + x ₃ = a ₁

x₂ + x ₃ + x₄ = a ₂

x ₃ = a ₃

x ₁ = a ₁ - a ₂ + x ₄

x ₂ = a ₂ - a ₃ + x ₄

x ₃ = a ₃

Taking x ₄ = 0, we get

x ₁ = a ₁ - a ₂,

x ₂ = a ₂ - a ₃

x ₃ = a ₃

and (a₁, a ₂, a ₃) = (a₁ - a ₂)(1,0, 0) + (a ₂ - a ₃)(1, 1, 0) + a ₃ (1, 1, 1) + 0(0, 1, 0)

Thus every vector (a₁, a ₂, a ₃) ∈ R³ is a linear combination of elements of S.

But (0, 0, 0) = 1(1,0, 0) + (-1)(1, 1, 0) + 0(1, 1, 1) + 1(0, 1, 0) which shows that S is L.D

Therefore S is not basis of R³.

Theorem 1:

Let V be a finitely generated vector space over F. Then the set S = { x ₁, x ₂, .....,x ₙ} is a basis of V iff every x ∈ V can be expressed as a unique linear combination of elements of S.

Proof:

Let S = { x ₁, x ₂, .....,x ₙ} be a basis of V. Then V = L(S). Therefore every x ∈ V can be expressed as a linear combination of elements of S.

If possible, let

x = a ₁ x ₁ + a ₂ x ₂+ ......+ a ₙ x ₙ, where a ₁, a ₂,... a ₙ ∈ F

x = b₁ x ₁ + b₂ x ₂+ ......+ bₙ x ₙ, where b ₁, b ₂,..... b ₙ ∈ F

Therefore

a ₁ x ₁ + a ₂ x ₂+ ......+ a ₙ x ₙ = b₁ x ₁ + b₂ x₂+ ......+ bₙ x ₙ

or ( a₁ - b₁)x ₁ + ( a₂ - b₂)x₂ +....+ ( a ₙ - b ₙ)x ₙ = 0

a₁ - b₁ = 0, a₂ - b₂ = 0, ....... , a ₙ - b ₙ = 0

since S is L.I

i.e. a₁ = b₁ , a₂ = b₂, ...... , a ₙ = b ₙ

Thus each element of V is uniquely expressible as a linear combination of elements of S.

Conversely suppose that x ∈ V is uniquely expressible as a linear combination of elements of S. Then V = L(S)

To prove that S is L.I

Let a ₁ x ₁ + a ₂ x ₂+ ......+ a ₙ x ₙ = 0

or a ₁ x ₁ + a ₂ x ₂+ ....+ a ₙ x ₙ = 0x ₁ + 0x ₂+ ......+ 0x ₙ

Therefore a₁ = 0 , a₂ = 0, ...... , a ₙ = 0

i.e. S = { x ₁, x ₂, .....,x ₙ} is L.I

Hence S is basis of V.

Theorem 2:

Every finitely generated vector space possesses a basis.

Proof:

Let V be a vector space generated by a finite set S = { x ₁, x ₂, .....,x ₙ} of non zero vectors in V.

If S is L.I. then it is basis of V. On the other hand, if S is L.D. ∃ some xₖ, 2 ≤ k ≤ n in S s.t. xₖ is a linear combination of preceding vectors x ₁, x ₂, ......,xₖ ₋₁

Since S generates V so each element of V is uniquely expressible as a linear combination of x ₁, x ₂, ......,xₖ, ....x ₙ.

Also, x ₖ is a linear combination of x ₁, x ₂, ......,xₖ ₋₁. It follows that each element of V is expressible as a linear combination of x ₁, x ₂, ......,xₖ ₋₁ , x ₖ ₊₁, .....,x ₙ.

Thus the subset S = {x ₁, x ₂, ......,xₖ ₋₁, x ₖ ₊₁, .....,x ₙ} of n- 1 elements of S also generated If S ₁ is L.I, then it is a basis of V.

On the other hand if S ₁ is L.D., then proceeding as above we can get subset S₂ of n - 2 elements of S which generates V. If S₂ is L.I., then S₂ is a basis of V. If not, then continuing this process, after a finite number of steps, we obtain a L.I. subset of S, which generates V. It may happen that we may be left with a single element generating V, which is L.I. and therefore it will become a basis of V.

Hence V possesses a basis.

Theorem 3:

Any two bases of a finitely generated vector space V have the same number of elements.

Proof.

Let V have the following two beset:

B ₁ = { x ₁, x ₂, .....,x ₘ}

B ₂ = {y ₁, y ₂, .......,y ₙ}

Since V is generated by B ₁ and so y ₁ ∈ V is a linear combination of elements of B ₁.

Therefore the set {y ₁, x ₁, x ₂, .....,x ₘ} is L.D.

Hence ∃ a vector x ₖ (≠ y ₁) which is a linear combination of the proceding vectors i.e. linear combination of y ₁ ,x ₁, ....,x ₖ₋₁

Let S ₁ = {y ₁, x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, .....,x ₘ}

obtained by deleting x ₖ from B ₁.

It is clear that each clement of V is linear combination of elements of S ₁.

Hence V is generated by S ₁.

Consequently, the element y ₂ ∈ B ₂ is a linear combination of elements of S ₁.

Therefore the set

{y ₂, y ₁, x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, .....,x ₘ} is L.D.

Therefore ∃ a vector xⱼ different from y₁ or y₂ which is linear combination of the proceding vectors i.e. of y ₂, y ₁, x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, .....,xⱼ ₋₁

Let S ₂ = {y ₂, y ₁, x ₁, x ₂, .....,x ₖ₋₁ , x ₖ ₊₁, ...

., xⱼ ₋₁, x ⱼ ₊₁, .....,x ₘ} clearly this set also generates V.

We continue to proceed in the above manner. Now if m ≺ n, then B₁ gets exhausted before B₂ and we will get.

S = { y ₘ, y ₘ ₋₁, ....., y ₁}

which is obtained by excluding α x from B₁ and including a y from B₂.

Thus Sₘ is a proper subset of B₂ generating V which is L.D.

Therefore the set B₂ becomes L.D. contradicting the L.I. of B₂

Hence m ≥ n Similarly, by changing the roles of the bases, we get n ≥ m

Hence m = n.

Theorem 4:

If a vector space V be generated by a set of n vectors, then every L.I. subset of V can have utmost n vectors.

Proof.

Suppose that S ₁ = { x ₁, x ₂, .....,x ₙ} is a set of generators of V and S₂ = {y ₁, y ₂, .....,y ₘ} is L.I. subset of V containing m vectors.

Then we have proved in previous theorem that m ≤ n, which shows that S₂ have utmost n vectors.

Theorem 5:

Every L.I subset of a finite dimensional vector space V(F) is either a basis of V or it can be extended to form a basis of V.

Proof.

Let S = { x ₁, x ₂, .....,xₘ} be L.I. subset of a finite dimensional vector space V(F). If S is not basis of V, suppose dim V= n, then V has a finite basis say

{ y₁, y₂, .....,yₙ}. It is clear that x ₘ is a linear combination of vectors y₁, y₂, .....,yₙ

Therefore the set { x ₘ, y₁, y₂, .....,yₙ} is LD. Since every super set of a L.D. set is L.D., and so the set. S = { x ₁, x ₂, .....,xₘ , y₁, y₂, .....,yₙ} is L.D.

Hence there exists a vector in S ₁, which is linear combination of its preceding vectors. This vector cannot be any of the x's since S is L.I. So it must be some y ₖ say.

Now each vector in V is linear combination of x ₁, x ₂, .....,xₘ , y₁, y₂, .....,yₙ and yₖ is a linear combination of vectors

x ₁, x ₂, .....,xₘ , y₁, y₂, .....,y ₖ ₋₁

Thesefore each element of V is expressible as a linear combination of elements of

S₂= {x ₁, x ₂, .....,xₘ , y₁, y₂, .....,y ₖ₋₁, y ₖ₊₁, ...yₙ}

Now, if S₂ is L.I. than it is a basis of V containing S.

Otherwise, the set S₃ can be obtained by excluding yⱼ (j > k) from S₂, generates V.

Now if S₃ is L.I. then it is a basis of V containing S. Continuing this process, till after a finite number of steps, either we get L.I. set which generates V and contains S or we are left with S itself generating V and S being L.I. will be a basis of V.

Hence, either S is already a basis or it can be extended to form a basis of V

Theorem 6:

Every set of (n + 1) vectors or more vectors in an n dimensional vector space V is L.D.

Proof:

Since dim V = n and so every basis of V will contain exactly n vectors.

Suppose S is any L.I. subset of V containing (n + 1) or more vector, then by previous theorem either S forms a basis or it can be extended to form a basis of V.

Therefore in each case the basis of V contains (n + 1) or more vectors, which is contrary to dim V = n

Hence S is L.D.

Theorem 7:

If V is an n-dimensional vector space then a set of n vectors in V forms a basis of V iff it is L.I. or iff V is generated by the set.

Proof:

Suppose a set of n vectors is L.I., then it must be a basis of V since dim V = n. For otherwise it can be extended to form a basis containing more than n elements which is not possible, since dim V = n.

Conversely suppose that the set

S = { x ₁, x ₂, .....,x ₙ} of n vectors of V forms a basis of V.

If S is L.I., then it will form a basis of V, othewise there will be a proper subset of S which will form a basis. This is not possible, since dim V = n.

Hence S is L.I.

Theorem 8:

Every subspace W of a finite dimensional vector space V(F) of dimenstion n is a finite dimensional space with dim m ≤ n.

Proof.

Let W be a subspace of a finite dimensional vector space V(F) with dim = n.

Let S = { x ₁, x ₂, .....,x ₙ} be a basis of V. Then

L(S)= V so that each element of V is a linear combination of elements of S.

Also W ⊆ V. Hence in particular each element of W can also be a linear combination of element of S. Also S is L.I. Therefore S is a basis of W or any subset of S is a basis of W (As every subset of L.I. is L.I).

In either case, the basis of W cannot have more than n vectors.

Hence m ≤ n.

Theorem 9:

Let W be a subspace of a finite dimensional vector space V(F).

Then dim V= dim W iff V= W.

Proof:

Let W be a subspace of a finite dimensional vector space V(F), then W ⊆ V.

Let W = V, then W is a subspace of V and V is a subspace of W.

Therefore,

W ⊆ dim V and dim V ≤ dim W

Hence dim V= dim W.

Conversely suppose that dim V = dim W

Let dim V = din W = n and let

S = { x ₁, x ₂, .....,x ₙ} be a basis of W then

S ⊆ W and L(S) = W and S is L.I

W ⊆ V

Therefore, x ᵢ ∈ W ⟹ x ᵢ ∈ V

i.e. S = { x ₁, x ₂, .....,x ₙ} is a L. I subset of V.

dim V = n ⟹ every L.I. subset of V containing n is a basis of V.

Therefore S is basis of V i.e. L(S) = V

Hence W = L(S) = V. i.e. W = V

LINEARLY DEPENDENT AND INDEPENDENT SETS

2022-06-30T00:35:00.004-07:00

LINEARLY DEPENDENT AND INDEPENDENT SETS

A finite se S = { α₁, α₂, α₃, .......,αₙ } of vectors in a vector space V is said to be linearly dependent ( denoted by K.D) if there exist α₁, α₂, α₃, ........αₙ, not all zero such that

a₁α₁ + a₂α₂ + a₃α₃ + ......+ aₙαₙ

If however

a₁α₁ + a₂α₂ + a₃α₃ + ......+ aₙαₙ = 0 ⟹ a₁ = 0, a₂ = 0,........, aₙ = 0 then the set S = { α₁, α₂, α₃, .......,αₙ } is said to be linearly independent. ( L.I)

From above definition of L.I of a finite set, we conclude that an infinite set of vectors of V is L.I if every finite subset is L.I because otherwise it is L.D

Example 1:

The set of vectors {(1,0,0) ,( 0,1,0) , (1,1,1) , (-1,1,-1)} of R³

Solution: Let

a₁(1,0,0) + a₂( 0,1,0) + a₃(1,1,1) + a₄(-1,1,-1) = (0,0,0)

where a₁, a₂, a₃, a₄ ∈ R.

⟹ a₁ + a₃ - a₄ = 0,

a₂ + a₃ + a₄ = 0 ⟹ a₃ = a₄, a₁ = 0

a₃ - a₄ = 0, a₂ = -2a

Taking a₄ = 1, a₃ = 1, a₂ = -2 we find that

The set is L.D

Example 2:

If F is the field of real numbers, prove that the vectors (1, 1, 0, 0), ( 0, 1, -1, 0) and (0, 0, 0, 3) in F⁴ are linearly independent.

Solution:

Let a₁, a₂, a₃ are real numbers.

Then for L.I we have

a₁(1, 1, 0, 0) + a₂( 0, 1, -1, 0) + a₃(0, 0, 0, 3) = (0, 0, 0, 0)

then a₁ + 0 + 0 = 0

a₁ + a₂ + 0 = 0

0 - a₂ + 0 = 0

0 + 0 + 3a₃ = 0

The only solution of above equation is a₁ = a₂ = a₃ = 0

Hence the given three vectors in F⁴ are linearly independent.

Example 3:

The set of vectors

{(1, 1, 1), ( -1, 0, 1) and (0, -2, 1)} in R³ is L.I

Solution:

Let a(1, 1, 1) + b( -1, 0, 1) + c(0, -2, 1) = (0, 0, 0) where a, b, c ∈ F

⟹ a - b = 0, a - 2c = 0, a + b + c = 0

⟹ a = b = c = 0

Hence the set is L.I

Example 4: The set { e₁, e₂,......,eᵢ,.....eₙ} of Rⁿ, Where eᵢ = (0, 0,....0, 1, 0, ....0) in which the ith coordinate is 1 and others are zero is L.I

Solution:

Let a₁e₁ + a₂e₂ + ......+ aₙeₙ = 0

⟹ a₁(1, 0, ...,0) + a₂(0, 1, 0,...,0) + ....+ aₙ(0,0,...1)

= (0,0,.....,0)

⟹ ( a₁, a₂, ....,aₙ) = (0, 0, ......,0)

⟹ aᵢ = 0 for i = 1,2,.....,n.

Hence the set is L.I

THEOREMS ON LINEARLY DEPENDENT AND INDEPENDENT SETS

Theorem 1:

A set consisting of a single non zero vector L.I

Proof:

Let S = {x} where x ≠ 0

and let ax = 0 for some scalar a.

Then ax = 0 ⟹ a = 0, since x ≠ 0

Therefore S is L.I

Theorem 2:

The set {0} is L.D

Proof:

For every non zero scalar a, we have a0 = 0

so, {0} is L.D

Theorem 3:

A set of vectors containing at least one zero vector is L.D

Proof:

Let S = { x₁, x₂, x₃, ...,xᵢ.....xₙ }, where at least one of them say xᵢ = 0. Then it is clear that

0.x₁ + 0.x₂ + 0.x₃ + ...+ 0.xᵢ +.....+ 0.xₙ

Thus ∑aᵢ xᵢ= 0, where aᵢ= 1 ≠ 0

Hence S is L.D

Theorem 4:

Every subset of a L.I set, is L.I

Proof:

Let S = { x₁, x₂, x₃, .....xₙ }, be L.I

and T = { x₁, x₂, x₃, .....xₖ }, ( 1 ≤ k ≤ n) be subset of S. Then ax₁ + ax₂ + .....a ₖ xₖ+ ....+ aₙ xₙ

⟹ a₁ = 0, a₂ = 0, ......,a ₖ = 0,.....,aₙ = 0

Let a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ = 0 for some scalar a₁ ,a₂ ......, a ₖ

Then a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ + 0 xₖ₊₁ + 0xₙ= 0

So a₁ = 0, a₂ = 0, ......,a ₖ = 0 Since S is L.I

Hence T is L.I

Theorem 5:

Every super set of a L.D set, is L.D

Proof:

Let S = { x₁, x₂, x₃, .....xₙ }, be L.D set of vectors and T = { x₁, x₂, x₃, .....xₖ, xₖ ₊ ₁, ....x ₙ } be super set of S.

Since S is L.D and so there exist scalar a₁ ,a₂ ......, a ₖ not all zero such that

a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ = 0

Therefore

a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ + aₖ₊₁ xₖ₊₁ + aₙxₙ= 0

Where

aₖ₊₁ = aₖ₊ ₂= ....... = a ₙ = 0

Thus there are scalar a₁, a₂,.....,a ₙ not all zero s.t

a₁ x₁ + a₂x₂ + .....+ a ₖ xₖ + a ₙ x ₙ = 0

Hence T is L.D

Theorem 6:

If a vector x is a linear combination of vectors x₁, x₂, x₃, .....xₙ , then the set { x₁, x₂, x₃, .....xₙ } is L.D

Proof:

Since x is linear combination of x₁, x₂, x₃, .....xₙ , so there exist scalar a₁ ,a₂ ......, a ₙ such that

a₁ x₁ + a₂x₂ + ....+ a ₙ x ₙ

or a₁ x₁ + a₂x₂ + ....+ a ₙ x ₙ + (-1)x = 0

Thus there exist scalar a₁ ,a₂ ......, a ₙ, a = 0 not all zero such that

a₁ x₁ + a₂x₂ + ....+ a ₙ x ₙ + ax = 0

Hence the set { x₁, x₂, x₃, .....xₙ } is L.D

Theorem 7:

If the set { x₁, x₂, x₃, .....xₘ } be L.I and the set { x₁, x₂, x₃, .....xₘ , x } be L.D , then x is a linear combination of vectors x₁, x₂, x₃, .....xₘ .

Proof:

Since the set { x₁, x₂, x₃, .....xₘ , x }is L.D, there exist scalar a₁ ,a₂ ......, a ₘ, a not all zero s.t

a₁ x₁ + a₂x₂ + ....+ aₘ xₘ + ax = 0 ......(1)

if a = 0, then

a₁ x₁ + a₂x₂ + ....+ aₘ xₘ + ax = 0

⟹ a₁ x₁ + a₂x₂ + ....+ aₘ xₘ = 0

⟹ a₁ = 0, a₂ = 0, ......, a ₘ= 0 since the set

{ x₁, x₂, x₃, .....xₘ } is L.I

Therefore a ≠ 0

so (1) can be written as

x = (- a₁/a) x₁ + (-a₂/a) x₂ + .....+ (-aₘ/a) xₘ

where a₁ /a stand for a₁a ⁻¹ etc

Hence x is linear combination of x₁, x₂, x₃, .....xₘ

Theorem 8:

A set {x₁, x₂, x₃, .....x ₙ} of non zero vectors is L.D iff there exist some vector xₖ (2 ≤ k ≤ n) of the which is linear combination of it's proceeding vectors.

Proof:

Let {x₁, x₂, x₃, .....x ₙ} be L.D. set of non zero vectors.

Consider the set {x₁}.

Since x₁ ≠ 0, the set {x₁} is L.I

Again consider { x₁, x₂ }. If it is L.I, then consider {x₁, x₂, x₃}. Proceeding in this way, let k be the first position integer (2 ≤ k ≤ n) for which the set {x₁, x₂, x₃, .....x ₖ} is L.D

Hence there exist scalar a₁, a₂, ....a ₖ not all zero such that a₁ x₁ + a₂x₂ + ....+ aₖ xₖ= 0

Here a ₖ ≠ 0 for otherwise, if a ₖ ≠ 0 , the set {x₁, x₂, x₃, .....xₖ₋₁ } will be L.D contradicting the definition of k .

Hence

xₖ= (- a₁/aₖ) x₁ + (-a₂/aₖ) x₂ + ....+ (-aₖ₋₁/aₖ) xₖ₋₁

where 1/aₖ= aₖ₋₁ is the multiplicative inverse of aₖ in the field and a₁/aₖ = a₁.aₖ₋₁ etc

This shows that aₖ is linear combination of preceding vectors x₁, x₂, x₃, .....xₖ₋₁

Conversely suppose that xₖ (2 ≤ k ≤ n) is a linear combination of it's preceding vectors

x₁, x₂, x₃, .....xₖ₋₁ so that

x = a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁= 0

or a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁ + aₖ xₖ= 0

where aₖ = -1 ≠ 0

This shows that {x₁, x₂, x₃, .....xₖ } is L.D

Since every super set of L.D set is L.D, it follow that {x₁, x₂, x₃, .....xₙ } is L.D

Theorem 9:

The set {x₁, x₂, x₃, .....x ₙ} of non zero vectors is L.D iff at least one of these vectors is linear combination of remaining ( n- 1) vectors.

Proof:

By above theorem, the set of non zero vectors {x₁, x₂, x₃, .....x ₙ} is L.D iff there exists aₖ (2 ≤ k ≤ n) which is linear combination of it's preceding vectors i.e.

xₖ = a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁

i.e

xₖ = a₁ x₁ + a₂x₂ + ....+ aₖ ₋₁ xₖ₋₁ + 0xₖ₊₁ + 0xₙ

i.e xₖ is linear combination of the remaining (n - 1) vectors.

LINEAR COMBINATION OF VECTORS

2022-06-29T22:21:00.003-07:00

LINEAR COMBINATION OF VECTORS

Let V(F) be a vector space. A vector x ∈ V is said to be linear combination of vectors x₁, x₂, x₃,.....xₙ ∈ V if x = a₁x₁ + a₂x₂ + .....+ aₙxₙ

where a₁, a₂, .......,aₙ ∈ F.

A linear combination of vectors is a sum of vectors where each vector is multiplied by a scalar coefficient (also known as a scalar multiple). The scalar coefficients are also known as weights or multipliers and they determine how much of each vector is included in the linear combination. The resulting vector is also a vector in the same space as the original vectors.

The mathematical notation for a linear combination of vectors is:

v = c1v1 + c2v2 + ... + cn*vn

where v is the resulting vector, c1, c2, ..., cn are the scalar coefficients, and v1, v2, ..., vn are the original vectors.

More about this.

A linear combination can also be represented using matrix notation. The vector v can be represented as a column vector, and the vectors v1, v2, ..., vn can be represented as column vectors in a matrix called the "coefficient matrix". The scalar coefficients c1, c2, ..., cn can be represented as elements in a column vector called the "weight vector".

The linear combination can then be represented as the matrix product of the coefficient matrix and the weight vector:

v = A * w

where A is the coefficient matrix and w is the weight vector.

Linear combinations of vectors have many applications in various fields such as physics, engineering, computer science and mathematics. For example, in physics, a linear combination of vectors can be used to represent the position of a particle in space, and in computer graphics, linear combinations of vectors are used to represent transformations of objects in 3D space.

Linear combinations of vectors are also used in the field of linear algebra, for example, to find the solution to a system of linear equations, or to find the eigenvectors of a matrix.

Example 1:

Express ( 1, 2, 3 ) as a linear combination of (1, 1, 1), (2, -1, 1) and (1, -2, 5) in R³(R).

Solution:

( 1, 2, 3 ) = a₁(1, 1, 1) + a₂(2, -1, 1) + a₃(1, -2, 5)

for a₁, a₂, a₃ ∈ R

= ( a₁ + 2a₂ + a₃, a₁ - a₂ -2a₃, a₁ - a₂ +5a₃)

then a₁ + 2a₂ + a₃ = 1, a₁ - a₂ -2a₃ = 2 ,

a₁ - a₂ +5a₃ = 3

Solving these equations we get

a₁ = 2, a₂ = -2/3 , a₃ = 1/3

Example 2:

The vector (2, -5, 4) cannot be expressed as a linear combination of ((1, -3, 2) and (2, -1, 1) in R³(R).

SMALLEST SUBSPACE CONTAINING A SUBSET

Let S be subset of a vector space V. Then, a subset U of V is called the smallest subspace containing S, if U is a subspace of V containing S and is itself contained in every subspace of V containing S. Therefore, the intersection of all subspaces of V containing S is also a subspace of V. This subspace is called the subspace of V generated or spanned by S and is denoted by [S].

If {Wᵢ}ᵢ∈I be a family of all subspaces of V containing S, then

[S] = ∩ᵢ∈I W ⊆ Wᵢ ∀ i ∈ I.

LINEAR SPAN

Let V(F) be a vector space and S be any non-empty subset of V. Then the linear span of set S is the set of all linear combination of finite subsets of elements if S and is denoted by L[S] or [S]. Thus,

L(S) = {x:x = ∑ αᵢxᵢ , αᵢ ∈ F, xᵢ ∈ S }

L(S) is also called the set spanned or generated by S.

Theorem 1:

The linear span L(S) of any subset S of a vector space V(F) is a subspace of V generated by S.

Proof. Let x,y be any two elements of L(S)

Then, x = α₁x₁ , + α₂x₂ + ....+ αₙxₙ = ∑ αᵢxᵢ

y = β₁y₁ + β₂y₂ + β₃y₃ ....+ βy =∑ βᵢyᵢ

where αᵢ βᵢ ∈ F and xᵢ yᵢ ∈ S

Now if a, b ∈ F, then

ax + by = ∑aαᵢxᵢ + ∑bβᵢyᵢ ∈ L(S)

Since ax + by is a linear combination of finite sets

x₁, x₂, ....,xₙ ; x₁, x₂, .......yₙ of elements of S.

Thus a, b ∈ F and x,y ∈ L(S) ⟹ ax + by ∈ L(S)

Hence L(S) is a subspace of V(F)

Also each element of S belongs to L(S),

because if xᵢ ∈ S, then xᵢ = 1xᵢ and this implies xᵢ ∈ L(S). Thus L(S) is a subspace of V(F) and S is contained in L(S).

Now if U is any subspace of V containing S, then each element of L(S) must be in U because U is to be closed under vector addition and scalar multiplication. Therefore L (S) will be contained in U.

Vector Subspace

2022-06-26T04:56:00.002-07:00

Vector Subspace

A non empty subset W of a vector space V over a field F is said to be a subspace of V if W itself forms a vector space over F w.r.t addition and scalar multiplication in V.

Let V be a vector space, then the zero space {0} and the whole space V are the trivial or improper subspace of V. All other subspace of V are called proper subspace of V.

Theorem 1:

A non empty set W of a vector space V is subspace of V if
(i) W is closed under addition
i.e x,y ∈ W ⟹ x + y ∈ W
(ii) W is closed under scalar multiplication
i.e α ∈ F, x ∈ W ⟹ αx ∈ W

Proof:

Suppose that W is a subspace of V, then

(i) W is closed under addition

i.e x,y ∈ W ⟹ x + y ∈ W

(ii) W is also closed under scalar multiplication

i.e α ∈ F, x ∈ W ⟹ αx ∈ W

Conversely suppose that W is a non empty subset of V such that conditions (i) and (ii) are satisfied.

Let x,y ∈ W

Now y ∈ W ⟹ (-1)y ∈ W by (ii)

⟹ -y ∈ W

Therefore x,y ∈ W ⟹ x, -y ∈ W

⟹ x + (-y) ∈ W by (i)

⟹ x - y ∈ W.

Thus W is subgroup of V under addition.

Since W is subset of V, axiom V₂) and V₃) of Vector space V will be satisfied.

Hence W is a vector space and so it is a subspace of V.

Theorem 2.

A non empty subset W of a vector space V over a field F is a subspace of V if α, β ∈ F and x,y ∈ W ⟹ αx + βy ∈ W.

Proof:

Suppose W is a subspace if V , then

α, β ∈ F and x,y ∈ W⟹ αx ∈ W , βy ∈ W

( since W is closed under scalar multiplication)

⟹ αx + βy ∈ W. (W is closed under addition)

Conversely suppose that

α, β ∈ F and x,y ∈ W ⟹ αx + βy ∈ W

Taking α = β = 1, we get

x,y ∈ W ⟹ 1x + 1y ∈ W

⟹ x + y ∈ W

Therefore it is closed under addition.

α ∈ F and x ∈ W ⟹ αx + 0y ∈ W, taking β = 0

Thus α ∈ F and x ∈ W ⟹ αx ∈ W

Therefore it is closed under scalar multiplication.

Hence W is a subspace of V by theorem 1.

Example 1:

Consider the following subset of R³ (R)

(i) W₁ = {(x₁ , x₂, 0) : x₁, x₂ ∈ R}.
(ii) W₂ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 0}.
(iii) W₃ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 1}.

Then W₁, W₂ are subspaces of R³, but W₃ is not a subspace of R³.

Solution:

(i) Let x = (x₁ , x₂, 0) , y = ( y₁, y₂, 0) ∈ W₁ and α, β ∈ R, then αx + βy = (αx₁ , αx₂, 0) + (βy₁, βy₂, 0)

= (αx₁ + βy₁, αx₂ + βy₂, 0) ∈ W₁

Hence W₁ is a subspace of R³

(ii) Let x = (x₁ , x₂, x₃), y = ( y₁, y₂, y₃) ∈ W₂

such that x₁ + x₂ + x₃ = 0 , y₁ + y₂ + y₃ = 0

Then αx + βy = (αx₁ + βy₁, αx₂ + βy₂, αx₃ + βy₃)

where α, β ∈ R

Now α(x₁ + x₂ + x₃) + β( y₁ + y₂ + y₃) = 0

or (αx₁ + βy₁) + (αx₂ + βy₂) + ( αx₃ + βy₃) = 0

Therefore αx + βy = ∈ W₂

Hence W₂ is a subspace of R³.

(iii) Let x = (x₁ , x₂, x₃), y = ( y₁, y₂, y₃) ∈ W₃

such that x₁ + x₂ + x₃ = 1 , y₁ + y₂ + y₃ = 1

If α, β ∈ R and α + β ≠ 1then

αx + βy = (αx₁ + βy₁, αx₂ + βy₂, αx₃ + βy₃)

but αx₁ + βy₁ + αx₂ + βy₂ + αx₃ + βy₃

= α(x₁ + x₂ + x₃) + β( y₁ + y₂ + y₃)

= α + β ≠ 1

Therefore αx + βy = ∉ W₃

Hence W₃ is not a subspace of R³.

Example 2:

In the vector space V(R) of all real valued functions, let W = {f ∈ V: f(-x) = f(x) ∀ x ∈ R}

be the set of even functions. Then W is a subspace of V.

Solution:

(af + bg)(-x) = (af)(-x) + (bg)(-x)

= af(-x) + bg(-x)

= af(x) + bg(x) (since f,g are even functions)

= (af)(x) + (bg)(x)

= (af + bg)(x)

Therefore af + bg is also even functions

Thus ∀ a, b ∈ R, f, g ∈ W

⟹ (af + bg) ∈ W

Hence W is a subspace of V.

Theorem 3:

The intersection of any two subspaces of a vector space V is also a subspace of V.

Proof:

Let W₁ and W₂ be two subspaces of a vector space V, Then clearly 0 ∈ W₁, 0 ∈ W₂

so that 0 ∈ W₁ ∩ W₂

Therefore W₁ ∩ W₂ is non empty.

Now let α, β ∈ F and x, y ∈ W₁ ∩ W₂

⟹ α, β ∈ F and (x, y ∈ W₁ and x, y ∈ W₂)

⟹ αx + βy ∈ W₁ and αx + βy ∈ W₂

since W₁ and W₂ are subspace of V

⟹ αx + βy ∈ W₁ ∩ W₂

Hence W₁ ∩ W₂ is a subspace V.

Theorem 4:

If { Wᵢ}ᵢ∈ᵢ be as arbitrary family of subspace of V, then ∩ᵢ∈ I Wᵢ is a subspace of V.

Proof:

Left as an exercise.

Note: If W₁ and W₂ are two subspaces of Vector space V, then W₁ ∪ W₂ need not be a subspace of V.

For example, if R be the field of real numbers

Then W₁ = {(x₁, x₂, 0): x₁, x₂ ∈ R} and

W₂ = {(0, x₂, x₃): x₂, x₃ ∈ R} are two subspaces of R³(R).

Now we consider the element x = (1, 2, 0) and y = (0, 3, 4) of W₁ ∪ W₂, then x + y = (1, 5, 4) ∉ W₁ ∪ W₂

Hence W₁ ∪ W₂ is not a subspace of R³(R).

Theorem 5:

Let W₁ and W₂ be two subspace of a vector space V, then W₁ ∪ W₂ is a subspace of V iff W₁ ⊆ W₂ or W₂ ⊆ W₁.

Proof:

Let W₁ and W₂ be two subspace of a vector space V.

also let W₁ ⊆ W₂ or W₂ ⊆ W₁ then either

W₁ ∪ W₂ = W₂ or W₁ ∪ W₂ = W₁

Hence W₁ ∪ W₂ is a subspace of W.

Conversely suppose W₁ and W₂ are two subspaces of V in such a way that W₁ ∪ W₂ is also a subspace.

If possible, let W₁ ⊆ W₂ and W₂ ⊆ W₁

Then ∃ x, y such that

x ∈ W₁ but x ∉ W₂ and y ∈ W₂ but y ∉ W₁

Now x ∈ W₁ ⟹ x ∈ W₁ ∪ W₂

y ∈ W₂ ⟹ y ∈ W₁ ∪ W₂

Since W₁ ∪ W₂ is a subspace of V and so

x + y = W₁ ∪ W₂ ⟹ x + y ∈ W₁ or x + y ∈ W₂

Now if x + y = W₁ , x ∈ W₁ ⟹ (x + y) - x ∈ W₁

⟹ y ∈ W₁

which is a contradiction

Again if x + y ∈ W₂, then

x + y ∈ W₂, y ∈ W₂ ⟹ (x + y) - y ∈ W₂

⟹ x ∈ W₂

which is again contradiction

Hence either W₁ ⊆ W₂ or W₂ ⊆ W₁

Example 1:

If V₃(F) = {(a, b, c): a, b, c ∈ F} is a vector space. Then a set W₃ = {(a, b, c): α₁a + α₂b + α₃c = 0 for fixed α, α, α ∈ F} I'd a subspace of V₃(F).

Solution:

Here W₃ is clearly a subset of V₃(F).

Also for x = (a₁, b₁, c₁), y = (a₁, b₂, c₂) ∈ W₃ such that

α₁a₁ + α₂b₁ + α₃c₁ = 0 ........(1)

α₁a₂ + α₂b₂ + α₃c₂ = 0 .........(2)

and α, β ∈ F, we have

αx + βy = α(a₁, b₁, c₁) + β(a₂, b₂, c₂ )

= (αa₁, αb₁, αc₁) + (βa₂, βb₂, βc₂ )

= (αa₁ + βa₂; αb₁ + βb₂; αc₁ + βc₂ )

Now α₁(αa₁ + βa₂) + α₂(αb₁ + βb₂) + α₃(αc₁ + βc₂ )

= α(α₁a₁, α₂b₁, α₃c₁) + β(α₁a₁, α₂b₂, α₃c₂)

= α0 + β0 [ by 1 and 2]

= 0

Hence αx + βy = W₃ and W₃ is subspace of V₃(F).

Example 2:

If α₁, α₂, α₃ are fixed elements of a field F, then the set W of all ordered triads (x₁, x₂, x₃) of elements of F such that

α₁x₁, α₂x₂, α₃x₃ = 0

is a subspace of V₃(F).

Solution:

Let x= ( x₁, x₂, x₃) and y = ( y₁, y₂ ,y₃) be any two elements of W.

Then x₁, x₂, x₃, y₁, y₂, y₃ are elements of F and are such that

α₁x₁, α₂x₂, α₃x₃ = 0 ........(1)

α₁y₁, α₂y₂, α₃y₃ = 0 .........(2)

Let α, β be any two elements of F, we have

αx + βy = α( x₁, x₂, x₃) + β( y₁, y₂ ,y₃)

= ( αx₁, αx₂, αx₃) + ( βy₁, βy₂ ,βy₃)

= ( αx₁ +βy₁, αx₂ + βy₂ , αx₃ + βy₃)

Now α₁( αx₁ +βy₁) + α₂(αx₂ + βy₂) + α₃( αx₃ + βy₃)

= α( α₁x₁, α₂x₂, α₃x₃) + β( α₁x₁, α₂x₂, α₃x₃)

= α0 + β0 [ by 1 and 2]

= 0

αx + βy = ( αx₁ +βy₁ , αx₂ + βy₂, αx₃ + βy₃)

Hence W is a subspace of V₃(F).

Example 3:

Let V(F) be any vector space. Then V itself and the subset of V consisting of zero vector only are always subspace of V.

Solution:

Left as exercise.

These two are called improper subspace. If V has any other subspace, then it is called a proper subspace. The subspace of V consisting of zero vector only is called the zero subspace.

Example 4:

Prove that U = {(a, b, c) : a, b, c are rationals} is not a subspace of R³(R).

Solution:

U will be a subspace of R³(R) iff :

(i) x, y ∈ U ⟹ x + y ∈ U and

(ii) α ∈ R, x ∈ U ⟹ αx ∈ U

Now let √2 ∈ R and x = (a, b, c) ∈ U where a, b, c are rational numbers. Then

αx = √2(a, b, c) = (√2a, √2b , √2c ) ∉ U.

Since √2a, √2b , √2c are not rational numbers.

Hence condition (ii) is not satisfied for U.

As such U is not subspace of R³(R).

Example 5:

If V = R³ , W ⊆ V. Show that W is not a subspace of V, if

W = {(a, b, c) : a² + b² + c² ≤ 1}.

Solution:

Let α is a any real number such that α ≻ 1 and x ∈ W such that x = (a, b, c). Then a² + b² + c² ≤ 1

Now αx = α(a, b, c) = (αa, αb, αc)

Since (αa)² + ( αb )² + (αc)² = a²( a² + b² + c² ) ≤ a ≻ 1

Therefore α ∈ R, x ∈ W but αx ∉ W

Hence W is not a subspace of V.

Vector spaces properties

2022-06-22T17:41:00.004-07:00

PROPERTIES

LET V(F) be a vector space, then

(i). a0 = 0, where a ∈ F, 0 on both sides represent zero vector.

PROOF:

We have a0 = a (0+0) = a0 + a0

or 0 + a0 = a0 + a0

⟹ a0 = 0 by right cancellation law

(ii). 0v = 0, where 0 on the left is scalar and on the right is a zero vector, v ∈ V

PROOF:

0.v = (0 + 0)v

or 0 + 0v = 0v + 0v

⟹ 0v = 0

(iii). a(-v) = -(av) = (-a)v, where a ∈ F, v ∈ V

PROOF:

we have v + (-v) = 0

therefore av + a(-v) = a0

or av + a(-v) = 0

therefore a(-v) is additive inverse of av

i.e. a(-v) = -av

Again (-a)v + av = [ -a + a]v

= 0v = 0

Therefore (-a)v is additive inverse of av

i.e. (-a)v = -av

Hence a(-v) = -(av) = (-a)v

(iv). av = 0 ⟹ a = 0 or v = 0

PROOF:

Let av = 0 and a ≠ 0

Then a exists in F , since a ∈ F

Therefore

a⁻¹ (av) = a⁻¹ 0

⟹ ( a⁻¹ a )v = 0

⟹ 1v = 0

⟹ v = 0

Again if v ≠ 0, then

av = 0 ⟹ av = 0v ⟹ a = 0

(v). a(v₁ - v₂) = av₁ - av₂, where a ∈ F , v₁, v₂ ∈ V

PROOF:

a(v₁ - v₂ ) = a[v₁ + (-v₂)]

= av₁ + a( -v₂)

= av₁ - av₂ (since a (-v₂) = - av₂)

(vi). av = bv ⟹ a = b, where v ( ≠0) ∈ F, v, v ∈ V

PROOF:

av = bv ⟹ av + (-bv) = bv + (-bv)

⟹ (a - b)v = 0

⟹ a - b = 0, since v ≠ 0

⟹ a = b

(vii). av₁ = av₂ ⟹ v₁ = v₂ where a ( ≠0) ∈ F, v₁, v₂ ∈ V

PROOF:

av₁ = av₂ ⟹ av₁ + [ -(av₂)] = av₂ + [ - av₂]

⟹ av₁ + a(-v₂) = 0

⟹ a( v₁ - v₂ ) = 0

⟹ v₁ - v₂ = 0, since a ≠ 0

⟹ v₁ = v₂

Vector space questions

2022-06-22T03:45:00.003-07:00

Question:

If X is the set of all real valued continuous (differentiable or integrable) functions defined on some interval [ 0,1]. Then to show that X is a vector space over the field R with addition and scalar multiplication defined as follows:

(𝔣 + g)x = 𝔣(x) + g(x) ∀ 𝔣,g ∈ X .......(1)

(α 𝔣) x = α 𝔣(x) ∀ α ∈ R ,𝔣 ∈ X, x ∈ [ 0, 1]

Solution. To show X be a vector space, we have to verify all the postulates of vector space. Let us verify ono by one.

V₁ : (X, +) is an abelian group.

Closure axiom for scalar multiplication :

For α ∈ E R . 𝔣 ∈ X ⟹ (α𝔣)(x) = α𝔣(x) ∈ X

⟹ α𝔣 ∈ X

V₂: For any α ∈ R: 𝔣,g ∈ X, we have

[ α (𝔣 + g) ](x) = α [ (𝔣 + g) ](x) ∀ x ∈ [ 0, 1]

= α [(𝔣(x)+ g(x) ]

= α (𝔣(x)+ αg(x)

= (α𝔣)(x)+ (αg)(x)

= (α𝔣+ αg)(x) ∀ x ∈ [ 0, 1]

Hence α (𝔣 + g) = α𝔣+ αg

V₃: For α, β ∈ R, 𝔣 ∈ X, we have

(α + β) 𝔣(x) = (α + β) 𝔣(x) ∀ x ∈ [ 0, 1]

= α𝔣(x) + β𝔣(x) as elements of R satisfy distributive law

= (α𝔣)(x) + (β𝔣)(x)

= (α𝔣 + (β𝔣)(x) ∀ x ∈ [ 0, 1]

Hence (α + β) 𝔣 = α𝔣 + β𝔣

V₄: For any α, β ∈ R, 𝔣 ∈ X, we have

[{(αβ) 𝔣}](x) = (α β) 𝔣(x)

= α [β 𝔣(x)] ( as elements of R are associative for multiplication)

= α [(β 𝔣)(x)]

= α [(β 𝔣)](x) ∀ x ∈ [ 0, 1]

Hence (αβ) 𝔣 = α (β 𝔣)

V₅: As 1 ∈ R, so for any 𝔣 ∈ X, we have

(1𝔣)x = 1𝔣(x) = 𝔣(x) ∀ x ∈ [ 0, 1]

( as 1 ∈ R is the multiplicative identity)

Hence 1𝔣 = 𝔣

Thus all postulate of vector space are satisfied showing that X(R) is a vector space.

Question.

Let S be set of all ordered pairs of real numbers. Define sums and scalar multiplication of pairs as follows :
(x₁ , y₁) + (x₂ , y₂) = (x₁ + x₂ y₁ + y₂)
and c (x₁ , y₁) = (cx₁ , y₁)
Show that S is not a vector space.

Solution.

Let a, b are any two real numbers and x = (x₁ , y₁) ∈ S. Then

(a + b)x = (a + b)(x₁ , y₁) = ((a + b) x₁ , y₁)

= (ax₁ + bx₁ ,y₁ )

and ax + bx = a(x₁ , y₁) + b(x₁ ,y₁ )

= (ax₁ , y₁) + (bx₁ ,y₁ )

= (ax₁ + bx₁, y₁ + y₁ )

= (ax₁ + bx₁, 2y₁ )

(a + b)x ≠ ax + bx ∀ x ∈ S

∴ Hence S is not a vector space.

Postulates of vector space

2022-06-20T04:49:00.004-07:00

Question:

If P(x) denotes the set of all polynomials in one indeterminate x over a field F then to show that P(x) is a vector space over F with addition defind as addition of polynomials and scalar multiplication defined as the product of polynomial by an element of F.

Solution:

Given P(x) = {p(x): p(x) = a₀ + a₁x + ·····+ aₙxⁿ +... = ∑ aₙxⁿ for aₙ, s ∈ F}

Addition of polynomials is defined as follows:

If P(x) =∑ aₙxⁿ ∈ P(x).q(x) =∑bₙxⁿ ∈ P(x).

then p(x) + q(x) = ∑ (aₙ + bₙ)xⁿ .......(1)

Scalar multiplication is defined as follows :

If p(x) =∑ aₙxⁿ ∈ P(x), α ∈ F

Then α p(x) =∑ (α aₙ) xⁿ .............(2)

Now to show that P(x) is a vector space, we have to verify all the postulates of vector space. Let us verify them one by one.

Closure axiom for addition:

For each p(x) . q(x) ∈ p(x) ⟹ p(x) + q(x) =∑ (aₙ + bₙ)xⁿ ∈ P(x)

(as if aₙ .bₙ ∈ F ⟹ aₙ + bₙ ∈ F as field is closed for +).

V₁: (i) For any p(x) = ∑ aₙxⁿ . q(x) = ∑bₙxⁿ.r(x) =∑ cₙxⁿ ∈ P(x).

We have

[p(x) + q(x)] +r(x) ={ ∑ aₙxⁿ + ∑bₙxⁿ} + ∑ cₙxⁿ

= ∑ (aₙ + bₙ)xⁿ + ∑ cₙxⁿ

= ∑ (aₙ + bₙ + cₙ)xⁿ

= ∑ (aₙ + (bₙ + cₙ))xⁿ

= ∑ aₙxⁿ + ∑(bₙ + cₙ)xⁿ

= p(x) + [ q(x) + r(x) ]

[as aₙ , bₙ , cₙ ∈ F ⟹ aₙ + (bₙ + cₙ) =aₙ + (bₙ + cₙ)]

Hence elements of P(x) arc associative for addition.

(ii) We know that

o̅(x) = ∑o̅xⁿ ∈ F(x)

∴ o̅ (x) + p(x) = ∑oxⁿ + ∑ aₙxⁿ ∀ p(x) ∈ P(x)

= ∑ (0 + aₙ)xⁿ

= ∑ aₙxⁿ (0 ∈ F is additive identity in F)

= p(x)

Similarly, p(x) + o̅(x) = p(x).

Hence o̅(x) is additive identity in P(x).

(iii) For each p(x) =∑ aₙxⁿ ∈ P(x), ∃ - p(x) ∑ (-aₙ)xⁿ ∈ P(x) such that

p(x) + (-p(x)) =∑ aₙxⁿ + ∑ (-aₙ)xⁿ

= ∑ [aₙx + (-aₙ)]xⁿ

(as -z is additive inverse of aₙ in F)

=∑ 0xⁿ = o̅(x)

Similarly, -p(x) + p(x) = o̅(x)

Hence additive inverse of each element in P(x) exists.

(iii) For any p(x). q(x) ∈ P(x), we have

p(x) + q(x) =∑ aₙxⁿ + ∑ bₙxⁿ

= ∑ (aₙ + bₙ)xⁿ

= ∑ (bₙ + aₙ)xⁿ

as addition is commutative in F

= ∑ bₙxⁿ + ∑ aₙxⁿ

= q(x) + p(x).

Hence addition is commutative in P(x). Thus (P(x), +) is an abelian group.

Closure axiom for scalar multiplication:

For a ∈ F . p(x) ∈ P(x) ⟹ a ap(x) = a∑ aₙxⁿ = ∑ (aaₙ)xⁿ ∈ P(x)

(as a ∈ F. aₙ ∈ F =aaₙ ∈ F, because F is closed for multiplication)

V₂ : For any a ∈ F , p(x).(x) ∈ P(x). we have

a (p(x) + q(x)) = a(∑ aₙxⁿ + ∑ bₙxⁿ)

= a (∑ (aₙ + bₙ)xⁿ)

= ∑ a(aₙ + bₙ)xⁿ

= ∑ (aₙ + bₙ)xⁿ

= ∑ (aaₙ + abₙ)xⁿ

(as elements of F satisfy the distributive law)

= ∑ (aaₙ)xⁿ + ∑(abₙ)xⁿ

= a∑aₙxⁿ + a∑bₙxⁿ

= ap(x) + aq(x)

V₃: For any α, β ∈ F, p(x) ∈ P(x), we have

(α + β) p(x) = (α + β)∑aₙxⁿ

= ∑((α + β)aₙ)xⁿ

= ∑(αaₙ + βaₙ)xⁿ

= ∑αaₙxⁿ + ∑βaₙxⁿ

= α∑aₙxⁿ + β∑aₙxⁿ

= α p(x) + βp(x)

V₄: For any α, β ∈ F, p(x) ∈ P(x), we have

(αβ) p(x) = (αβ)∑aₙxⁿ

= ∑((αβ)aₙ)xⁿ

= ∑(α(βaₙ))xⁿ

(as elements of F are associative for multiplication)

= α(∑(βaₙ)xⁿ)

= α(β∑aₙxⁿ)

= α (βp(x))

V₅: As 1 ∈ F, so for any p(x) ∈ P(x), we have

1p(x) = 1 ∑aₙxⁿ = ∑(1aₙ)xⁿ = ∑aₙxⁿ = p(x)

(as 1 is multiplication identity of F)

Hence showing that P(x) is a vector space.

Vector spaces

2022-06-19T02:07:00.004-07:00

Vector Spaces

INTRODUCTION

We are familiar with two terms as the sum of vectors and multiplication of a vector by a scalar. The set of vectors forms an abelian group with respect to addition. For any three vectors u, v and w we have

(i) Closure law : u + v is a vector

(ii) Associative law : (u + v) + w = u + (v + w)

(iii) Existence of identity: u + 0 = 0 + u = u

(iv) Existence of inverse : u + (-u) = (-u) + u = 0

(v) Commutative law : u+v=v+u

- u is the additive inverse of u, is the negative of the vector u, which has the same magnitude as u but is oriented in the opposite direction. 0 is the zero vector or null vector. The multiplication of vector u by a scalar a is a vector au. Also we have

(vi) a(bu) = (ab)u where a, b are scalars.

(vii) a (u + v) = au + av

(viii) (a + b)u = au +bu

We shall axiomatize the above properties by defining two operations namely internal binary operation and external binary operation.

Let V be a non-empty set of vectors and F be a non- empty set of scalars called a field.

The mapping o : V x V ⟹ V

defined by u o v = u + v ∈ V, ∀ u,v ∈ V is internal binary operation called vector addition.

The mapping o' :F x V ⟹ V defined by

a o' u = a.u, ∀ a ∈ F , u ∈ V is external binary operation in V over F called scalar multiplication.

Now we shall give the abstract definition of a vector space.

VECTOR SPACE

Let F be a field whose elements celled scalers. Then the system (V, +, · ) consisting of

(i) a non-empty set V of vectors,

(ii) an internal binary operation + in V called addition of vectors

(iii) an external binary operation (·) in V over F called scalar multiplication, is said to be a vector space or linear space V over F if the following axioms are satisfied :

V₁) The system (V, +) is an abelian group i.e. (i) v₁ + v₂ ∈ V, ∀ v₁ v₂ ∈ V

(ii) ( v₁+ v₂) + v₃ = v₁ + (v₂+ v₃) ∀ v₁, v₂, v₃ ∈ V

(iii) ∃ an element 0 ∈ V such that

v+0 = 0+v= v, ∀ v ∈ V,

The additive identity 0 is called zero vector or null vector.

(iv) v₁ + (-v₁) = (-v₁) + v₁ = 0

(v) v₁ + v₂ = v₂ + v₁

V₂) (i) ∀ α ∈ F, v ∈ V, α v ∈ V

(ii) Scalar multiplication is associative i.e.

∀ α,β ∈ F, v ∈ V,

α (βv) = (αβ)v

(iii) 1. v = v ∀ v ∈ V, where 1 is the unit element of F.

V₃) (i) Multiplication by a scalar is left distributive w.r.t. vector addition i.e.

α(v₁+v₂) = αv₁ + αv₂, ∀ a ∈ F, v₁,v₂ ∈ V

(ii) multiplication by a vector w.r.t. scalar addition is right distributive i.e.

(α+β)v = αv+βv ∀ α,β ∈ F, v∈V

The vector space V over a field F is denoted by V (F) or simply by V.

Example 1. The set of all vectors in three dimensional space over the field R of real numbers is a vector space.

Example 2. The field C of complex numbers is a vector space over the field R of real numbers.

Solution.

V₁) The system (C, +) is an abelian group.

V₂) (i) ∀ a ∈ R, u ∈ C, au ∈ C.

(ii) Scalar multiplication is associative i.e.

∀ α, β ∈ R, u=x+iy ∈ C

α (β u) = α (β x + iβ y) = (αβ x + i αβ y) = (αβ)(x +iy) = ( αβ)u

(iii) 1u =u ∀ u∈C, where 1 is the unit element of R.

V₃) (i) ∀ a ∈ R, u₁,u₂ ∈ C

α (u₁ + u₂) = α u₁ ₊ α u₂

(ii) ∀ a, β ∈ R, u = C

(α + β)u =α u +β u

Hence C is a vector space over R.

Example 3.

Let F be a field and H be a subfield of F. Then F forms a vector space over H w.r.t. addition in F as vector addition and the scalar product of α ∈ H and v ∈ F as the product of α and v.

Solution.

V₁) (F, +) is an abelian group.

V₂) (i) ∀ α, β ∈ H, v ∈ F, α v ∈ F (since α ∈ F as H ⊆ F)

(ii) since the multiplication in F is associative and so

∀ α, β ∈ H, v ∈ F

α ( β v) = ( α β)v

(iii) 1 ∈ H, v ∈ F

1.v = v

V₃) distributive laws hold in field F.

Hence F forms a vector space over H.

Thus the field R of real numbers will form a vector space over the field Q of rational numbers.

Example 4. Let F be a field. Then the set

Fⁿ = {(a₁, a₂, ....... aₙ) : aᵢ ∈ F}

forms a vector space over F w.r.t. addition and scalar multiplication defined as follows.

If x = (a₁, a₂, ....... aₙ), y = (b₁, b₂, ....... bₙ) ∈ Fⁿ then x + y = (a₁ + b₁, a₂ + b₂.......... aₙ + bₙ) and if a is any element of F, then

ax = (aa₁, aa₂, .......aaₙ).

Solution.

V₁) The system (F , +) is an abelian group, since

(i) ∀ x = (a₁, a₂, ....... aₙ), y = (b₁, b₂, ....... bₙ) ∈ Fⁿ

x + y = (a₁ + b₁, a₂ + b₂....... aₙ + bₙ)∈ Fⁿ

(ii) ∀ x = (a₁, a₂, ..... aₙ), y = (b₁, b₂, .... bₙ) , z= (c₁, c₂, ....cₙ) ∈ Fⁿ

(x + y) + z = (a₁ + b₁, a₂ + b₂...... aₙ + bₙ) + (c₁, c₂, ....cₙ)

= (a₁+b₁)+c₁, (a₂+b₂)+c₂, ......(aₙ+bₙ)+cₙ

=a₁+(b₁+c₁), a₂+(b₂+c₂), .......aₙ+(bₙ+cₙ)

=x+y+z

(iii) 0=(0,0,.....,0) ∈ Fⁿ is additive identity in Fⁿ, since ∀ x ∈ Fⁿ

x+0=x

(iv) To each x = (a₁,a₂,a₃ ... aₙ) ∈ Fⁿ, ∃

-x = (-a₁, - a₂, .... - aₙ) ∈ Fⁿ s.t

x + (-x) = ( 0, 0, ......0)

(v) x + y= ( a₁ + b₁, a₂ + b₂, ........aₙ + bₙ)

= (b₁ + a₁, b₂ + a₂, ....bₙ +aₙ)

= y +x, ∀ x,y ∈ Fⁿ

V₂) ∀ a ∈ F, x ∈ Fⁿ

ax = (aa₁, aa₂, .......aaₙ) ∈ Fⁿ

(ii) a,b ∈ F, x ∈ Fⁿ

a(bx) = a(ba₁, ba₂, .....baₙ)

=(aba₁, aba₂, ......abaₙ)

=(ab)x

(iii) ∀ x ∈ Fⁿ

1.x = x where 1 is the unit element of F,

V₃) (i) ∀ a ∈ F, x,y ∈ Fⁿ

a(x + y)= a ( a₁ + b₁, a₂ + b₂, ........aₙ + bₙ)

= ( aa₁ + ab₁, aa₂ + ab₂, ........aaₙ + abₙ)

= ax + ay

(ii) ∀ a,b ∈ F, x ∈ Fⁿ

(a + b)x =( ( a + b)a₁, (a + b)a₂, ........(a + b)aₙ)

= ( aa₁ + ba₁, aa₂ + ba₂, ....., aaₙ + baₙ)

= ax + bx

Hence Fⁿ is a vector space over F.

Fⁿ is called an n-dimensional vector space over the field F.

Example 5. (i) The set Qⁿ of all ordered n tuples of rational numbers is the rational vector space over the field Q of rational numbers.

(ii) The set Rⁿ of all ordered n tuplcs of real numbers is real vector space over the real field R. Rⁿ is also called an n-dimensional real coordinate space.

(iii) The set Cⁿ of all ordered n tuples of complex numbers forms the complex vector space over the complex field C.

Example 6. A vector space consisting of a single clement, namely, the zero vector 0, is called zero space or null space and is denoted by {0}.

Whole numbers

2022-05-09T09:26:00.004-07:00

Whole numbers:مکمل اعداد

In our daily routine we meet such situations that we count objects like:

How many people are there in crowd?

How many students are there in this school?

How many books are there in the library?

How many cakes are in the box?

How many eggs are there in the crate?

To answer these questions we associate different objects with numbers for example single object with number 1 two objects with numbers 2 and so on.It means we denote them by different symbols as below.

1,2,3,4,5,6,7,8,9,......

These symbols are called numerals.

Natural numbers: قدرتی اعداد

1,2,3,4,5,6,.... These are numbers that we use for counting the different objects so these are called counting numbers or natural numbers. We note that the natural numbers are starting form 1 so we said that the smallest natural number is 1. If we think which is the largest natural number then we have no answer because the largest natural number does not exist.

Example : مثالیں

If a person think that the largest natural number is 100 ( suppose )

And if we add 1 in 100 then it will be the next natural number. If we suppose the largest natural number is 1000000000000......... and we add 1 in it then it will be the next natural number. So no one tell or judge the largest natural number.

Whole Numbers: مکمل اعداد

In the starting of natural numbers if we add 0 before 1 then these numbers are known as whole numbers. The set of whole Numbers is denoted by W.

0,1,2,3,4,5,6,7,8,9,........

The smallest number or we say that the whole numbers are starting with 0 and we cannot tell the largest number of whole Numbers like natural numbers.

Representing Whole Numbers on Number Line:

نمبر لائین پر مکمل اعداد کا اظہار

We draw a line and Mark a point with 1st whole number which is 0.

On this line we Mark points at equal distance and each point we write 1,2,3,4,5,6,7,8,9,... respectively.

Properties of Whole Numbers:

مکمل اعداد کی خصوصیات

1- There is no whole number on left side of zero (0).

2- zero (0) is the smallest whole number.

3- Each number on number line is successor of its previous number, mean each number is 1 more than its previous number or we say that each number is predecessor of its next number.

4- Each whole number on number line is greater than the each whole number on its left,

e.g 6 is greater than 2, 5 is greater than 3, and 8 is greater than 0.

Addition of Whole Numbers:

مکمل اعداد کی جمع

When we add whole numbers we write them in vertical column and each digit is place according to its place value.

Example

Subtraction of Whole Numbers:

مکمل اعداد کی تفریق

When we subtract whole numbers we write them in vertical column and each digit is place according to its place value.

Examples:

Laws of Addition of Whole numbers:

Commutative law: قانون مبادلہ

Suppose two whole numbers x and y if we add them like x+y or y+x the result remains unchanged. Means the result x+y = y+x

Example: 7+5 = 12 if we change their order like 5+7 = 12 so the result is not changed by changing their order.

Associative Law : قانون تلازم

If we add three whole numbers 1st of all we add two numbers and the result is added to 3rd whole number. For example there are three numbers 4,5 and 7

(4+5)+7 = 4+(5+7)

9+7 = 4+12

16 = 16

So (4+5)+7 = 4+(5+7) , This law is called associative law of Addition.

Note:

Commutative and associative laws do not hold in subtraction and division.

These laws only hold in addition and multiplicative.

Additive Identity: جمعی ذاتی عنصر

If we observe the following examples

1+0 = 1 and 9+0 = 9 and 7+0 = 7

We note that the sum of zero and a whole number is always the whole number itself so zero (0) is known as the additive Identity.

Laws of Multiplication

Commutative law :

Suppose two whole numbers x,y then there multiplication is x×y = y×x this is called Commutative law w.r t multiplication.

Example:

If there are two whole numbers 4 ,5 then

4×5 =20 if we change their order then 5×4 = 20 , Hence the result is same after changing their order. So 4×5= 5×4

Associative law:

Suppose three whole numbers like 3,5, and 7 then according to associative law

( 3×5)×7 = 3×(5×7)

15×7 = 3×35

105 = 105

In associative law 1st we multiply two numbers and then their result is multiply by 3rd number.

Distributive law of multiplication over addition:

ضرب کا قانون تقسیمی بلحاظ جمع

Consider three whole numbers 3,4,5 then the distributive law of multiplication over addition of Whole numbers as

3×(4+5) = (3×4) + ( 3×5)

3×9 = 12 + 15

27 = 27 L.H.S = R.H.S

Distributive law of multiplication over subtraction:

ضرب کا قانون تقسیمی بلحاظ تفریق

Consider three whole numbers 3,4,5 then the distributive law of multiplication over subtraction of Whole numbers as

3×(4-5) = (3×4) - ( 3×5)

3×(-1) = 12 - 15

-3 = -3 L.H.S = R.H.S

Multiplicative Identity:

ضربی ذاتی عنصر

Observing the following examples

1×4 = 4 or 7× 1 = 7 we note that the product of any whole number with 1 is always the whole number itself. So 1 is called multiplicative Identity.

Cube and cube roots

2021-04-20T05:42:00.012-07:00

Cube and cube roots

Learning Outcomes

After reading the post readers will be able to,

Find the cube of a number.
Know about the perfect cubes.
Know about the cube root and how to find the cube root of a giving number.

Cubes:

Cube of a number means to multiply a number by itself three times. Suppose y be any number then,

y×y×y = y³

For example

2×2×2 = 2³

3×3×3 = 3³

4×4×4 = 4³

5×5×5 = 5³

6×6×6 = 6³

7×7×7 = 7³

And so on.

Perfect cubes:

Perfect cube is a number that is the result of multiplying (product) an integer by itself three(3) times. It can also be define as an integer to the third power of another Integer.

Example: show that 8 is a perfect cube.

Solution: 8 = 2×2×2 = 2³ so 8 is a perfect cube of 2.

Example: show that 27 is a perfect cube.

Solution: 27 = 3×3×3 = 3³

So 27 is a perfect cube of 3.

Example: Is 216 a perfect cube?

Solution:

216 = 2×2×2×3×3×3 = 2³ × 3³

= (2×3)³ = 6³ so 216 is a perfect cube of 6.

Example: show that 343 is a perfect cube.

Solution:

343 = 7×7×7 = 7³ so 343 is a perfect cube of 7.

Example: find the cube of 5.

Solution: (5)³ = 5×5×5 = 125 Ans.

Example: find the cube of 3.

Solution: (3)³ = 3×3×3 = 27 Ans.

Example: find the cube of (1.2).

Solution: (1.2)³ = (1.2)×(1.2)×(1.2) = (1.44)×(1.2) = 1.728 Ans.

Cube Roots:

In math a cube root of a number is denoted by y⅓ is a number such that a³ = y i.e a = y⅓.

The symbol of cube root is ∛.

Example: find the cube root of 125.

Solution: Factorization of 125 is

125 = 5×5×5 = 5³

∛(125) = ∛(5×5×5) = (5³)⅓ = 5 Ans.

Example: find cube root of 9261.

Solution: Factorization of 9261 is

9261 = 3×3×3×7×7×7 = 3³ × 7³

∛(9261) = ∛(3³×7³) = (3³×7³)⅓ = (3³)⅓ × (7³)⅓ = 3×7 = 21

Properties of cubes of numbers:

1) Cube of a positive (+) number is positive (+). For example 3³ = 27

2) Cube of a negative (-) number is negative (-). For example (-4)³ = -64

3) Cube of an even numbers is even. For example 6³ = 216

4) Cubes of odd numbers is odd. For example 7³ = 343

5) Cube of distributive property under multiplication is

(5×8)³ = 5³ × 8³

and cube of distributive property under division is

(5/8)³ = 5³/8³

6) Cube number of the perfect cubes

2³ = 8 ,

6³ = 216 ,

8³ = 512 so 8, 216 and 512 are perfect cubes.

Questions:

Q1 : Is 512 perfect cube?

ANS: first we factorize 512 as

512 = 2×2×2×2×2×2×2×2×2

512 = 2³ × 2³ × 2³

512 = 8×8×8

512 = 8³

So 512 is a perfect cube.

Q2: 81 is a perfect cube ?

ANS: first we factorize 27 as

81 = 3 ×3×3×3

81 = 3³ × 3¹

So 81 is not a perfect cube.

Question 3: is 6859 perfect cube?

Answer: first we factorize it as

6859 = 19×19×19

6859 = 19³

So 6859 is a perfect cube.

Question 4: 729 is a perfect cube?

Answer: first we factorize it as

729 = 9×9×9

729 = 9³

So 729 is a perfect cube.

Question 5: (1000/125) is a perfect cube?

Answer: first we factorize it as

(1000/125) = (10×10×10)/(5×5×5)

(1000/125) = 10³/5³

(1000/125) = (10/5)³

So (1000/125) is a perfect cube.

Example: find the cube of (1.4)

Solution:

Taking cube of (1.4)

(1.4)³ = (1.4) × (1.4) × (1.4)

(1.4)³ = (1.96) × (1.4)

(1.4)³ = 2.744 Ans.

Example: find cube of (0.8)

Solution: Taking cube of (0.8)

(0.8)³ = (0.8) × (0.8) × (0.8)

(0.8)³ = (0.64) × (0.8)

(0.8)³ = 0.512 Ans.

Example: find the cube root of 729.

Solution:

Taking cube root of 729.

∛729 = ∛(3×3×3×3×3×3)

∛(729) = ∛(3³) × ∛(3³)

∛(729) = (3³)⅓ × (3³)⅓

∛(729) = 3×3 = 9 Ans.

Example: find the cube root of 15625.

Solution:

Taking cube root of 15625.

∛(15625) = ∛(5×5×5×5×5×5)

∛(15625) = ∛(5×5×5) × ∛(5×5×5)

∛(15625) = ∛(5³) × ∛(5³)

∛(15625) = (5³)⅓ × (5³)⅓

∛(15625) = 5×5 = 25 Ans.

Example: Find the cube root of 13824.

Solution:

Taking cube root of 13824.

∛(13824) = ∛(2×2×2×2×2×2×2×2×2×3×3×3)

∛(13824) = ∛(2³) × ∛(2³) × ∛(2³) × ∛(3³)

∛(13824) = (2³)⅓ × (2³)⅓ × (2³)⅓ × (3³)⅓

∛(13824) = 2×2×2×3

∛(13824) = 24 Ans.

Frequently asked questions. (FAQs)

Are cubes and cube roots same?

we understand this by a example, 216 is called the cube of 6. While on the other hand, the cube root of a number is the reverse or opposite process of the cube of a number. The cube root of a number is written as ∛ . Considering the same example, 2 is called the cube root of the number 8.

What is the cube of √ 3?

The cube root of 27 is 3 and of 8 is 2. Cube root of any number y is a number x, such as x³= y. So, to find the cube root of 3 we have to determine a number, which when multiplied three times, gives the number 3, such as x³ = 3 or x = ³√3.

What is Cube Root of a Number?

The cube of a number is the value of the third exponent or power three of the number. For example, the cube of 3 is 3³= 3 × 3 × 3 = 27. While the cube root is the reverse process or opposite process of the cube of a number and is denoted by ∛. For example, ∛27, that is, the cube root of 27 = 3 because when 3 is multiplied three time with itself, it gives 27.

What is Cube Root Used For?

The cube root is used to solve the cubic equations. Cube roots are also used to find the dimensions of a cube if its volume is given.

How do you find cube root of a number?

We define the cube root as, "Cube root is the number which needs to be multiplied thrice or multiplied three times to get the original number." Now, let us look at the cube root formula, where z is the cube root of x. ∛x = z.

How we find the cube root of 64?

The cube root of 64 is the number which when multiplied by itself three times which gives the product as 64. Since 64 can be expressed as 2 × 2 × 2 × 2× 2× 2. So, the cube root of 64 = ∛(2 × 2 × 2 × 2 × 2 × 2) = 4.

What is a example of cube root ?

The cube root of number is a value when we multiplied by itself three times or thrice produces the original value. For example, the cube root of 64, denoted as ³√64, is 4, because when we multiply 4 by itself three times we get 4 x 4 x 4 = 64 = 4³ or 2⁶

Find the cube root of 4?

The cube root of 4 is a number, when we multiplied by itself three times or thrice it gives the product as 4. So 4 can be expressed as 2×2. The cube root of 4 = ∛(2 × 2) = 1.587.

What is the fast way to find cubes of given numbers?

Cube of a number means, a number that is multiplied by itself thrice or three times, then the product is called the cube of that given number.

1³=1×1×1=1

2³=2×2×2=8

3³=3×3×3= 27 and so on..

4³= 64

6³= 216

7³= 343

8³= 512

9³= 729

10³= 1000

What does cube root look like?

The cube root ( ∛ ) of a number is written with a small number 3 which is called the index, just outside and above the radical symbol.

This small or little 3 distinguishes cube roots from square roots which are written without a small number outside and above the radical symbol.

What is the cube root symbol?

The cube root of a number is the factor that we multiply by itself three times or thrice to get that number. The symbol for cube root is ∛ .

Have a nice day.

Square and square root

2021-04-15T03:50:00.013-07:00

Square and square root

Learning Outcomes:

After reading this post readers will be able to,

Define what is a perfect square.
Identify a number is a perfect square or not.
Recognize the properties of perfect square.
Find the square root of the nature numbers by factorization method and division method.
difference between square and square root

Introduction:

We know that the area of a square can be calculated by multiplying length of square by itself as

Area of square = length × length

Suppose a square has a length "x" then

Area of square = (x) × (x) = (x)²

It means x² is a area of square whose length is x or simply we can say that x² is the square of x . i.e The square of x = x²

Thus the square of a number can be defined as below:

The product of a number with itself is called square of it.

Perfect Square:

Any natural number is called a perfect Square, if it is the square of any natural number. To make it clear lets find the square of some natural numbers.

1² = 1 × 1 = 1

2² = 2 × 2 = 4

3² = 3 × 3 = 9

4² = 4 × 4 = 16

5² = 5 × 5 = 25

6² = 6 × 6 = 36

7² = 7 × 7 = 49

8² = 8 × 8 = 64

9² = 9 × 9 = 81

10² = 10 × 10 = 100 and so on.

Hence 1 is the square of 1.

4 is the square of 2.

9 is the square of 3.

16 is the square of 4 and so on.

To check a number is a perfect Square or not:

To check a number is perfect Square or not, write the number as a product of its prime factors of all factors can be grouped in pairs then the given number is a perfect square.

Example: check the following numbers are perfect square or not.

1) 3969

2) 6084

3) 3873

Solution:

1) prime factors of 3969 = 3×3 ×3×3 × 7×7

We note that each factor form a pair. So 3969 is perfect square.

2) prime factors of 6084 = 2×2 × 3×3 × 13×13

Each factor of 6084 from a pair. So it is a perfect square.

3) prime factors of 3872 = 2×2 × 2 × 2×2 × 11×11

We note that 2 is a factor which can't be paired with any equal factor. So 3872 is not a perfect square.

Properties of perfect squares:

1) The square of an even number is even:

As we know the natural numbers are divided into two groups, even numbers and odd numbers.

See the following squares,

2² = 2×2 = 4

4² = 4×4 = 16

6² = 6×6 = 36

8² = 8×8 = 64

10² = 10×10 = 100

12² = 12×12 = 144

We note that the squares of all even numbers are even numbers.

2) Square of an odd number is odd:

See following examples,

1² = 1×1 = 1

3² = 3×3 = 9

5² = 5×5 = 25

7² = 7×7 = 49

9² = 9×9 = 81

So the squares of odd numbers are also odd numbers.

3) The square of a proper fraction is less than itself:

To Square a proper fraction we multiply the numerator and denominator by itself.

For example (2/5)

We take square of it as (2/5)²

= (2/5) × (2/5) = (2×2)/(5×5) = 4/25

Now we compare the fraction 2/5 with its square 4/25

2/5 , 4/25

Take L.C.M of 5 and 25 which is 25 and multiply it to both the fractions

(2/5)×25 , (4/25)×25

50 , 20

So 50 > 20

so it is observe that the square of a proper fraction is less than itself.

so 2/5 > 4/25

Similarly (1/3)² = 1/3 x 1/3 = 1/9

so 1/3 > 1/9

4) Square of a decimal less than 1 is smaller than the decimal:

To find the square of decimal, we use following method.

(0.4)² = (0.4) × (0.4) = 0.16

Is 0.16 is smaller than 0.4 or greater ? Certainly 0.16 is smaller than 0.4

so 0.16 < 0.4

it means square of a decimal is always smaller than the given decimal.

Square Root:

Process of finding square root is an opposite operation of squaring a number.

To understand this we find some perfect squares.

2² = 4 ( 2 squared is 4)

5² = 25 ( 5 squared is 25)

7² = 49 ( 7 squared is 49)

The above equations can be read as

2 is the square root of 4.

5 is the square root of 25 and

7 is the square root of 49.

Similarly we can find the square root of any square number. For this we use a symbol ( √ ) which represent a square root. For example,

√(y²) = y where "√" is called radical sign. Here y² is called radicand.

Square root by prime Factorization:

As we know that:

The square root of 4 is , √4 = √(2²) = 2

The square root of 9 is , √9 = √(3²) = 3

The square root of 25 is , √25 = √(5²) = 5

And so on. But in large perfect squares it is difficult for us to guess their square roots. So we used an other method which is called prime factorization method. To understand this method see following example.

Suppose the number is 36

36 = 2×2×3×3

Take square root on both side.

√(36) = √(2×2×3×3)

Write them in pair of prime factors of a perfect square.

√36 = √(2×2) × √(3×3)

√36 = √2² × √3²

So √36 = 2 × 3 = 6 Ans.

Finding square roots of fractions:

As we know that there are three types of common fractions.

1) proper fraction

2) Improper fraction

3) Compound fraction or mixed fraction

Example: square root of a common fraction (144/256)

Solution:

Ist we take square root of (144/256).

Example: An improper fraction (144/81), find the square root.

Solution:

Square roots of decimals:

In decimals first of all we change them into common fractions and then find square root. After finding square root we write the answer in decimal form again. To understand this we see an example.

Example: Find the square root of 0.64.

Solution:

First we change decimal into fraction as 0.64 = 64/100

Now we take square root of it as √(64/100)

Factorization of 64 = 2×2×2×2×2×2

Factorization of 100 = 2×2×5×5

Square root by division method:

In previous example we have learnt the process to find square root of natural numbers by Prime factorization method but now we learn the square root of natural numbers by division method.

Example: By division method find the square root of 324.

Solution:

Example: By division method find the square root of 585225.

Solution:

Example: By division method find the square root of (4096/15129).

Solution:

Have a nice day.

Exponents

2021-04-04T02:47:00.008-07:00

Exponents

Learning Outcomes:

After reading the post readers will be able to:

1) Recognize base, exponent and value.

2) Know about the product law of exponents.

aˣ × aⁿ = aˣ⁺ⁿ
aⁿ × bⁿ = (ab)ⁿ

3) Quotient law.

aˣ ÷ aⁿ = aˣ⁻ⁿ
aⁿ ÷ bⁿ = (a/b)ⁿ

4) power law. (aˣ)ⁿ = aˣⁿ

5) For zero(0) exponent: a⁰ = 1

6) For exponent as negative Integer. a⁻ⁿ = (1/aⁿ)

7) Concept of power of integer that is (-a)ⁿ when "n" is even or odd integer.

Exponents or Indices:

Base, Exponent and Value:

We know that the repeated multiplication of n number can be written in a short by using exponent.

For example

5×5×5 can be written as 5³. We read it as 5 to the power of 3 where 5 is the base and 3 is the exponent or index.

Note: The exponent of a number indicates that, how many times a number is multiplied with itself.

Similarly as above, 12×12 can be written as 12² . We read it as 12 to the power of 2 where 12 is the base and 2 is the exponent.

So from above examples we can conclude that if a number "s" is multiplied with itself n-1 times then the product will be sⁿ ,i.e.

Sⁿ = s×s×s×s×.......×s (n-1 times multiplication of s with itself)

We read it as "s to the power of n" or "nth power of s" where "s" is the base and "n" is the exponent.

Example: Express in exponential form. (-5) × (-5) × (-5)

Solution: (-5) × (-5) × (-5) = (-5)³

Example: Express in exponential form. 3×3×3×3×3×3×3

Solution: 3×3×3×3×3×3×3 = (3)⁷

Example: Express in exponential form. (1/7)×(1/7)×(1/7)×(1/7)

Solution: (1/7) × (1/7) × (1/7) × (1/7) = (1/7)⁴

Example: (-3/8) × (3/8) × (-3/8)

Solution: (-3/8) × (-3/8) × (-3/8) = (-3/8)³

Example: Identify the exponent and base 34⁵

Solution: 34⁵

Base = 34

Exponent = 5

Example: Identify exponent and base of (-3/8)⁷

Solution:

(-3/8)⁷

Base = (-3/8)

Exponent = 7

Example: Identify exponent and base (-x/z)ᵅ

solution (-x/z)ᵅ

Base = (-x/z)

exponent = a

Example: write in simplest form (-2)⁵

Solution: (-2)⁵

= (-2)×(-2)×(-2)×(-2)×(-2)

= -32 Ans.

Example: (2/3)³

Solution: (2/3)³

= (2/3)×(2/3)×(2/3)

=8/27 Ans.

Example: write in simplest form (-1/4)⁴

Solution: (-1/4)⁴

= (-1/4)×(-1/4)×(-1/4)×(-1/4)

= 1/ 256

Note: [ (-) × (-) = (+) ] , [ (-) × (+) = (-) ] and [ (+) × (+) = (+) ]

Laws of exponents OR Indices:

Exponents are used to solve many problems so it is important that we understand the laws for working with exponents. Now we discuss these laws one by one.

Product Law:

1) when bases are same but exponents(power) are different:

See the example to understand

2³ × 2² = (2×2×2) × (2×2)

= (2×2×2×2×2)

= 2⁵

From this example we can notice that same result can be obtained by adding the exponents of two numbers.

2³ × 2² = 2³⁺² = 2⁵

Similarly,

(-3/7)² × (-3/7)⁵ = (-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)

= (-3/7)⁷

Again using the short method to find result

(-3/7)² ×(-3/7)⁵ = (-3/7)²⁺⁵ = (-3/7)⁷

Form above we can deduce the following law:

While multiplying two(2) numbers with the same base we add their exponents but the base remains unchanged i.e. for any number "b" with exponents x and n, this law is written as

(b)ˣ × (b)ⁿ = (b)ˣ⁺ⁿ

2) when bases are different but exponents are same:

As we know

2³ × 5³ = (2×2×2) × (5×5×5)

= (2×5) × (2×5) × (2×5)

= (2×5)³

Similarly,

(-1/7)³ × (3/7)³

= [ (-1/7)×(-1/7)×(-1/7) ] × [ (3/7)×(3/7)×(3/7)]

= (-1/7 × 3/7)(-1/7 × 3/7)( -1/7 × 3/7)

= ( -1/7 × 3/7)³

We can deduce the law from above examples

While multiplying two(2) rational numbers having the same exponent the product of two(2) bases I written with the given exponent. Suppose two rational numbers are "x" and "y" with exponent "n" then xⁿ × yⁿ = (xy)ⁿ

Example: Simplify (-1/4)² × (2/3)²

Solution:

(-1/4)² × (2/3)² = [ (-1/4) × (2/3)]²

= [ -2/12 ]² = [ -1/6 ]² Ans.

Example: (-3/2)³ × (-3/2)⁴

Solution : (-3/2)³ × (-3/2)⁴

= (-3/2)³⁺⁴ = (-3/2)⁷ Ans.

Quotient Law:

1) when bases are same but exponents are different

See the following examples to understand

2⁷ ÷ 2³ = 2⁷/2³

= (2×2×2×2×2×2×2)/ (2×2×2)

= 2×2×2×2 = 2⁴

Find the same quotient by another method

2⁷/2³ = 2⁷⁻³ = 2⁴

Similarly,

(-3/7)⁵ ÷ (-3/7)²

= [ (-3/7)×(-3/7)×(-3/7)×(-3/7)×(-3/7)] ÷ [(-3/7)×(-3/7)]

(-3/7) ×(-3/7)×(-3/7) = (-3/7)³

Short method to find Quotient is

(-3/7)⁵ ÷ (-3/7)² = (-3/7)⁵⁻² = (-3/7)³

We can suggest another law from above examples

The division of two rational numbers with same base can be performed by subtracting their exponents. Let suppose "b" is the base of any two rational numbers with exponents "x" and "n" such that" b" is not equal to zero and "x" is greater than "n" then,

bˣ ÷ bⁿ = bˣ⁻ⁿ

2) when bases are different but exponents(power) are same

As we know

(2/3)⁴ = (2/3)×(2/3)×(2/3)×(2/3)

= (2×2×2×2)/ (3×3×3×3)

= 2⁴/3⁴ = 2⁴ ÷ 3⁴

Similarly,

(x/y)⁵ = (x/y)×(x/y)×(x/y)×(x/y)×(x/y)

= ( x⁵ / y⁵) or x⁵ ÷ y⁵

This law can be written as

If two rational numbers "a" and "b" where "b" is not equal to zero and "n" is there exponent, then

aⁿ ÷ bⁿ = (a/b)ⁿ

Example: 9⁸ ÷ 3⁸ simplify it.

Solution: 9⁸ ÷ 3⁸ = (9/3)⁸ = 3⁸

Example: (-3/11)⁷ ÷ (-3/11)⁴

Solution: (-3/11)⁷ ÷ (-3/11)⁴

= (-3/11)⁷⁻⁴ = (-3/11)³

Example: (14)¹¹ ÷ (63)¹¹

Solution: (14)¹¹ ÷ (63)¹¹ = (14/63)¹¹ = (2/9)¹¹

Power law:

As we studied that bᵅ × bⁿ = bᵅ⁺ⁿ. Let we use this law to simplify an expression (7⁴)².

(7⁴)² = 7⁴ × 7⁴ = 7⁴⁺⁴ = 7⁸ is the same as 7⁴ˣ²

We solve another expression by using same law

[ (-1/2)⁷ ] = (-1/2)⁷ × (-1/2)⁷

= (-1/2)⁷⁺⁷ = (-1/2)¹⁴ is the same as (-1/2)⁷ˣ²

From above examples we deduced that the base remains same with a new exponent equal to the product of the two (2) exponents, i.e.

(bᵅ)ⁿ =bᵅˣⁿ= bᵅⁿ

Zero exponent:

By quotient Law , anything divided by itself is 1 as below.

4²/4² = (4×4) / (4×4) = 1

This can also b written as

4²/4² = 4²⁻² = 4⁰ = 1

Similarly,

(-5)⁴ / (-5)⁴ = 1

Or it can be written as also

(-5)⁴ / (-5)⁴ = (-5)⁴⁻⁴ = 1

So any non zero rational number with zero exponent is equal to 1. Suppose "z" be any non zero rational number with zero exponent then

z⁰ = 1

Negative Exponents:

Look at the given pattern below.

10² = 10 × 10

10¹ = 10

10⁰ = 1

10⁻¹ = 1/10

10⁻² = 1/10 × 1/10

......... so on

and

10⁻ⁿ = 1/[10×10×10×........×10(n-1 times multiplication)] =1/10ⁿ

In general it can be written as

a⁻ⁿ = 1/aⁿ

Example: Express in a single exponent. (3⁴)⁵

Solution:

(3⁴)⁵ = 3⁴ˣ⁵ = 3²⁰

Example: [(-2/3)³]²

Solution: [(-2/3)³]² = (-2/3)³ˣ² = (-2/3)⁶

Example: [(1/7)⁵]⁶

Solution:

[(1/7)⁵]⁶ = (1/7)⁵ˣ⁶ = (1/7)³⁰

Example: change following negative exponent into positive exponent.

1) (-2/5)⁻⁴

2) (a/-b)⁻⁶

Remember that,

When we multiply a negative number by itself it gives us positive result because minus(-) time minus is plus(+).

For example

(-3)×(-3) = (+9) and (-7)×(-7) = (+49)

Now see an other pattern given below

(-2)² = (-2)×(-2) = (+4)

(-2)³ = (-2)×(-2)×(-2) = (-8)

(-2)⁴ = (-2)×(-2)×(-2)×(-2) = (+16)

(-2)⁵ = (-2)×(-2)×(-2)×(-2)×(-2) = (-32) ....so on.

From above examples it is clear that a negative number with an odd exponent gives a negative result. So it can b explain as,

Let "x" be any positive rational number and "n" be any non zero Integer, then according to this law:

1) If "n" is an even integer, then (-x)ⁿ is positive.

2) If "n" is an odd integer, then (-x)ⁿ is negative.

Example: (2/5)⁻³ × (2/5)³ + (3/5)⁵ × (3/5)⁻⁵

Solution:

(2/5)⁻³ × (2/5)³ + (3/5)⁵ × (3/5)⁻⁵

= (2/5)⁻³⁺³ + (3/5)⁵⁺⁽⁻⁵⁾

= (2/5)⁰ + (3/5)⁰ = 1+1 = 2 Ans

Example: (-2/7)⁵ × (-2/7) × [(-2/7)²]

Solution:

(-2/7)⁵ × (-2/7)⁻² × [(-2/7)²]⁻¹

= (-2/7)⁵⁺⁽⁻²⁾ × (-2/7)²ˣ⁽⁻¹⁾

= (-2/7)³ × (-2/7)⁻²

= (-2/7)³⁺⁽⁻²⁾

= (-2/7)³⁻² = (-2/7)¹ = -2/7 Ans.

Have a nice day.

Algebra

2021-03-29T07:37:00.008-07:00

Algebra

Learning Outcomes:

After reading this article or post readers will be able to,

Explain the definition of algebra.
Know that about sentence and about statement.
Understand the difference between like and un-like terms.
Addition and subtraction of algebraic expressions.
Simplify the given algebraic expressions grouped with brackets.
Recall basic formulas.
How to factorize expressions of different types.
Recognize simultaneous linear equations with one and two variables.
Different methods to solve simultaneous linear equations.

Algebra:

The word algebra, which is taken from the Arabic word Al Jabar. A famous Muslim mathematician Al khawarzmi wrote a famous book, the name of book is Al Jabar Wal Muqabala in 820 AD, in that book he described the method of solving difficult and complex mathematical problems.

Algebra is an important branch of math which provides us the solution of many complex mathematical problems in an easy way. Especially when we showing or represent a quantity by a symbol without knowing its numerical value.

Relationship between Algebra and Arithmetic:

All of us well aware of natural numbers, which are 1,2,3,4,.... And use of the basic operations like ( +,-, ×,÷ ) in arithmetic.

In algebra we use letters a,b,c,d,e,.....,z in addition of numbers to generaliz the arithmetic which help us to express a quantity without knowing its numerical value.

We shall explain it with the help of following example.

1+2 = 3 if x=1, y=2 then z= 3

2+3 =5 if x=2, y = 3 then z=5

3+4=7 if x=3 , y=4 then z=7

....... so on. x+y = z

In above examples x+y = z is giving us a general form which is representing all given Arithmetical statements. So algebra is a general form of the arithmetic.

x + y = z shows the sum of two(2) numbers represented by x and y is equal to the number represented by z.

Sentence:

A group of words that gives us a complete sense is called sentence.

For example,

Goats give milk.
Birds are flying in the air.
He writes very well.

These are examples of a sentence.

Statement:

A statement is also a sentence that may be true or false.

For example,

a) Banana is a fruit. ( True )

b) Mumbai is in America. ( False )

c) Thank you. (neither true nor false)

Here (a) and (b) are statement but (c) is not.

Open statement:

A statement from which we cannot decide that, it is true statement or false statement so such type of statement is called open statement. For example,

2 + ∆ = 6

We try to guess the value of ∆ to make the statement true.

2 + 3 = 6 (false) , 2 + 4 = 6 (true)

Here 4 is the required number which is making the statement true or 4 is satisfying the statement.

Variable and constant:

We learnt in previous post that in algebra a letter of the alphabet (x,y,z etc) is used to represent a number of values. So these letters are called variable, and numbers 0,1,2,3,4,... have definite fixed values are called constants.

Algebraic Expressions:

Such a expressions in which the numbers or variables or both ( variables and numbers) are connected by operational signs (+,-,×,÷) are called algebraic expressions.

For example: 4 , 2x , 5x+3y , a+b , x+y-z etc.

Algebraic terms:

The operational signs plus( + ) and minus (-) separate and connect the parts of an algebraic expression. These parts are called or known as the algebraic terms of the algebraic expression.

For example: in an expression x + 2y + 5 , there are three terms x, 2y , 5

Constant term:

In above expression x + 2y + 5 , the term 5 is a fixed and remains unchanged and only values of x and y vary. So term 5 is called a constant term.

Index or Exponential form:

We know that 2×2×2 = 2³ and 5×5×5×5 = 5⁴. These forms are called index or Exponential forms and can be read as power of 2 and power of 5.

In algebra we can write an index / exponential form as:

y × y = y² or a×a×a = a³

In y², variable y is called base and power 2 is called exponent. We can read it as y raise to the power 2.

Coefficient:

The multiplying factor (mean any Integer) of a variable is called its coefficient. For example in 7y , 7 is the coefficient of y.

Like and un like terms:

The terms of the same kind which are only differ by their coefficients are called like terms. Such terms can be reduced to a single term by subtracting and adding.

For example,

(1) x + 3x + 4x = 8x , (2) 5xy - 2xy -xy = 2xy

And the terms having different variables or same variables with different exponents are called unlike terms. For example:

(1) x,y,z (2) ab, ac ,bc

Addition and subtraction of algebraic expressions:

Please visit my previous post.

Simplification:

As we know that brackets are used to indicate the order for performing operations. The four (4) kinds of brackets are,

A) _____ is called bar or we called it vinculum.

B) ( ) is called curved brackets or round brackets or parentheses.

C) { } is called braces or curly brackets.

D) [ ] is called square brackets or box brackets.

Sometimes in algebra we cannot Simplify an expression into a single term within the brackets. For example in 3x - ( x + y ) we cannot Simplify the expression ( x + y ). For such situation

1) First expand brackets.

2) Simplify the whole expression as given below,

3x - ( x + y) = 3x - x - y = 2x - y

Hence 2x - y is the simplest form of the above algebraic expression.

-ve ( negative) sign before the brackets means, change the signs of all the terms with in brackets.

For example. - (x + y - z ) = -x -y +z

Example: [ 5x - { 3b + ( 6a - 2a + b)}]

Solution:

Example: [ 2a + { c - a + ( a + 2b + c )}]

Solution:

Evaluation:

The process of finding the numerical value or absolute value of an expression by using numbers in place of variables is called evaluation.

The absolute value or numerical value of an expression varies according to the given value of variables. For example, if a= 2 and b= 1 then the numerical value of a+b is 2+1 = 3 and if a=4 and b=2 then the numerical value of a+b = 4+2 =6.

Example: If a=3 and b = 4 then solve or prove that (a+b)² = a² + 2ab + b²

Solution:

Basic algebraic formulas:

1) ( x + y )² = x² + y² + 2xy

Example: solve (107)² by using formula.

Solution:

(107)² = (100 + 7 )²

= (100)² + (7)² + 2(100)(7)

= 10000 + 49 + 1400

= 11449 Ans.

2) (x - y )² = x² + y² - 2xy

Example: Evaluate (87)² by using formula.

Solution:

(87)² = (90 - 3 )²

= (90)² + (3)² - 2(90)(3)

= 8100 + 9 - 540

= 7569 Ans

3) x² - y² = (x - y )(x + y )

Example: evaluate 107 × 93 by using formula.

Solution:

107 × 93 = (100 + 7)(100 - 7)

= (100)² - (7)²

= 10000 - 49

= 9951 Ans.

Factorization:

Factors of an expression are the expressions whose product(multiplication) is the given expression. The process of expressing the given expressions as the product of its factors is called Factorizing or Factorization.

1) Ka + Kb + Kc (type 1)

Example: 2x - 4y + 6z factorize it.

Solution: 2x - 4y + 6z = 2(x -2y + 3z)

Example: x² - xy + xz factorize it.

Solution: x² - xy + xy = x ( x - y - z )

Example: factorize 3x² - 6xy

Solution: 3x² - 6xy = 3x ( x - 2y )

2) ac + ad + bc + bd (type 2)

Example: 3x + cx + 3c + c² factorize it

Solution: 3x + cx + 3c + c²

= ( 3x + cx ) + ( 3c + c² )

= x ( 3 + c ) + c ( 3 + c )

= (x + c ) ( 3 + c )

Example: 2x²y - 2xy + 4y²x - 4y²

Solution:

2x²y - 2xy + 4y²x - 4y²

= 2y ( x² - x + 2yx - 2y )

= 2y [ x (x - 1 ) + 2y ( x -1 )]

= 2y ( x- 1 )(x + 2y )

3) x² + 2xy + y² and x² - 2xy + y² (type 3)

Example: 9a² + 30 ab + 25b² factorize

Solution:

9a² + 30 ab + 25b²

= (3a)² + 2(3a)(5b) +(5b)²

= (3a + 5b )²

Example: 16x² - 64x + 64 factorize

Solution:

16x² - 64x + 64

= 16 ( x² -4x + 4 )

= 16 [ (x)² -2(2)(x) + (2)² ]

= 16 ( x - 2 )²

Example: 8x³y + 8x²y² + 2xy³ factorize

Solution:

8x³y + 8x²y² + 2xy³

= 2xy ( 4x² + 4xy + y² )

= 2xy [ (2x)² + 2(2x)(y) + (y)² ]

= 2xy ( 2x + y )²

4) x² - y² (type 4)

Example: 25x² - 64 factorize it

Solution:

25x² - 64

= (5x)² - (8)²

= ( 5x + 8) ( 5x - 8)

Example: factorize 16y²b - 18bx²

Solution:

16y²b - 81bx²

= b ( 16y² - 81x² )

= b [ (4y)² - (9x)² ]

= b ( 4y + 9x ) ( 4y - 9x )

Example: factorize ( 3x - 5y )² - 49z²

Solution:

( 3x - 5y )² - 49z²

= ( 3x - 5y )² - (7z)²

= ( 3x - 5y + 7z) ( 3x - 5y - 7z )

Example: By using formula evaluate (677)² - (323)²

Solution:

(677)² - (323)²

= (677 + 323) ( 677 - 323)

= 1000 × 354

= 354000

5) x² + 2xy + y² - z² and x² - 2xy + y² - z² (type 5)

Example: a² - 2ab + b² - 4c²

Solution:

( a² - 2ab + b² ) - 4c²

= ( a - b )² - (2c)²

= ( a - b - 2c ) ( a- b +2c )

Example: 4a² + 4ab +b² - 9c²

Solution:

4a² + 4ab +b² - 9c²

= (2a)² + 2(2a)(b) + (b)² - 9c²

= (2a+b)² - (3c)²

= ( 2a + b - 3c)( 2a + b + 3c )

Cubic formulas:

1) ( x + y )³ = x³ + y³ + 3xy(x + y )

Example: solve ( 3a + 4b )³

Solution: ( 3a + 4b )³

= (3a)³ + (4b)³ + 3(3a)(4b)(3a + 4b)

= 27a³ + 64b³ + 36ab(3a + 4b)

= 27a³ + 64b³ + 108a²b + 144ab²

2) (x - y )³ = x³ - y³ - 3xy ( x - y )

Example: ( 2a - 3b )³

Solution:

( 2a - 3b )³

= (2a)³ - (3b)³ - (3)(2a)(3b)(2a - 3b )by formula

= 8a³ - 27b³ - 18ab (2a - 3b )

= 8a³ - 27b³ - 36a²b + 54ab²

Simultaneous linear equations:

If there are two or more linear equations consisting of same set of variables are satisfied simultaneously by the same values of the variables then these equations are called simultaneous linear equations.

Recognizing simultaneous linear equations in one(1) and two(2) variables:

As we know that a linear equation is an algebraic equation in which each term is either a variable or constant or the product of variable or a constant. The standard form of linear equation consisting of one (1) variable is,

ax + b , ∀ a,b ∈ R

Similarly a linear equation in two variables is of the form ax + by = c where a,b and c are constants. Two (2) linear equations considered together from a system of linear equations. For example x + y = 3 and x - y = 2 is a system of two linear equations with two(2) variables x and y. This system of two (2) linear equations is known as the simplest form of linear system which can be written in the general form as,

a₁x + b₁y = c₁

a₂x + b₂y = c₂

Solution of simultaneous linear

equations:

There are many methods to solve simultaneous linear equations that is

1) Method of equating the coefficients.

2) Method of elimination by substituting.

3) Method of cross Multiplication.

Have a nice day.

Polynomials

2021-03-24T02:43:00.008-07:00

POLYNOMIALS

Learning Outcomes:

After reading the post the reader will be able to:

What are the constant(مستقل مقدار), variable(متغیر), literal(حرفی مقدار) and algebraic expression.
Polynomial
Degree of a polynomial(کثیر رقمی کا درجہ)
Coefficient(عددی سر) of a polynomial
Recognize polynomial in one, two and more variables(متغیرات).
Recognize polynomials of various(مختلف) degrees (for example linear, quadratic, cubic and biquadratic polynomials).
Add, subtract and multiply polynomials.
Divide a polynomial by a linear polynomial.

ALGEBRAIC EXPRESSIONS: (الجبری جملے)

An algebraic expression is made up of symbols (علامات) and signs of algebra. Algebra helps us to make general formula (جنرل فارمولا) because algebra is linked with arithmetic.

For example, x² + 2 x + 1 and √x -(1/√x) , "where x is not equal to zero" are algebraic expressions.

Recall Constant, Variable, Literal (حرفی مقداریں)and Algebraic Expression.

Constant: A symbol that has a fixed numerical value (مخصوص عددی قیمت) is called a constant.

For example in 2x+5, 2 and 7 is a constant term.

Variable: Variable is a symbol, usually a letter that is used to represent a quantity that may,have an infinite (uncountable) number of values are also called unknowns. For example, in 4x+ y + 3z; so x, y and z are variables.

Literal: The alphabets that are used to represent constants (مستقل مقدار) or coefficients are called literals. For example, in ax² + bx + c; so a, b and c are literals (حرفی مقداریں) whereas x is a variable.

Algebraic Expression: An expression which connects (جوڑنا) variables and constants by algebraic operations (الجبری عوامل) of addition, subtraction, multiplication and division is called an algebraic expression.

A few algebraic expressions(الجبری جملے) are given below:

(1) 24 (ii) x + 2z (iii) 4x - y + 7 (iv) {- 2 /x}+ y (v) 3y + 7z -6

Note:

[Algebra was introduced by Muslim Mathematician named Al Khawaraizmi(الخوارزمی) {780 - 850}. He was also considered as the father of modern Algebra]

POLYNOMIAL: (کثیر رقمی)

Definition: A polynomial expression or simply a polynomial is an algebraic expression consisting(مشتمل) of one or more terms in each of which the exponent of the variable is zero(0) or a positive integer(مثبت صحیح عدد).

For example, 13, -1x, 5x + 2y, x² — 3x + 6 are all polynomials. The following algebraic expressions(الجبری جملے) are not polynomials. 1/x , x⁻², x² - x⁻³ +5, x³ + y⁻⁴ , x/y.

Degree of a polynomial:(کثیر رقمی کا درجہ)

Degree of a Polynomial is the highest power(بڑی قوت) of a part (term) in a polynomial. Degree of a term in a polynomial is the sum of the exponents(قوت نما) on the variables in a single term. The degree of 2x³y⁴ is 7 as 3 + 4 = 7

Coefficient of a Polynomial:

In a term the number multiplied by the variable (عددی سر) is the coefficient of the variable. In 4x + 6y, 4 is coefficient of x and 6 is coefficient of y.

Recognition (پہچان) of Polynomial in one, two and more Variables:

a) Polynomials in one Variable

(ایک متغیر والی کثیر رقمی)

Consider the following Polynomials:

(1) x⁴ + 4 (ii) x² + x + 1 (iii) y³ +y² - y + 1 (iv) y² - y + 8 , In these polynomials (i) and (ii) x is the variable and in polynomials (iii) and (iv) y is the variable. All these polynomials are polynomials in one(1) variable.

b) Polynomials in two Variables:

(دو متغیر والی کثیر رقمی)

Consider the following Polynomials:

(i) x² + y + 2 (ii) x²y + xy + 6 (iii) x²z + xz + z (iv) x²z + 8, In polynomials (i) and (ii) x, y are the variables(متغیرات). In these above polynomials (iii) and (iv) x, z are the variables. All these polynomials(کثیررقوم) are in two variables.

c) Polynomials in more Variables:

Similarly x³yz + xy²z + xy + 7 is a polynomial in three variables x, y and z.

Recognition of polynomials of various degrees:

a) Linear Polynomials: (یک درجی کثیر رقمی)

Consider the following polynomials:

(i)x + 2 (ii) x (iii) x + 2y (iv) x + z, In all these polynomials the degree of the variables x, y or z is one. Such types of polynomials are linear (یک درجی) polynomials.

b) Quadratic Polynomials: (دو درجی کثیر رقمی)

Let us write a few polynomials in which the highest exponent or sum of exponents is always 2. (i) x² (ii) x² - 3 (iii) xy + 1

In the first two polynomials x is the variable and its degree(درجہ) is 2. In the 3rd Polynomial x, y are the variables and sum of their exponents is 1 + 1= 2. Its degree is also 2. Therefore above polynomials of the type (i), (ii) and (iii) are quadratic polynomials.

c) Cubic polynomials:(سہ درجی کثیر رقمی)

Consider the following examples of polynomials:

(I) 5x³ + x² - 4x +1

(ii) x²y + xy² + y - 3

The degree of each one of polynomial is 3. These polynomials are called cubic polynomials.

d) Biquadratic polynomials:(چہار درجی)

We take a few polynomials of 4 degree.

(i) x⁴ + x²y² + x³y + y³ + 3

(ii) y² + y⁴ + y³ + y +9

These polynomials are biquadratic(چہار درجی) polynomials.

Operations on polynomials:

Addition of polynomials ( algebraic expressions)

If P(x) and Q(x) are polynomials then their addition is represented by P(x) + Q(x). In order to add two or more than two polynomials first we write the polynomials in ascending or descending order and like terms each in the form of columns. Finally we add the coefficients of like terms.

Example: Find the sum of ( 3x³ + 5x² - 4x ) , ( x³ - 6 + 3x² ) and ( 6 - x² - x )

Solution:

Subtraction of polynomials ( algebraic expressions)

The subtraction of polynomials P(x) and Q(x) is represented as P(x) - Q(x). If the sum of two polynomials is zero then P(x) and Q(x) are called additive Inverse of each other.

Same as addition method we write the polynomials in ascending or descending order and then we change the sign of every term of the polynomial which is to be subtracted.

Example: solve ( 5x⁴ + x - 3x² - 9 ) - ( 2x³ - 4x² + 8 - x )

Solution:

Multiplication of polynomials:

Multiplication of the polynomials is explained by following examples

Example: Solve ( 4x² ) × ( 5x³ )

Solution:

Example: Find product of following polynomials.

( 3x² + 2x - 4 ) and ( 5x² - 3x + 3 )

Solution:

Example: Solve ( 2x - 3 ) × ( 5x + 6 )

Solution:

Division of polynomials:

The division is reverse process of Multiplication. The division of polynomials is explained by following examples.

Example: ( -8x⁵ ) ÷ ( -4x³ )

Solution:

Example: ( x³ - 2x + 4 ) ÷ ( x + 2 )

Solution:

Note that if a polynomial is completely (exactly) divisible by another polynomial then the remainder is zero (0).

Overall Summary:

An expression (sentence) which Connects or links variables and constants by algebraic operations of addition subtraction, multiplication and division is called an algebraic expression.

Constants are algebraic symbols that have a fixed value and these values are unchange able.

A symbol(sign) in algebra which can assume different numerical values is called a variable.

A literal (حرفی مقدار) is a value that is expressed as itself. For example, the number 45 or the word "speed" are both literals.

An algebraic expression which has finite (countable ) number of terms and the exponents of variables are whole numbers, is called polynomial.

A polynomial is either zero (0) or an be written as the sum of a finite number of non-zero terms.

In a polynomial coefficient is a number or symbol(sign) multiplied with a variable in an algebraic term.

The polynomials of degree one(1) are called linear polynomials.

The polynomials of degree two(2) are called quadratic polynomials.

The polynomials of degree three(3) are called cubic polynomials.

The polynomials of degree four(4) are called biquadratic polynomial.

Have a nice day.

Number system

2021-03-23T00:47:00.007-07:00

NUMBER SYSTEM

Learning Outcomes:

After reading the article readers will be able to:

Recognize base of a number system.
Number system with base 2,5,8 and 10
Understand:
Binary number system or base 2
Number system with base 5
Octal number system or base 8
Base 10 or decimal number system
Conversion (تحویل) of binary system, base 5 and base 8 to decimal system and vice Versa.
Addition, subtraction and Multiplication of base 2,5 and 8.

NUMBER SYSTEMS: (عددی نظامات)

We know that any number can be formed with the help of 10 digits i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8 9. These numbers are called numerals (علامات) and these numerals are known as 'Arabic numerals'.(عربی علامات)

Base of a Number System:

The number of digits involved (استعمال) in a number system is called the base of that number system.

If a number system (عددی نظام) involves only two digits 0, 1, then base (اساس) is 2. A number system, in which 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 are used, is a system with base 10 (اساس دس کا نظام).

Similarly, a number system (عددی نظام ) in which five digits 0, 1, 2, 3 and 4 are used is a system with base 5.

Number System with Base 2, 5, 8 and 10:

(a) Number System with Base 2:

A number system formed by two digits 0, 1 is called Binary system (اساس دو کا نظام) and its base is 2. This system is not used in everyday(روزمرہ) life apparently(بظاہر). But it is very important number system because it is used in all types of computers(کمپیوٹر). Because computer stores(جمع کرنا) information in the form of binary numbers so the binary system is of primary importance in the modern age(جدید دور) of computer.

(b) Number System with Base 5:

This number system(عددی نظام) involves digits 0, 1 2, 3 and 4. The largest digit (سب سے بڑا عدد) in base 5 system is 4.

(c) Number System with Base 8:

The number system(عددی نظام) with base 8 is called octal system. In this system(نظام) eight digits 0, 1, 2, 3, 4, 5, 6 and 7 are used. The largest (سب سے بڑا) digit in octal system is 7.

(d) Decimal Number System:

Decimal number system (اعشاری نظام ) is the most popular number system in the world. In this system (نظام) ten digits (0 to 9) are used. Every number can be expressed (ظاہر کرنا) as the sum of multiples (اضعاف) of powers of 10 and 10 is called its base.

CONVERSIONS: (تحویلات)

The above discussed (بیان کیا گیا) number systems are all place value number systems. The numbers used in these systems can be converted (تبدیل کرنا) from one system to another system. The method of successive division (مسلسل تقسیم) is used to convert a number from one system to another system. The division (تقسیم) is performed by the base of the system in which it is being converted.

Conversion from decimal number system (اعشاری نظام) to other number systems:

A) Conversion from decimal (base 10) to binary system:

Example: Convert 15 into binary system or base 2.

Solution:

We write the numbers from bottom to top.

Example: change 541 into base 2 or binary system.

Solution:

Thus 541 = (1000011101)₂

B) Conversion from decimal system (base 10) to a number with base 5:

Any number of decimal system can be converted to equivalent no. with base 5 as

Example: Convert 17 to an equivalent no. with base 5

Solution:

Example: Convert 89651 to an equivalent no. with base 5

Solution:

Thus 89651 = ( 10332101)₅

C) Conversion from decimal to base 8 or Octal system:

Example: convert 824 to an equivalent no. with base 8(Octal system)

Solution:

Hence 824 = (1470)₈

Example: Convert 4837 to an equivalent number with base 8 (Octal system).

Solution:

Result, 4837 = (11345)₈

Conversion (تحویل) from other Number system to decimal (اعشاری) number system:

A) Conversion from binary system to decimal system:

We can understand this conversion by following examples

Example: change (1101)₂ into equivalent number in decimal system.

Solution:

(1101)₂ =( 1×2³) + (1×2²)+(0×2¹)+(1×2⁰)

(1101)₂ = 8+4+0+1 = 13

B) From base 5 to decimal system (base 10)

Example: change (413242)₅ into equivalent decimal system.

Solution: (413242)₅ = (4×5⁵)+(1×5⁴)+(3×5³)+(2×5²)+(4×5¹)+(2×5⁰)

(413242)₅ = (4×3125)+(1×625)+(3×125)+(2×25)+(4×5)+(2×1)

(413242)₅ = 12500+625+375+50+20+2 = 13572

C) Conversion from Octal system (base 8) to decimal system:

Example: convert (126)₈ to decimal system.

Solution: (126)₈ = 1×8²+2×8¹+6×8⁰

(126)₈ = (1×64)+(2×8)+(6×1)

(126)₈ = 64+16+6 = 86

Example: Convert (424002)₈ to decimal system.

Solution: (424002)₈ = (4×8⁵)+(2×8⁴)+(4×8³)+(0×8²)+(0×8¹)+(2×8⁰)

(424002)₈ = (4×32768)+(2×4096)+(4×512)+(0)+(0)+(2×1)

(424002)₈ = 131072+8192+2048+0+0+2 = 141314

Addition, Subtraction and Multiplication:

A) Addition base 2:

There are only two digits 0 and 1 used in the binary number system.

When we add the numbers and if the sum is greater than 1 then divide the sum by 2, write the remainder and carry quotient to next digit.

Following examples are helpful to understand the addition process of binary number system.

Example: add these two numbers of binary system. (111)₂ and (10)₂

Solution:

Example: solve: (10110111)₂ + (100011)₂

Solution:

B) Subtraction of binary number system:

Example: solve (101)₂ - (11)₂

Solution:

Example: find (10011)₂ - (1101)₂

Solution:

C) Multiplication of base 2:

Example: solve (11)₂ × (10)₂

Solution:

Example: multiply (11011011)₂ by (10101)₂

Solution:

Base 5 (Octal system) number system:

A) Addition of base 5

In addition if the sum of two or more digits is greater than 5 then divide the sum by 5 and write the remainder and carry the quotient to the next digit.

Example: Find sum of (4)₅ and (3)₅

Solution:

Example: solve (12433)₅ + (31243)₅

Solution:

B) Subtraction of base 5:

Example: solve (3421)₅ - (2143)₅

Solution:

C) Multiplication of base 5:

Example: solve (23)₅ × (14)₅

Solution:

Example:

Multiply (421)₅ by (234)₅

Solution:

Base 8 OR Octal number system:

In Octal system there ae eight Numbers 0,1,2,3,4,5,6,7.

A) Addition in base 8 OR Octal system:

Example: Solve (6)₈ + (7)₈

Solution:

Example: (64)₈ + (44)₈

Solution:

Example: Solve (255636)₈ + (143576)₈

Solution:

B) Subtraction in base 8 OR Octal number system:

Example: Solve (14)₈ - (6)₈

Solution:

Example: (604)₈ - (247)₈

Solution:

Example: (455122)₈ - (216634)₈

Solution:

C) Multiplication in base 8 OR Octal number system:

Example: solve (36)₈ × (43)₈

Solution:

Example: (446)₈ × (213)₈

Solution:

Addition(جمع), Subtraction (تفریق) and Multiplication (ضرب) of numbers with different bases:

For this first of all we convert all the numbers into decimal system OR base 10 and perform the given operations. Then the answer is converted into base 2, base 5 and base 8 as required.

Example: solve (100111)₂ + (4123)₅ + 567 and give answers in base 2 ,base 5 and base 10.

Solution: We convert (100111)₂ and (4123)₅ into base 10 OR decimal system.

pic

example. solve (777)₈ - (2343)₅ - (1000111)₂ and give your answer in binary system (base 2).

Have a nice day.

Information Handling

2021-03-20T04:41:00.006-07:00

INFORMATION HANDLING

Learning Outcomes:

After reading this post Readers will be able to:

1- Know about the Information Handling

2- How to collect Data and represent Data?

3- Types of Data

4- Difference between grouped and ungrouped Data

5- Know about Horizontal and vertical bar graphs

6- About pie graph and how to draw pie graph

7- What is frequency and frequency distribution

8- How to draw histogram which represent frequency table

9- How to calculate Mean, weighted mean, median and mode

Information Handling

Introduction:

Information Handling is a branch of statistics which deals with the collection, analysis(تجزیہ), explanation (وضاحت) and presentation of data. The word statistics has been deduced from the Latin word Statisticum collegium that means council of state(ریاستی کابینہ) and the Italian word statista which means politician(سیاستدان).

In the world around us there are a lot of questions and situations (حالات) that we want to understand, describe, explore and access(رسائی ).

For example

How many hospitals are there in out country?

How many children were born during the last ten (10) years?

How many doctors will be required in the next five years?

To know about such things, we collect information and present it in a manageable way( منظم انداز) so that useful conclusions (نتائج ) can be drawn. The branch of statistics that deals with this process is called information Handling (معلومات داری).

Data:

A data is a set of information and facts (حقائق ) which is represented in the form of figures. We can collect data by several ways, depending upon the quantity of the data and reason for it's collection.

OR Data means facts or groups of information that are normally the results of measurements(پیمائش ), observations and experiments. These results help us in reviewing our past performance(کارکردگی ) and future planning. For example the government of a state prepares it's budgets (بجٹ) and development plans on the basis of a collected data about the resources and population.

Presentation of data:

After the collection of data the most important step is its presentation (ظاہر کرنا) that provides basis to draw conclusions. Data can be represented in the form of tables (جدول) and different kind of groups.

Ungrouped data:

The data that is collected in a raw form and it provides us information about individuals(انفرادی ). Data in such a form is called ungrouped data (غیر گروہی مواد).

Example:

11 players of a cricket team enhanced the score of the team in a one day match as given in following table.

Grouped data:

The data which provides us the information about groups is called the grouped data (گروہی مواد). For example we can represent the data of the above example in groups as,

Number of players who scored between 1-10=1

Number of players who scored between 11-20=2

Number of players who scored between 21-30=2

Number of players who scored between 31-40=5

Number of players who scored between 41-50=1

We can also represent (ظاہر کرنا) the above data by using a table as given below.

Graph:

A graph is a drawing that shows the relationship( تعلق) between numbers and quantities. We use graph to display a data in a simple, attractive and comprehensible (قابل فہم) way.

Types of graph:

There are two types of graph.

1- Bar graph

2- pie graph

Bar graph:

A graph in which data is represented by a number of rectangular bars (مستطیلی بار) is called a bar graph. A bar has a uniform width (چوڑائی ) and an equal distance from the other bars. we use two rays to draw a bar graph which are called axis and marking on axis is called it's scale. A graph with horizontal bars (افقی بار) is called horizontal bar graph and graph with vertical bars (عمودی بار) is called vertical bar graph.

Vertical bar graph:

We an understand this type of bar graph by an example.

John got the following marks out of 100 marks

Solution:

Choose a suitable scale (مناسب سکیل) to draw a graph.

Draw x-axis and y-axis with the common point O.

Indicate marks according to the scale(سکیل), along y-axis

Along x-axis mark each month (ہر ماہ) after a suitable distance.

Draw coloured rectangular bars (رنگ دار مستطیلی بار) according to the scale for each month.

Pic

How to read bar graph:

1- The bar graph represent in general the marks of John that he got in six(6) tests.

2- January is the month in which John got the maximum marks.

3- September is the month in which John got the least marks.

4- The ratio of the marks of October to the marks of November is 80:60 or 4:3

5- The month of best performance is January and that of worst performance is September.

Horizontal bar graph:

Observe the following example

John traveled to help the poor as given in the following table.

Solution:

Choose a suitable scale (مناسب سکیل) to draw a graph.

Draw x-axis and y-axis with the common (مشترک) point O

Along y-axis mark each day after a suitable distance (مناسب فاصلہ)

Draw coloured rectangular bars according to the scale(سکیل کے مطابق) for each day.

How to read this bar graph:

1- The bar graph is representing in general the distance that John traveled in five(5) days.

2- Monday is the day in which John traveled the maximum distance.

3- Tuesday is the day in which John traveled the least distance.

4- the ratio (نسبت) of distance of Monday to the distance of Tuesday is 920:640 or 23:16

5- The day of maximum(زیادہ) traveling is Monday and that of least traveleing is Tuesday.

Pie graph:

The representation of a numerical data in the form of disjoint sectors (قطعہ) of a circle is called a pie graph.

A pie graph is generally used for the comparison (موازنہ) of some numerical facts classified in different classes. In this graph the central angle measure 360 degree which is subdivided into the ratio (نسبت) of the size of the groups to shown through this graph.

Example:

Read the following pie graph which is showing the favourite fruit (پسندیدہ پھل) of 900 boys?( When pie graph is giving)

1- How many boys , favourite fruit is mango ?

2- How many boys , favourite fruit is banana?

3- How many boys , favourite fruit is apple ?

4- How many boys , favourite fruit is grapes ?

5- which fruit is the most favourite of the boys?

Solution:

We use the following formula (فارمولا) to solve it

Required quantity = (given angle/360)× total quantity

1- Mango is favourite fruit = (210/360)×900 boys = 525 boys

2- Banana is favourite fruit = (60/360) × 900 boys =150 boys

3- Apple is favourite fruit = (50/360) × 900 boys = 125 boys

4- Grapes as a favourite fruit = (40/360)×900 boys = 100 boys

EXAMPLE:

It is compulsory (ضروری) for each student to take part in different games. Out of 1800 students in school 750 play cricket, 200 students play badminton, 400 students play hockey and 450 students play football. In order to represent (ظاہر کرنا) their comparison(موازنہ), draw a pie graph. (When data is giving)

Solution:

Total students = 1800

Formula:

Required angle = (no. Of students play a game ÷ total students) ×360

Angle in pie graph for badminton = (200 / 1800)×360 = 40 degree

Angle in pie graph for cricket = (750/1800)×360 =150 degree

Angle in pie graph for hockey = (400/1800)×360 = 80 degree

Angle in pie graph for football = (450/1800)×360 = 90 degree

Construction of pie graph:

First of all Draw a circle of suitable radius.

Draw an angle of 40 degree for badminton ,150 degree for cricket, 80 degree for hockey and the remaining angle will be 90 degree which represent the football.

Frequency and frequency distribution

Frequency:(تعدد)

The number of times a value occur in a data is called frequency of that value.

Example:

15 students obtained marks out of 10 in a test as following

5,3,7,10,7,9,3,7,5,4,6,8,7,5,2

In this data the frequency of 3 marks is 2.

The frequency of 5 is 3.

The frequency of 7 marks is 4

Frequency distribution:(تعددی تقسیم کا جدول)

To write a data in the form of a table in such a way that the frequency (تعدد) of each class can be observed at once is called its frequency distribution(تعددی تقسیم کا جدول).

Construction of frequency distribution table:

We can understand this by an example,

There are 50 students and they got marks as following in a 60 marks test.

25,33,34,26,35,28,29,36,30,54,30,39,36,37,39,40,37,34,27,41,37,41,38,42,48,51,40,51,43,40,41,39,48,51,53,41,37,52,28,46,44,37,39,52,51,40,45,46,43,53

Make a frequency distribution table taking six(6) classes of equal size by tally marks.

Solution:

In data we find lowest value which is = 25

Highest value is data = 54

Total classes = 6

Size of class interval = ( highest value - lowest value) / (number of classes)

Size of class interval = (54-25)/6

Size of class interval = 29/6

Size of class interval = 5 ( approximately)

Example:

The number of units consumed electricity by 31 households are as following

727,773,859,711,860,747,862,738,774,852,791,836,834,752,828,792,908,839,752,715,880,838,

852,816,751,834,818,835,831,778,837.

Make a frequency distribution table with 10 classes.

Solution:

Lowest value in data = 712

Highest value in data = 908

Number of classes = 10

Class interval size = (908-711)/10

Class interval size = 197/10 = 19.7 = 20 (approximately)

Construction of Histogram:

We are familiar with pie and bar graph. Another common graphic way to presenting data is by means of histogram. A histogram is similar (same عکس) to bar graph but it is constructed for a frequency table.

In histogram the values of the data are represented along the horizontal axis and the frequency are shown by bars perpendicular to the vertical axis. Bars of equal width(چوڑائی ) are used to represent individual (انفرادی ) classes of frequency table.

To draw a histogram from grouped data the following steps are followed.

1- Draw x-axis and y-axis .

2- Mark class boundary of the classes along x-axis.

3- Mark frequency along y-axis.

4- Draw bar for each class so that the height of the bar drawn for each class is equal to the frequency of the class.

We can understand it by a example,

Example:

The detail of distances travelled (سفر کے تفصیل) daily by the residents of a locality are given below. Construct a histogram(کالمی نقشہ) for the following frequency table.

Solution:

The frequency distribution table is:

Histogram: (کالمی نقشہ)

Calculation of measures of central tendency:

Mean (average):

If x₁ ,x₂,.......,xₙ given quantities. Then their average or mean can be calculate by the following formula. ( X denotes Mean )

X = (x₁ + x₂ +......+xₙ) / n

X = (addition of all values)/ (number of values)

A student scores in eight papers are 58,72,65,85,94,78,87,85. Find the mean score.

X = ( 58+72+65+85+94+78+87+85) /8

X = 624/8 = 78

Hence the mean score is 78.

Weighted Mean: (اوزانی اوسط)

When all values of given data have same importance then we use Mean(اوسط). But when different values have different importance then these values (اقدار) are known as weights.

If x₁,x₂,x₃,.......,xₙ have the weights w₁,w₂,w₃,.....,wₙ then

Weighted mean = Xw = (x₁w₁ + x₂w₂ + x₃w₃ + .........+ xₙwₙ ) / (w₁+w₂+.......+wₙ)

Example:

The following data describes the marks of a student in different subjects and their weights assign to these subjects are also given:

Find its weighted mean:

Solution:

Weighted mean = [ 4(74) + 3(78) + 5(74) + 6(90) ] / (4+3+5+6)

Weighted mean = ( 296+234+370+540)/(18)

Weighted Mean = 1440/18 =80 marks

Median : (وسطانیہ)

If some data is arranged in descending or ascending order then the median of the data is :

(a) The middle value of data , if it consists of odd number of values.

(b) If the data consists even values then the Median is the mean of two middle values.

Example:

Find the median if the weights in kg of 9 students are under as

29,32,45,27,30,47,35,37,33

Solution:

First we arrange these values in ascending order

27,29,30,32, 33, 35,37,45,47

The middle value is 33 So Median is 33 kg

Example:

Find the median if the weights in kg of 10 students are under as

29,32,45,27,30,47,35,37,33,31

Solution:

First we arrange these values in ascending order

27,29,30,31,32, 33, 35,37,45,47

so there are two values in the centre

Median = (32+33)/2

Median = 32.2 ans

Mode: (عادہ)

Mode is the value that occurs (آنا) most frequently(سب سے زیادہ) in a data. In case no value is repeated (دہرانا) in data then the data (مواد) has no mode. When two or more values (اقدار) occur with the same greatest frequency , then each is a mode.

Example: Find the mode (عادہ) of the following data.

2,1,8,6,3,2,9,0,5,2,7,5,1,2

Solution:

Since 2 is repeated 4 times so this data has mode i.e 2

Example: find the mode of following data.

6,7,3,9,0,1,4,2

Solution:

This data has no mode because no value is repeated in this data.

Example: Find the mode of given data.

2,2,2,1,3,4,5,5,5,6,7

Solution:

In this data 2 is repeated 3 times and 5 is also repeated 3 times so this data has two modes i.e 5 and 2.

Note:

1- A data can have more than one mode(مواد کا ایک سے زیادہ عادہ ہو سکتا ہے).

2- A data may or may not have mode.

Have a nice day

Demonstrative Geometry

2021-03-16T08:02:00.006-07:00

DEMONSTRATIVE GEOMETRY
Learning Outcomes:

In this post we discuss the following topics,

Demonstrative Geometry

Types of assumptions

The parts of propositions

What is geometrical theorem

Examples of Axioms

Postulates

Examples of postulates

What is corollary

Converse of a theorem

Demonstrative Geometry: (اثباتی جیومیٹری)

Demonstrative geometry is a branch (شاخ) of mathematics in which theorems (مسائل ) on geometry are proved through logical reasoning. It demonstrates the truth (سچائی ) of mathematical statements concerning geometric figures.

ریاضی کی وہ شاخ جس میں جیومیٹری کے مسائل کا حل منطقی استدلال سے معلوم کیا جاے۔اس میں جیومیٹری کی اشکال سے مدد لی جاتی ہے اور ریاضی کے مسائل کی سچائی کو ثابت کیا جاتا ہے۔

Basics of reasoning

Basics of reasoning in mathematics are:

(استدلال کی بنیادیں)

Basic Concepts: (بنیادی تصورات) Some concepts are accepted without defining (بغیر بیان کے) them for example point, line or plane.

بعض اوقات ایسے تصورات ہوتے ہیں جن کو تعریف کے بغیر قبول کر لیا جاتا ہے۔مثلا نقط۔خط اور مستوی

Assumptions: (مفروضے) Some statements (بیانات) are accepted true without proofs. These are called basic assumptions.

کچھ بیانات ایسے ہوتے ہیں جن کو بغیر ثبوت کے درست تسلیم کر لیا جاتا ہے۔ ہم انہیں بنیادی مفروضے کہتے ہیں۔

Types of Assumptions: (مفروضوں کی اقسام )

Axioms: (اصول متعارفہ)

An axiom is a self-evident truth which needs no proof(جیسے ثبوت کی ضرورت نہ ہو) or demonstration and can be taken for granted.

اصول متعارفہ ریاضی کی تقریبا تمام شاخوں میں مشترک ہوتا ہے اور ایسے بغیر ثبوت کے تسلیم کیا جاتا ہے۔

Examples of Axioms (مثالیں)

i. A whole (total ) is always greater (بڑا) than its part or a part cannot be equal to the whole.

ii. Things which are equal (برابر) to the same thing are equal.

iii. If equals be subtracted from equals, the differences (فرق) are equal.

iv. Doubles and halves (نصف) of equal are equal.

(کل ہمیشہ جزو سے بڑا ہوتا ہے یا جزو کبھی کل سے کے برابر نہیں ہوتا )

( اشیاء جو ایک ہی چیز کے برابر ہوں آپس میں بھی برابر ہوتی ہیں)

( اگر برابر میں سے برابر تفریق کیے جایں تو فرق بھی برابر ہوتے ہیں )

( برابر کے دوگنا اور ان کے نصف آپس میں برابر ہوتے ہیں )

Postulates: (اصول موضوعہ)

A postulate is that elementary statement (بیان) which we have to assume while making a demonstration.

( ایسے مفروضے جنہیں ابتداء میں ہی صحیح تسلیم کر لیا جاتا ہے۔اور ان کا تعلق ریاضی کی کسی مخصوص شاخ سے ہو تا ہے۔)

Examples of Postulates : ( مثالیں )

1- A straight line (سیدھی لاین) may be drawn from one point to any other point in the same plane.

2- We can produce a infinite(لا متناہی) straight line (سیدھی لاین) to any length in a straight line in either direction.

3- We can cut off (کاٹنا) a straight line of any length from a given straight line either from it or by producing it.

The magnitude (مقدار) of an angle does not depend upon the length of its arms.

Parts of Propositions : (مسلہ کے اجزا)

A proposition is a declarative sentence (بیانیہ فقرہ) that is either true or false. The various parts of propositions are,

Enunciation: (بیان مسلہ)

It is the statement of a geometrical truth(سچائی ) which we are going to prove.

ii. Given: (معلوم)

For the sake of convenience (آسانی) and clarity (وضاحت), we first of all put down what is given to us or what is assumed. (فرض کرنا)

iii. To Prove: (مطلوب)

In this part,we put down what we are going to prove (ثابت کرنا) or establish.

iv. Construction: (عمل)

In this part of the proposition (مسلہ), we note down all the additional lines or figures (اشکال) which must be drawn so that, we may be able to arrive at the required result.

v. Proof: (ثبوت)

In this part, we establish (ثابت کرنا) the truth with a suitable line of reasoning.

Meaning of a Geometical Theorem, Corollary and Converse of a Theorem

Geometrical Theorem: (ہندسی مسلہ)

A theorem is that kind of proposition in which we establish (ثابت کرنا) a geometrical truth by means of reasoning with the help of a geometrical figure(ہندسی اشکال).

(ہندسی مسلہ ایک ایسا بیان ہے جیسے سچائی استدلال اور ہندسی اشکال کی مدد سے ثابت کیا جاتا ہے)۔

Examples of Theorems (مثالیں)

i. The sum of the interior angles (اندرونی زاویے) of a triangle is 180°.

ii. If two angles of a triangle are congruent(متماثل), the sides opposite these angles are congruent.

Corollary: (نتیجہ صریح)

A corollary is also a proposition(مسلہ), the truth of which can be immediately drawn from the theorem that has already been proved(پہلے سے ثابت شدہ).

For example, a theorem in geometry is "The angles opposite two congruent sides (متماثل اطراف) of a triangle are also congruent". A corollary to that statement is that in an equilateral triangle(مساوی الاضلاع مثلث) all angles are congruent.

Converse of theorem: (عکس مسلہ)

When the 'Given معلوم' of one proposition becomes the 'To Prove مطلوب ' of the other proposition and vice versa, the two propositions are called converse (عکس) of each other.

( اگر کسی مسلہ کے معلوم کو مطلوب کے ساتھ اور مطلوب کو معلوم میں بدل دیا جاے تو نیو مسلہ کو پہلے مسلہ کا عکس مسلہ کہتے ہیں )

For example the converse of Pythagoras theorem (مسلہ فیثا غورث) is "If the sum of the squares of two sides is equal to the square of the third side of the triangle(مثلث), triangle is a right triangle(قایمہ زاویہ مثلث)".

Pythagoras theorem

Theorems

Theorem 1:

If a straight line(خط مستقیم) stands on another straight line, the sum of measures(پیمائش ) of two angles so formed is equal to two right angles.

(اگر ایک خط مستقیم دوسرے خط مستقیم پر واقع ہو تو اس طرح جو دو متصلہ زاوے بنتے ہیں ان کا مجموعہ دو قایمہ زاویہ کے برابر ہوتا ہے۔)

Corollary: If two straight lines intersect(قطع کرنا) one another, the four angles equal to four right angles. (اگر دو خطوط ایک دوسرے کو قطع کریں تو اس طرح بننے والے چار زاویوں کی مقدار کا مجموعہ چار قایمہ زاویہ کےبرابر ہوتا ہے۔)

Theorem 2:

If the sum of measures of two adjacent angles (متصلہ زاویے) is equal to two right angles, the external arms (بیرونی بازو) of the angles are in a straight line.

corollary 1: if two straight lines (دو سیدھی لاین) cut one another, the sum of the measures of the angles so formed is equal to four (4) right angles.

Corollary 2: When any number of straight lines (چند سیدھے خطوط) meets at a point, the sum of the consecutive angles (مسلسل تمام زاویے) so formed is equal to four right angles.

Theorem 3:

If two lines intersect (قطع) each other then the opposite vertical angles(متقابلہ راسی زاویے) are congruent.

Theorem 4:

If two sides of a triangle are congruent (متماثل) then the angles opposite (متقابلہ زاویے) these sides are congruent.

Corollary 1: in an equilateral triangle (مساوی الاضلاع مثلث)all angles are congruent.

Corollary 2: The bisector (ناصف) of the vertex angle (زاویہ راس )of an isosceles triangle (مساوی الساقین) is also the perpendicular bisector of the base of the triangle.

Theorem 5:

The sum of measures(پیمائش کا مجموعہ) of the three angles of a triangle is 180⁰.

Corollary 1: Each angle of an equilateral triangle (مساوی الاضلاع مثلث) is 60⁰.

Corollary 2: In a right angled triangle the acute angles (حادہ زاویے) are complementary.

Have a nice day.

Financial Arithmetic

2021-03-14T07:14:00.007-07:00

Learning Outcomes:

After reading this post the readers are able to understand

1- what is partnership?

2- what is inheritance?

3- what is banking and it's other related terms like

PLS saving account,

Current deposit account,

PLS term deposit account,

Foreign currency account,

Negotiable instruments like

Cheque, demand draft,pay order, online banking, ATM , debit card, credit card,

4- readers are able exchange or conversion of currencies.

5- Readers are able to calculate profit ,loss and markup and also discount.

6- Readers are able to know the types of finance like

Overdraft, Running finance, Demand finance and leasing also.

7- Readers are able to calculate percentage %.

8- Readers are able to well known about insurance and its types like,

Life insurance

Vehicle insurance and about income tax.

Let's start...

Partnership: (شراکت)

A business (کاروبار) in which two or more persons run the business and they are responsible(ذمہ دار) for the profit and loss is called the partnership.

If the partners( شراکت دار) start the business and close it together with same or different investment capital, this partnership is called a simple partnership( سادہ شراکت داری).

If the partners contribute (شامل) different capitals for different time periods or at least one partner contributes two or more capitals for different time periods, then this partnership is called a compound partnership ( مرکب شراکت داری). In this case the profit or loss (نفع اور نقصان) is divided in the ratio of monthly investments.

Example:

Saud & Ammar start a business with capitals of Rs 56000 and Rs 64000 respectively. After 1 year they earned a profit of Rs 22500. Caculate the share of each one.

solution:

Example:

Tahir start a business with a amount of Rs 15000. after five (5) months Umer also joined him with an investment of Rs 30000. At the start of nine (9) months Usman joined them by investing Rs 45000. At the end of year they earned a profit of Rs 406000. Find the share of every one.

solution:

Inheritance: (وراثت )

When a person dies, then the assets (اثاثہ) left by him are called inheritance and it is distributed among his legal inheritors ( ورثہ) according to Islamic Sharjah Law (شریعت اسلام). In Islam the principles (اصول)of distribution of inheritance are given below.

First of all his/her funeral expenses(تجہیزوتکفین کے اخراجات) and all his/her all debt be paid.

Then execute the will upto one third (1/3 ) of his/her property if asked for.

Then distribute (تقسیم) the remaining inheritance accordingly.

We can understand the procedure(طریقہ) of inheritance with the help of following examples.

Example:

A person left his property (جایداد) of Rs 640000. A debt (قرض) of Rs 40000 was due to him and Rs 5000 was spent on his burial(تدفین). Distribute the amount between his widow ( share 1/8), one (1) daughter and two (2) sons according to the Islamic law.

Solution:

BANKING: (بینکنگ)

It is a business activity of accepting (وصول کرنا) and safeguarding (حفاظت کرنا ) the money and then earn a profit by lending out this money.

Define Commercial Bank deposits:

(کمرشل بنک ڈیپازٹس کی تعریف )

The function of bank which accepts deposits, provides loans and other services to the clients (لوگ۔گاہک) is known as commercial banking.

Types of a Bank Account:

There are four types of bank accounts.

1- PLS Saving Bank Account: (نفع و نقصان شراکتی سیونگ اکاونٹ)

It is an account on the basis (بنیاد) of profit and loss sharing. The bank uses the deposits in some business (کاروبار) and gives the share in profit and loss to the account holder at the end of specified period(مخصوص وقفہ). This account is meant to encourage (حوصلہ افزاٸ) the saving habits among the persons having small income means. Zakat (زکوٰۃ ) is deducted on notified balance on first Ramadan each year.

2- Current Deposit Account: (کرنٹ ڈیپازٹ کاونٹ)

This account is usually opened by businessmen (کاروباری آدمی ) who have a number of deposits and withdrawal regularly. It is a running account and no interest (نفع) is paid on its balance. In this account amount can be deposited and withdrawn (نکلوانا) at any time during banking hours without any notice. No Zakat is deducted (منہاکرنا) on this account.

3- PLS Term Deposit Account:

(نفع و نقصان شراکتی ٹرم ڈیپازٹ اکاونٹ)

This account is free of interest(سود). PLS term deposit holder(کھاتہ دار) shares profit and loss on the rate determined by the bank after every six months. The rate of profit on fixed deposits is comparatively (نسبتاً ) higher than saving deposits.

4-Foreign Currency Account: (فارن کرنسی اکاونٹ)

A foreign currency account (کھاتہ) is the account maintained in a commercial bank in the currency other than Pakistani currency. Usually foreign currency accounts are maintained in Dollars ڈالر, Pounds پاونڈ , Euro یورو etc. Foreign currency accounts are exempted (چھوٹ) from Zakat and taxes. Rate of profit in this account is very low(بہت کم).

Describe negotiable instruments(نیگوشی ایبل انسرومنٹس)

like cheque, demand draft and pay order:

Negotiable Instrument:

It is a document (کاغزات) which can be transferred from one person to another. It is payable either to the order of the bearer (مالک) or to his agent(نماندہ) as the case may be. This document is entitled to receive that amount which is mentioned (ذکر) in it.

Cheque: (چیک)

A cheque is a written order that instructs a bank to pay the specific(مخصوص) amount from a specified account to the holder of the cheque. A crossed cheque (کراس چیک) has to be deposited in the specified account.

Demand Draft: (ڈیمانڈ ڈرافٹ)

It is a method used by individuals(انفرادی) to make transfer payments from one bank account to another. The bank receives (وصول کرنا) the money in advance before it issues the draft. A very small fee is charged (فیس وصول کرنا) by the bank to prepare it.

Pay order: (پے آرڈر)

It is a document which instructs (ہدایت نامہ) a bank to pay a certain amount to a third party. Pay order issued by the bank on the request of its customer (گاہک). It is issued (جاری کرنا) on the receipt of full amount for which a pay order is issued by the bank. It can be encashed (رقم حاصل کرنا) from any other bank.

Online Banking

Explain online Banking:(آن لاین بینکنگ کی وضاحت)

The use of Internet by banks to assist (سہولت) their customers through on-line banking. It allows customers to perform banking transactions (منتقلی) such as money withdrawal, pay utility bills and transfer funds from their account to another account. A good online bank will offer its customers (گاہک) just about every service traditionally available through a local branch.

Transactions through ATM (Auto Teller Machine): (اے ٹی ایم)

An automated teller machine (ATM) is electric devices (الیکٹرانک مشین) that allows a bank's customer, to draw cash and check their account balances (کھاتہ کی رقم) without any need for a humane teller. The transactions are as given below.

Withdraw money,

make deposits,

print a statement,

check account balances(بیلنس ) and trasfer money (رقم) between accounts.

Debit Card: (ڈیبٹ کارڈ)

It is a plastic payment card that provides card holder electronic access (رسائی ) to his bank account at anytime and anywhere. It is a facility provided (مہیا کرنا ) to the customers to perform different transactions. It is a smarter and secured (محفوظ) way to make quick payments at the time of purchase (خرید کرنا) of different goods from traditional or online market.

Credit Card (Visa and Master): (کریڈٹ کارڈ)

It is a thin plastic card which can be used to buy articles(خریداری اشیاء ). Visa and Master cards are used worldwide (ساری دنیا) for making payments. These are not the names of cards but are the names of global credit card companies(کمپنیاں). Credit card holder is charged an annual fee(سالانہ فیس).

Conversion of Currencies: (کرنسیوں کا تبادلہ)

A foreign currency exchange rate (تبادلہ کی شرح) is a price that represents how much it costs to buy the currency of one country using the currency of another country.

Convert Pakistani Currency to well-known international (بین الاقوامی)currencies:

Currency conversion rates are not permanent (مستقل) but these change day by day. We use these currency rates to convert(تبدیل) Pakistani currency to differen international currencies. (rate of US $ is equal to Rs. 157)

Example:

Mr. Naeem wants to exchange( تبدیل کرنا) Pakistani Rupees 50000 to US dollars. How many US dollars will he receive? (1US dollars = 157 rupees)

Solution:

Amount to be changed = 50000 rs

Rate of 1 US dollar = 157 rs

Number of US dollars = 50000 / 157 = 318.47 US dollars Ans.

Example:

Change 75000 Pakistani Rupees in UK pounds ( 1UK pound = 218 rupees)

Solution:

Amount to be changed = 75000 rs

Rate of UK one UK pound = 218 rs

Number of UK Pounds = 75000/218 = 344.03 UK Pounds Ans.

Profit / Markup

Profit: (منافع)

When we deposit (جمع کروانا) money into a bank, the bank uses our money and in return pays an extra amount(اضافی رقم) along with actual deposit. The extra money which the bank gives for the use of our amount is called profit (منافع) on the deposit.

Markup: (مارک اپ)

When we borrow (قرض) money from bank to run a business, the bank in return receives some extra amount alongwith the actual money (اصل رقم) given. This extra money which the bank receives (وصول کرنا) is known as markup.

Principal amount: (اصل رقم)

The amount we borrow (ادھار لینا) or deposit (جمع کروانا) in the bank is called principal amount.

Profit / Markup rate:

The rate (شرح) at which the bank gives share to its account holders (کھاتہ دار) is known as profit / markup rate. It is expressed in percentage.

Period:(مدت)

The time for which a particular (خاص) amount is invested in a business is known as period.

How to calculate profit/markup: (منافع اور مارک اپ معلوم کرنا)

For calculation of profit/markup we use following formula.

Profit/markup = principal × time × rate

OR I = P × R × T

Now we solve some examples by using this formula.

Example:

Younis borrow Rs 65000 from bank at the rate of 5% for two (2) years. Find or calculate the amount of markup (مادک اپ) and the total amount to be paid.

Solution:

Example:

A student purchase a computer by taking loan from the bank on simple interest. The student took loan of Rs 25000 at the rate of 10% for 2 years. Find markup to paid and the total amount to be paid back.

Solution:

Calculate Principal Amount: (اصل زر معلوم کرنا )

The formula used in previous(سابقہ) examples, we will use same formula for principal amount as,

Profit/markup = principal × time × rate

Now

Principal amount = profit or markup/ ( Rate × time)

Example:

Find the principal amount is taken to bring in Rs 640 as profit at the rate of 4% in two (2) years?

Solution:

Example:

A man got some loan on which he has to pay Rs 3500 as markup at the rate of 10% for 3.5 years. Find the amount of loan.

Solution:

Calculate (profit/markup) rate: (منافع یا مارک اپ معلوم کرنا)

Profit/markup = principal × time × rate

Rate = markup / (principal amount × Time)

Example:

At what annual rate percent (%) of markup (مارک اپ) would the principal amount ( اصل زر) Rs 68000 become Rs 86360 in three (3) years ?

Solution:

Calculate the period (Time): (مدت معلوم کرنا)

Profit/markup = principal × time × rate

Period (time) = markup / (principal × rate)

Example:

A man got loan from bank at the rate of 3% per year for some period. In how much period ( کتنے عرصہ میں) his loan of Rs 65000 will become Rs 68900.

Solution:

Types of Finance:

Overdraft (OD): ( اوور ڈراٹ)

It is a borrowing facility (سہولت) provided by a bank to account holder to withdraw (نکلوانا) some amount in excess of his original account balance. In other words if there is no amount left in an account and the bank does not send a cheque back due to lack (کمی) of funds in the drawer's account, then this is called Overdraft.

Running Finance: (گردشی سرمایہ)

Running Finance is very similar (ایک جیسا)( to overdraft. The aim of running finance is to give a chance (موقعہ) -to the customers to withdraw more money that they actually have. Therefore it can be considered (تصور کرنا) as a credit facility which is meant for a credit limit with a variable interest rate. Usually (عام طور پر) the running finance is granted for a period of 1 year.

Demand Finance: ( ڈیمانڈ فنانس)

One can think of demand as a person's willingness (مرضی) to go out and buy a certain product. For example market demand is the total of what everybody ( ہر آدمی) in the market wants and is willing to pay for. To meet (پورا کرنا) these requirements banks have demand finance. Demand is a type of loan (قرض) that may be called in by the bank (or lender مقروض) at any time. It may be either short (مختصر)term or long term.

Leasing: (ٹھیکہ)

A lease is a contractual agreement (معاہدہ) between the lessee (user)(اثاثہ کرایے پر لینے والا) to pay the lessor (owner) for the use of an asset. It means the user rents (کرایہ) the land or goods rented out by the owner. The ownership of the leased asset during the leased period (اٹھیلہ کا دورانیہ) known as term remains with the lessor. Hire purchase is a method (طریقہ) of buying goods in which payments of purchase price is spread over specific term by payment of an initial deposit known as the down payment(ڈاون پے منٹ). It is explained with the help of examples.

Example:

A company gets house on lease for six (6) years. According to agreement( راضی نامہ) the company paid Rs 1000000 as down payment (ڈاون پیمنٹ) and shall pay Rs 20000 per month as rent. After three (3) years company shall increase the rent 3% . Find the total amount the lesser (owner) would get.

Solution:

Percentage: (فی صد)

The definition of percentage is ' out of hundred or per hundred (ہر سو پر)'. Percentage is denoted by %.

Profit and Loss: (نفع و نقصان )

The term profit can b understand as if the selling price (S.P)(قیمت فروخت) is higher than the cost price ( C.P)(قیمت خرید) then the profit occur. It can be written as

Profit (نفع)= sale price - Cost price Or

Profit = S.P - C.P

And if the cost price is higher (زیادہ) than the selling price then loss occur and it can be written as

Loss (نقصان) = Cost price - Sale price. Or

Loss = C.P - S.P

Percentage Profit and Percentage Loss: (نفع و نقصان فیصد میں)

Percentage Profit and Percentage Loss will be find by using following formulas.

Percentage Profit(نفع فیصد) = ( profit / cost price) × 100

Percentage loss (نقصان فیصد)= ( loss / cost price) × 100

Example:

John bought a motor cycle for 50000 rupees and sold it for 56000 rupees. Find his percentage profit?

Solution:

Example:

Aslam bought a piece of land worth Rs. 300000 and sold it for Rs. 240000. Find his Profit or loss Percentage?

Solution:

Discount:(چھوٹ)

The meaning of discount is to reduce (کمی) the price is an article from its market price which is also called list price or regular price. After reduction (کمی کے بعد) the amount is known as the sale price. The discount is the amount that we saved (بچت) in buying an article.

Discount (چھوٹ)= Market price - sale price

Example:

Alam bought some articles of worth Rs. 2500. He was allowed 15 percent (%) discount on his purchase. Find or calculate sale price of the said articles.

Solution:

Example:

The market price of an article is Rs. 1700. The sale price of that article is Rs 1360. Find the discount in percentage (%).

Solution:

Insurance: (بیمہ)

Insurance is a system of protecting or safeguarding (تحفظ) against risk or injuries. It provides financial protection for property(جایداد), life, health etc. against specified contingencies such as death, loss or damage and involving payment of regular premium(قسط) in return for a policy guaranteeing. The contract(معاہدہ) is called the insurance policy. The party bearing the rish is the insurer (بیمہ کمپنی) or assurer and the party Whose risk is covered is known as insured or assured.

There are many types of insurance including life, health, property and so.. We discuss only two types of insurance.

1- Life insurance
2- Vehicle insurance

Life insurance: (زندگی کا بیمہ)

Life insurance is an agreement (معاہدہ) between the policy owner and the insurance company for an agreed time period. Insurance company agrees to pay back a sum equal to original amount (اصل زر) and the profit at the end of agreed period or on the death or critical illness of the policy owner. In return the policy owner agrees (راضی) to pay regular installments of premium.

Example:

Hamza got a life insurance policy of Rs. 500000. Rate of annual premium is 4.5% of the total amount of the policy whereas the policy fee is at the rate of 0.25% . Calculate the annual premium of policy.

Solution:

Vehicle Insurance: (گاڑی کا بیمہ)

The vehicle insurance provides a protection against risks (خطرات) to the vehicle. The amount of policy in this depends upon the actual value (اصل قیمت) of the vehicle.

Example:

Abu Bakar got his motorcycle insured for one year. The price of his motorcycle is Rs. 50000 and the rate of insurance (انشورنس کی شرح) is 4.5 %. Find the amount of premium.

Solution:

Income Tax: ( انکم ٹیکس)

Income tax imposed (لاگو ہونا ) on the annual income of a person whose income exceeds a certain limit (خاص حد) which is determined by the government. The rules for income tax are amended (تعین کرنا ) by the government from time to time.

Exempt Income: (چھوٹ کی رقم )

Tax exempt Income is money on which a person does not have to pay tax (ٹیکس). In other words it is income (آمدنی) which is not subject to income tax.

Taxable income: (قابل ٹیکس آمدن )

Taxable income is the difference (فرق) of annual income and exempted income.

Taxable income = Annual Income - exempted income

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Ratio and Proportion

2021-03-11T05:10:00.006-08:00

Learning Outcomes

In this post the reader will be able to understand the following topics.

1- What is the Ratio?

2- Equivalent Ratios.

3- Common friction, numerator, denominator, antecedent and Consequent.

4- About proportion and it's two types

-Direct proportion and

-Indirect proportion.

5- Compound proportion

Ratio:

The numerical comparison between the two(2) quantities of the same kind is called ratio.

Now we make it clear with an example:

John and Noni bought a book for 60 dollars. John paid 40 dollars and Noni paid 20 dollars. Now we can compare the amount paid by John and Noni to find the relation between the two amount. This comparison can be done by the following two (2) ways or methods.

A) By finding their difference.

( 40 dollars - 20 dollars) = 20 dollars

The difference of 20 dollars does not state the comparison with the other amount. Therefore it is not proper way to show the comparison between two quantities.

B) By writing them into a fraction.

Amount paid by Noni / Amount paid by John = 20/40 = 1/2

This fraction is showing that for every 1 dollar of Noni ,John paid 2 dollars. This fraction shows the clear comparison between these two amounts. Therefore it is a better way to show the comparison between the two amounts. It can be written by putting a colon (:) between two quantities.

Noni John

1 : 2

And read as 1 is to 2 but it can be read as 2 is to 1 , because by changing the order of elements of a ratio the value is also changed. Therefore 1:2 and 2:1 are two different ratios.

In general we can write it as x:y is not equal to y:x . A ratio can also be written for more than two(2) quantities, i.e

3:4:5 or x:y:z etc

Antecedent and Consequent:

If we have a ratio 2:4 , than the first element of this ratio is called antecedent and the second element is called consequent.

2 is called Antecedent and 4 is known as consequent.

Example:

Write the quantities in the form of x:y:Z

500 grams, 800 grams and 1 kilogramme

500 : 800 : 1000 (1kg= 1000gms)

Each is divided by 100 than

5 : 8 : 10

Equivalent Ratios:

As we know that a ratio is another form of a common fraction. So there are same rule for finding the equivalent fractions can be used to achieve equivalent Ratios.

Rule : When both the elements of a ratio are multiplied or divide by same number the value of a ratio is not changed.

Example

Relationship between ratio and fraction:

In fact ratio is the simplest form of common fraction(کسر عام), in which numerator denotes Antecedent and the denominator denotes consequent.

Note:

A ratio has no unit. It is just a number that indicates how many times one quantity is greater than the other.

Proportion:

The relation (تعلق) of equality of two ratios is called proportion.

For example, 1:2 = 2:4 and 2:4 = 3:6 or so on..

Further we mak it clear with the following example:

John bought two cups at the rate of 25 dollar per cup. How much did John pay for two cups?

Cost of 1 cup = 25 dollars

Cost of 2 cups = 25×2 = 50

Now we can write it as

Cups cost

1:2 = 25:50

From the above two equivalent Ratios are indicating the relation between two different quantities of cups and their prices. We can say that the two ratios are proportional to each other which are denoted by a symbol :: and can be written as 1:2 :: 25:50 and read as

1 is to 2 is proportional(تناسب) to 25 is to 50.

In above proportion the first and fourth elements are called extreme of a proportion and the second and third elements are called mean of a proportion, It can be shown as below.

Now we solve some examples of proportion.

So if there are four quantities a,b,c and D are in proportion then these are written as a:b :: c:d

Actually it is a relationship between two Ratios a:b and c:d

Proportion is of two (2) kinds:

1- Direct proportion
2- Indirect proportion

Direct proportion:

Direct proportion is a relationship in which one quantity increases or decreases in a same proportion by increasing or decreasing the other quantity.

Example:

A washerman irons 2 shirts in 10 minutes. How many shirts can the washerman iron in one hour ?

Inverse proportion:

Inverse proportion is a relationship in which one quantity increases by decreasing the other quantity and same quality decreases by increasing the other quantity.

Example:

A project can be completed by 150 working in 40 days. But project manager call 30 more workers after 16 days. Now in how many days will the remaining work be completed?

Compound proportion:

The relationship between two (2) or more proportion is known as compound proportion.

We try to understand compound proportion with the help of examples.

Example:

If 35 workers dig 805 cm³ of earth in 5 hours, how much of the earth will 30 workers dig in 6 hours ?

Example:

8000 rupees are sufficient for s family which have 4 members for 40 days. For how many days 15000 rupees will be sufficient for a family of 5 members?

Example:

If 4200 men have sufficient(خوراک) food for 32 (thirty two) days at a rate of 12 hectogram per person ,how many men may leave(چھوڑ جانا) so that the same food be sufficient for 42 ( forty two) at a rate of 16 hectogram per man ( person).

Solution:

As the number of days increases the number of persons decreases. So it is an inverse proportion.

As the quantity of food increases the number of persons decreases. So it is also inverse proportion.

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Simplification

2021-03-08T04:36:00.006-08:00

Learning Outcomes:

After studying this post, readers will be able to:

1- Know that the following four(4) kinds of brackets

______ vinculum, (ونی کلم۔بار)

( ) parentheses or curved Brackets or round brackets, (خم دار بریکٹس)

{ } braces or curly brackets, (کرلی بریکٹس۔بریسس)

[ ] square brackets or box brackets, (باکس بریکٹس۔ مربعہ نما بریکٹس)

2- Readers Know the order of preference as,

______ , ( ), { } and [ ] , to remove (simplify) them from an expression.

3- Readers are able to recognize BODMAS rule to follow the order (ترتیب) in which the operations, to simplify mathematical expressions, are performed.

4- Simplify mathematical expressions involving fractions (کسر عام)and decimals(کسر اعساریہ) grouped with brackets using BODMAS rule.

اوپر بریکٹس کی چار اقسام بیان کی گیں ہیں۔ہم انکو اسی ترتیب کے ساتھ حل کرتے ہیں۔ مطلب سب سے پہلے بار والی یا ونی کلم والی رقم حل کریں گے اور اس کے بعد خم دار بریکٹس پھر کرلی برکٹس اور آخر میں باکس بریکٹس کو حل کرتے ہیں۔

Introduction:

We have already learnt simplification of an expression involving two or more of four operations but sometimes it becomes more important to simplify some operations before the others in an expression. For this purpose, we use brackets which indicate the order of simplification. There are four types o f brackets.

1- ________ is called a bar or viniculum

2- ( ) is called a round or curved brackets or parentheses.

4- [ ] is called box brackets or square brackets.

Some Rules

There are some rules to simplify (حل کرنا) an expression containing brackets.

Rule 1: When an expression (حسابی جملہ)contains only addition and Subtraction: work from left to right within the brackets.

Rule 2: When an expression contains only multiplication(ضرب) and division: work from left to right within the brackets.

Rule 3: When an expression contains (مشتمل) any three or all four operations: use the BODMAS rules.

BODMAS Rule:

Every one of us know that addition, subtraction, multiplication and division are the four (4) basic operations (بنیادی عوامل) of math. An expression which contains all these operations we use simplification rule(اختصار کا قانون). This rule tells us that we performed this simplification rule in the following order but after removing (Simplify) the brackets.

BODMAS

BO for Brackets Of
D for Division
M for Multiplication
A for Addition
S for Subtraction

So this simplification rule is called BODMAS Rule.

ہم جانتے ہیں کہ ریاضی میں چار بنیادی عوامل ہیں۔ جمع ۔تفریق۔ضرب اور تقسیم ۔ اس قانون کےمطابق یہ چاروں عوامل بریکٹس ہٹانے کے بعد ہم جسں ترتیب کے ساتھ حل کرتے ہیں وہ ترتیب ذیل میں درج ہے۔ سب سے پہلےتقسیم کا عمل پھر ضرب کا عمل پھر جمع کا عمل اور آخر میں تفریق کا عمل کیا جاتا ہے۔

Some Examples:

Example 01:

Example 02:

Example 03:

Example 04:

Example 05:

Example 06:

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Integers

2021-03-06T07:19:00.007-08:00

Integers:

( Learning outcome )

After studying this post, Readers Will be able to Know that ;

1- The natural numbers(قدرتی اعداد) 1, 2, 3, 4,5,6,......., are also called positive(مثبت) integers and the corresponding negative numbers (منفی) —1, —2, —3, -4, -5,...... ..., are called negative integers, and zero `0' is an integer which is neither positive(نہ ہی مثبت) nor negative(نہ ہی منفی).

2- Recognize integers. (صحیح اعداد کی پہچان)

3- Represent integers on a number line (نمبر لاین)

4- Know that on the number line any number lying to the right of zero (0) is positive (مثبت)and to the left of zero(0) is negative(منفی). Readers also know that to the right of another number is greater(بڑا), to the left of another number is smaller(چھوٹا).

5- Readers know that every positive integer is greater than a negative integer.

6- Readers Know that every negative integer is less than a positive integer.

7- Arrange a given list of integers in ascending (ترتیب صعودی)and descending order(ترتیب نزولی).

8- Define absolute or numerical value (عددی قیمت) of a number as its distance from zero on the number line which is always positive.

9- Arrange the absolute or numerical values of the given integers in ascending (تریب صعودی) and descending order.

10- Use number line to display:

(A) sum(جمع) of two or more given negative integers(منفی صحیح اعداد),

(B) Difference (فرق)of two given positive integers(مثبت صحیح اعداد),

11- Add two integers (with like signsایک جیسی علامت والے ) in the following three steps:

a) Take absolute values of given integers(صحیح اعداد),

b) Add the absolute values,

c) Give the result(نتیجہ) with the common sign,

12- Add two integers (with unlike signs مختلف علامت والے) in the following three steps:

a) Take absolute values of given integers,

b) Subtract (تفریق) the smaller absolute value from the larger,

c) Give the result (نتیجہ) with the sign of the integer with the larger absolute value.

13- Recognize subtraction as the inverse(الٹ) process of addition.

14- Subtract one integer from the other by changing (تبدیل کرنا) the sign of the integer being subtracted and adding according to the rules for addition of integers.

15- Readers recognize that :

The product (حاصل ضرب)of two integers of like signs is a positive integer.

The product (حاحل ضرب) of two integers of unlike signs is a negative integer.

Recognize that division (تقسیم) is the inverse process of multiplication.

Recognize (پہچان) that on dividing(تقسیم) one integer by another.

-if both the integers have like signs the quotient( حاصل تقسیم) is positive.

-if both the integers have unlike signs(مختلف علامات) the quotient is negative.

16- Know that division of an integer by zero '0 is not possible(نا ممکن).

Integers:

Positive Integers: مثبت صحیح اعداد

Every one is familiar with the numbers like 1,2,3,4,5,.... These numbers can be added(جمع), subtracted(تفریق), multiplied(ضرب) and divided (تقسیم) as the rules expressed in previous classes. These numbers are known as natural numbers(قدرتی اعداد). These numbers are also called or known as positive Integers.

Negative Integers: منفی صحیح اعداد

All of us know that the natural numbers which are starting form 1 , All the natural numbers are greater than zero (0). If we think is there any number which is less than zero (0)? We understand this by an example. If we subtract 4 from 7 then the result is 3 i.e. 7-4= 3. But what is the result if 4 - 7 = ???

So remember that there is an other type of numbers which are called negative numbers because if we subtract 4 - 7 then the result is -3. So there are two types of numbers ( by sign) positive numbers and negative numbers, further we can understand these numbers by a number line

Hance zero (0) is known as starting point(نقطہ آغاز), so these positive and negative numbers with zero are called integers.

Examples:

5-8 = -3 , 3-4 = -1 , 3-5 = -2 and so...

These numbers -1, -2 ,-3 are called negative integers(منفی صحیح اعداد).

We know that 1 is less than 2 and 0 is less then 1. In the same way 0 is greater than -1 and -1 is greater than -2.

So again we understand this, the whole numbers(مکمل اعداد) 0,1,2,3,4.... together with negative numbers ( -1,-2,-3,-4......) are called integers.

We can write them as;

Note:

Zero (0) is also an integer but it is neither positive integer nor a negative Integer.

Integers are also known as directed numbers because these numbers also represent the direction(سمت) as well as measurement(پمائش).

Ordering of Integers: صحیح اعداد کی ترتیب

We note that on a number line, the integers that lies to the left of another Integers is always(ہمیشہ) smaller and the integer that lies to the right of the same integer is always greater. For example zero (0) lies to the left of all positive Integers on the number line. So

0 ≺ 1,2,3,4,5,....

And the integer zero (0) lies to the right of all negative integers. So

0 > -1,-2-3,-4,......

We can write the above statements as

.....-5 < -4 < -3< -2< -1 < 0 < +1< +2 < +3< +4........

Absolute or Numerical value of an Integer: صحیح اعداد کی عددی قیمت

Numerical or absolute (عددی قیمت) value of a number is distance from zero (0) to that number on the number line. If we find the absolute or Numerical value of 4 and -4 ,we note that 4 and -4 both the numbers are at equal distance(فاصلہ) from zero (0). So both have equal absolute or Numerical value that is 4. We write this as

| +4 | = 4 and | -4 | = - ( -4 ) = 4

Now we solve some examples:

Find numerical values of following

-6 ,-8 , 12

So | -6 | = - ( - 6 ) = 6 numerical value of -6 is 6

| -8 | = - ( -8 ) = 8 numerical value of -8 is 8

| 12 | = 12 numerical value of 12 is 12

Addition of Integers: جمع

The Integers with like sign and unlike sign are added in the following way

Subtraction of Integers: تفریق

Subtraction is the opposite process of addition, the integers are subtract from each other as following,

Multiplication of Integers: ضرب

We should know that when signs (+,-,×,÷) are multiply each other then what will be the result

Two like signs multiply each other then

(+) × (+) = +

(-) × (-) = +

When opposite signs multiply each other then

(+) × (-) = -

(-) × (+) = -

Now we multiply the integers as following

Division of Integers: تقسیم

In division we follow the same rule of multiplication for signs.

Examples,

Division of an integer by zero (0) is impossible(ناممکن).

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Factors and Multiples

2021-03-03T19:05:00.004-08:00

Factors and Multiples (عاد اور اضعاف)

Student learning outcomes:

Define a factor عادas a number which divides the dividend completely مکمل طور پرleaving zero (0) remainder.
Define a multiple اضعاف as a dividend into which a factor can divide.
Define even numbers as the numbers which are completely divisible by 2 or Multiples of two (2)
Define odd numbers طاق اعداد as the numbers which are not completely divisible by 2 or the numbers which are not Multiples of two (2).
Define prime numbers مفرد اعداد(2,3,5,7,11,13,17,.....), the numbers which have only two factors, 1st is 1 and other is itself.
Define composite numbers مرکب اعداد ( 4,6,8,9,10,12....), the numbers which have more than two factors.
Students are able to Know that 1 is neither prime nor composite as it has only one factor which is 1 itself.
Students are able to Know that 1 is a factor of every number.
Students are able to Know that 2 is only even prime number where as all other prime numbers are odd.
Test by inspection whether the numbers 2,3,4,5,6,8,9,11,12,15,25 can divide a given number completely.
Define prime factorization مفرد تجزی , the process of factorizing a number into its prime factors.
Recognize the index notationقوت نماٸ شکل.
Factorize تجزیa given number and express ظاہر کرنا its factors اجزا in the index notation(power form).
Define HCF (highest common factor) عاداعظمas the greatest number which is common factor of two or more numbers.
Find HCF( highest common factor) of two or more than two numbers by prime factorization and Long division method.
Define LCM ( least common multiple) ذواضعاف اقلas the smallest number which is a common multiple of two or more numbers.
Find LCM( least common multiple) of two or more than two numbers by prime factorization and Long division method.

Solve real life problems related to HCF (highest common factor) and LCM(least common multiple).

Factors:عاد

We know that if a number is divided ( تقسیم کیا جاے) by another number and the remainder (باقی) is zero (0) then the 1st number is said to be divisible (قابل تقسیم )by 2nd number.

For example:

From the above examples it can be seen that the number 18 is divisible by 1,2,3,6,9 and also by 18. These numbers are known as the factors (عاد) of the number 18, i.e, the factors of 18 are 1,2,3,6,9 and 18.

Similarly we can find the factors of any other numbers as giving

The factors of 12 = 1,2,3,4,6,12

The factors of 15 = 1,3,5,15

The factors of 4 = 1,2,3,6,7,14,21,42

So a factor of a number can be defined as(عاد کی تعریف)

A number that divides(تقسیم کرنا) the given number completely(پورا) is called a factor of the given number.

Note:

Every number has at least two factors(ہر نمبر کے کم از کم دو عاد ہوتے ہیں) except one (1)(سواے ایک کے). One has only 1 factor which is one (1) itself.

Multiples: اضعاف

All the numbers which are divisible (قابل تقسیم )by another number are called the multiples of that number.

For example:

Multiples of 2= 2,4,6,8,10,.......

Multiples of 3 = 3,6,9,12,.......

Multiples of 5 = 5,10,15,20,25,30,....

In the given example as we say that the factors of 18 are 1,2,3,6,9,18, we can also say that 18 is a multiple of each of the number 1,2,3,6,9,18.

The multiples of

2 = 2,4,6,8,10,12,14,16,18,20,....

The multiples of 3 = 3,6,9,12,15,18,21,....

The multiples of 9 = 9,18,27,36,.....

Thus a number is said to be a multiple(اضعاف) of each of its factor. It can also be noticed that the multiple of a number is either greater(بڑا) than or equal to the number itself.

Types of Natural numbers: قدرتی اعداد کی اقسام

Natural numbers are classified(جماعت بندی) in two ways, either even (جفت) and odd (طاق)or prime(مفرد) and composite(مرکب).

Even numbers: جفت اعداد

The numbers which are divisible (قابل تقسیم) by two (2) are called even Numbers. We can also say that all multiples of two (2) are even numbers. It mean 2,4,6,8,10,12.... are all even numbers because all numbers are multiple of two (2). The set of even Numbers is denoted(ظاہر کرنا) by captital letter (بڑے حروف)E.

E = { 2,4,6,8,......}

Odd numbers: طاق اعداد

The numbers which are not divisible by two (2) are called odd(طاق) numbers. It can also be said that the numbers which are not Multiples(اضعاف) of two (2) are odd numbers, i.e 1,3,5,7,9,11,... are all odd numbers. The set of odd numbers is denoted by the capital letter O.

O = { 1,3,5,7,....}

Prime numbers:مفرد اعداد

A number having exactly two factors(ایسے اعداد جن کے صرف دو عاد ہوں), one (1) and the number itself is called the prime number, i.e 2,3,5,7,11,13,17,.... are all prime numbers. The set of prime (مفرد) numbers is denoted(ظاہر کرنا) by the capital letter P.

P = { 2,3,5,7,11,.... }

Note:

The number two (2) is only even prime number where as all other prime numbers are odd.

Composite Number: مرکب اعداد

A number having factors other than two (2) (ایسے اعداد جن کے دو سے زیادہ عاد ہوں)and itself is called a composite Number or we can say that the numbers having more than two factors are composite numbers,i.e 4,6,8,9,10,12,... are all composite numbers because each of these numbers has more than two factors(عاد). The set of composite numbers(مرکب اعداد) is denoted (ظاہر کرنا) by the capital letter C.

C = { 4,6,8,9,10,12,....}

Note :

The numbers 1 is neither prime (نا ہی مفرد) nor composite (نا ہی مرکب) Number because it has only one factor which is 1 itself.

Examples:

Factors of 56 = 1,2,4,7,8,14,28,56

Factors of 121 = 1,11,121

Factors of 36 = 1,2,3,4,6,9,12,18,36

Multiples of 3 = 3,6,9,12,15,18,........

Multiples of 9 = 9,18,27,36,45,54,.......

Multiples of 12 = 12,24,36,48,60,........ e.t.c

Factorization: تجزی

The process (عمل) of writing a number into its factors(عاد) is called factorization.

We know that a natural number can be expressed as the product (حاصل ضرب) of its factors as giving below.

32 =( 2 )× (2) ×( 2) × (2)×(2)

42 = 2 × 3 × 7

So if a number which can be expressed as the product of prime factors (مفرد اجزاے ضربی) is called prime factorization.(مفرد تجزی)

The prime factors of a number can be written in any order like

42 = 2×3×7

42= 3×2×7 or

42 = 7×2×3

42 = 7×3×2

But often write them in ascending order(ترتیب صعودی).

The prime factors (مفرد عاد) of a number can be expressed by using a tree like diagram is called Factor Tree Method.

24 can be factorize by factor tree method

Index Notation: قوت نماٸ شکل

Look at the prime factors(مفرد عاد) of following numbers.

49 = 7×7

81 = 3×3×3×3

32 = 2×2×2×2×2

125 = 5×5×5

So we can write(لکھنا) the prime factors of above given Numbers as.

49 = 7×7 = 7² ( square of 7)

81 = 3×3×3×3 =3⁴ (3 to the power of 4)

32 = 2×2×2×2×2 =2⁵ (2 to the power of 5)

125 = 5×5×5 = 5³ ( 5 to the power of 3 or cube of 5)

Highest common factor: عاداعظم

To find HCF there are two methods

1- prime factorization method

2- Long division method

HCF by prime Factorization Method:

Lets consider the numbers 72, 48 and 132 and try to find their HCF by prime Factorization Method.

HCF by long Division Method:

Let's try to find HCF by long division method of the numbers 928 and 324.

The last divisor( آخری تقسیم کرنے والا) is the highest common divisor of two numbers.

So the HCF (عاداعظم) of 928 and 324 is 4.

Least common multiple (LCM): (ذواضعاف اقل)

The least common multiple ( LCM ذواضعاف اقل) of two or more than two numbers is the smallest (چھوٹا)number which is the multiple of each of the given numbers.

Look at the multiples of 4 and 6 which are given as

Multiples of

4 = 4,8,12,16,20,24,28,32,36,40,....

Multiples of 6 = 6,12,18,24,30,36,42,.......

We note that 12, 24, 36,.... Are the common Multiples of 4 and 6 but the smallest among (درمیان) then is 12, Hance 12 is known as the least common multiple of 4 and 6.

There are two methods to find the least common multiple (LCM).

1- Prime factorization method ( مفرد تجزی کا طریقہ)

2- Division Method (تقسیم کا طریقہ)

LCM by prime Factorization Method:

To understand this method we can solve an example. Lets find LCM of 36, 48 and 56 by prime Factorization method.

So LCM of 36,48 and 56 = 2⁴ × 3² ×7

LCM = 16 ×9 × 7

LCM = 1008

LCM by division method:

To find LCM by division method which is simpler than the prime factorization method. To understand this we solve the following example.

LCM of 24, 36, 54 and 81 by this method.

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Representing Sets through Venn diagrams

2021-02-24T03:10:00.005-08:00

Representing Sets through Venn diagrams: وین ڈایگرامز

( venn diagrams were introduced by John venn a British philosopher )

A Venn diagram is a simple closed figure which shows the sets and their relationships.

In Venn diagrams we represent the universal set by a rectangle and the other sets are represented by a closed circle into the inner area of the rectangle. (which represent the universal set).

وین ڈایگرامز میں یونیورسل سیٹ کو مستطیل بکس میں ظاہر کیا جاتا ھے۔اور باقی سیٹوں کو سادہ بند دائرہ کی شکل میں یونیورسل سیٹ کے اندر ظاہر کیا جاتا ھے۔

Now we represent the different set operations in Venn diagrams.

Union of Sets:سیٹوں کا یونین

If there are two sets, S and D.

S ={1,2,3} , D ={ 1,2,3,4}

By observing elements of these both sets it is clear that set S is subset of set D and the union of these both sets is

{ 1,2,3,4}

In Venn diagram we draw both the sets in a universal set and draw the set S in inner area of set D because S is subset of set D.

ہم دونوں سیٹوں کو یونیورسل سیٹ کے اندر اس طرح ظاہر کریں گے کہ سیٹ ایس کو سیٹ ڈی کے اندر بنائیں گے کیونکہ سیٹ ایس سیٹ ڈی کا تحتی سیٹ ہے۔

The Venn diagram of this example can be drawn as

Pic.

In Venn diagrams we shade the region on the basis of our result.

If set D is subset of set S then Venn diagram is represented by (in union)

اگر سیٹ ڈی سیٹ ایس کا تحتی سیٹ ہو تو وین ڈایاگرام کچھ اس طرح سے ہو گی۔

Pic.

If two sets are Overlapping sets: اگر دو سیٹ متراکب سیٹ ہوں

In overlapping sets both sets have some common elements but not the elements of both sets are same.

متراکب سیٹ ایسے سیٹ ہوتے ہیں جن کے چند ایک ارکان ایک جیسے ہوں پر سارے ارکان ایک جیسے نہ ہوں ۔متراکب سیٹ کہلاتے ہیں۔

Example, A={ 1,2,3,4,5 } ,B ={ 2,4,6,8}

In union the result is { 1,2,3,4,5,6,8 }

So we shade the area of both sets

Pic.

If two sets are disjoint sets: اگر دو سیٹ غیر مشترک سیٹ ہوں۔

Disjoint sets have no common elements e.g A ={ 1,2,3,4} ,C ={ 6,7,8 }

In union the result of these sets is

{ 1,2,3,4,6,7,8 }

So we shade the area of both sets.

غیر مشترک سیٹ ایسے سیٹ ہوتے ہیں جن کا کوئی بھی رکن ایک جیسا نہ ہو

Pic.

Intersection of Sets: سیٹوں کا تقاطع

سیٹوں کے تقاطع میں ہم سیٹوں کے مشترکہ ارکان جواب میں لکھتے ہیں۔

If A and B are two sets and A is subset of B ,it means all elements of A is present in set B.

By taking intersection the result is set A, so Venn diagram of these sets as

Pic.

If set B is subset of set A then by taking intersection of these sets the result is set B.

The Venn diagram is

Pic.

If set A and set B is overlapping sets متراکب سیٹthen taking the intersection of these sets the result is the common elements of A and B.

The Venn diagram is

Pic

If set A and set B is disjoint sets غیر مشترک سیٹ then taking the intersection of these sets the result is a empty set so in Venn diagram we shade no area of these both sets.

جب دو سیٹ غیر مشترک ہوں تو ان کا تقاطع لینے سے جواب خالی سیٹ آ تا ہے۔ اس کا مطلب ہم کوئی علاقہ شیڈ نہیں کریں گے۔

The Venn diagram is

Pic

Difference of two sets: دو سیٹوں کا فرق

If set A and set B are two sets and set A is subset of set B, so when we take difference A - B ,the result is a empty set { }. In Venn diagram we shade no area.

The Venn diagram is

Pic

If set B is subset of set A and by finding A - B we get the elements of set A which are not present in set B so we shade the area of set A.

Venn diagram is

Pic

If set A and set B is overlapping sets متراکب سیٹ . If we find A - B the result is that elements of set A which are not present in set B.

If we find B - A then the result is that elements of set B which are not present in set A.

The Venn diagram in both conditions as

If set A and set B are disjoint sets غیر مشترک سیٹ then by taking A - B , the result is the elements of set A.

So we shade the area of set A. If we find B - A the the result is the elements of set B so we shade the area of set B.

The Venn diagram is

pic

Complement of a set: سیٹ کا کمپلینٹ

If U is a universal set and other set A , the complement of set A is A' = U - A

In result we get the elements of universal set which are not present in set A so in Venn diagram we shade the area of set U and leave the area of set A.

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Have a nice day.

Subset

2021-02-21T03:24:00.009-08:00

In this content we discuss the following terms:

Subset
Types of subset
Proper subset
Improper subset
Super set
Power set
Disjoint sets
Overlapping sets
Universal set
Difference between sets
Complement of a set
Union of sets
Intersection of sets
Commutative law in sets
Associative law in sets
Distributive law in set
De Morgan's law

Subset:تحتی سیٹ

If A and B are two sets i.e

A={1,2,3,4} , B ={1,2,3,4,5,6}

In these sets we can easily observe that all the elements of set A is present in the set B so we said that set A is subset of set B.

Subset can be define as " If every element of a set A is also the element of set B then the set A is called the subset of set B". And it is denoted by a symbol A ⊂ B.

اگر کسی سیٹ کے تمام ارکان کسی دیے گئے سیٹ میں موجود ہوں تو وہ اس دیئے گئے سیٹ کا تحتی سیٹ کہلاتا ھے۔

If we read this D ⊂ G

set D is subset of set G.

Note:

ہر سیٹ اپنے آپ کا بھی تحتی سیٹ ہوتا ہے۔

Every set is also a subset of itself.e.g

set G is subset of set G, so G ⊂ G

Set D is subset of set D , so D ⊂ D

Types of Subset:

There are two types of Subset

1- Proper subset

2- Improper subset

واجب تحتی سیٹ اور غیر واجب تحتی سیٹ

Proper Subset: واجب تحتی سیٹ

If D and F are two sets, and all the elements of set D is also the element of set F but there is at least one element of set F is not present in set D, then set D is called proper subset of set F.

اگر ایک سیٹ کے تمام ارکان کسی دوسرے سیٹ میں موجود ہوں اور دوسرے سیٹ کے کم از کم ایک رکن ایسا ہو جو پہلے سیٹ میں موجود نہ ہو تو پہلا سیٹ دوسرے سیٹ کا واجب تحتی سیٹ ہو گا۔

It is denoted by D ⊂ F.

Example:

A={1,2,3,4,5,6} , S={2,4,6}

So set S is subset of set A and written as S ⊂ A.

Improper subset: غیر واجب تحتی سیٹ

If there are two set like H and T and both set contains same elements and equal number of elements, in other words both are equal sets OR set H is subset of set T and set T is subset of set H then we say set H is improper subset of set T and set T is improper subset of set H.

ایسے دو سیٹ جن کے تمام کے تمام ارکان ایک دوسرے میں موجود ہوں۔ یا دو مساوی سیٹ آ پس میں ایک دوسرے کے غیر واجب تحتی سیٹ ہوں گے۔

Super Set: سپر سیٹ

If A ={1,2,3,4} and

B={1,2,3,4,5,6,7,8,9,10}

In this example set A is subset of set B and it is denoted as A ⊂ B.

But the set B is the Super Set of set A and it is denoted as B ⊃ A.

Set B contains all elements of set A and also other elements.

اگر ایک سیٹ دوسرے سیٹ کا واجب تحتی سیٹ ہے تو دوسرا سیٹ پہلے سیٹ کا سپر سیٹ کہلاتا ہے۔

Note:

1- Each set is an improper subset of itself.

2- Empty set has no proper subset of itself.

Because empty set has only one subset which is empty set and it is improper subset.

3- A singleton set has only one proper subset of itself.

4- All the subsets of a set in proper subset except itself.

Power set: پاور سیٹ

A set which contains all the possible subsets of a given set is called power set.

ایسا سیٹ جو دیے گئے سیٹ کے تمام ممکنہ تحتی سیٹوں پر مشتمل ہو

Example:

D ={ 1,2}

The Power set of D can be written as

P(D) = { {} , {1} ,{2}, {1,2} }

In is example first three subsets are proper subset and 4th is improper subset of set S.

Short cut formula to fine that a set contains who many subsets of itself:

Which is 2ⁿ ( n is number of elements the set)

For example a set which contains four elements and how many subsets of this set have, we use the formula.

2ⁿ = 2⁴ = 2×2×2×2. = 16

Disjoint Sets: غیر مشترک سیٹ

Two sets are said to be Disjoint if there are no common element between them.

If we take intersection between these two sets then the result is empty set.

ایسے دو سیٹ جن کا کوئی بھی رکن ایک جیسا نہ ہو غیر مشترک سیٹ کہلاتے ہیں۔

Example:

A={1,2,3} , D ={5,7,8} in this example we see there is no common element between them so set A and set D in Disjoint set.

Overlapping Sets: متراکب سیٹ

Two sets are said to be Overlapping Sets if there is at least one element is common between them but not all the elements are common. In other words they are not subset of each other.

ایسے دو سیٹ جن کے کم از کم ایک رکن آ پس میں ملتا ہو پر سارے ارکان ایک جیسے نہ ہوں متراکب سیٹ کہلاتے ہیں۔

Examples of Overlapping Sets:

S ={1,2,3,4} ,H ={2,4,6,8} are overlapping Sets

F={3,5,7,8,9} , G ={8,12,14,15} are overlapping Sets

Universal Set: یونیورسل سیٹ

Universal Set is that which contains all the possible elements of the sets under consideration.

Universal Set is denoted by U.In other words universal Set in the super Set of all the given set which are under consideration.

ایسا سیٹ جو زیر غور یا زیر بحث دیئے گے سیٹوں کے تمام ارکان پر مشتمل ہو یونیورسل سیٹ کہلاتا ہے

Example:

U={1,2,3,4,5,6,7,8,9,10,11,12}

A={1,3,5}

C={2,4,6,8}

D={3,4,5,6,7}

In this example set U in universal Set of set A,set C and set D.

Difference of two sets:دو سیٹوں کا فرق

If S and D are two sets then S difference D is the set which contains all the elements of set S which are not the elements of set D. It is denoted by

S - D or S/D.

And D difference S is the set which contains all the elements of set D which are not the elements of set S. It is denoted by D - S or D/S.

Note:

S - D is not equal to D - S

Complement of a set: سیٹ کا کمپلینٹ

If A is a set and U is its universal Set then the difference set U/A is called the complement of set A and it is denoted by A' .Set A' ( complement of A) contains all the elements of universal Set U which are not the elements of set A.

یونیورسل سیٹ سے کسی بھی سیٹ کا کا فرق اس سیٹ کا کمپلینٹ کہلاتا ہے۔

Example:

U={1,2,3,4,5,6}

D={2,4,6}. Then

D' = U/D ={ 1,3,5 }

Operations on set

Union of two sets: دو سیٹ کا یونین

If A and B are two sets them the union of set A and set B can be represented by A B and the result is a set which contains all the elements of set A or set B or both A and B.

دو سیٹ کا یونین پہلے سیٹ کے ارکان یا دوسرے سیٹ کے ارکان یا دونوں سیٹوں کے تمام ارکان پر مشتمل ہوتا ہے۔

Example:

A ={1,2,3,4} , B {2,4,6,7}

So A ∪ B ={1,2,3,4} ∪ { 2,4,6,7}

A ∪ B ={ 1,2,3,4,6,7} ans.

If we find B ∪ A then

B ∪ A ={2,4,6,7} ∪ { 1,2,3,4}

B ∪ A ={1,2,3,4,6,7}

So we observe the result that

A ∪ B = B ∪ A

This is called Commutative law w.r.t union

قانون مبادلہ بلحاظ یونین کہلاتا ہے۔

Intersection of two sets: دو سیٹ کا تقاطع

If A and B are two sets them the intersection of set A and set B can be represented by A B and the result is a set which contains all the common elements of set A and B.

دو سیٹ کا تقاطع ان دونوں سیٹوں کے مشترکہ ارکان پر مشتمل ہوتا ہے

Example:

A ={1,2,3,4} , B {2,4,6,7}

So A ∩ B ={1,2,3,4} ∩ { 2,4,6,7}

A ∩ B ={2,4} ans.

If we find B A then

B ∩ A ={2,4,6,7} ∩ { 1,2,3,4}

B ∩ A ={2,4} ans.

So we observe the result

A ∩ B = B ∩ A

This is called Commutative law w.r.t Intersection.

قانون مبادلہ بلحاظ تقاطع کہلاتا ہے۔

Associative Law w.r.t Union and Intersection:

قانون تلازم بلحاظ یونین اور تقاطع ۔

If S,D snd F are three sets then the Associative law can be written as

S ∪ (D ∪ F ) = (S ∪ D) ∪ F Associstive law w.r.t union

S ∩ ( D ∩ F) = ( S ∩ D) ∩ F Associative law w.r.t intersection

Note:

To find union / intersection of three sets first of all we find union or intersection is any two sets and then the union or intersection of the third set with the resultant set.

Distributive Laws:

قانون تقسیمی

If Z, C and V are three sets then Distributive law of union over intersection is

Z ∪ ( C ∩ V ) = ( Z ∪ C ) ∩ (Z ∪ V )

And Distributive law of intersection over union is

Z ∩ ( C ∪ V ) = (Z ∩ C ) ∪ ( Z ∩ V )

De Morgan's Laws: ڈی مارگن کے قوانین

If H and K are the sub set of a universal set U then

(H ∪ K)´ =H´ ∩ K´ and (H ∩ K )´ = H´ ∪ K´

( ' ) symbol represent the compliment of a set.

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disjoint set examples
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disjoint set Venn diagram
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overlapping sets definition
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Have a nice day.

SET

2021-02-18T03:30:00.017-08:00

Definition of set:

A set can be define as " collection of objects or elements or numbers is called Set but these objects/elements/numbers/ things are distinct and well defined.

مختلف اور واضح اشیاء کے اجتماع یا اکٹھ کو سیٹ کہتے ہیں۔

Examples of set:

Dinner set,Water set,Tea set,Geometry set,Set of flowers,Set of natural numbers,Set of prime numbers, etc

The idea of set is given by a mathematician whose name is George Cantor جارج کینٹر who was born in Russia on 3rd March 1848. He died in Germany on 6 Jan. 1918. He published an article in 1873 which give birth of set theory.

Set Notations:( سیٹ لکھنے کا طریقہ)

For recognition every thing is given or represented by a name like Hydrogen is denoted by H, Oxygen is by O similarly Set is also represented by English alphabets (us captital letter to give the name of a set) and the elements or things of a set is written within the curly bracket and these elements of a set is separated by commas.

Examples:

Set of Natural numbers can be written as

N={1,2,3,4,......}

Set of whole Numbers can be written as

W={0,1,2,3,4,5,.........}

Set of pet animals can be written as

P={cow,goat,horse,sheep}

Set of flowers can be written as

F={rose, Lilly,gladia, jasmine}

Now we discuss some important set :

N= set of natural numbers ={1,2,3,4.........}

Natural numbers are start from 1 and continue......

W=set of whole Numbers ={0,1,2,3,4.......}

If we add 0 before natural numbers then it becomes whole Numbers.

E=set of even numbers ={ 0,2,4,6,8,10,.....}

The numbers which are divisible by 2 are called even numbers.

O=set of odd numbers ={1,3,5,7,9,......}

The numbers which are not divisible by 2.

P=set of prime Numbers ={2,3,5,7,11,13,....}

The numbers which have two factors, Ist is 1 and other is itself

C= set of compound numbers ={4,6,8,9,10,12,....}

The numbers which have more then two factors .

Z=set of integers ={.....-3,-2,-1,0,1,2,3,4,5.......}

Q=set of rational numbers ={ p/q , p,q ∈ Z ∧q not equal to zero}

Expressing a Set:

(سیٹ کو ظاہر کرنے کا طریقہ)

There are 3 ways to expressing a Set.

1- Descriptive form

2- Tabular form

3- Set builder form

Descriptive form:

(بیانیہ طریقہ)

If we describe a set in words OR describe with the help of a statement then it is called descriptive form.

اگر سیٹ کو اس کی خاصیت کے مطابق الفاظ میں لکھا جائے تو اس کو بیانیہ طریقہ کہتے ہیں۔

Examples:

P= set of solar planets.

S= Set of colours of rainbow

P= Set of prime Numbers

Z= Set of integers

W= Set of whole Numbers

N= set of natural numbers

F= Set of four vegetables

D= Set of five games

H= Set of pet animals , etc

Tabular Form :

(اندرانی طریقہ)

If we write the numbers or elements of a set in curly brackets ( braces) and each number/element is separate by comma ( ,) then is known as Tabular form,

Tabular Form is also known as Roster Form.

اگر کسی سیٹ کے ارکان کو {} میں "٫" کی مدد سے الگ الگ کر کے لکھا جائے تو اس کو اندرانی طریقہ کہتے ہیں۔

Examples:

Set of natural numbers can be written in Tabular Form as

N ={1,2,3,4,5...........}

Set of whole Numbers can be written in Tabular/Roster form as

W ={0,1,2,3,4,5.........}

Set of english alphabets can be written in Tabular/Roster form as

X ={ a,b,c,d,...........,x,y,z}

Set Builder Form:

(ترقیم سیٹ ساز)

If we write a set by using common property of all it's members/elements then it is called set builder form.

اگر کسی سیٹ کے تمام ارکان کو ان کی مشترکہ خصوصیت کی بنیاد پر لکھا جائے تو اس طریقہ کو ترقیم سیٹ ساز کہتے ہیں۔

Examples:

Set of natural numbers can be written in set builder form as

N={ x/x ∈ N }

Set of whole numbers can be written in set builder form as

W = {x/x ∈ W }

Set of english alphabets can be written in set builder form as

X={ x/x ∈ English alphabets }

To write a set into Set Builder Form you should know some important symbols:

< less than

≤ less than or equal to

> greater than

≥ greater than or equal to

( if the open side of symbol is on right hand then it is LESS THAN and if the open side of the symbol is on left hand then it is known as GREATER THAN)

I such that

∈ belongs to

∧ and

∨ or

Types of Sets

1- Finite sets:

متناہی سیٹ

A set which have countable / finite number of elements is called finite set.

ایسا سیٹ جس کے ارکان کی تعداد محدود ہو

Examples of finite set:

S ={ 1,2,3,4,5}

W ={ 2,6,8,9,0,1}

A= A set of first five natural numbers

C = The set of vowels

H = students of seventh class

2- Infinite Set:

غیر متناہی سیٹ

A set which have uncountable/unlimited/infinite number of elements is called an infinite set.

ایسا سیٹ جس کے ارکان کی تعداد لا محدود ہو

Examples:

A ={1,2,3,4.............}

S= { 1,3,5,7,11,........}

K = a set of whole Numbers

M = a set on integers

F= a set of compound numbers

etc.

In a finite set we can easily tell/find the last element of the set but in infinite set we cannot tell/find the last element or number of that set.

3- Empty set:

خالی سیٹ

Empty means nothing, so empty set is that set which has no element or containing nothing.

Empty set is also known as NULL set.

Empty set is denoted by a letter ∅ ( faii) فائی which is Greek letter.

It is also represented by empty brackets like { } .

ایسا سیٹ جس میں کوئی رکن نہ ہو

Examples of empty set:

A = a set of horns of an ass.

A = a set of 200 feet tall boys.

Singleton Set :

یک رکنی سیٹ

Such a set which have only be element is called singleton Set.

ایسا سیٹ جس میں ایک ہی رکن ہو

Examples :

A ={ 1}

D ={2/3}

H= a set of whole Numbers less then 1

G= a set of prime Numbers which is also even numbers

Or G= {2}

Equal and Equivalent Sets

Equal Set:

مساوی سیٹ

Two sets are said to be equal Set if the number of both sets are equal in counting and the elements of both sets are same.

For example A={1,2,3} , B={3,2,1}, we can write A=B

In these both sets we see that each set contain three elements and these three elements are also same.

ایسا سیٹ جس کے ارکان کی تعداد بھی برابر ہو اور ارکان بھی ایک جیسے ہوں۔

Equivalent Set :

مترادف سیٹ

Two sets are said to be equivalent if the number of elements of both sets are equal in number but it is not necessary that the elements of both sets are same.

Two sets are equivalent if and only if one to one correspondence can be established b/w them.

ایسا سیٹ جس کے ارکان کی تعداد برابر ہو لیکن سیٹ کے ارکان کا ایک جیسا ہونا ضروری نہیں۔

ایسا سیٹ جس میں ایک سے ایک کی مطابقت قائم ہو جائے۔

Example. A={1,2,3} ,B ={ a,b,c}

These two sets are equal in number of elements but the elements are not same.

ایک مساوی سیٹ کو ہم مترادف سیٹ کہہ سکتے ہیں لیکن ایک مترادف سیٹ ،مساوی سیٹ ہو بھی سکتا ہے اور نہیں بھی ۔

Non Equivalent Set :

غیر مترادف سیٹ

The two sets are called non equivalent if one to one correspondence cannot established between these two sets.

Example:

A={1,2,3,4} , B={a,c,d,f,h,g} These two sets are non equivalent sets.

List

set
set of rational numbers is denoted by
set of rational and irrational numbers
set of whole numbers
set of rational numbers
set of irrational numbers
set of prime numbers
set builder form
sets
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roster form description of set
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tabular minimization
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infinite set
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empty set
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empty set is a subset of every set
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singleton set definition
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which set contains non equivalent numbers

Easy mathematics

BASES AND DIMENSIONS

BASES AND DIMENSIONS

(a) Bases.

(b). Dimension of a vector space.

LINEARLY DEPENDENT AND INDEPENDENT SETS

LINEARLY DEPENDENT AND INDEPENDENT SETS

THEOREMS ON LINEARLY DEPENDENT AND INDEPENDENT SETS

LINEAR COMBINATION OF VECTORS

LINEAR COMBINATION OF VECTORS

SMALLEST SUBSPACE CONTAINING A SUBSET

LINEAR SPAN

Vector Subspace

Vector Subspace

Theorem 1:

A non empty set W of a vector space V is subspace of V if(i) W is closed under additioni.e x,y ∈ W ⟹ x + y ∈ W(ii) W is closed under scalar multiplicationi.e α ∈ F, x ∈ W ⟹ αx ∈ W

Theorem 2.

A non empty subset W of a vector space V over a field F is a subspace of V if α, β ∈ F and x,y ∈ W ⟹ αx + βy ∈ W.

(i) W₁ = {(x₁ , x₂, 0) : x₁, x₂ ∈ R}.(ii) W₂ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 0}.(iii) W₃ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 1}.

Theorem 3:

The intersection of any two subspaces of a vector space V is also a subspace of V.

If { Wᵢ}ᵢ∈ᵢ be as arbitrary family of subspace of V, then ∩ᵢ∈ I Wᵢ is a subspace of V.

Let W₁ and W₂ be two subspace of a vector space V, then W₁ ∪ W₂ is a subspace of V iff W₁ ⊆ W₂ or W₂ ⊆ W₁.

If V₃(F) = {(a, b, c): a, b, c ∈ F} is a vector space. Then a set W₃ = {(a, b, c): α₁a + α₂b + α₃c = 0 for fixed α, α, α ∈ F} I'd a subspace of V₃(F).

Let V(F) be any vector space. Then V itself and the subset of V consisting of zero vector only are always subspace of V.

Prove that U = {(a, b, c) : a, b, c are rationals} is not a subspace of R³(R).

W = {(a, b, c) : a² + b² + c² ≤ 1}.

Vector spaces properties

LET V(F) be a vector space, then

(i). a0 = 0, where a ∈ F, 0 on both sides represent zero vector.

(ii). 0v = 0, where 0 on the left is scalar and on the right is a zero vector, v ∈ V

(iii). a(-v) = -(av) = (-a)v, where a ∈ F, v ∈ V

(iv). av = 0 ⟹ a = 0 or v = 0

(v). a(v₁ - v₂) = av₁ - av₂, where a ∈ F , v₁, v₂ ∈ V

(vi). av = bv ⟹ a = b, where v ( ≠0) ∈ F, v, v ∈ V

(vii). av₁ = av₂ ⟹ v₁ = v₂ where a ( ≠0) ∈ F, v₁, v₂ ∈ V

Vector space questions

If X is the set of all real valued continuous (differentiable or integrable) functions defined on some interval [ 0,1]. Then to show that X is a vector space over the field R with addition and scalar multiplication defined as follows:

Let S be set of all ordered pairs of real numbers. Define sums and scalar multiplication of pairs as follows : (x₁ , y₁) + (x₂ , y₂) = (x₁ + x₂ y₁ + y₂) and c (x₁ , y₁) = (cx₁ , y₁) Show that S is not a vector space.

Postulates of vector space

If P(x) denotes the set of all polynomials in one indeterminate x over a field F then to show that P(x) is a vector space over F with addition defind as addition of polynomials and scalar multiplication defined as the product of polynomial by an element of F.

Closure axiom for addition:

Closure axiom for scalar multiplication:

Vector spaces

Vector Spaces

VECTOR SPACE

Whole numbers

Cube and cube roots

Cube and cube roots

Learning Outcomes

Find the cube of a number.Know about the perfect cubes.Know about the cube root and how to find the cube root of a giving number.

Cubes:

Example: show that 8 is a perfect cube.

Example: show that 27 is a perfect cube.

Example: Is 216 a perfect cube?

Example: show that 343 is a perfect cube.

Example: find the cube of 5.

Example: find the cube of 3.

Example: find the cube of (1.2).

Cube Roots:

The symbol of cube root is ∛.

Properties of cubes of numbers:

Q1 : Is 512 perfect cube?

Q2: 81 is a perfect cube ?

Question 3: is 6859 perfect cube?

Question 4: 729 is a perfect cube?

Question 5: (1000/125) is a perfect cube?

Example: find the cube of (1.4)

Example: find the cube root of 15625.

Example: Find the cube root of 13824.

Frequently asked questions. (FAQs)

What is the cube of √ 3?

What is Cube Root of a Number?

What is Cube Root Used For?

How do you find cube root of a number?

How we find the cube root of 64?

What is a example of cube root ?

Find the cube root of 4?

What is the fast way to find cubes of given numbers?

What does cube root look like?

A non empty set W of a vector space V is subspace of V if
(i) W is closed under addition
i.e x,y ∈ W ⟹ x + y ∈ W
(ii) W is closed under scalar multiplication
i.e α ∈ F, x ∈ W ⟹ αx ∈ W

(i) W₁ = {(x₁ , x₂, 0) : x₁, x₂ ∈ R}.
(ii) W₂ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 0}.
(iii) W₃ = {(x₁ , x₂, x₃) : x₁ + x₂ +x₃ = 1}.

Let S be set of all ordered pairs of real numbers. Define sums and scalar multiplication of pairs as follows :
(x₁ , y₁) + (x₂ , y₂) = (x₁ + x₂ y₁ + y₂)
and c (x₁ , y₁) = (cx₁ , y₁)
Show that S is not a vector space.

Find the cube of a number.
Know about the perfect cubes.
Know about the cube root and how to find the cube root of a giving number.

Define what is a perfect square.
Identify a number is a perfect square or not.
Recognize the properties of perfect square.
Find the square root of the nature numbers by factorization method and division method.
difference between square and square root