Abstract Math: ∑£¢∞§¶•ªº–≠‘“πˆ¨¥†®´åß∂ƒ©˙∆˚¬…æ÷≥≤µ˜∫√ç≈

2011-12-31T11:15:00.000-08:00

Closed sets under addition.

BASIC SETS WHICH ARE CLOSED UNDER ADDITION:

if a and b are two elements in G then a+b is in G.

Natural Numbers: Are closed under addition.

a. The natural numbers: {1,2,3,....}. Adding two natural numbers, you get another natural number.
If a and b are in N then a+b is also in N.
Example: 1 and 2 are in N and 1+2 = 3 is in N. Closed under addition.

Whole Numbers: are closed under addition.

b. The whole numbers: {0,1,2,3,.....} Adding two whole numbers, you get another whole number.

If a and b are in W, then a+b is in W.

Example: 0 and 1 are in W and 0+1 = 1 is in W. Closed under addition.

Integers: Are closed under addition.

c. The integers: {...-3,-2,-3,0,1,2,3,...}. Adding two integers, you get another integer.

if a and b are in Z then a+b is also in Z.

Example: -2 and 8 are in Z and -2+ 8 = 6 is in Z. Closed under addition.

Rational Numbers: Are closed under addition.

Adding two rational numbers, you get another rational number.

If a and b, b,c, d ≠ 0, are elements in Q, than a/b +c/d are also in Q.

Example -1/2 + 1 = 1/2 is in Q. Closed under addition.

Irrational Numbers: Aren't closed under addition.

Adding two irrational numbers, you don't necessarily get an irrational number.

If a and b are in I then a+b is in I, is not satisfied.

Example: √2 + -√2 = 0 but zero is not an irrational number. Not closed under addition.

Real Numbers: Are closed under addition.

Adding two real numbers, you get another real number.

If a and b are in R then a+b is in R.

Example: √2 + 0 = √2 is in R. Closed under addition.

Complex Numbers: Are closed under addition.

(a + bi) + (c + di) = (a+c) + (b+d)i

Example: (3 + 3i) + (2 + 5i) = (3+2) + (3+5)i = 5+8i.

SETS WHICH ARE CLOSED UNDER MULTIPLICATION:

if a and b are two elements in G then a•b is in G.

SETS WHICH ARE CLOSED UNDER SUBTRACTION:

if a and b are two elements in G then a•b is in G.

SETS WHICH ARE CLOSED UNDER DIVISION:

if a and b are two elements in G then a•b is in G.

SUBGROUPS

2011-12-30T20:42:00.001-08:00

Check the link below. It might make it easy for anyone to understand the meaning of subgroups.

Subgroups
http://dogschool.tripod.com/subgroups.html

CAN LINEAR ALGEBRA DESCRIBE THE WORLD?

2011-12-23T12:09:00.000-08:00

Background

1. The Singular case in Linear Algebra.

When you have a system of equations in two dimensions. If the two lines are
parallel, the two lines have the same slope and they do not intercept. The
tow lines are increasing or decreasing, depending on their slopes, at the same
rate of change.
If we use a table of values from a linear equation, we can see this rate of change by observing the values of y and of x, as they increase or decrease. Sometimes the lines are increasing by two, by four, etc. If increasing by two then the line have a slope of two, if increasing by four, the line have a slope of 4 and so on.
In linear Algebra when two lines do not intercept the system is said to be singular. This word here
"singular." Was the one giving me the most trouble. Finally, I understand it with perfection.
In 3D, you not only have the case when two lines are parallel (actually two parallel planes). You also have other cases. Thesingular case increases with the number of dimensions increases. This is only a natural consequenceas with most things in existence, the level of complexity increases with as we move from a simple case to the more robust case.

2. The Vector case.

When I was in college I learned about vector addition and scalar multiplication. Back then I did not
know what was going on. I used to get so confused, when they talked about Linear Combinations and Unit vectors.
Well finally I have begun to understand that if you have any vector whatsoever in 2D for example, that is a point in the
coordinate plane. that is a (2,4) as an example, you not only have the coordinates of the point, but also, and very but very elegantly, a line from the origin to the point. Thus, we can thing about any point in two, three, fourth, etc. points in space forming a unique line, or vector from the origin. What a revelation!!!!

HOW TO WRITE IT

2011-04-22T10:06:00.000-07:00

At the Orsini

HOW TO WRITE IT

Writing an Essay on a math problem (one of the many ways out there)

INTRODUCTION
Write an introduction about the problem your are going to solve. You should do this introduction after you had finished writing the section on evaluation.

HEURISTICS:

1. Analysis:
Investigate the problem, learn as much about it, formulate a plan to solve it.

2. Synthesis:
Carry out your plan and find the solution.

3. Evaluation:
Check your solution using a different plan and compare it to your work in both analysis and synthesis.

CONCLUSION:
Write this section after writing your introduction.

Try this method with any problem you want to solve. Use college ruled lined paper. Fold the paper in have to write in two sections (front and back.) Leave a line between notes.

PROBLEM SOLVING HEURISTIC

2011-04-22T09:47:00.000-07:00

HEURISTICS: (This format can be used to solve any problem whatsoever)

1. Analysis:
Investigate the problem, learn as much about it, formulate a plan to solve it.

2. Synthesis:
Carry out your plan and find the solution.

3. Evaluation:
Check your solution using a different plan and compare it to your work in both analysis and synthesis.

Try this method with any problem you want to solve. Use college ruled lined paper. Fold the paper in have to write in two sections (front and back.) Leave a line between notes.

HOW TO SOLVE IT : FROM A BOOK BY G. POYLA

2011-04-21T23:31:00.000-07:00

INTRODUCTION
I just finished reading "How to solve it by G. Poyla."
What I learned.

SCHEMA 1:
Analysis or reflection.
When presented with a math problem, do first an analysis of the whole problem. Figure out how you are going to solve the problem. You can work forward or backward. If you work forward you will be working from the data to the unknown. If you are going to work backward you are going to work from the unknown to the data. Sometimes you can work forward and backwards at the same time.
After you have reflected on all you know and can do from your analysis, including doing a little research and had decided on some plan of attack. Your analysis is complete. Once you'd decided what you are going to try---use a formula, draw a table, make a drawing, do some computation, solve an equation, work forward, work backward, etc. Now you are ready to move into action and actually solve the problem--on paper.

SCHEMA 2
Synthesis or action:
Use a pencil and paper and carry out your plan to find the answer to your problem. After finding the answer you are ready to move to further reflection.

SCHEMA 3
Evaluation or further reflection.
Here you must restate how you solved the problem and whether or not you can now solve the problem in a different way.

CONCLUSION
Doing the Analysis, Synthesis and Evaluation, completes the circle, that is, completes the whole enchilada.

THINKING ABOUT MATH

2011-03-19T14:34:00.000-07:00

Last time I was writing about equivalence. After one year I have not changed my idea. The idea of equivalence stands. I have been extremely busy working on teaching math, thus the little time to write. I have been paying careful attention on how to write a doctoral dissertation. That is, how to conduct research and how to write an acceptable paper. I have found out the process simple enough. I even constructed a simple version of a dissertation myself. I have found out I can do my own research on anything and write a decent paper. I have created a small template of it. It all seems to simple to me. Lately I have been working on lesson plans. I have gotten here and there on abstract algebra but I have not been able to advance a step. I have reviewed my own calculus notes to not much surprise. It is abstract algebra which worries me everyday. I wish I had all the time in the world to dedicate all my attention to it.

UNIVERSAL SETS?

2010-09-10T20:00:00.000-07:00

Can we use the idea of a set when solving word problems?

Two sets are the equal if they contain the same elements. Two sets are equivalent if they contain the same number of elements.

It is possible to think of two equal sets when solving problems. One set A and one set B. In set A is the unsolved problem, while in the set B will be the solved problem.

This should be the way in which we should think when solving problems in general. It is a matter of putting the pieces together.

ANTECEDENTS VERSUS CONCLUSIONS

2010-08-21T21:02:00.000-07:00

It has been a long time since I last posted but, I have been pretty busy like always.

Einstein stated that the most simple formulas explain the most phenomena in the the universe or something like that.

What can be more simple than an "If P then Q statement."

Say you wanted to prove anything. You can proceed in this way.

Say you wanted to prove that a set A is equal to a set B.

What can you do?

According to tradition we must prove that A is a subset of B and then prove that B was a subset of A.

Let us put this into an "if p than q" statement:

Say if A is a subset of B and B is a subset of A [this is our P statement] then A = B [this is our Q statement.]

Now we have a reason to prove mathematical statements of this sort in this fashion.

Previously my teacher will say if you want to prove that two sets A and B are equal, you must prove that A is a subset of B and that B is a subset of A and you will be done.

But now I say if you want to prove that two sets A and B are equal, you must clearly see the

whole argument first and then prove that they are equal. Start by defining the mathematical statement for the antecedent P and then by defining the mathematical statement for the conclusion p and then you will see the whole mathematical argument more clearly.

Take for example 2x - 5 = 7

then 2x = 12

then x = 6.

say x = 6

then 2x = 12

then 2x -5 = 7

Where is my mathematical argument?

Say if (2x - 5 = 7 then x = 6 and if x = 6 then 2x -5 = 7 ) Then (x = 6 and 6 = x.) and my understanding is wholly, but my teachers never taught me in this way.

Restating:

If A [2x - 5 = 7] is a subset of B [ x = 6] and B [ x = 6 ] is a subset of A [2x - 5 = 7] then A = B.

Clearly in this situation the two sets are equal.

But more is to be said into this, we must say that the proof is by deduction as opposed by induction. Each statement in the argument is deduced from the previous statement. Deduction is from the particular to the general while induction is from the general to the particular.

One is to conclude from this example that the more proofs we construct in this way that we will have no other option than to conclude by induction that knowledge becomes more easily acquired if proceed in this fashion above. Induction in this sense is more like by "Experimentation." That, is the more we experiment the more we understand about our subject matter.

2009-07-08T23:38:00.000-07:00

After having explored some modular tables, my mind is empty. I think I need to recharge my batteries.

DEFINITIONS

2009-06-24T10:01:00.000-07:00

RINGS:
note: most elements in most rings do not have inverses?

WHAT ARE PROPER SUBGROUPS?

2009-05-25T20:15:00.000-07:00

The question asks for the definition of proper subgroups. A subgroup of a group has the same number of elements of the group whose elements belong to, but a proper subgroups contains less number of element the main group contains. It is said that the order of the subgroup must divide the order of the main group. In the case the order of the subgroup and the order of the main group are the same then the set is simply called a subgroup of the group.

CLICK HERE FOR ORIGINAL PICTURE

THERE ARE SUBGROUPS? REALLY?

2009-05-24T01:57:00.000-07:00

Accordingly, a subgroup is a group itself. It can be the group itself or some other group with a less order and with all the elements contained in the main group and the order of the subgroup must less or equal the order of the main group. That is, the order of the subgroup must divide the order of the group. The important thing to remember here is that subgroups are themselves groups satisfying the four operations of groups. Fair enough for now.

GROUP EXPLORER

THIS ONE IS EASY OR SO IT SEEMS TO ME: THE ORDER OF A GROUP.

2009-05-24T01:44:00.000-07:00

The order of a group is the number of elements in the group. The order of one of the elements in a group can be tricky to find unless we know the the group itself and the factorization of the group is also known--other wise forget it. I think we are talking about infinite groups here as opposed to finite ones. This one concept is easy enough for me to understand, however I can't wait to start finding the order of some groups and some elements of groups because I never understood before what my subtle professor was talking about during that unforgettable lecture on the order of groups and the order of the elements in a group.

AND NOW INTO THE TRIVIAL GROUP

2009-05-24T01:28:00.000-07:00

The trivial group contains only one element. In the case of addition we have the zero group {0} and in multiplication we have the multiplicative group {1}.

BUT THEN WHAT IS A QUASIGROUP?

2009-05-24T00:28:00.000-07:00

It is said that a quasigroup need not have an identity nor be associative. But what kind of monsters might quasigroups be? To name an example we can think of LATIN SQUARES in which multiplying any row or column you get the same product. This concept I have never hitherto contemplated and I can only now see the beauty of it. check it out.

NOW WHAT IS ACTUALLY A SEMIGROUP?

2009-05-23T23:38:00.000-07:00

As I read, a semigroup is nothing other than an associative groupoid. It is said that a semigroup does not have to have an identity. Now I am curious to know what is a groupoid and why there does not need to exist an identity element in a semigroup and what are they or how they look. As far as inverses is concern there need to be none. Thus the set of integers will not qualify here. for every element in the integers with the ecception of zero has an inverse and as we saw earlier, the set of the integers is a ring and futhermore an integral domain. So semigroups must be sets with no identity and no inverses. I am thinking of the set of natural numbers. Natural numbers have no inverses and we know that the associative propety will apply for addition. In the natural numbers there exist no identity for addition because the identity for addition is zero and zero only belongs to the whole nubmers, yet something else must a a semigroup with no inverses and no identity property. I will have to wait to see more on examples of semigroup to see their truly importance, but for now I am satisfy with how much I have undertood so far about groups, rings, semigroups, and on idempotents. My worry so far is knowing more about groups. As far as I can understand a semigroup with an identity is a groupoid, and the natural numbers under the operation of addition while it seems to be a semigroup is not a monoid.

WHAT IS THEN A MONOID?

2009-05-23T23:13:00.000-07:00

To my understanding a monoid is a semigroup. In a semigroup the only property missing from a group is the inverse property. The elements in a monoid, it is said, need not have inverses. But I do not know why at this point. I will need to read more to see what kind of monoids are out there. The monoid, I know, must have an identity, as there is always an identity in every group. In a monoid as in a group, we have the associative property, as well as, the closure property. Thus so far as I can understand a monoid is a semigroup but as the need for not necessarily carrying inverses is not yet clear to me. I suppose that if in a monoid there are inverses, this is only a bonus. What makes a monoid different from a group is simply that a group has inverses for sure while in a monoid inverses can be absent.

THEN WHAT IS A GROUP?

2009-05-23T22:57:00.001-07:00

Simply stated a group is a set satisfying the operations of closure, associative, identity and the inverse property.

SEE ORIGINAL

RESEARCH
GROUP THEORY AN ONLINE COURSE
DEFINITION OF A GROUP

IDEMPOTENTS AND NILPOTENTS

2009-05-23T22:25:00.000-07:00

GROUPS, RINGS, FIELDS AND INTEGRAL DOMAINS

THE ASSUMED OPERATION FOR Z (THE SET OF INTEGERS) IS ADDITION. I AM ONLY PROVIDING THE TABLES FOR MULTIPLICATION for Zn.

DISCUSSION

QUESTION: WHAT IS A RING?

ANALYSIS: RING

RING: (properties)

Addition properties
1-5 AN ABELIAN GROUP

1. Associative property of addition
2. commutative property of addition
3. identity property of addition
4. inverse property of addition
5. distributive property

Multiplication properties

ASSOCIATIVE RING 1-6
6. associative of multiplication (non associative rings also exist: octonions, Sloane's,A037292, )

OPTIONAL CONDITIONS FOR RINGS:

COMMUTATIVE RING: 1-7
7. commutative of multiplication
(non-commutative rings: Real-valued matrices, quaternions)

RING WITH AN IDENTITY: 1-8 APPLY
8. multiplicative identity ( Some authors required 8 for it to be a ring)
(with no multiplicative identity: Even-valued integers, ) There might be an identity for multiplication but sometimes there is none but if it exists it must be unique. Z6 is a commutative ring with an identity. in Z6 only 1 and 5 have multiplicative inverses. These two elements form form the group of units of Z6. Z6 is a typical ring.

RING WITH INVERSES 1-9 BECOMES A FIELD
9. multiplicative inverses (the integers Z have no multiplicative inverses) might not exist but if they do, they must be unique

SYNTHESIS: RING

For a ring is enough to be an Abelian group 1-5. 6 is the associative property of multiplication. Thus 1-6 will make it into an associative ring. 1-5 will make it into a non-associative ring. 7 will make it into a commutative ring. 8 will make it a ring with multiplicative identity, be this ring associative, non-associative, commutative or non commutative. 9 will make it a ring with multiplicative inverses. In the case 1-8 applies and ab = ba = 0 only if a = 0 or b = 0 will make it into an integral domain. An integral domain has no zero divisor; that is, if ab=ba=0 for a ≠ 0 and b ≠ 0. Finally in a ring -ab=a(-b)=-ab and also 0a = a0 =0 these two properties work as a result of 5. You can use 5 to prove both properties.
Z is a ring without multiplicative inverses. Most elements in most rings do not have inverses.

QUESTION: WHAT IS A FIELD?

FIELD:
1-9 must satisfied.
Q is a field. R is a field. C is a field. Zn is a field for n = prime. Zn is a commutative ring with identity. When n is composite Zn has zero divisors, and thus is not even an integral domain.

QUESTION: WHAT IS A SKEW FIELD?

ANALYSIS: SKEW FIELD

if 7 does not apply it is called a SKEW FIELD

SYNTHESIS: SKEW FIELD

A Skew field is therefore a field whose element are does not commute under multiplication.

QUESTION: WHAT IS AN INTEGRAL DOMAIN?

ANALYSIS:
An integral domain is a commutative ring 1-7 with an identity 1-8 in which ab=0 iff a = 0 or b = 0. An integral domain has no zero divisors, that is if ab=ba=0 iff a ≠ 0 and b ≠ 0. Z is an integral domain. Z is commutative. Z has an identity. Z has no inverses. Z has no zero divisors.

SYNTHESIS:
The set of integers is an integral domain. The set of integers is a commutative ring. It has an identity 1. The elements in the set has no inverses and no zero divisors.

INTEGRAL DOMAIN:
2Z has no zero divisors, but it is not an integral domain because it lacks an identity
GROUP:
A UNIT:
AN INVERSE:
A ZERO DIVISOR:
IDENTITY:
PRIME:
COMPOSITE:

EXAMPLES OF RING:
The Natural Numbers is not even a ring because addition does not have inverses.
Z, Q, R, and C are all commutative rings. They all have multiplicative identities. They are all integral domains. Q,R and C are fields. These are but a few of many fields.

ANALYSIS AND SYNTHESIS: MODULAR MATHEMATICS (CLOCK ARITHMETIC)

QUESTIONS: WHAT IS AN IDEMPOTENT? WHAT IS A NILPOTENT?

ORDINARY MULTIPLICATION TABLE:

THE SET OF Z (INTEGERS) FORM A FIELD
IT IS A COMMUTATIVE RING WITH IDENTITY 1
(INTEGER MODULO 0?)
THERE ARE NO ZERO DIVISORS. IN FACT, THE SET OF INTEGERS IS AN INTEGRAL DOMAIN, AS IT HAS NO ZERO DIVISORS. This integral domain is a ring only and not a field for in this system division is impossible. This is an example of ring that is not a field.

In integer modulo 0, the divisor is 0 itself. This means we would have to divide by zero to find out answers. However, division by zero is not allowed in ordinary arithmetic. Thus, I am let to conclude I must divide by 1. Dividing by 1 let me to the desired multiplication. For example, 5•5 = 25 and 25 divided by 1 is 25 and this seems to work whenever I multiply any non-negative integer by any other non-negative integer including multiplication by zero. I notice that there is only one idempotent in this system. that is one number satisfying the equation x•x = x, namely 1•1 = 1. Any other observation at this point will have to wait until I can synthesize any further analysis. I must mention that the rational numbers is a field because division is possible there and the integers is only a ring as division does not exists withing the set of integers. This is the reason why the integers does not form a field under multiplication.

INTEGER MODULO 1: (1 is neither prime nor composite)

THERE ARE NO ZERO DIVISORS.
THIS IS NOT AN INTEGRAL: There is no identity 1 ≠ 0 ;1•a = a = a•1
ACTUALLY THE IDENTITY HERE IS 0 AS 0•0 = 0 =0• 0 BUT THIS IS ONLY AN EXCEPTION, ALL OTHER UNITS ARE EQUAL TO 1.
0 ALSO BEHAVES AS A UNIT, BUT THIS IS ALSO AN EXCEPTION
THIS IS THE TRIVIA RING R= 0.

How interesting can be a clock with only one number? The number zero.

Here we could only multiply 0•0 to get 0 and nothing more. According to the rule of idempotent elements we have x•x = x or as I noted above 0 • 0 = 0. Also x exp n = 0 for some n > 0 in this case n=2 we get that 0 is also a nilpotent. There are no nonzero zero divisors. That is, for any a and b in the system ab = 0 for some a ≠ 0 and some b ≠ 0.
We will see the appearance of zero divisor in integer modulo 4 where 2•2 = 0, since 2•2 is 4 in regular multiplication but when dividing 4 by 4 we get 1 remainder 0; thus making (4 = 0 modulo 4) in the sense of explaining the meaning of clock arithmetic.
In integer modulo 1 we are supposed to divide by 1 after multiplying. Example 0•0 =0 and 0÷1 is zero (see ordinary multiplication above with 1 as a divisor only that in ordinary multiplication the elements in the set are infinite.) Also notice that 1 is neither a prime nor a composite number. the following integer modulo is a prime number. Integer modulo with prime numbers give rise to fields. I will explain this later.

INTEGER MODULO 2 ( 2 is a prime number)

THIS IS A FIELD: A COMMUTATIVE RING WITH AN IDENTITY
THERE ARE NO ZERO DIVISORS; this is an integral domain
1 HAS AN INVERSE AS 1•1=1= 1•1; ab = 1 = ba.
THE SET OF UNITS {1} IS THE SET OF NONZERO ELEMENTS THAT HAS A INVERSE
1 IS THE IDENTITY as 1•a = a = a•1

We still have no zero divisors. Two is a prime number. 1 and 0 are both idempotent. 0 is the only nilpotent. Notice there is no need to divide in this modulo by two as the product 1•1 = 1. In the following modulo we will use the concept of division as 2 • 2 is 4. Thus we will say 4 ÷ 3 = 1 or say 4 = 1 (mod 3).

INTEGER MODULO 3 (3 is a prime number)
THERE ARE NO ZERO DIVISORS.
As 3 is prime this is a field
1 has an inverse, 2 has an inverse
This field is an integral domain (including addition)
1 is the identity

Notice that 2•2 = 4 and 4 = 1 (mod 3); this is indeed the first time we need of division to find our answer.

So far as before we have that 0 • 0 = 0 and 1 • 1 = 1. Thus as before 0 is a idempotent as well as a nilpotent. The element 1 is only an idempotent. There are no zero divisors.

INTEGER MODULO 4 (4 is a composite number)
THE FIRST ZERO DIVISOR APPEARS, NAMELY: 2.
THIS IS NOT AN INTEGRAL DOMAIN AS IT HAS A ZERO DIVISOR
1 has an inverse, 3 has an inverse
{1,3} form a group of units for Z4
THIS IS NOT A FIELD AS 4 IS NOT PRIME
THIS IS A COMMUTATIVE RING WITH AN IDENTITY =1 (including addition)

Check this one out. This is the first case in which we encounter zero divisors. That is a•b = 0 yet a ≠ 0 nor b ≠ 0 that is 2• 2 = 4 = 0 (mod 4).
0 is a neilpotent, 1 is an idempotent, 0 is also an idempotent. Also this is the first time I distinguish 2 to be both a zero divisor and a nilpotent at the same time.

2 • 2 = 4 = 0 (mod 4)
2 • 3 = 6 = 2 (mod 4)
3 • 2 = 6 = 2 (mod 4)
3 • 3 = 9 = 1 (mod 4)

Notice that : The remainder is the answer of every product when divided by 4.

INTEGER MODULO 5 (5 is a prime number)
NO ZERO DIVISORS.
THIS IS A FIELD AS 5 IS PRIME
THIS IS AN INTEGRAL DOMAIN AS IT DOES NOT HAVE ANY ZERO DIVISORS
1, has an inverse, 4 has an inverse
{1,4} is a group of unit of Z5
1 IS THE IDENTITY FOR THIS FIELD (including the operation of addition of addition)

Notice that there are no zero divisors in the table. The only idempotents are 0 and 1 and the only nilpotent is 0.

INTEGER MODULO 6
THE SECOND SET OF ZERO DIVISORS APPEAR {2,3,4}
MAKING Z6 NOT AN INTEGRAL DOMAIN AS IT HAS ZERO DIVISORS
1 AND 6 ARE THE ONLY ELEMENTS HAVING INVERSES {1,6} FOR A GROUP OF UNITS OF Z6
Def. a is a unit iff ab=1=ba and 1≠0.
THIS IS A COMMUTATIVE RING WITH IDENTITY 1 IF WE CONSIDER BOTH: ADDITION AND SUBTRACTION FOR Z6

Zero divisors appeared in integer modulo 4. Here in integer modulo 6, zero divisors make their appearance again. From the table, 3 and 4 and zero divisors. 0,1,3, and 4 are idempotents. 0 appears to be the only nilpotent. Impotant is the fact that no zero divisor in the modulo is a also a nilpotent element.

INTEGER MODULO 7 (7 IS PRIME)
NO ZERO DIVISORS.
THIS IS A FIELD AND AN INTEGRAL DOMAIN (including addition)
1 is the identity (multiplication)
one is the only element that has inverse

The first thing to notice here is that there is no zero divisors. The only nilpotent is 0. There are two idempotents: 1, and 0.

INTEGER MODULO 8 (8 IS COMPOSITE)
THE THIRD SET OF ZERO DIVISORS APPEAR.
THIS IS NOT AN INTEGRAL DOMAIN AS IT HAS ZERO DIVISORS
NOT A FIELD AS 8 IS NOT PRIME
ONLY A COMMUTATIVE RING WITH IDENTITY 1
1,3,5,7, HAVE INVERSE
{1,3,5,7,} FOR A GROUP OF UNITS OF Z8

Notice that there are 3 zero divisors {2,4,6}. There are also two neilpotents {0,4}. There are also two idempotents {0,1}. Important too is the fact that 4 is both a nilpotent and a zero divisor.

INTEGER MODULO 9 (9 Is a composite number)
THE FOURTH SET OF ZERO DIVISORS APPEAR: {3, 6}. THIS TWO ZERO DIVISORS ARE AMAZINGLY ENOUGH BOTH NILPOTENT ELEMENTS.
NOT AN INTEGRAL DOMAIN AS IT HAS ZERO DIVISORS

As before 0 and 1 are both idempotents. 0 is also, a nilpotent. 3 is also a nilpotent. 6 is also a nilpotent. Notice that 3 and 6 are both zero divisors and nilpotents.

INTEGER MODULO 10 (10 IS NOT PRIME)
THERE ARE ZERO DIVISORS {2,4,5,6,8}. NON OF THIS ZERO DIVISORS IS A NILPOTENT. THE ONLY NILPOTENT IS 0. THE IDEMPOTENT ARE {0,1,5,6}. NOTICE THAT 6 IS BOTH A ZERO DIVISOR AND AN IDEMPOTENT. SO IS 5 ALSO.

INTEGER MODULO 11
THERE ARE NO ZERO DIVISORS
2 IDEMPOTENTS {0,1}
ONE NILPOTENT {0}
NOTICE THAT THE TABLE LOOKS VERY CLEAN.

IDEMPOTENT
NILPOTENT

USC MATH DEPARTMENT LINK AND RINGS

2009-05-18T23:34:00.000-07:00

You can find the mathworld link at the USC math department related links.

CLICK ON THE LINK BELOW: GO TO ALGEBRA. THEN TO RING THEORY: THEN TO DIVISION ALGEBRA.

YOU WILL FIND ALL ABOUT RING THEORY THERE. ENJOY!

http://math.usc.edu/mathematics/relatedlinks.html

MAKING SENSE OF MATHEMATICAL RINGS AND FIELDS

2009-05-17T13:06:00.000-07:00

https://www3.imsa.edu/programs/e2k/curriculum/68reflections.mod/teaching

Last night was a very productive night. As I went over my old math notes, I came across a lecture on rings. This has been a lecture I'd never understood completely. As I read on, I begun to understand what a ring was. First I understood that for a set to be a ring, certain math properties must apply withing the set. But it all was so confusing until I begun to analyze more carefully. I ask to myself what is this about and how can I understand it? Let's take a look at a most original set of numbers: the set of integers:

..........-1, 0, 1, 2, ............

I found out that this set is a commutative ring and by definition an integral domain as it does not contain any zero divisors. zero divisors came into existence in the study of modular mathematics.

An example of a zero divisor would be 2•2 = 0. Here 2 is a zero divisor as 0 divided by 2 will give us 2 in the modular set : (0,1,2,3) this is called the set of all integers mod4. To give an example if we multiply 2•0 we get 0. If we multiply 2•1 =2 ; 2•2 = 4 but there is no four in this set so 4 becomes 0. Similarly 2•3 = 6 but there is no 6 so 2. Therefore clearly 2 is a zero divisor in the sense that 2 is not zero and yet when multiplying 2•2 we get 0. This makes 2 a zero divisor. Of importance is the fact that in the set of integers there are no zero divisors. The only way we can get a•b = 0 is if a or b is zero. This makes the set of all integer an integral domain. An integral domain is a ring. In particular the set of integers is a commutative ring. The fact that the set of integers has no inverse element such that a• -a = 1 deprives the integers from being a mathematical field unlike the field of the rational numbers where 2/5 • 5/2 = 1. The inverse element in this case is the reciprocal fraction, something clearly missing in in integers which are void of fractions. Thus, finally I got to understand fields and rings.

WHAT IS A COMMUTATIVE RING?

THE SET MUST BE A COMMUTATIVE GROUP: That is the following properties for addition must apply:

CLOSURE PROPERTY OF ADDITION (Magma or groupoid)
1) The result of adding any two members of the set, must give rise to another member in the set. For example: in the set of all integers (a ring) adding 3 and 4 give 7. Clearly, 3, 4, and 7 are integers. Thus adding any two integers will give rise to another integer, i.e; the sum of any a and b is in the set of all the integers. This property is called the closure property. Let a and b be in the set then a+b is in the set.

ASSOCIATIVE PROPERTY OF ADDITION (Semigroup)
2) letting a, b and c be in the set will ensure that (a+b)+c =a+(b+c), that is the associative property applies. (This makes it a semi-group)

IDENTITY PROPERTY OF ADDITION (Monoid)
3) Given any a there exist an element 0 such that a+0 =a. It must be true that in the set you must find a zero element. Clearly we can find the zero in the set of the integers as by definition the set of the integers are the positive and negative natural numbers including zero.

INVERSE PROPERTY OF ADDITION (group)
4) This property says that given any a in the set, there must also exist an -a, and this is clearly true in the case of the integers as every positive integers has its own corresponding opposite negative integer. Example: 4+( -4) = 0

COMMUTATIVE PROPERTY OF ADDITION (Abelian Group)
5) Given any a and b we have that a+b = b+a and this is the commutative property. that is given any two integers we can commute them and get the same answer.

THE SET MUST BE A COMMUTATIVE UNDER MULTIPLICATION TO A COMMUTATIVE RING, OTHER WISE IT IS SIMPLY A RING. The following properties must apply:

CLOSURE PROPERTY OF MULTIPLICATION
1) Given any elements a and b then a•b must be an element in the set. Clearly given any two integers their product is also an integer. Example: 2•3 = 6.

ASSOCIATIVE PROPERTY OF MULTIPLICATION (Seudoring)
2) Given any elements in the set a, b and c; the associative property of multiplication must apply. This clearly is the case in the set of the integers. That is, (a•b)c = a(b •c) This is clearly true when multiplying any three integers. Example: (3•2)•5 = 3•(2•5) = 30.

IDENTITY PROPERTY OF MULTIPLICATION (Ring)
3) Given any element a in the set, a•1 =a = 1•a

COMMUTATIVE OF MULTIPLICATION (Commutative Ring)
4) Given any elements in the set a and b then a•b = b•a: Example -2• 3 = 3• -2 = -6

ZERO PROPERTY OF MULTIPLICATION (Integral Domain)
5) Given any a and b, a•b = 0 or b•a = 0 if and only if a = 0 or b = 0.

INVERSE PROPERTY OF MULTIPLICATION (Field)
6) For any a then a• -a = 1 = -a• a. (This is the definition for a ring to be a field.)

So you see these is pretty neat stuff considering that most people in the world have no idea whatsoever about abstract algebra. But as I always said: Understanding is directly proportional to the way in which material is presented to the world. Seeing what is possible and not possible in the world makes way lot of sense when we consider that not all teachers are philosophers nor all philosopher teachers.

AT VENICE BEACH CALIFORNIA THINKING ABOUT NOUMENA AND ABOUT PHENOMENA AND OTHER MENTAL INCURSIONS

2009-05-15T21:03:00.000-07:00

TAKING A NAP AT VENICE BEACH.

IN THIS STATE OF ABANDONMENT

2009-05-14T22:59:00.000-07:00

In this state of solitude of abandonment I might say,
only an abacus would make me happy.