Math Rescue

Algebra: Inverse Functions, Finding Inverses

2015-04-27T11:16:00.000-05:00

I'm so sorry this post is a week late. Life got busy. :-)

Now that we understand the definition of an inverse function and how to test them, we can learn how to find them. In this post, I'll share my favorite method of finding the inverse. How do you like to do it?

Step 1: Rewrite with the variables y and x. Use y in place of f(x), g(x), etc.

This is straight-forward. If you are given a function in terms of different variables, such as f(b) and b, change them to y and x. Use y for the dependent variable (usually f(x) or g(x), etc.) and x for the independent variable (usually already x, but may be another letter).

Step 2: Swap the independent and dependent variables.

Swap x and y. Shown below.

Step 3: Solve for y.

Using this new equation, solve for y. Shown below.

Step 4: Replace y with the inverse notation of the original function.

In this case, we replace y with f^(-1)(x) because the original function was f(x). If the original function were, say, g(b), then we would replace x with b, and y with g^(-1)(b). Shown below.

Step 5: Test the inverse.

Using the method we discussed in the previous post, test the inverse you found. Those steps are shown below.

Our answer passes the test of the theorem we learned about, so it is correct.

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Let's try another example. This one is a little more complicated, and will show you why it is important to test your answer at the end!

Steps 1-3: Rewrite in terms of y and x; swap the variables; solve for y.

All three of the first steps are shown below. Notice the last step of solving for y. When you take the square root of a variable, you must include both the positive and negative outside of the square root. Therefore, we actually have two answers for the inverse, this time.

Step 4: Rewrite with inverse notation of the original function.

Step 5: Test the inverse(s).

We must test both of our answers. First, we will test the positive version, shown below.

This answer passes the test. Now, we must test the negative version, shown below.

This answer passes the first portion of the test, but fails the second portion. Remember, both parts of the theorem must hold true in order for the two functions to be considered inverses. Therefore, g^(-1)(x) = - sqrt(x-1) is NOT an inverse of g(x) = x^2 + 1! We simply throw out this answer and only give the positive version, shown below.

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That's how I like to find the inverse of a function. :-) If you have another method you prefer, I'd love to hear all about it in a comment!

That concludes this algebra series on inverse functions. If you have a request for my next series, leave it as a comment on this, or any, post, or send it to me via e-mail (find my e-mail on the Contact page). I'd love to hear from you!

Algebra: Inverse Functions, Testing Inverses

2015-04-15T13:40:00.000-05:00

Remember that theorem I gave you in the Introductory post for this series?

We can use this theorem to test whether two functions are inverses of each other.

To test whether these two functions are inverses, we need to look at the two composite functions in the theorem above.

Step 1: Evaluate the first composite function in the theorem above, (f o g)(x).

To form a composite function, locate the variable (x) in the first function (circled in red in the image above). Insert the entire second function into this variable (circled in green, above, with arrow showing the insertion point).

Now, evaluate this new function, shown below.

If the first composite function equals x, continue to Step 2.

Otherwise, the two functions are NOT inverses.

In this example, the first composite function, (f o g)(x), does equal x. Therefore, we continue to Step 2.

Step 2: Evaluate the second composite function in the theorem above, (g o f)(x).

Form this composite function as before. This time, the entire first function (circled in red above, with arrow showing insertion point) is inserted into the variable placeholder of the second function (circled in green).

As before, evaluate this new function, shown below.

If the second composite function equals x, the functions ARE inverses.

If EITHER of the two composite functions do not equal x, the functions ARE NOT inverses!

In this example, the two functions are inverses of each other.

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Here's another example.

Begin with Step 1 as before.

The first composite function cannot be reduced any further than (3x^2 - 1), as shown in the image above. Therefore, it does not equal x. And therefore, the functions f(x) and g(x) are NOT inverses.

There is no need to proceed to Step 2. According to the theorem, both composite functions must equal x for the two functions to be inverses of each other.

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The composite functions need not be evaluated in the order given. You could evaluate the second composite function first. Sometimes, one of the two is easier to reduce than the other. If you get stuck trying to reduce one of them, try the other.

In the next, and final, post of this series, we'll discover how to find the inverse of a given function. :-) I hope you'll stick around for it!

Interesting Solution

2015-04-14T14:47:00.000-05:00

A reader, Richard Odland, contacted me with a different approach to puzzle I shared some time ago, "A Corny Question" on this post. (My solution can be found on the follow-up, Solutions post.)

I found his approach interesting, and received his permission to share it with you all!

Here is a link to a Google Document with his solution. :) Let us know what you think of it!

And thank you, Richard!

Algebra: Inverse Functions, Introduction

2015-04-01T14:00:00.000-05:00

Think of a function as a machine. It takes an input (x-value, independent variable), performs some operation (equation) on it, then spits out an output (y-value, dependent variable).

The inverse of a function is the undo button in the world of functions. It takes an output, performs an operation, and gives back the input that would have went into the original function machine.

There is a way to test whether two functions are the inverses of each other. It is actually a mathematical theorem, which says:

We will use this information in the next two posts, which make up this series on inverse functions. In the first, we will discuss how to use the above theorem to test whether two functions are the inverses of each other. In the second, we'll see how to find the inverse of a given function.

I hope you'll stay tuned!

Early April Mini-Series Preview!

2015-03-25T14:00:00.000-05:00

The next little series coming up in early April will be on finding and testing the inverse of a function. :-) Stay tuned!

If you want more updates, you can find the Math Rescue blog on Google+ and Twitter!

Algebra: Multiplying Polynomials, Method 3

2015-03-18T14:00:00.000-05:00

If you missed the introductory post to this series, you should go read it.

This is the last method I'll be sharing for how to multiply polynomials. I call this one the "Numerical" method because it is set-up like a standard multiplication problem with numbers. I do not find this method particularly quick (like "The Table" method), but it is organized and easy to understand because it builds upon a skill with which the majority of students are already comfortable.

Our first example is an easy one, as usual. :-)

Step 1: Write the polynomials in a column, aligning them at the right, with like terms stack (introduce missing terms with a zero coefficient), like you would a standard multiplication problem. Write the longer polynomial on top.

In this example, there are no missing terms. But, if there were, I'd introduce them (as in previous methods), with zero coefficients. Like terms should be aligned. Notice in the image above that the term (-2x) in the first polynomial is aligned in the same "column" as the (+x) in the second polynomial.

Step 2: Multiply the last term of the bottom polynomial by each term of the top polynomial, from right to left. Write these products below the solid line, aligned with the term from the top polynomial. This is just like a standard multiplication problem with numbers.

In this step, we are working with the last term of the bottom polynomial (circled in red in the image above). Make sure you include the sign along with the term when multiplying! First, this term is multiplied by the last term of the top polynomial (circled in green). The product is written below the term used from the top polynomial (underlined in green).

Continue multiplying the last term (red circle) of the bottom polynomial by each of the terms in the top polynomial, from right to left. These are shown in purple and blue, respectively.

Step 3: Multiply the next to last term of the bottom polynomial by each term of the top polynomial, from right to left. Write these products below the products from Step 2, aligning them one term from the right side (leave a space below the far-right term of the first row of products). This is just like a standard multiplication problem with numbers.

Continue in this same manner until every term of the bottom polynomial has been used. The number of rows in the "products" section (below the solid line) should be equal to the number of terms in the bottom polynomial.

Now, we are multiplying across the top polynomial using the second to last term of the bottom polynomial (circled in red in the image above). These products are aligned below the products from Step 2, but moved one "space" to the left. This ensures that like terms are aligned in the products! This should look very familiar to you as a standard multiplication problem.

In this example, the multiplying is complete. Every term of the bottom polynomial has been multiplied through the top polynomial. Notice, there are two rows of product polynomials below the solid line, which is equal to the number of terms in the bottom polynomial (2).

Step 4: Draw another solid line below the rows of product polynomials. Add these polynomials together. Like terms are added together. The final product polynomial is yielded. (If there are any terms with a zero coefficient, rewrite the product polynomial and remove them.)

This step is where like terms are combined to form the final product polynomial. Simply add the terms in each column (example, boxed in green, above). The final product polynomial is boxed in red.

This method can be difficult to explain, but it really does work exactly like multiplying numbers with multiple digits that is taught in elementary/primary school.

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And now for an example with missing terms and longer polynomials. :-)

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I hope you have found this series helpful. As always, if you have any questions, comments, or requests for future topics, leave a comment below. :-)

Algebra: Multiplying Polynomials, Method 2

2015-03-16T14:00:00.000-05:00

Read the introduction to this series here.

This second method is my favorite way to multiply polynomials. I call it "The Table" method or "The Box" method. Even way up in Calculus III in university, my homework and test pages would have little boxes in the margins wherever I had to multiply polynomials. Many students find this method easier to follow and more organized. While tutoring, my students rarely missed terms when using this method. However, tables can be confusing in their own right, so this method may not be for everyone!

Let's begin with a simple example.

Step 1: Set up a table with one polynomial across the top (one term per column) and the second polynomial down the left side (one term per row) (including all +/- signs!).

***Important! Introduce any missing terms with zero coefficients (in red on first polynomial, below). Doing this will ensure your like terms align properly when you multiply. With enough practice, you will be able to skip this, but it is very helpful when you first begin using this method!***

In the image below, you can see I have set up my table with the first polynomial across the top and the second down the left-hand side. I included the missing term of the first polynomial, giving it a coefficient of zero (in red).

Step 2: For each cell in the table, multiply the term for that column by the term for that row.

So, begin by multiplying x times x^2, which yields x^3. This cell is marked with a green circle. Continue multiplying in this way until every cell of the table is filled, as shown in the image below. Watch the signs on each term carefully!

Step 3: Beginning with the first cell, write out the product polynomial by combining like terms on the diagonal as you go.

Notice that like terms are collected on diagonals, illustrated with the green and yellow highlights in the image below.

As you write out your answer (boxed in red above), simply add/subtract like terms along each diagonal, working from the top left corner to the bottom right.

You can help ensure you do not miss terms by marking off the cells you have already written as you go.

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Here is another, slightly more complicated example, worked out in steps below:

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As you can see, this method is very compact, even with large polynomials.

There is one more method I'll be sharing with you. Look for it Wednesday, March 18th!

Algebra: Multiplying Polynomials, Method 1

2015-03-13T12:00:00.000-05:00

Read the introduction to this series here.

I call this first method "Term-by-Term." While tutoring, I found that most students attempted to multiply polynomials in this way. When the polynomials are small (two binomials, for example), this method is fast and easy to do. However, when the polynomials are large, it is cumbersome and lengthy.

Step 1: Multiply the first term of the first polynomial by each term in the second polynomial (including +/- signs!).

In the image below, first the operation shown by the red arrow is performed, then the operation shown by the blue arrow. The resulting products (underlined in red and blue) are written out as terms in the product polynomial.

Step 2: Multiply the second term of the first polynomial by each term in the second polynomial (including +/- signs!).

In the image below, the two operations are shown in green and purple. These products, underlined, are tacked onto the products from the previous step in the product polynomial.

Continue multiplying in this way until every term from the first polynomial has been used.

Last Step: Combine like terms and rewrite in standard polynomial format (with exponents in descending order).

In this example, there are no like terms to combine, so we simply need to write the terms in the correct order (shown boxed in red below).

You see, when multiplying two binomials, this method works very well. Some students may find it difficult to keep track of which terms they are currently multiplying. My suggestion is to try drawing the arrows, as I've done, while you multiply one term across the second polynomial. Then, erase those arrows and draw new ones for each successive term.

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Let's try multiplying two trinomials with this method...

The images below show each step.

And the final answer...

Hopefully, this example illustrated how cumbersome this method becomes with long polynomials. Each step takes up more and more space on the page. In the end, there can be dozens of terms to sort through and combine.

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The method next up in this series is much more conducive to multiplying longer polynomials. Look for it next Monday, March 16!

Algebra: Multiplying Polynomials, Introduction

2015-03-11T14:22:00.000-05:00

Many of the students I tutored were confused any time they had to multiply polynomials. Over the few years I tutored, I developed three methods for performing this arithmetic. It never failed that at least one of these three methods would "click" with someone who previously struggled to understand how this works.

I'll be sharing these three methods with you in a series of posts, in order to give each method enough attention. My names for the methods are as follows:

My personal favorite is "The Table" method -- it is most intuitive to me. Even up through Calculus III in university, my homework and tests would have a little table scrawled in the margins where ever I was required to multiply polynomials. :-)

I hope you find at least one of these methods helpful. I will also give a brief description of when each method may be preferred over another.

Please stick with me through this series. Let me know if you have any questions or comments, or if you have a request for a future post.

Find me on Google+ and Twitter!

Site Update: New Look!

2015-03-11T10:36:00.004-05:00

I've updated the look of the site with a new layout and theme. I hope this makes my posts more readable.

Also, the Math Rescue Blog has gone social!

Find me on GooglePlus!

Follow me on Twitter!

I look forward to connecting with you!

Notice: Update to "Trigonometric Identities" Post

2015-03-10T16:25:00.000-05:00

I was recently informed of two small errors in my Trigonometric Identities "cheat sheet" given in my blog post titled "Trigonometry: Proving Trigonometric Identities," originally posted on March 6, 2012.

I have since corrected those errors. I have updated the original post with a notice to that effect. I also corrected the errors in the accompanying Google Document "cheat sheet" I shared on that post. If you downloaded this "cheat sheet" before 03/10/2015, please re-visit the link given in the original post and download the new version.

I am sorry for an inconvenience.

Algebra: Dividing Polynomials, Part 2 of 2

2014-07-05T12:59:00.000-05:00

In Part 1 of this series, I explained standard long division of numbers, and we covered a very basic polynomial division problem. Today, we will cover three situations you may encounter: (1) when there is a "missing term" in the dividend, (2) when the leading co-efficient of the dividend is not 1, (3) when there is a remainder.
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Situation #1: How to set up the problem when there is a "missing term" in the dividend.

Look at the problem above. Do you see the where the "missing term" should be? If you look at the exponents of the dividend, you'll see that there is no term for x^2. In order to properly divide these polynomials, we need to reintroduce the missing term. We can do that without changing the value of the dividend by using a co-efficient of zero, as shown circled in red below.

Now that the problem is set up in a familiar way, you can solve it as we did the problem in the Part 1. The steps are shown in the image above.

You should be aware that there may be multiple "missing terms", as in the polynomial (x^5 - x^2 + x). In such a case, introduce enough terms that all exponents are covered. In this example, you would write (x^5 + 0x^4 + 0x^3 - x^2 + x + 0) under the division bar.
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Situation #2: The dividend's leading co-efficient is not 1.

The leading co-efficient in a polynomial is the co-efficient on the term with the highest exponent. In the polynomial (2x^3 - x^2 - 3x + 2), the leading co-efficient is 2 because it is on the term with the highest exponent, 3. All of the problems we have dealt with so far have had a leading co-efficient of one. Let's use the polynomial I've given you already to work another division problem.

The first step of this problem is just like the others we have solved. Begin by dividing the first term of the dividend (green circle) by the first term of the divisor (orange box) to get the first term of the quotient (red box). Then, multiply this first term of the quotient by the whole divisor (underlined in purple). This new polynomial (underlined in blue) is written below the dividend and subtracted. Do you see how this step is just like the first step of the other problems we solved? The only, very minor difference, is the initial division of the first terms of the dividend and divisor.

Now, you can finish solving the problem as you would any other, shown below.

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Situation #3: The remainder is not zero.

Consider the problem I have worked below. You will notice that at the end, I am left with (-5x+1) (red box). I cannot divide (-5x) by the first term of the divisor, (x^2) because (x^2) is larger. What do we do with this remainder?

Your instructor should explain how to represent this remainder. Two representations are given below, although the second looks more professional and is my preference.

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I hope this series has been helpful to you. If you have any questions about this topic or something you would like me to cover in the future, leave them below. Thanks!

Algebra: Dividing Polynomials, Part 1 of 2

2014-06-28T11:08:00.001-05:00

Dividing polynomials may appear more difficult to the student who is either unfamiliar/unpracticed in regular long division or is not comfortable with the rules of exponents. This topic will be covered in a two-part series. Today, we'll cover the basics of long division with numbers and a very simple polynomial division problem. In part 2, we will review three more complicated types of problems you may encounter.

First, let's review regular long division. If you are already familiar with this, you can skip ahead to the next portion of this post. Unfortunately, many kids get through third grade (or whenever they teach this now) without a good handle on the division process. Then, when those kids get to Algebra class, they struggle Big Time with dividing polynomials. The great news is that it is never too late to learn AND this is a useful skill, even in Calculus!

Let's define some terms. In a division equation, the number being divided is called the dividend. The number that is dividing is the divisor. The result of dividing the dividend by the divisor is called the quotient. If the divisor does not divide evenly into the dividend, there will be a remainder, which is usually expressed as a fraction.

Below is a typical long division problem being worked in steps. The explanation is given below the image.

We divide the divisor into the first number of the dividend (red circles). In this case, we say "2 goes into 5, 2 times." So, we write the step 1 quotient (bold circle) above the number in the dividend we used: we write 2 above the 5. [NOTE: When dividing polynomials, the divisor will never be larger than the first "term"/"place" of the dividend.] Then, we write the product of 2 and 2 (4, red square) below the 5 and subtract. We get 1 and then we "bring down" the next place in the dividend, in this case, a 0, making 10.

In step 2, our new dividend is 10 (green circle). We divide and see that "2 goes into 10, 5 times." So, we write the step 2 quotient (bold circle) above the next place of the original dividend (above the 0, which is what we "brought down" to make 10). Then, we write the product of 2 and 5 (10, green square), below the 10 and subtract. We get 0 and then bring down the next place in the original dividend, which is a 4.

We continue this process, which takes about 5 steps, seen in the image above, until we have a number left that is not divisible by our divisor. That is, 1 is less than 2. Therefore, we call 1 our remainder (bold blue circle). We write the remainder (blue circle) next to the quotient and show it over the divisor (purple circle) as a fraction. The entire number 25,215 1/2 (bold pink box) is the answer (quotient) to the division problem.

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We can apply these same general steps when we divide polynomials. You should be familiar with exponent rules. If you aren't, check out this website that will give you a good overview.

Let's begin with a pretty basic example. Divide (x^3 + 5x^2 + 7x + 2) by (x + 2). After we set up the problem, our first step is to divide the first term of the divisor (blue box) into the first term of the dividend (blue circle). Here, that means we are dividing x^3 by x. According to the rules of exponents, if the base is the same (x is the base here), you simply subtract the exponents. Since 3 - 1 = 2, x^3 divided by x is x^2. We write this first step quotient (red circle) above the first term of the dividend (blue circle).

Then, we multiply the first step quotient (red circle) by the entire divisor (underlined in green). We get x^3 + 2x (underlined in orange), which we write below the dividend. Then, we subtract terms. I suggest that you write (x^3 + 2x) in parentheses as I have done so that when you subtract, you remember to subtract both terms. This will be helpful when some of the terms may be negative (we will cover that in the next post of this series). After subtracting, we have 3x^2. We then bring down (arrow) the next term of the dividend (with its sign) to set up step 2.

You can see step 2 in the image below. This time, we divide the first term of the divisor (blue box) into the first term of the new dividend at the bottom (blue circle). That is, divide 3x^2 by x, which yields 3x (red circle). Then multiply this second step quotient (red circle) by the entire divisor (underlined in green), yielding the product (3x^2 + 6x), underlined in orange. We write this below the second step dividend we used, and subtract terms. This leaves us with (x + 2) and we can move onto step 3.

Look at the image of step 3 below. It follows steps 1 and 2 exactly. Can you follow along?

After subtracting at the end of step 3, we are left with a remainder of zero. That means our divisor divides evenly into our dividend, and our final quotient is x^2 + 3x + 1.

You've just walked through a basic polynomial division problem with me. Join me next time for part 2 of this series where we will look at the following situations: (1) when a term is "missing," (2) when the leading co-efficient is not 1, and (3) when there is a remainder.

As always, feel free to leave any questions in the comment section below. If you have ideas for future posts, feel free to leave them here as well!

Site Update

2014-06-23T13:43:00.003-05:00

Wow, guys! It's hard to believe it's been nearly a year since I've posted anything here. But, I promise, I have a great excuse. ;-)

I had a baby in early April. :-) I spent the first four months of that pregnancy feeling nauseous most of the day, especially when I was on the computer. Since then, life has just gotten busy for me.

However, I am getting back into my studies and hoping to bring more content to you all. I've worked through my Trigonometry book and have a few ideas for posts in that subject. And now I've started on my Calculus book.

As always, if there is a specific topic you would like me to cover, don't be too shy to ask! You can do so in the comments on any post or by e-mailing me at

m a t h r e s c u e {at} h o t m a i l . c o m

(without the spaces). I'd love to tackle any problem or topic you've got in mind. :-)

Math Brain Teasers: The Solutions

2013-07-17T13:03:00.000-05:00

As promised, here are my processes of finding the solutions to the two math brain teasers presented in the previous post. Did you find the answer by another means? I'd love to hear about it in the comments below!

Coin Counting
Jones gives Smith as many coins as Smith has. Smith then gives Jones as many coins as Jones has remaining. Now Jones has 36 coins, and Smith has 42 coins. How many coins did each of them have to begin with?

These word problems are easiest to solve by setting up math equations. First, we need to define a couple of variables: let j = the number of coins Jones started with, and let s = the number of coins Smith started with.

Now, we can begin to build equations. Consider how many coins Jones has at the end: 36 coins. Two things happen to get him from j coins to 36 coins. (1) Jones gives Smith as many coins as Smith has. How many coins is that? We've defined Smith's starting amount as s coins. After this transaction, Jones has (j - s) coins remaining. (2) Smith gives Jones as many coins as Jones has remaining (after giving Smith s coins). We've determined Jones has (j - s) coins left. Our final Jones equation looks like this:

Next, we need to write the Smith Equation. Smith winds up with 42 coins. Two things happen to get him from s coins to 42 coins. (1) Jones gives Smith as many coins as Smith has. Smith has s coins. Jones gives him s coins. Therefore, Smith now has (s + s) coins. (2) Smith gives Jones as many coins as Jones has remaining. We already determined that Jones has (j - s) coins remaining after giving Smith s coins. Therefore, Smith gives Jones (j - s) coins. Now Smith has (s + s) - (j - s) coins. Our final Smith equation looks like this:

Finally, we need to solve this system of equations. I like the elimination method.

A Corny Question
A merchant has 21 sacks of grain: seven full, seven half-full, and seven empty. He wants to divide them equally among his three sons. How can he do this--without transferring any portion of grain from sack to sack--so that each son will have not only an equal quantity of grain, but also an equal number of sacks?

This one stumped me as I tried to consider the full, half-full, and empty sacks of grain. I finally made progress by defining a weight for a full sack of grain. I chose an arbitrary 10 pounds. Next, I tried using equations to solve this puzzle, but didn't get very far--I struggled with defining variables, having too many variables, and then going 'round in circles with equations that weren't working. Finally, I used trial and error, and I was able to come up with the correct answers.

Using my arbitrary 10 lbs for a full sack of grain, we see that the merchant has a total of 10(7) + 5(7) + 0(7) = 105 lbs of grain among the 21 sacks. Therefore, each son needs to receive 105 / 3 = 35 lbs of grain and 21 / 3 = 7 sacks. I set up a table to begin giving the sons sacks of grain.

First, I started by giving the first son (boxed in green) the maximum number of full sacks he could get. Dividing 35 by 10 gives 3 and 1/2. So Son1 can get 3 full sacks, which is 30 lbs of grain. In order to give him 35 lbs of grain, he needs only 1 half-full sack. Then, in order to give him 7 sacks, he needs 7 - 3 - 1 = 3 empty sacks. His totals are perfect: 35 lbs of grain and 7 sacks. There are enough sacks of grain for Son2 to get the same as Son1: 3 full, 1 half-full, and 3 empty.

For Son3 (boxed in orange above), there is only 1 full sack of grain left (7 - 3 - 3 = 1). That gives him 10 lbs of grain. In order to make 35 lbs of grain, he needs 25 more pounds (35 - 10 = 25) or 5 half-full sacks (25/5 = 5). That is perfect because there are exactly 5 half-full sacks left (7 - 1 - 1 = 5). Now he has 1 full sack and 5 half-full sacks. In order to have 7 sacks total, he needs 1 empty sack. It's perfect again! There is exactly 1 empty sack left (7 - 3 - 3 = 1). We can check the first row by totaling the number of each type of sack and the total weight.

Therefore, solution 1 is as follows: Son1 and Son2 each get 3 full, 1 half-full, and 3 empty sacks; Son3 gets 1 full, 5 half-full, and 1 empty sacks.

We need to check for other possible solutions. Let's try giving Son1 (boxed in purple) one less than the maximum number of full sacks, that is only 2 full sacks. Then, in order to have 35 lbs of grain, he needs 15 more lbs or 3 half-full sacks (15/5 = 3). In order to have 7 sacks total, he needs 2 empty sacks (7 - 2 - 3 = 2). Let's give this same set-up to Son2.

For Son3 (boxed in blue), there are 3 full sacks left (7 - 2 - 2 = 3), giving him 30 lbs of grain. There is 1 half-full sack left (7 - 3 - 3 = 1), which gives him 35 lbs--perfect! Finally, there are 3 empty sacks remaining (7 - 2 - 2 = 3), which is the exact number he needs in order to have his 7 sacks total (7 - 3 - 1 = 3). Once again, we can check our solution by totaling the number of each type of sack and the total weight.

Solution 2, then, is as follows: Son1 and Son2 each get 2 full, 3 half-full, and 2 empty sacks; Son3 gets 3 full, 1 half-full, and 3 empty sacks of grain.

We can try for more solutions. In the third row (brown box), you see that I gave Son1 only 1 full sack this time. Then, he needed 5 half-full sacks to make 35 lbs and 1 empty sack to make 7 sacks. I can't give the same to Son2 because there are not enough half-full sacks (5 + 5 = 10 > 7). I can't give Son2 only 2 full sacks because then he would need 3 half-full sacks, and there are not enough for that either (5 + 3 = 8 > 7). If I give Son2 3 full sacks and continue on with the process, I find the solution is the same as Solution 1, just re-arranged a bit.

Lastly, I can try giving Son1 no full sacks (not shown in image). Then, I have to give him 7 half-full sacks in order for him to get 35 lbs of grain (35/5 = 7). Then he receives no empty sacks because he has his total of 7 sacks. After that, there are no more half-full sacks to give. There is no way to give any son 35 lbs of grain in only full and empty sacks because 10 does not evenly divide into 35. Therefore, there are no other solutions.

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I hope you found these puzzles entertaining, as I did. Did you solve both correctly? Let me know in the comments! I would also love to read any alternate methods of solving them that you might have come up with--just post in the comments below!

Math Brain Teasers: The Puzzles

2013-07-16T11:14:00.000-05:00

Any time I come across a math-y brain teaser, I just have to try to solve it. That's what happened when I read through the latest issue of Reader's Digest (August 2013) called "Fun and Games." I thought it would be fun to share these little teasers with you. Tomorrow, I will reveal the answers along with my process of finding the solutions, so make sure you subscribe to e-mail updates or check back then! These puzzles are the property of Reader's Digest.

Coin Counting

Jones gives Smith as many coins as Smith has. Smith then gives Jones as many coins as Jones has remaining. Now Jones has 36 coins, and Smith has 42 coins. How many coins did each of them have to begin with?

A Corny Question

A merchant has 21 sacks of grain: seven full, seven half-full, and seven empty. He wants to divide them equally among his three sons. How can he do this--without transferring any portion of grain from sack to sack--so that each son will have not only an equal quantity of grain, but also an equal number of sacks?

Discrete: Propositional Equivalences

2013-05-01T10:47:00.000-05:00

Today in Discrete Math, we are looking at problems dealing with propositional equivalences. We will cover logical equivalences and propositional satisfiability.

The most fool-proof, though time-consuming, way to verify logical equivalences is by construction of truth tables. This method is sufficient for small compound propositions, but quickly grows out of hand for longer propositions. Also, it requires little thought (which is a large part of its appeal), but reasoning through equivalences will make you more familiar with the laws than truth tables will. Your professor may even disallow the use of truth tables as “proof” of your assertions.

In problem 1 only, we will use a truth table. The process requires little instruction, as it is so simple.

Begin the table with a column for each proposition and enough rows for every combination of truth values for those propositions. In general, 2^n rows are needed to write a truth table for a compound proposition with n base propositions (plus 1 heading row). In this example, there are 3 base propositions, so we will start with 3 columns and 2^3 + 1 = 9 rows. Fill in the heading row and the 8 rows below with every possible combination of truth values for the three propositions (below).

Begin with the left side of the equivalence (part a). Add a column to the table for every operator, including every negation, conjunction, and disjunction. Here, we have two operators (two disjunctions), so we need to add two columns to our table (below). Determine the truth values along each column. Recall that a disjunction is only truth if both propositions are false (red arrows). It is easier to first fill in all the false values (red circles), then add in the truths afterward. Blue arrows and circles refer to the last column, where the only false value appears at the bottom.

Next, create a second truth table using the right side of the equivalence (below).

The final step is to compare the far right columns of each truth table. If the truth values are exactly the same,

then the two compound propositions are equivalent. These are, as shown below.

Part (b) of this problem is worked the same way. Try it on your own. :)

Another use of truth tables is for determining whether a compound proposition is a tautology. A tautology is a compound proposition that is always true. Therefore, the truth table of a tautology will have only T entries in the far right column.

Reasoning through logical equivalences will better help you learn the laws and basic equivalences. Let’s try an example.

Problem 2 is actually asking that you verify that the contrapositive of a conditional statement is equivalent to the original statement. To show that two compound propositions are logically equivalent (without the use of truth tables), we need to show that both sides are true for exactly the same combinations of truth values of the propositional values. Alternately, we can show that both sides are false for the same combinations.

I will call the conditional statement side A (or the left) and the contrapositive side B (or the right). First, we must determine which route we will take, either when side A is true or when it is false. You should recall that a conditional statement is true in every case except the one in which (T -> F), that is, when p is T but q is F.

We have two choices: (1) go through each of the three cases in which A is T and determine if B is also T OR (2) consider the only case in which A is F and see if B is F as well. Obviously, the second option is less work, so let’s do that!

Considering the case when side A is F, we will assume that p is T and q is F. Then notq is T (the opposite of q) and notp is F (the opposite of p). Then side B is the situation where (T -> F), which, as noted earlier, is F. Therefore, side B is F when side A is F. Furthermore, this is the only case where side B is F (as we know from the truth table of a conditional statement). Therefore, side A and B are false for exactly the same set of truth values of p and q, and thus, the two sides are logically equivalent.

There, that wasn’t so hard, was it?

Our final example deals with propositional satisfiability. A compound proposition is satisfiable if there is at least one set of truth values for the propositions that makes it true. This (or these) set(s) of truth values are called the solution to the compound proposition. If the compound proposition is always false (the opposite of a tautology), then it is unsatisfiable.

To determine if the compound proposition is satisfiable, we need to find one set of truth values that make the entire proposition true. If we cannot find such a set, we know the proposition is unsatisfiable. Of course, this can be done with truth tables, but we will use our reasoning this time.

Look across the entire compound proposition. Notice that this long string is made up of three conjunctions (red) of four “pieces” (each section in parentheses can be considered a piece; in green below). A conjunction (or set of conjunctions) is only true if every piece is true.

The first piece is a basic conditional statement. It is true in three cases as noted below. Keeping these three cases in mind, we will look at each individually. First, we will assume the truth values of that case. Then, we will systematically go through each of the pieces to determine if they are also true in that case. Once we get to a false “piece,” we can discard that case and move to the next, beginning again with the piece on the left.

We begin by assuming the case truth values. In case (1), that is the case where both p and q are true. The second “piece” contains notq, which would be false (the opposite of q). That makes the second piece entirely false. At this point, we do not need to check the other two “pieces.” With conjunctions, one false “piece” makes the entire thing false.

Now we move on to case (2), below, where p is F and q is T. We must again check the second “piece,” in green, which we find to be true this time. So now we go on to the next “piece,” in blue. Since p is assumed F, notp is T, and thus the third “piece” is true as well. Now the final “piece,” in orange below: we have determined that notp is T and notq is F. This makes the final piece false. That spoils the entire case, and we can mark it off.

We follow the same procedure for the third case. We again return to the “beginning” of the compound proposition and work our way across, piece by piece. This time, the third “piece” fails us. And since every case that makes the first “piece” true does not make every other “piece” true, we have determined there is no set of truth values that will make this entire compound proposition true. Therefore, it is unsatisfiable. To check your work, you can use a truth table.

That’s all for this installment of Math Rescue. I hope you learned something and will continue to return for more helpful posts! Leave a comment below with any topic suggestions, questions, or feedback. Thanks for reading!

Discrete: Applications of Propositional Logic, Part 2

2013-04-15T12:00:00.000-05:00

We are re-visiting Applications of Propositional Logic today for Part 2, which includes Boolean searches and logic circuits. Make sure you have already read through Part 1, as information there will come in handy for this section.

Problem 3 is about Boolean searches on the internet. When searching on the internet (on Yahoo! or Bing, for example), you can use Boolean search terms to narrow your sea of results. The connectives used are AND, OR, and NOT. If you wanted to search for information about universities in Mexico, you could type “university AND Mexico,” but this would return results for universities in New Mexico also (because “Mexico” is in the name). To further narrow your search, you could instead search for “(university AND Mexico) NOT New,” which would exclude results that contain the word “new.”

Note that Google, the word “NOT” is not used. Instead, a minus sign (“-“) is placed before an excluded term.

To include a search term, use the conjunction AND; to exclude, use NOT. To include either of two search terms, use the connective OR.

Lastly, we will look at logic circuits, which are sometimes used in the design of computer hardware. Instead of words, we use symbols to create logic circuits. The symbol set consists of gates: the “inverter” (NOT gate), the OR gate, and the AND gate. Notice the difference in the shape of the AND and OR gates (shown below).

Here is problem 4.

Begin by reading the statement from left to right. Here, we first need p and r to go through NOT gates, then through an OR gate (green). Beginning at the top left, we construct the two NOT gates that then feed into a single OR gate (green). Next, we need a q that feeds first through a NOT gate, then into an AND gate (red) along with the result of the previous OR gate.

Below that q, we need another p that again feeds through a NOT gate. Then we need another q and r, that go to an OR gate (blue). The product of this OR gate (blue) joins up with the p that has gone through the NOT gate and together, they go through an AND gate (purple). The products of the two AND gates (red and purple) come together and go through an OR gate (orange).

You can check the work by following through the circuit and writing out the products. You should end up with the same statement as given in the problem.

I hope this series has helped you learn more about propositional logic and has allowed you to practice what you have already learned by applying your knowledge to various types of problems. The best way to improve your skills with this topic is to practice (as with any math, right?).

Don’t forget to leave comments about what you are working on, what you would like me to cover in the future, or how you are doing in your studies! I love to read them, and will try to respond to each of you. I am always excited to receive topic suggestions. Thanks for reading!

Discrete: Applications of Propositional Logic, Part 1

2013-04-15T11:31:00.000-05:00

I would first like to offer my apologies to my readers for my absence as of late. Life has gotten in the way, as I moved to a new house, acquired a new pet, planted a late garden, and have been without internet for nearly a month. I am hoping the internet situation will soon be remedied, but until then, I hope you will not abandon this blog for inactivity.

Growing weary of Algebra, Calculus, and Number Theory, I decided to jump into a new subject today: Discrete Math. We begin with propositional logic; more specifically, the applications of such. I assume you have already learned the base definitions (proposition, implication, etc.) of propositional logic and turn my attention to the applications in my textbook

Part 1 of this topic includes translating English sentences and testing the consistency of system specifications.

Problem 1 is an example of how to translate English sentences into propositional logic.

In this case, the propositions are defined for us. After reading the sentence a couple of times, a key phrase should instinctively pop out at you: “only if.” You should recall that this phrase is one of many ways that a conditional statement (or implication) is expressed. When you see this phrase, or any other equivalent, underline or add parentheses to define the first and second propositions, generic p and q. Here is a list of English phrases that are used to express the conditional statement p -> q:

It is very important that you pay attention to the order that p and q appear in these phrases. Many times q appears first, but comes last in the propositional statement!

Back to our problem, you can see below that I’ve underlined the first proposition, “p,” in green and the second, “q,” in blue. The first proposition is simple, but the second is a compound proposition we will have to digest.

Let’s take a closer look at that compound proposition “q.” When breaking down a compound proposition, look for key words such as “and” and “or.”These words highlight conjunctions (and’s) and disjunctions (or’s). “And” appears twice in this proposition (I’ve boxed them in blue). Once again, we can separate everything between the conjunctions into individual propositions (here, underlined in orange, purple, and red).

The orange proposition is exactly proposition “r” given in the problem. The purple proposition is similar to the proposition “m” given in the problem, but there is one key difference: the word “not.” In fact, the portion underlined in purple is the negation of proposition “m,” and it is written as shown or with a bar over the “m.” Lastly, the red proposition is, again, the negation of the listed proposition “b.”

Putting all of these together, as shown above, with conjunctions between each (if “or” had been present, we would use a disjunction instead), we get the second part of the sentence, generic proposition “q.” Since this is a compound proposition, we enclose it in parentheses.

We now return to the original statement. Notice that the first proposition, generic “p,” is exactly the given proposition “g,” and the second is what the compound proposition we just put together. Since this is a conditional statement, we connect the two with an arrow pointing right, as shown below in the final answer.

Translating English sentences into propositional logic requires that you find key phrases, such as the conditional statement phrases we considered earlier and words like “and,” “or,” and “not.”

Problem 2 (below) is an example of how to determine the consistency of a set of system specifications.

A set of system specifications are consistent if they do not contain conflicting requirements. To determine if system specs are consistent, perform the following steps:

You can walk through step (2) by reasoning through truth values or by the use of a truth table. We will do both in this example.

Now back to our problem. Step (1) is to translate these system specifications into propositional logic. This works exactly as Problem 1 did, except this time, we need to define our own propositions. Define propositions by finding individual statements in the list and assigning them a letter. You can use generic letters (p, q, r), but it is easier if you choose a letter with some connection to the statement.

Each statement is underlined in a different color (above). When a statement is repeated (that is, it has already been assigned a color), I have underlined it in the same color. In this way, we see there are really only three propositions at work here (three colors). They are defined explicitly below.

Use the information from Problem 1 to write each statement using propositional logic. The statements are as follows:

Step (2), then, is where we determine if the specs are consistent. Let’s use reasoning first. In order for all four specs to be true, the final spec must be true. In order for (4) to be T, s must be F (because spec (4) is the negation of s). Now, assuming s is F, the only way for spec (1) to be T is for r to also be F (check the truth table for conditional statements). Now assume that both r and s are false. Since r is F, the only way for spec (3) to be true is if i is also F (again, using the truth table for conditional statements). Since s and i are both false, spec (2) is T. When all three propositions are false, all of the system specifications are true. Therefore, the system is consistent.

Alternately, you could look for a set of truth values that makes all specs true by using a truth table. Begin by creating a column for each proposition (towards the left, below) and listing every combination of truth values for those statements. Since there are 3 statements, there will be 2^3 = 8 possibilities or rows.

Next, create a column for each system specification and fill in its truth values according to the truth tables for that operator (here, only conditional statements and a negation are used). Pay careful attention to the order of the propositions. A conditional statement is only false if p is T and q is F (generic p, q). Look at the first spec. It deals with propositions r and s, the first two columns. Look for the cases where r is true, but s is false (I’ve placed a red arrow on those rows.). First, fill in those entries as F, then the rest as T. Fill in the remaining conditional statements similarly (blue and green arrows and circles). The last spec is a simple negation of s. Fill this column in by writing F every time a T appears under s, and writing T every time an F appears.

The system is consistent if there exists a set of propositional truth values that makes ALL specifications true. Look, therefore, for a row in the set of specs in which every spec is true. Here, that occurs at the bottom, when all three propositions are F (in purple, above). If there didn’t exist a row of all true specs, then the system would be inconsistent.

Thus we conclude Part 1 of Applications of Propositional Logic. Part 2, coming next week, will include information on Boolean searches and logic circuits!

New Video! Trigonometry

2012-12-06T11:17:00.000-06:00

I've uploaded a new YouTube video in the Trigonometry category: Evaluating Base Angles, the Hand Method.

Click here to go to the video.

Click here to go to the original blog post.

Algebra: Using Graphing Utilities to Solve Word Problems

2012-10-08T14:51:00.002-05:00

Proper use of a graphing utility is the most important skill you should develop. Once you master the calculator (or other utility), you can perform any task and complete any problem. You then only need the mathematical skills necessary to set up the information for solving. You can avoid many tedious steps, solving problems the easy way, if you just know how to use your calculator.

In my College Algebra textbook, the very first section covers how to use your graphing utility to model word problems. We have already discussed the Calculate Menu and How to Input Functions—this post will cover how to interpret and solve word problems using your graphing utility of choice.

In this post, I’ll be using a program called Microsoft Mathematics for graph images, but will be instructing you with button and menu names from the TI-84 Plus Silver Edition. Any graphing calculator will perform these tasks, though you may need to consult your manual (or find it online) if you have a different calculator and cannot find the buttons or menus I mention here.

Let’s jump right in with our first word problem.

First, I like to underline/circle/box (or highlight with various colors) pertinent information, to help me sort through the words and get straight to the facts. Underline the question or directive (red below). Circle plain information or direct substitutions (blue below). Box indirect substitutionary information; i.e. information that will help you create an equation (green below).

The second step is to draw a picture representing the problem, if possible.

Let’s consider the information we know. The blue, boxed information below is what we gather directly from the problem. Next, we must obtain equations. Since we are told the tank is a right circular cylinder, and since we are given a volume, it is reasonable to assume we need the formula for the volume of a right circular cylinder, which I’ve labeled “Equation A.” Consider the portion we boxed in green earlier—this sentence tells us a relationship between the radius and the height of the tank. With this information, we form “Equation B.”

We have a volume to substitute (given), but then we are left with Equation A containing two variables, r and h. Equation A and B form a system of equations with two variables. The best way to solve this is to rewrite Equation B in terms of one of the variables and then substitute that into Equation A and solve for the other variable. This method is called “Substitution.”

First, solve Equation B for h (you could solve for r and would obtain the same final answer). Let’s call this Equation C (purple box below).

Substitute Equation C for “h” in Equation A. Also, replace V with the volume given in the problem, 40,000 cubic feet. Simplify the right side and now we have Equation D.

This is the equation we need to solve. You’ll notice, however, that this is a cubic function that you cannot solve by hand (you could if the left side was zero). Therefore, we have to use a graphing utility to solve. There are a few ways of doing this.

1) Input each “side” of Equation D as a separate function in the Y= menu, then graph the functions and use the Table menu to locate their intersection points.
2) Input each “side” of Equation D as a separate function in the Y= menu, then use the Calculate menu to find their points of intersection.
3) Re-arrange Equation D so that one side is equal to zero, input the new function into the Y= menu and find the roots, or x-intercepts, of that equation.

Method (2) is the easiest, in my opinion, so that is what we’ll do. If you’ve seen my blog post on Graphing Functions, then you already know how to input each “side” of Equation D into your calculator as two separate functions. And if you’ve seen my blog post on the Calculate Menu, you already know how to find the intersection of two functions.

When you graph the two functions,

they should look something like this:

(Note: The blue horizontal line is Y1 and the green curve is Y2.)

Below, I have circled the intersection points we need to find in red.

Use the Calculate Menu (2^nd, TRACE; 5: Intersect) to determine these points. If you need help, see my post on using the Calculate Menu here. Remember, these “x-values” are the radius values. Notice that a radius cannot be negative, so we can dismiss our first answer. We can find the corresponding height values by simply inserting each radius value into Equation B from earlier in the problem.

And that’s it! Many problems are solved using this method. Anytime you have an equation you cannot solve, you can input each side into the calculator as separate functions (if there is only one variable throughout) and then simply find the intersection points of the functions.

Another way you might use your calculator in Algebra is if you need to fill in a table of values. Instead of calculating each answer individually, you can input the equation as a function in your calculator. To find x-values that correspond to given y-values, use the first option in the Calculate Menu (1: value). To find y-values that correspond to given x-values, use the Table Menu (2^nd, GRAPH).

I hope this post has helped you learn to use your calculator when solving algebra word problems.

Number Theory: Finite Induction

2012-09-07T13:00:00.000-05:00

Today, I want to give you some step-wise instruction sets for using (finite) mathematical induction to verify formulas and establish/verify inequalities. I assume you know and understand the Well-Ordering Principle, the First Principle of Finite Induction, and the Second Principle of Finite Induction. These should be covered somewhere in the very beginning of your Number Theory textbook (or perhaps in the Appendix). Throughout this post, all variables are assumed to be integers.

You will find proofs requiring the use of mathematical induction throughout many classes and textbooks. Today, we are going to focus on two types of problems: those that require you to verify/establish a formula and those that asks you to obtain/establish a particular inequality.

HOW TO VERIFY FORMULAS USING THE FIRST PRINCIPLE OF INDUCTION

Step 1 – “basis for induction” – Verify the formula for the least element.

Step 2 – “inductive hypothesis” – Assume the formula is true for some arbitrary element, usually k.

Step 3 – “induction” – Prove the formula is true for the next consecutive element, that is, k + 1.

Let’s try a few examples.

The first two steps are simple. In the first, you want to verify that the formula works for the least element. You will be given a range for which the formula works—here, it is integers greater than or equal to one. So, 1 is our least element. We verify the formula is accurate for n = 1. In step 2, you state your hypothesis, that is, the assumption that the formula is true for some arbitrary n value. I like to use n = k for my arbitrary value, but you can use any variable you’d like. We let this k be greater than 1 because we already verified the formula works for n = 1. Step 3 is where students have the most trouble.

In this step, you have to actually prove that the formula is true for the next consecutive element, which is k + 1. It is very helpful to write out what you hope to accomplish, which I usually precede by writing “want to show.” Start with the left side of the formula and group the terms together that appeared in Step 2. We already know that these first k terms “work” in the formula. After this, it is only a matter of manipulation.

Combining the fractions is the next logical step (be sure to get a common denominator first); then just see where simplifying will take you. Eventually, you get to a point where you can’t simplify anymore, and that usually means it’s time to factor. How do you know which factors to check for? Look back at your stated goal—you see that (k + 1) appears in the numerator. I simply divided (long division of polynomials) the numerator by (k + 1) and then rewrote the second factor to better resemble the formula.

You should end with a statement that summarizes the induction process. That is, since the formula works for n = 1 and works for k + 1, we know the formula works for all integers greater than or equal to 1. Many professors will dock points if you do not summarize your proof with a concluding statement, so don’t forget to do that!

All problems like this work similarly. More difficult problems will simply require more manipulation in the third step. The first two steps are always about the same—the only difference may be the number that requires verification in the first step. Remember that you always want to verify the least element. It is usually 1, but not always, so pay attention to that.

Here is another one worked out, but with some of the steps consolidated.

Again, Steps 1 and 2 are pretty much the same.

Just as before, we group the first k terms that appeared in Step 2 and use the formula for them. The rest is pretty straight-forward, I think. In the “factoring” step, I factored (k + 1) from the numerator first, then factored (2k + 1) from what was left. How did I know to factor those out? I considered my stated goal and saw that these factors appeared there. If you try to factor something that appears in your goal and do not get the answer you need, then you should check your work—you probably made a mistake earlier (or in your factoring work).

HOW TO VERIFY FORMULAS USING THE SECOND PRINCIPLE OF INDUCTION

The Second Principle is very similar to the First.

Step 1: Verify the formula for all necessary least elements.

Step 2: Assume the formula is true for all elements less than some arbitrary element, k.

Step 3: Prove the formula is true for the arbitrary element k.

Here is an example.

Notice that the nth number is defined in terms of the previous two. So the 3rd number is defined in terms of numbers a1 and a2; the fourth in terms of a2 and a3; etc. This is why n is limited by 3. If n = 2, there is no number defined for a(n-2) = a0. This changes our first step—we must verify the formula for the least necessary elements. In this case, we verify the first two elements. In other cases, you may have to verify the least 3 or 4 elements. It all depends on how the nth number is defined.

Step 2 also shows us a difference between the two Principles. Notice that I am assuming the formula is true for all integers between our least element (1) and k, not including k. We must assume this because if we only assume it works for k, we don’t know if the formula is true for a(k-1) or a(k-2), but we must use those in order to make our proof! After making the assumption, note the numbers that you will need in the proof, which all depends on how that nth number is defined. In this case, I need a(k-1) and a(k-2).

Begin by defining what you want to show, which is simply the formula with n = k. Use the definition given in the problem to start with, then replace a(k-1) and a(k-2) by the formula definitions from Step 2 (underlined in blue). Now, simply manipulate the numbers until you reach your goal. In this instance, I multiplied through the parentheses, then simplified (green and red underlines). Notice how I simplified the red portion: the rules of exponents will come up over and over again in Number Theory! After simplifying, I can factor the power of 2 from the first two terms (leave the last term alone). After that, I end up with 10, which is 5 * 2, and this 2 can simplify with the power of 2 I already have. And I am left with what I wanted to determine.

As before, you should end your problem with a conclusion of some sort. The Second Principle is essentially the same as the First. Since the formula is true for all values up to and including an arbitrary value k, we can say the formula is true for all numbers on which it is defined, which is n >= 1 in this case.

HOW TO USE INDUCTION TO ESTABLISH AN INEQUALITY

Step 1: Verify the inequality for any necessary least elements.

Step 2: Assume the inequality is true for some arbitrary element (obtain equation A).

Step 3: Note the inequality desired for the next consecutive element (obtain equation B).

Step 4: Manipulate Eq. A so that the left side of Eq. A matches the left side of Eq. B (obtain equation C).

Step 5: Connect Eq. B and Eq. C by an inequality (obtain equation D).

Step 6: Prove Eq. D is true and extrapolate to form statements that prove Eq. B.

I know, this sounds very complicated. It will make more sense as we go through an example!

These problems are long and can get a little confusing, which is why I label pertinent equations (A, B, C, etc). I encourage you to do this as well as you work.

In Step 1, we verify that the inequality is true for the least element, n = 1. It is true. In Step 2, we assume the inequality is true for an arbitrary value k, which is greater than 1 (we already verified 1) and form our first equation, Eq. A, by inserting k for n into the inequality. Simple so far. In Step 3, we determine what we want to prove. We want to show that the inequality works for the next consecutive element, that is, k + 1. To obtain Eq. B, we simply insert (k+1) in place of n in the original inequality.

Here is where everything is complicated a bit. The purpose of the next few steps is to obtain a new inequality that will allow us to prove Eq. B. You are probably familiar with the Squeeze Theorem which works on a similar principle—if a < b and b < c, then we can definitively say that a < c. Think of Eq. B as our “a < c.” We now need two inequalities that link together that will allow us to conclude that Eq. B is true.

In Step 4 (above), we are going to form Eq. C that will be our “a < b” statement. Begin with Eq. A and manipulate it until the left side of Eq. A matches the left side of Eq. B. In order to do this, you need the items found in the blue box. You need to be very familiar with exponential rules and factorials in order to do these problems. After simplifying, we have obtained Eq. C. The left side matches the left side of Eq. B. Remember that whatever you do to one side of an equation, you must do the same to the other—the right side of Eq. C no longer matches Eq. A and in fact, doesn’t match any of our equations.

Step 5 is where we form our “a < b < c” statement in order to figure out our “b < c” statement. I like to think of opening up Eq. B; prying it apart in the middle and leaving a big gaping hole. Notice that we have the “a” and “c” portions of our “a < b < c” statement in place now. Equation C fits in this hole (green). It “links” up with the left side because of Eq. C. The portion underlined in red (“b < c”) is the portion that is “missing”… we have not previously established it. This portion is labeled Eq. D.

Step 6 is the proof of Eq. D. Begin by simplifying the right side. Again, you need to be familiar with factorials here—rewrite the right side. The easiest way to prove an inequality is to move all relevant terms to one side (so that 0 is on the other) and then prove that the terms sum/produce a positive or negative number (depending on the sign). Here, I chose to subtract the right side from the left, but you could subtract the left side from the right. Now all we need to do is simplify and prove that the left side is negative (or is equal to 0).

Always be on the lookout for common factors. If I hadn’t rewritten the factorial on the right side, I would not have any common factors. So, factor out anything that is in common and simplify what remains. Again, be on the lookout for common factors—I found another at the bottom of this photo—and factor them.

Now we need to show that this inequality is true—that the left side really is negative. Consider the limitations in place at the beginning of the problem—we let k > 1. From here, we can multiply both sides by -2, which reverses the inequality; so -2k < -2, which means -2k is negative (obvious, yes?). If k > 1, then 2k > 2. Three (3) is the least possible value that 2k could be equal to; if 2k = 3, then (2k)! = 3! = 6; therefore, (2k)! >= 6. I simplified this by saying (2k)! is at least 2. Either way, we know it is positive. And again, since k > 1, k+1 is definitely greater than 2 and is therefore positive as well.

So, we have a negative multiplied by two positives. This will always yield a negative result and so, we know that our final statement is true. Therefore, Eq. D is true (as that’s where we began in Step 6). Since Eq. D is true, our linkage of inequalities (“a < b < c”) is also true. Since this is true, we can say that the left side is less than or equal to the right side and therefore the inequality from Step 3 (Eq. B) is true, which is what we wanted to show.

We essentially began with “a < b” (Eq. C). We found a “c” such that “b < c” (Eq. D) and proved it. Therefore, we could safely say that “a < b < c” and thus, “a < c” (Eq. B).

I know this is confusing, but it will get easier with practice. Just keep working at it. Labeling the equations is a huge help, too.

I hope you now feel more comfortable with mathematical (finite) induction. It is a method of proof you will come across throughout even a short mathematical career/study track.

Calculator Help: The Calculate Menu – Part 2

2012-09-03T12:30:00.000-05:00

I am using a TI-84 Plus Silver Edition graphing calculator for demonstrations; consult your calculator's manual for help finding menus and buttons.

In the previous post, we learned how to use the value, zero, minimum, and maximum commands found in the Calculate menu on a graphing calculator (click here to view that post). Today, we will look at the last three commands: intersect, derivative, and integral. In an Algebra class, you will probably use the intersection command frequently, but you will not use the derivative and integral commands until you reach Calculus. Even then, many students do not realize the advantage those two little commands offer. Let's get started.

To begin, refer to the first set of instructions in the previous post linked above to set up your calculator with our function and window, so that you can see the graph I'll be working with throughout this post.

HOW TO USE THE "INTERSECT" COMMAND

The "intersect" command allows you to determine the point of intersection of two functions. For this section, input the following function into the second slot in the Y= menu.

Press GRAPH to graph both functions.

Navigate to the Calculate menu (press 2ND then TRACE) and choose option 5: intersect. Take notice of a few things. The calculator has placed a cursor on the graph of the first function at a random point (circled in blue below) and has opened with the first prompt (circled in red): "First curve?"

We want to know the intersection point located in the bottom right, circled in green in the photo above. To answer the first prompt, move the cursor close to the intersection point and press ENTER.

Since there are only two functions graphed here, the calculator automatically jumps to the second equation—in the upper left corner of the screen, the second equation is displayed and the cursor has "jumped" from the first curve to the second. Since this point is close to the intersection, we can press ENTER here without moving the cursor. If it were not very close to the intersection we desire to find, we would arrow closer to it. Press ENTER again to skip through the guess screen and obtain the intersection answer.

There are a couple of things to note here. If there were many functions graphed on the screen, we could toggle through them by pressing the UP or DOWN arrow key in order to identify which curves we want to find the intersection for. Also, the cursor must be near the intersection you desire to find when you press ENTER for each curve. The calculator will find the intersection point closest to the cursor points you entered. Use the "intersect" command to find the other two points of intersection for these two curves. You should get the following answers: (0.89m -2.66) and (-1.47, 4.42).

The "intersect" command is also helpful for finding all x-values that yield a particular y-value. For example, if we want to know at what points our first function is equal to -10, we can do the following:
Change the second function to y = -10.

Then press GRAPH. Use the "intersect" command to find the three points of intersection of the two curves. You should get the following answers: (-2.25, -10), (1.49, -10), and (4.76, -10).

HOW TO USE THE "DERIVATIVE" COMMAND

Clear the second function from your calculator, but keep the first one and graph it in the same window we've been using. The "derivative" command allows you to determine the derivative of the function at any given x-value in the window you have set. Open the Calculate menu and choose option 6: dy/dx. You can choose an x-value in one of two ways: move the cursor along the curve to the point you desire or type in the x-value you want to evaluate. Let's evaluate the derivative at x = 5. When you type a number, an input line will appear automatically.

Press ENTER after you input the value and the calculator will return the value of the derivative at that point. We would round this answer to dy/dx = 28.

This command is very helpful when you first learn about derivatives. You will have to find derivative values manually in the beginning, but this command allows you to check those answers.

HOW TO USE THE "INTEGRAL" COMMAND

I cannot input the "integral" symbol here, but the final Calculate command is the integral. The integral over an interval is the area under the curve (between the curve and the x-axis) over that interval. To practice, let's find the integral of this function over the interval (1, 5). With the same function graphed on your screen, open the Calculate menu and choose the last option. The calculator will place a cursor on the screen (circled in blue below) and opens with the first prompt (circled in red below): "Lower Limit?"

The lower limit we want to consider is x = 1. You can enter limits one of two ways: move the cursor to the point on the graph or directly input the x-value. I prefer to directly input the x-value because it is not usually possible to place the cursor exactly on the desired limit. Type "1" and an input line will automatically appear at the bottom of the screen.

Press ENTER to set the lower limit. You can probably guess the next prompt: "Upper Limit?" As for the lower limit, enter the upper limit, which in this case is 5.

Press ENTER (there is no guess screen this time) and the calculator does two things. It shades the integral for a visual representation and also gives the numerical value, which is -69.33 in this case.

In order for this command to work, you must choose limit values that are contained in the window you have set. This is a fantastic way to check your integration answers for Calculus class!

There you have it. We have now covered all of the commands found in the Calculate menu. I hope you now have the ability and confidence to use these commands to help you perform well in your math classes. Most math teachers/professors require you to show your steps on homework, but these commands allow you to very easily check your answers before turning in your work. If you have any questions or comments, please leave them below!

Calculator Help: The Calculate Menu – Part 1

2012-08-30T13:59:00.000-05:00

I am using a TI-84 Plus Silver Edition for demonstration today; consult your manual for help locating similar menus and buttons on your device.

Before tackling this post, make sure you know how to input functions into your graphing calculator. If you need help with that or just need a reminder, review my blog post called Calculator Help: Input Functions. The very first section in my College Algebra textbook is called "Using Graphing Utilities." The applications portion of the exercise set requires students to find local extrema and intersections of functions. These two tasks are accomplished via the "Calculate" menu on most graphing calculators (TI-83 through TI-89 most specifically). This menu contains many other useful commands and I want to discuss all of them. Today, we'll begin by looking at the first four commands (value, zero, minimum, and maximum). Next time, we will cover commands 5 through 7 (intersect, derivative, and integral).

To find the Calculate menu, press 2ND then TRACE, which is located just below the graphing screen.

The Calculate menu contains seven commands: value, zero, minimum, maximum, intersect, derivative, and integral.

For practice, input the following function into your Y1 slot. Make sure all other function slots are empty and that all stat plots are turned off. (Refer to my post Calculator Help: Statistics for instructions on turning on/off stat plots.)

Use the following window values to obtain the graph seen below.

Now that we have a graph to work with, let's look at each of the Calculate commands individually.

HOW TO USE THE "VALUE" COMMAND

This command requires the input of an x-value and it will determine the y-value at that point. Let's try it out! In the Calculate menu, choose the first command 1: value. The calculator then prompts you for the x-value you want to evaluate. For this command to work, the x-value you enter must be within the viewing window.

Let's evaluate the function at x = 1 (which is in the viewing window of (-4, 8), so it is a valid input). Type in "1" and press ENTER.

The calculator will place an "x" mark on the point it located for us (circled in blue on the photo below) and identified the y-value that corresponds to the input (circled in red on the photo below).

Notice that the evaluated function is written at the top of this screen. If we input multiple functions, we can cycle through them after inputting an x-value by pressing the UP and DOWN arrow keys.

The "value" command is very helpful when you need to evaluate a function at a few points. For more than a few points, it is more helpful to consult the table, which we will look at in another post someday.

HOW TO USE THE "ZERO" COMMAND

This command confuses many students. The "zeroes" of a function are the x-values at which the function crosses the x-axis. This occurs whenever the y-value of the function is zero, hence the name. You may also see these values referred to as "x-intercepts." This command will NOT tell you the y-value at which x = 0 (which is the y-intercept). To find the y-intercept, use the "value" command we covered above and input "x = 0." To find the x-intercepts (a.k.a. zeroes) of a function, use the "zero" command. With the same function input into the calculator, return to the Calculate menu and choose the second command 2: zero.

The calculator places an "x" mark on the function (circled in blue on the photo above) and asks a question, "Left Bound?" (circled in red). The calculator is prompting you to move the cursor to a point on the graph that is "left bound" from the zero. Let's try to find the zero farthest to the left (circled in green above). To determine which side of the axis is left bound, place your finger on the zero and move it left, following the graph. From this point, left bound is below the x-axis. Use the LEFT arrow key to move the cursor to some point below the x-axis, left of the zero we want to identify. You can stop the cursor at any point left of the desired zero—there is no precise place to move it to. I usually stay fairly close to the axis while making sure I am definitely below it (below the axis, y-values are negative; above the axis, y-values are positive—use this to be sure you are below the axis in this case).

When you have positioned the cursor left of the desired zero, press ENTER. The calculator places a filled right-pointing arrow on the graph (circled in purple on the photo below) to mark where you have "saved" your cursor position and now prompts you to move the cursor to a point on the graph that is "right bound" from the zero (circled in red).

From the zero, right bound is above the x-axis, so move the cursor somewhere close, but above, the x-axis, then press ENTER.

Above, you see the next prompt that appears. The calculator is asking if you want to enter a guess for the answer. Entering a very close guess will save the calculator a bit of time finding the answer. You do not have to enter a guess. To continue from the guess screen without entering a guess, simply press ENTER. You can enter a guess by keying in an x-value or moving the cursor to the zero you are looking for. You can help the calculator find the zero faster also by choosing left and right bound points that are very close to the actual zero. The larger the interval you choose, the longer it will take the calculator to find your answer.

We were trying to find a point at which the y-value is zero, but here, it is giving us a point with a y-value that is not zero. The x-value for this zero is an estimate—it is a decimal. Therefore, the calculator evaluated the function at this estimated point. The y-value given is, however, very, very close to zero.

Try finding another zero; say the one in the middle. This time, if you begin at the zero and move left, you will be moving above the x-axis and if you move to the right, you will move below the x-axis. It is very important that, when entering left bound and right bound points, you pay attention to where the graph is located. Left bound will not always be below—as in this case, it is above! You should find this zero to be at the point (0.65, 0).

HOW TO USE THE "MINIMUM" COMMAND

The "minimum" command is used to find a local minimum point, which is the point that forms a "valley" in the graph. With the same function and window set in the calculator, navigate to the Calculate menu and choose command 3: minimum. Visually locate the minimum. This is the point we want to numerically identify.

You will notice that the prompts for this command are the same as the prompts for the previous "zero" command. The first prompt is "Left Bound?" Move the cursor using the LEFT or RIGHT arrow key to a point that is left of the minimum we want to identify. Anywhere left of the point is sufficient, but remember that the farther your cursor is set from the point, the longer the calculator will take to find the answer.

Press ENTER when you've chosen a point left of the minimum. The calculator will place a solid right-pointing arrow on the screen to mark your spot and bring up the next prompt: "Right Bound?" As before, move the cursor to the right of the minimum and press ENTER. The calculator will once again request a guess, which you can skip by pressing ENTER again.

My calculator estimated the minimum to be at the point (3.36, -24.75) as you can see below.

HOW TO USE THE "MAXIMUM" COMMAND

The "maximum" command is used to find a local maximum point, which is the point that forms a "hill" in the graph. It works exactly as the "minimum" command works. You will move the cursor to the left of the maximum, press ENTER, move the cursor to the right of the maximum, press ENTER again, and then press ENTER once more to move past the guess screen. Here is my answer of (-0.69, 8.60):

That covers the first four commands in the Calculate menu. These commands are very helpful when solving problems that involve complicated formulas without "pretty" (whole) numbers. They also help when you learn to find these values by hand, as you can check your answers! In the next post, we will cover commands 5 through 7, so stay tuned.

Trigonometry: Evaluating Base Angles - The Hand Method

2012-08-10T14:26:00.000-05:00

Notice: On 12/6/2012, I uploaded a YouTube video that corresponds to this blog post. You can find it by following this link.

Most Trigonometry students are encouraged (and sometimes required) to memorize the "base angle" evaluations for the three fundamental trigonometric functions. These five core angles (in degrees) are 0, 30, 45, 60, and 90. The three fundamental trig functions are sine, cosine, and tangent. So students spend hours attempting to memorize 15 values... or they don't memorize them and do poorly on the test.

Of course, a calculator will give you these values quite easily. Memorizing them is useful though because (1) these angles show up frequently in introductory trigonometric and calculus courses and (2) having the answers memorized saves time on exams and homework. But memorization is a chore and is difficult for most students. That's why I really love this "hand method" of remembering (or rather, determining) these values. I did not create this--I learned it while taking Pre-Calculus my junior year of high school, and I've been using it ever since!

Each finger on the left hand with the palm facing towards you (thumb up, pinkie down) represents one of the five base or core angles. Visualize your hand as the first quadrant--the pinkie is in the 0 degree position and the thumb is in the 90 degree position. Then it is easy to see that the ringer finger is the 30 degree finger, the middle is the 45 degree finger, and the pointer represents the 60 degree finger. Here is a visual with angles in degrees and radians. Knowing the fingers in both measurements will save you the time it would take to convert all radian measures into degrees!

Once you have the fingers identified, we can begin evaluating the base angles, those found in the first quadrant. We will learn later in the post how to evaluate angles that are constant multiples of these base angles.

First, let's look at evaluating the cosine or sine of one of these core angles. In my visual below, I've chosen to evaluate at 60 degrees. Step 1 is to identify the finger that corresponds to the angle. Here, the pointer finger represents 60 degrees. Step 2 is to lower that finger, just fold it down, as in the image below. To find the cosine of this angle, simply count the number of fingers still standing above the lowered finger. Then, take the square root of this number and divide by 2. This is the cosine of the angle. To find the sine of this angle, count the number of fingers still standing below the lowered finger, which is 3 in this case. Take the square root of this number and divide by 2.

Easy enough, right? Evaluating tangent is simple too. You only have to do one extra thing. After lowering the required finger, flip your hand over so that the pinkie is on top and the thumb is on bottom. The fingers still represent the same angle values. Avoid confusion by lowering your finger first, then flipping your hand. After flipping your hand over, to find the tangent of the angle, count the number of fingers above (which is 3 here) and the number of fingers below (which is 1). Take the square root of the number above and divide by the square root of the number below.

See, I told you it was just as easy! Only one extra step, flipping the hand over. You can quickly evaluate any of these base angles at any of the three fundamental trig functions using this method. I've compiled a few for you to try. Evaluate these trig functions at the given angles using the hand method. Don't convert between radians and degrees--practice learning the fingers in both units. After you've evaluated all of them, use a calculator to check your answers (your calculator will give decimal approximations; enter your answer to see if the decimals match). Make sure your calculator is in the right mode for the type of angle you are evaluating!

You can also see the answers quickly by clicking here.

It is also possible to evaluate the trig functions at any constant multiple of the five core angles using this method. There is just an extra step involved, and that is to determine the sign (positive or negative) the answer will have. You know that all angles can be found on the unit circle. The unit circle contains 360 degrees (or 2pi radians). Angles between 0 and 90 degrees are in quadrant I of the standard x,y-plane; angles between 90 and 180 are in quadrant II; angles between 180 and 270 are in quadrant III; and angles between 270 and 360 are in quadrant IV.

But there are angles greater than 360 and angles less than 0. We discussed coterminal angles in another post. Review How to Find a Coterminal Angle in that post.

The first step to evaluating a trig function at a multiple of one of the five core angles is to determine which core angle is multiplied. To do this, find the angle that is coterminal with the given angle. For example, given the angle -405, the angle coterminal with this is 315 degrees. This angle is a multiple of 45 degrees: 315 = 7 * 45. Find this by trial and error--divide the angle by each of the five base angles. Whichever base angle gives a whole quotient, that is the base angle you are looking for.

The second step is to evaluate the trig function at the base angle you found.

The third step is to determine the quadrant of the original angle. Since -405 is coterminal with 315, and since 315 lies in the fourth quadrant, I know that -405 lies in quadrant IV.

Now, you can determine the sign of your final answer. Refer to the diagram below:

This little phrase will help you remember which trig functions are positive in which quadrants: All Students Take Calculus. "All" stands for, well, all. That is, in quadrant I, all of the trig functions are positive. "Students" stands for sine; so only the sine function is positive in the second quadrant. "Take" stands for tangent; so only the tangent function is positive in the third quadrant. And "Calculus" stands for cosine; so only cosine is positive in the fourth quadrant.

In my example, -405 degrees lies in the fourth quadrant where only cosine is positive. If I'm evaluating cosine at this particular angle, the answer is positive; if I'm evaluating sine or tangent at this angle, the answer is negative.

Here are a few examples:

In each instance, the steps are the same. First, identify the base angle by finding the angle that is coterminal to the given angle. Divide the given angle by each of the core angles to determine which angle to evaluate. Then determine which quadrant the angle is in and use it to determine the sign of the final answer.

Practice this method of angle evaluation enough, and it will be very helpful to you in future classes. After trigonometry, you will probably use it in calculus and maybe other subjects as well. You can impress your teachers or classmates with the speed at which you can answer evaluation questions. Of course, it only works for multiples of the five core angles, but you will see that these angles appear frequently in homework and test questions because trig functions evaluated at these angles yield nice numbers.

Edit: This post was edited on 8/16/2012 to correct a mistake in the "Examples" photo.

Edit #2: This post was edited on 12/6/2012 to add the notice at the top.