## Square Roots Part 1

Note:  Square roots are pretty tricky to teach and learn because the tendency is to seek answer-getting methods.  Patience at the onset, allowing for full development of conceptual understanding is key.  Do not revert to tricks and quick "gets" when first learning square roots.  Always revert back to the question they ask and how you know if you've answered that question.

square roots part 1

Square Roots

Part 1

Introduction: Square roots are consistently among the most misunderstood topics in developmental math. Similar to exponents, students must possess both procedural fluency but also a solid conceptual foundation and the ability to read and understand what square roots mean, in order to be proficient with them. It is often the case that problems with square roots do not lend themselves to a correct first step, but rather, offer many equally viable methods of approach.

Square Roots Ask a Question: What number squared is equal to the radicand? The radicand is the number inside the square root symbol (radical). This expression asks, what number times itself (squared) is 11?

$\sqrt{11}$

This is a number. It is not 11. It turns out this number is irrational and we can never actually write what it is more accurately than this.

Big Idea: The area of a square is calculated by squaring a side (multiplying it by itself). Since all sides of a square are equal, this about as easy of an area to calculate as possible. A square root is giving us the area of a square and asking us to find out how long a side is.

$\overline{)42}$

For example, this square has an area of forty-two. Instead of writing out the question, “How long is the side of a square whose area is forty-two?” we simply write, $\sqrt{42}.$

The majority of the confusion with square roots comes back to this definition of what a square root is. To make it as clear as possible, please consider the following table.

 English Math How long is the side of a square that has an area of 100? $\sqrt{100}$ How long is the side of a square that has an area of 10? $\sqrt{10}$

These two numbers were chosen because students inevitably write $\sqrt{100}=\sqrt{10}.$

 English Math Answer How long is the side of a square that has an area of 100? $\sqrt{100}$ 10 How long is the side of a square that has an area of 10? $\sqrt{10}$ 3.162277660168379…

Key Knowledge: In order to be proficient with square roots we need to know about perfect squares. A perfect square is a number that is the product of a number squared. Sixteen is a perfect because four times four is sixteen.

The reason you need to know perfect squares is because square roots are asking for numbers squared that equal the radicand. So if the radicand is a perfect square, we have an easy ‘get,’ that is, simplification.

For example, since 42 = 16, and the square root of sixteen $\left(\sqrt{16}\right)$ asks for what number squared is 16, the answer is just four.

Let’s take a look at the first twenty perfect squares and what number has been squared to arrive at the perfect square, which we will call the parent.

 Perfect Square “Parent” 1 1 4 2 9 3 16 4 25 5 36 6 49 7 64 8 81 9 100 10 121 11 144 12 169 13 196 14 225 15 256 16 289 17 324 18 361 19 400 20

You should recognize these numbers as perfect squares as that is a key piece of knowledge required!

Pro-Tip: When dealing with square roots it is wise to have a list of perfect squares handy to help you familiarize yourself with them.

How to Simplify a Square Root: To simplify a square root all you do is answer the question it is asking.

The best way to go about that is to see if the radicand is a perfect square. If so, then just answer the question. For example:

Simplify $\sqrt{256}$

Since this is asking, “What number squared is 256?” and 256 is a perfect square, 162, the answer to the question is just 16.

$\sqrt{256}=16$

What if we had something like this:

$\sqrt{{x}^{2}}$

If you’re confused by this, revert back to the question it is asking. This is asking, “What squared is x $–$ squared?” All you have to do is answer it.

$\sqrt{{x}^{2}}=x$

What if the radicand was not a perfect square?

If you end up with an ugly square root, like $\sqrt{48},$ all you have to do is factor the radicand to find the largest perfect square.

List all factors, not just the prime factors. In fact, the prime factors are of little use because prime numbers are not perfect squares. And again, we are looking for perfect squares because they help us answer the question posed by the square root.

48

1, 48

2, 24

3, 16

4, 12

6, 8

Pro-Tip: When factoring, do not skip around. Check divisibility by all of the numbers in order until you get a turn around. For example, after 6, check 7. Seven doesn’t divide into 48, but 8 does. Eight times six is forty eight, but you already have that pair. That’s how you know you’re done!

In our list we need to find the largest perfect square. While four is a perfect square, sixteen is larger. So we need to use three and sixteen like shown below.

$\sqrt{48}=\sqrt{16}\cdot \sqrt{3}$

The square root of three is irrational (square roots of prime numbers are all irrational), but the square root of sixteen is four. So rewriting this we get:

$\begin{array}{l}\sqrt{48}=\sqrt{16}\cdot \sqrt{3}\\ \sqrt{48}=4\cdot \sqrt{3}\end{array}$

See Note 1 and Note 2 below for an explanation of why the above works.

Fact:

That means that Let’s see if it is true.

${\left(4\sqrt{3}\right)}^{2}=4\sqrt{3}×4\sqrt{3}$

Because we can change the order in which we multiply, we can rearrange this and multiply the rational numbers together first and the irrational numbers together first.

$4\sqrt{3}×4\sqrt{3}=4×4×\sqrt{3}×\sqrt{3}$

The square root of three times itself is the square root of nine.

$4×4×\sqrt{3}×\sqrt{3}=16×\sqrt{9}$

The square root of nine asks, what squared is nine. The answer to that is three.

$16×\sqrt{9}=16×3$

So,

Note 1: $4\sqrt{3}$ cannot be simplified because the square root of three is irrational. That means we cannot write it more accurately than this. Also, the product of rational number and an irrational number is irrational. So, $4\sqrt{3}$ is just written as “four root three.”

Note 2: We can separate square roots into the product of two different square roots like this:

$\sqrt{75}=\sqrt{25\cdot 3}$ figure a.

or

$\sqrt{75}=\sqrt{25}\cdot \sqrt{3}$ figure b.

If we consider the question being asked, what number squared is seventy five, we can see why this works. What number squared is seventy five is the same as what number squared is twenty five times three,” (figure a). The number squared that is twenty five times the number squared that is three is the same as the number times itself that is twenty five times three.

For example:

But also: $\sqrt{64}=\sqrt{4}\cdot \sqrt{16}$

And this simplifies to:

$\sqrt{64}=2\cdot 4=8$

What we will see in a future section is that square roots are actually exponents, exponents are repeated multiplication and the order in which you multiply does not matter. This allows us to manipulate square root expressions in such a fashion.

Let us work through two examples. Before we do, let us define what simplify means in the context of square roots. Simplify with square roots means that the radicand does not contain a factor that is a perfect square and that all terms are multiplied together.

Simplify: $9\sqrt{8}$

What is the nine doing with the square root of eight? It is multiplying by it. We cannot carry out that operation. However, eight, the radicand, does contain a perfect square, four. Do not allow the fact that 9 is also a perfect square confuse you. This is just 9, as in 1, 2, 3, 4, 5, 6, 7, 8, 9. The square root of eight cannot be counted. It is asking a question, remember?

$9\sqrt{8}=9\sqrt{4}\cdot \sqrt{2}$

Pro-Tip: When rewriting radical expressions (square roots), write the perfect square first as it is easier to manipulate (you won’t mess up as easily).

$\begin{array}{l}9\sqrt{4}\cdot \sqrt{2}\\ 9\cdot 2\cdot \sqrt{2}\\ 18\sqrt{2}\end{array}$

Example 2:

Simplify $8\cdot \sqrt{\frac{32}{{x}^{4}}}$

The eight is multiplying with the radical expression. Just like we could separate the multiplication of square roots, we can also separate the division, provided it is written as multiplication by the reciprocal. So, let’s consider these separately, to break this down into smaller pieces that are easier to manage.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{\sqrt{32}}{\sqrt{{x}^{4}}}$

Let’s factor each square root, looking for a perfect square. Note that x2 times x2 is x4.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{\sqrt{16}×\sqrt{2}}{\sqrt{{x}^{4}}}$

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=8×\frac{4×\sqrt{2}}{{x}^{2}}$

Notice that 8 is a fraction 8/1.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=\frac{8}{1}×\frac{4×\sqrt{2}}{{x}^{2}}$

Multiplication of fractions is easy as π.

$8\cdot \sqrt{\frac{32}{{x}^{4}}}=\frac{32\sqrt{2}}{{x}^{2}}$

Summary: Square roots ask a question: What number squared is the radicand? This comes from the area of a square. Given the area of a square, how long is the side?

To answer the question you factor the radicand and find the largest perfect square.

Time for some practice problems:

1.7 Square Roots Part 1 Practice Set 1

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{125}=\\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{27}=\\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{162}=\\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{75}=\\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{45}=\end{array}$ $\begin{array}{l}6.\text{\hspace{0.17em}}\sqrt{4{x}^{2}}=\\ 7.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{4{x}^{4}}=\\ 8.\text{\hspace{0.17em}}\sqrt{\frac{4}{25}}=\\ 9.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{98}=\\ 10.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}=\end{array}$ $\begin{array}{l}11.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{4}\cdot \sqrt{8}=\\ 12.\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\sqrt{8}=\\ 13.\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\sqrt{9}=\\ 14.\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\sqrt{\frac{1}{4}}=\\ 15.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{300}=\end{array}$

1.7 Square Roots Part 1, Practice 2

Simplify problems 1 through 8.

1.      $\sqrt{24}$ 2. $\sqrt{8{x}^{3}}$

3. $\sqrt{200}$ 4. $\sqrt{27}$

5. $\sqrt{7x}$ 6. $4\sqrt{{a}^{2}b}$

7. $3x\sqrt{98{x}^{2}}$ 8. $\frac{1}{3}\cdot \sqrt{\frac{9}{{x}^{2}}}$

9. Show that ${a}^{2}b=\sqrt{{a}^{4}{b}^{2}}$ 10. Why is finding perfect squares appropriate

when simplifying square roots?

## Where the First “Law” of Logarithms Originates – Wednesday’s Why E6

rule 1

Wednesday’s Why $–$ Episode 6

In this week’s episode of Wednesday’s Why we will tackle why the following is “law” of exponents is true in a way that hopefully will promote mathematical fluency and confidence. It is my hope that through these Wednesday’s Why episodes that you are empowered to seek deeper understanding by seeing that math is a written language and that by substituting equivalent expressions we can manipulate things to find truths.

${\mathrm{log}}_{2}\left(M\cdot N\right)={\mathrm{log}}_{2}M+{\mathrm{log}}_{2}N$

Now of course the base of 2 is arbitrary, but we will use a base of two to explore this.

The first thing to be aware of is that exponents and logarithms deal with the same issue of repeated multiplication. There connection between the properties of each are tightly related. What we will see here is that the property of logarithms above and this property of exponents below are both at play here. But it is not so easy to see, so let’s do a little exploration.

Just to be sure of how exponents and logarithms are written with the same meaning, consider the following.

${a}^{b}=c↔{\mathrm{log}}_{a}c=b$

Let us begin with statement 1: ${2}^{A}=M$

We can rewrite this as a logarithm, statement 1.1: ${\mathrm{log}}_{2}M=A$

Statement 2:
${2}^{B}=N$

We can rewrite this as a logarithm, statement 2.1: ${\mathrm{log}}_{2}N=B$

If we take the product of M and N, we would get
2A ·2B. Since exponents are repeated multiplication,

2A ·2B = 2A + B

This gives us statement 3: $M\cdot N={2}^{A+B}$

Let us rewrite statement 3 as a logarithmic equation.

$M\cdot N={2}^{A+B}\to {\mathrm{log}}_{2}\left(M\cdot N\right)=A+B$

In statements 1.1 and 2.1 we see what A and B equal. So let’s substitute those now.

$\begin{array}{l}{\mathrm{log}}_{2}\left(M\cdot N\right)=A+B\\ {\mathrm{log}}_{2}\left(M\cdot N\right)={\mathrm{log}}_{2}M+{\mathrm{log}}_{2}N\end{array}$

It took a little algebraic-juggling to get it done, but hopefully you can now see that this is not a law or a rule, but a property of repeated multiplication, just like all of the properties of exponents are consequences of repeated multiplication.

Let me know what worked for you here and what did not. Leave me a comment.

Thank you again.

## How to Teach Exponents – Part 1

Exponents are one of the most difficult topics to teach because once you understand how they work, it seems so obvious.  And once you understand something to the point of it being obvious, your memory of how difficult it was to learn and what caused problems is almost entirely erased.

To “Get It,” with exponents a balance between conceptual understanding and procedural efficiency must be struck.  This balance is probably more important with exponents than any other topic in Algebra 1.  This is because sometimes one method or technique of simplifying an exponential expression will be fluid and pain-free.  Another problem the same method will lead to confusion and head-ache.

In order for students to be well versed in exponents they must see multiple ways of approaching each problem.  They must understand where all of the “rules” come from, why they’re true and how they’re related to repeated multiplication.

In the video below I will discuss some of my specific points of emphasis with respect to exponents as well as some general math-teaching tricks you can use.

So, if you’d like to use my PowerPoint, feel free to download it here.  Make it your own, change graphics, add bellwork, whatever you decide.  The only thing I ask is you share where you found it and let me know how it went.

Keep in mind, this might be more than one day for your class, depending on class duration, aptitude and other factors.  If this material could not be covered in one day I would strongly advise creating some quality homework that forces them to think about what has been learned so far.  Homework is not just practice, it is for learning!

As always, thank you for reading.

## Why Do Some Fractions Repeat as Decimals?

There's a video of this material at the bottom of the page.

If you'd like a PowerPoint that covers this material, email me at thebeardedmathman@gmail.com

Why some rational numbers repeat as fractions

Why Some Rational Numbers are Non-Terminating Decimals

I’d like to explore the relationship between non-terminating, but repeating, decimals and their rational equivalents. The topic is worth exploration because it can help provide insight as to why something like 0.99999999999999999… is equal to 10, not just approximately 10, kind of. The moment you stop writing the 9s, though, it is no longer equal to ten. Through this exploration I hope you’ll see why these things occur.

Beyond that, I think it is very cool to explain something that has bugged me for years. (For the sake of simplicity, we will only consider proper fractions that are fully reduced.)

Why 1/3 is 0.3333333333 … repeating infinitely is easily enough seen by dividing 1 by three with long division. You end up with a loop of 10 divided by 3, which always has a remainder of one. But why does such a nice and easy rational number end up with a non-terminating decimal equivalent, when some other, ugly numbers, like 7/32 are terminating?

For the sake of clarity, to convert a fraction into a decimal you divide the numerator by the denominator. So, $¼$ would be 0.25 as shown below:

$\frac{1}{4}=4\begin{array}{c}\hfill 0.25\\ \hfill \overline{)\begin{array}{l}1.00\\ \underset{_}{-8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}20\\ \underset{_}{-20}\end{array}}\end{array}$

I am going to make a few points that will be pulled together to show why some rational numbers are non-terminating decimals.

Point One:

The first big thing to note is that, as you probably already know, 0.25 is the same as $\frac{25}{100}$ . This ratio is equal: $\frac{1}{4}=\frac{25}{100}$. Similarly, $0.157=\frac{157}{1,000}.$ However, $0.\overline{13}\ne \frac{13}{100}$.

Point Two:

Decimals are equivalent to fractions with denominators that are a power of ten. Consider the following decimal, 0.111111… repeating.

The first decimal, 0.1 is $\frac{1}{10}$. The second decimal, 0.11 is $\frac{1}{10}+\frac{1}{100}$, the third, 0.111 is $\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}$, and so on.

Then $0.1111=\frac{1}{10}+\frac{1}{100}+\frac{1}{1,000}+\frac{1}{10,000}$ .

This means that decimals, rewritten as fractions, all have denominators that are an exact power of ten.

So, $0.1111=\frac{1}{10}+\frac{1}{100}+\frac{1}{1,000}+\frac{1}{10,000}$, which is $0.1111=\frac{1}{{10}^{1}}+\frac{1}{{10}^{2}}+\frac{1}{{10}^{3}}+\frac{1}{{10}^{4}}.$

Let’s see what that would look like for a terminating decimal, like $\frac{1}{8},$ which is 0.125.

$0.125=\frac{1}{10}+\frac{2}{100}+\frac{5}{1,000}$. If you’re unsure of this being true, I’ll get common denominators and we will see the sum is $\frac{125}{1,000}.$

$\frac{1}{10}\cdot \frac{100}{100}+\frac{2}{100}\frac{10}{10}+\frac{5}{1,000}$

$\frac{100}{1,000}+\frac{200}{1,000}+\frac{5}{1,000}=\frac{125}{1,000}.$

Decimals are fractions with a denominator that’s an exact power of ten.

Point Three:

Let us consider a different method of converting fractions into decimals, to shed light on why non-terminating decimals spring forth from some rational numbers.

Let us convert one-fifth into a decimal by setting up a ratio where the denominator must be a power of ten.

$\frac{1}{5}=\frac{x}{10}$

The power of ten is easy here because five divides into ten evenly.

Changing the denominator into ten by multiplying by two over two:

$\frac{2}{2}\cdot \frac{1}{5}=\frac{x}{10}$

Give us:

$\frac{2}{10}=\frac{x}{10}$, so .

Let’s see how this works for fraction like $\frac{1}{8}.$

$\frac{1}{8}=\frac{x}{{10}^{?}}$

The smallest power of ten that eight divides into is 1,000. The smallest that 16 divides into is 10,000, and the smallest that 32 divides into is 100,000. We will explore how to see that in a moment, but let’s finish 1/8th first.

$\frac{1}{8}=\frac{x}{1000}$

$\frac{125}{125}\cdot \frac{1}{8}=\frac{x}{1000}$

$\frac{1}{8}=\frac{125}{1000},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{8}=0.125$.

If we look at this in terms of factors and exponents, we can see an interesting occurrence. The column on the left will be a denominator of two and the column on the right will be the smallest power of ten we can use, set up as a ratio. Then we’ll factor these and gain valuable, hopefully, insight.

 Un-factored Factored Denominator 2 Denominator 10 Denominator 2 Denominator 10 $\frac{1}{2}$ $\frac{x}{10}$ $\frac{1}{{2}^{1}}$ $\frac{x}{{\left(2\cdot 5\right)}^{1}}$ $\frac{1}{4}$ $\frac{x}{100}$ $\frac{1}{{2}^{2}}$ $\frac{x}{{\left(2\cdot 5\right)}^{2}}$ $\frac{1}{8}$ $\frac{x}{1000}$ $\frac{1}{{2}^{3}}$ $\frac{x}{{\left(2\cdot 5\right)}^{3}}$ $\frac{1}{16}$ $\frac{x}{10,000}$ $\frac{1}{{2}^{4}}$ $\frac{x}{{\left(2\cdot 5\right)}^{4}}$

To easily see how to make $\frac{1}{16}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{10,000}$, let’s look at the factored form.

$\frac{1}{{2}^{4}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{{\left(2\cdot 5\right)}^{4}}$

Do you see that if to make 24 = ${\left(2\cdot 5\right)}^{4}$, we would need to multiply it by 54?

$\frac{{5}^{4}}{{5}^{4}}\cdot \frac{1}{{2}^{4}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{{\left(2\cdot 5\right)}^{4}}$

Another way to see this by setting up an equation, for the denominators:

${2}^{4}n={2}^{4}×{5}^{4}$

Solving by dividing by n we arrive at:

$n={5}^{4}.$

Then:

$\frac{1}{16}\cdot \frac{{5}^{4}}{{5}^{4}}=\frac{x}{10,000}$

$\frac{1}{16}=\frac{625}{10,000}$

 Un-factored Factored Denominator 2 Denominator 10 Denominator 2 Denominator 10 $\frac{1}{2}$ $\frac{5}{10}$ 0.5 $\frac{1}{{2}^{1}}\cdot \frac{{5}^{1}}{{5}^{1}}$ $\frac{{5}^{1}}{{\left(2\cdot 5\right)}^{1}}$ $\frac{1}{4}$ $\frac{25}{100}$ 0.25 $\frac{1}{{2}^{2}}\cdot \frac{{5}^{2}}{{5}^{2}}$ $\frac{{5}^{2}}{{\left(2\cdot 5\right)}^{2}}$ $\frac{1}{8}$ $\frac{125}{1000}$ 0.125 $\frac{1}{{2}^{3}}\cdot \frac{{5}^{3}}{{5}^{3}}$ $\frac{{5}^{3}}{{\left(2\cdot 5\right)}^{3}}$ $\frac{1}{16}$ $\frac{625}{10,000}$ 0.0625 $\frac{1}{{2}^{4}}\cdot \frac{{5}^{4}}{{5}^{4}}$ $\frac{{5}^{4}}{{\left(2\cdot 5\right)}^{4}}$

Let’s consider how this would work for a number that’s not a perfect power of 2, like 20.

$\frac{1}{20}=\frac{x}{10}$

$\frac{1}{\left({2}^{2}\cdot 5\right)}=\frac{x}{\left(2\cdot 5\right)}$

Keep in mind, we’re not just trying to get the denominators equal, we want them to be equal and a perfect power of 10. (So a denominator of 20 will not work, because decimals are all powers of ten.)

If you see the factors of 20 have 22, which tells us we are going to have a denominator of 102, 100.

$\frac{1}{\left({2}^{2}\cdot 5\right)}\cdot \frac{5}{5}=\frac{x}{\left(2\cdot 5\right)}\cdot \begin{array}{c}\\ 2\cdot 5\end{array}$

$\frac{5}{100}=\frac{x}{100}$

And 1/20th is 0.05, so we’re golden!

By using this new method, we can see that we must change the denominator of our original rational number into a power of ten in order for it to be written as a decimal (since decimals are all powers of ten).

The Pay-Off

Given these three points I will show you why some decimals are non-terminating. Short-story long here, perhaps, but the purpose is not for answer-getting, but to explore things mathematically and discover understanding in things previously mysterious.

Let’s take 1/3 for starters. To rewrite one-third as a decimal, accurately, without truncating the decimal expansion, we need to write multiply three by itself so it will be a power of 10.

$\frac{1}{{3}^{n}}=\frac{x}{{10}^{n}}$

This is impossible. The easiest reason why is that the powers of three end in, 1, 3, 7, 9 only. There is no power of three that ends in zero, and all powers of ten end in zero.

It turns out that all rational numbers, with denominators that have a prime factor other than 2 or 5 suffer such a fate. Seven ends in 1, 3, 7 or 9, only. Eleven only ends in 1.

Decimals compare a whole number to a power of ten. There is no way to multiply the prime numbers, other than 2 and 5, by a power that will result in a whole number.

Let us consider another method of explanation of why 3, for example, will never be a power of 10.

${3}^{x}={10}^{x}$

Taking the logarithm of both sides gives us:

$\mathrm{log}{3}^{x}=\mathrm{log}{10}^{x}$

$x\mathrm{log}3=x\mathrm{log}10$

Because log10 = 1,

$x\mathrm{log}3=x$

Solving for x

$x\mathrm{log}3-x=0$

$x\left(\mathrm{log}3-1\right)=0$

Log3 $–$ 1 cannot equal zero because 101 is not 3, so x must be zero.

Plugging that value into our original equation we see that only power of 3 and 10 that yields equal value is zero.

${3}^{0}={10}^{0}$

Conclusion:

I hope you really see the reason why rational numbers become non-terminating decimals is because of the nature of what decimals are. They are powers of 10. There are not any whole number multiples of prime numbers, other than two or five that produce a power of ten. So you cannot covert a seven into a power of ten, other than the power of zero.

Again, the purpose of this exploration was just that, to explore the math behind something we just gloss over and hopefully make connections and other such things.

This is not an attempt at a proof, but rather a justification and explanation. Let me know what works and what doesn’t work for you.

## Exponents Part 2

two

Exponents Part 2

Division

In the previous section we learned that exponents are repeated multiplication, which on its own is not tricky. What makes exponents tricky is determining what is a base and what is not for a given exponent. It is imperative that you really understand the material from the previous section before tackling what’s next. If you did not attempt the practice problems, you need to. Also watch the video that review them.

In this section we are going to see why anything to the power of zero is one and how to handle negative exponents, and why they mean division.

What Happens with Division and Exponents?

Consider the following expression, keeping in mind that the base is arbitrary, could be any number (except zero, which will be explained soon).

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3\cdot 3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 3. Note that 3 is just 31.

$\frac{{3}^{5}}{{3}^{1}}$

If we write this out to seek a pattern that we can use for a short-cut, we see the following:

$\frac{{3}^{5}}{{3}^{1}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3}$

If you recall how we explored reducing Algebraic Fractions, the order of division and multiplication can be rearranged, provided the division is written as multiplication of the reciprocal. That is how division is written here.

$\frac{{3}^{5}}{{3}^{1}}=\frac{3}{3}\cdot \frac{3\cdot 3\cdot 3\cdot 3}{1}$

And of course 3/3 is 1, so this reduces to:

$\frac{{3}^{5}}{{3}^{1}}=3\cdot 3\cdot 3\cdot 3={3}^{4}$

The short-cut is:

$\frac{{3}^{5}}{{3}^{1}}={3}^{5-1}={3}^{4}$

That is, if the bases are the same you can reduce. Reducing eliminates one of the bases that is being multiplied by itself from both the numerator and the denominator. A general form of the third short-cut is here:

Short-Cut 3: $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$

This might seem like a worthless observation, but this will help articulate the very issue that is going to cause trouble with exponents and division.

$\frac{{3}^{5}}{{3}^{1}}={3}^{5}÷{3}^{1}$ .

But that is different than

${3}^{1}÷{3}^{5}$

The expression above is the same as

$\frac{{3}^{1}}{{3}^{5}}$

This comes into play because

$\frac{{3}^{1}}{{3}^{5}}={3}^{1-5}$,

and 1 $–$ 5 = -4.

Negative Exponents?

In one sense, negative means opposite. Exponents mean multiplication, so a negative exponent is repeated division. This is absolutely true, but sometimes difficult to write out. Division is not as easy to write as multiplication.

Consider that 3-4 is 1 divided by 3, four times. 1 ÷ 3 ÷ 3 ÷ 3 ÷ 3. But if we rewrite each of those ÷ 3 as multiplication by the reciprocal (1/3), it’s must cleaner and what happens with a negative exponent is easier to see.

$1÷3÷3÷3÷3\to 1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}$

This is classically repeated multiplication. While one times itself any number of times is still one, let’s go ahead and write it out this time.

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot {\left(\frac{1}{3}\right)}^{4}$

This could also be written:

$1\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\cdot \frac{1}{3}\to 1\cdot \frac{{1}^{4}}{{3}^{4}}$

The second expression is easier, but both are shown here to make sure you see they are the same.

Since 1 times 14 is just one, we can simplify this further to:

$1\cdot \frac{{1}^{4}}{{3}^{4}}=\frac{1}{{3}^{4}}.$

Negative exponents are repeated division. Since division is hard to write and manipulate, we will write negative exponents as multiplication of the reciprocal. In fact, if instructions say to simplify, you cannot have a negative exponent in your final answer. You must rewrite it as multiplication of the reciprocal. Sometimes that can get ugly. Consider the following:

$\frac{b}{{a}^{-5}}$

To keep this clean, let us consider separating this single fraction as the product of two rational expressions.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{1}{{a}^{-5}}$

The b is not a problem here, but the other rational expression is problematic. We need to multiply by the reciprocal of $\frac{1}{{a}^{-5}}$, which is just a5.

$\frac{b}{{a}^{-5}}=\frac{b}{1}\cdot \frac{{a}^{5}}{1}={a}^{5}b$.

This can also be considered a complex fraction, the likes of which we will see very soon. Let’s see how that works.

$\frac{b}{{a}^{-5}}$

Note: ${a}^{-5}=\frac{1}{{a}^{5}}$

Substituting this we get:

$\frac{b}{\frac{1}{{a}^{5}}}$

This is b divided by 1/a5.

$b÷\frac{1}{{a}^{5}}$

Let’s multiply by the reciprocal:

$b\cdot {a}^{5}$

Now we will rewrite it in alphabetical order (a good habit, for sure).

${a}^{5}b$

Let us consider one more example before we make our fourth short-cut. With this example we could actually apply our second short-cut, but it will not offer much insight into how these exponents work with division.

This is the trickiest of all of the ways in which exponents are manipulated, so it is worth the extra exploration.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}$

As you see we have four separate bases. In order to simplify this expression we need one of each base (2, x, y, z), and all positive exponents. So let’s separate this into the product of four rational expressions, then simplify each.

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}\to \frac{2}{{2}^{-2}}\cdot \frac{{x}^{-2}}{x}\cdot \frac{{y}^{-5}}{{y}^{3}}\cdot \frac{z}{{z}^{-5}}$

The base of two first:

$\frac{2}{{2}^{-2}}\to 2÷{2}^{-2}$

We wrote it as division. What we will see is dividing is multiplication by the reciprocal, and then the negative exponent is also dividing, which is multiplication by the reciprocal. The reciprocal of the reciprocal is just the original. But watch what happens with the sign of the exponent.

First we will rewrite the negative exponent as repeated division.

$2÷\frac{1}{{2}^{2}}$

Now we will rewrite division as multiplication by the reciprocal.

$2\cdot {2}^{2}={2}^{3}$

Keep in mind, this is the same as 23/1.

We will offer similar treatment to the other bases.

Consider first $\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Negative exponents are division, so:

$\frac{{x}^{-2}}{x}=\frac{{x}^{-2}}{1}\cdot \frac{1}{x}$

Notice the x that is already dividing (in the denominator) does not change. It has a positive exponent, which means it is already written as division.

$\frac{{x}^{-2}}{1}\cdot \frac{1}{x}\to \frac{1}{{x}^{2}}\cdot \frac{1}{x}=\frac{1}{{x}^{3}}$

This is exactly how simplifying the y and z will operation.

$\frac{{2}^{3}}{1}\cdot \frac{1}{{x}^{2}\cdot x}\cdot \frac{1}{{y}^{5}\cdot {y}^{3}}\cdot \frac{z\cdot {z}^{5}}{1}$

Putting it all together:

$\frac{2{x}^{-2}{y}^{-5}z}{{2}^{-2}x{y}^{3}{z}^{-5}}=\frac{{2}^{3}{z}^{6}}{{x}^{3}{y}^{8}}$.

Short-Cut 4: Negative exponents are division, so they need to be rewritten as multiplication by writing the reciprocal and changing the sign of the exponent. The last common question is what happens to the negative sign for the reciprocal? What happens to the division sign here: $3÷5=3×\frac{1}{5}$. When you rewrite division you are writing it as multiplication. Positive exponents are repeated multiplication.

${a}^{-m}=\frac{1}{{a}^{m}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{{a}^{-m}}={a}^{m}$

This is the second to last thing we need to learn about exponents. However, a lot of practice is required to master them fully.

To see why anything to the power of zero is one, let’s consider:

${3}^{5}$

This equals three times itself five total times:

${3}^{5}=3"#x22C5;3\cdot 3\cdot 3\cdot 3$

Now let’s divide this by 35.

$\frac{{3}^{5}}{{3}^{5}}$

Without using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}=\frac{3\cdot 3\cdot 3\cdot 3\cdot 3}{3\cdot 3\cdot 3\cdot 3\cdot 3}=1$

Using short-cut 3, we have this:

$\frac{{3}^{5}}{{3}^{5}}={3}^{5-5}$

Five minutes five is zero:

${3}^{5-5}={3}^{0}$

Then 30 = 1.

Τhe 3 was an arbitrary base. This would work with any number except zero. You cannot divide by zero, it does not give us a number.

The beautiful thing about this is that no matter how ugly the base is, if the exponent is zero, the answer is just one. No need to simplify or perform calculation.

${\left(\frac{{3}^{2x-1}\cdot {e}^{\pi i}}{\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}}\right)}^{0}=1$

Let’s take a quick look at all of our rules so far.

 Short-Cut Example ${a}^{m}\cdot {a}^{n}={a}^{m+n}$ ${5}^{8}\cdot 5={5}^{8+1}={5}^{9}$ ${\left({a}^{m}\right)}^{n}={a}^{mn}$ ${\left({7}^{2}\right)}^{5}={7}^{10}$ $\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$ $\frac{{5}^{7}}{{5}^{2}}={5}^{7-2}={5}^{5}$ ${4}^{-3}=\frac{1}{{4}^{3}}$ ${a}^{0}=1$ 50 = 1

Let’s try some practice problems.

Instructions: Simplify the following.

1. ${\left({2}^{8}\right)}^{1/3}$ 2. $3{x}^{2}\cdot {\left(3{x}^{2}\right)}^{3}$

3. $\frac{5}{{5}^{m}}$ 4. $\frac{{5}^{2}{x}^{-3}{y}^{5}}{{5}^{-3}{x}^{-4}{y}^{-5}}$

5. $7÷7÷7÷7÷7÷7÷7÷7$ 6. $9{x}^{2}y÷9{x}^{2}y$

7. $9{x}^{2}y÷\left(9{x}^{2}y\right)$ 8. ${\left({x}^{2}\cdot 2{x}^{6}\right)}^{2}\cdot {\left({x}^{2}\cdot 2{x}^{6}\right)}^{-2}$

9. ${\left({a}^{m}\right)}^{n}\cdot {a}^{m}$ 10. $\frac{{\left(3{x}^{2}+4\right)}^{2}}{{\left(3{x}^{2}+4\right)}^{3}}$

## Exponents Part 1

exponents part 1

Exponents Part 1

One of the biggest things to understand about math is how it is written. The spatial arrangement of characters is syntax. Syntax, in English, refers to the arrangements of words to convey meaning.

Exponents are just a way of writing repeated multiplication. If we are multiplying a number by itself repeatedly, we can use an exponent to tell how many times the number is being multiplied. That’s it. Nothing tricky exists with exponents, no new operations or concepts to tackle. If you’re familiar with multiplication and its properties, exponents should be accessible.

That said, it is not without its pitfalls. A balance between conceptual understanding and procedural short-cuts is needed to avoid those pitfalls. The only way to strike that balance is through a careful progression of exercises and examples. An answer-getting mentality will lead to big troubles with exponents. People wishing to learn how exponents work must seek understanding.

Let’s establish some facts that will come into play with this first part of exponents.

1.      Exponents are repeated multiplication

To simplify simple expressions with exponents you only need to know a few short-cuts, but to recall and understand, we need more. These facts are important.

With an exponential expression we have a base, the number being multiplied by itself, and the exponent, the small number on the top right of the base which describes how many times the base is being multiplied by itself.

${a}^{5}$

The number a is the base. We don’t know what a is other than it is a number. It’s not a big deal that we don’t know exactly what number it is, we still know things about this expression.

Five is the exponent, which means there are five a’s, all multiplying together, like this: $a\cdot a\cdot a\cdot a\cdot a.$

Something to keep in mind is that this expression equals another number. Since we don’t know what a is, we cannot find out exactly what it is, but we do know it’s a perfect 5th power number, like 32. See, 25 = 32.

What if we had another number multiplying with a5, like this:

${a}^{5}\cdot {b}^{3}$

If we write this out, without the exponents we see we have 5 a’s and 3 b’s, all multiplying together. We don’t know what a or b equals, but we do know they’re multiplying so we could change the order of multiplication (commutative property) or group them together in anyway we wish (associative property) without changing the value.

${a}^{5}\cdot {b}^{3}=a\cdot a\cdot a\cdot a\cdot a\cdot b\cdot b\cdot b$

And these would be the same:

$\left(a\cdot a\right)\left(a\cdot a\cdot a\right)\left(b\cdot b\cdot b\right)$

$\left(a\cdot a\right)\left[\left(a\cdot a\cdot a\right)\left(b\cdot b\cdot b\right)\right]$

$\left(a\cdot a\right){\left[ab\right]}^{3}$

${a}^{2}{\left[ab\right]}^{3}$

This is true because the brackets group together the a and b, making them both the base. The brackets put them together. The base is ab, and the exponent is 3. This means we have ab multiplied by itself three times.

Keep in mind, these are steps but exploring how exponents work to help you learn to read the math for the intended meaning behind the spatial arrangement of bases, parenthesis and exponents.

Now, the bracketed expression above is different than ab3, which is $a\cdot b\cdot b\cdot b$.

${\left(ab\right)}^{3}\ne a{b}^{3}$

Let’s expand these exponents and see why this is:

${\left(ab\right)}^{3}\ne a{b}^{3}$

Write out the base ab times itself three times:

$\left(ab\right)\left(ab\right)\left(ab\right)\ne a\cdot b\cdot b\cdot b$

The commutative property of multiplication allows us to rearrange the order in which we multiply the a’s and b’s.

$a\cdot a\cdot a\cdot b\cdot b\cdot b\ne a\cdot b\cdot b\cdot b$

Rewriting this repeated multiplication we get:

${a}^{3}{b}^{3}\ne a{b}^{3}$

The following, though, is true:

$\left(a{b}^{3}\right)=a{b}^{3}$

On the right, the a has only an exponent of 1. If you do not see an exponent written, it is one. If we write it out we see:

$\left(a\cdot b\cdot b\cdot b\right)=a\cdot b\cdot b\cdot b$

In summary of this first exploration, the base can be tricky to see. Parenthesis group things together. An exponent written outside the parenthesis creates all of the terms inside the parenthesis as the base. But if numbers are multiplying, but not grouped, and one has an exponent, the exponent only belongs to the number just below it on the left. For example, $4{x}^{3},$ the four has an exponent of just one, while the x is being cubed.

Consider: ${\left(x+5\right)}^{3}.$ This means the base is x + 5 and it is multiplied by itself three times.

Repeated Multiplication Allows Us Some Short-Cuts

Consider the expression:

${a}^{3}×{a}^{2}.$

If we wrote this out, we would have:

$a\cdot a\cdot a×a\cdot a$.

(Note: In math we don’t use colors to differentiate between two things. A red a and a blue a are the same. These are colored to help us keep of track of what’s happening with each part of the expression.)

This is three a’s multiplying with another two a’s. That means there are five a’s multiplying.

${a}^{3}×{a}^{2}={a}^{5}$

Before we generalize this to find the short-cut, let us see something similar, but is a potential pitfall.

${a}^{3}×{b}^{2}$

If we write this out we get:

$a\cdot a\cdot a×b\cdot b$

This would not be an exponent of 5, in anyway. An exponent of five means the base is being multiplied by itself five times. Here we have an a as a base, and three of those multiplying, and a b as a base, and two of those multiplying. Not five of anything.

The common language is that if the bases are the same we can add the exponents. This is a hand short-cut, but if you forget where it comes from and why it is true, you’ll undoubtedly confuse it with some of the other short-cuts that follow.

Short-Cut 1: If the bases are the same you can add the exponents. This is true because exponents are repeated multiplication and the associative property says that the order in which you group things does not matter (when multiplying).

${a}^{m}×{a}^{n}={a}^{m+n}$

The second short-cut comes from groups and exponents.

${\left({a}^{3}\right)}^{2}$

This means the base is a3, and it is being multiplied by itself.

${a}^{3}×{a}^{3}$

Our previous short cut said that if the bases are the same, we can add the exponents because we are just adding how many of the base is being multiplied by itself.

${a}^{3}×{a}^{3}={a}^{3+3}={a}^{6}$

But this is not much of a short cut. Let us look at the original expression and the outcome and look for a pattern.

${\left({a}^{3}\right)}^{2}={a}^{6}$

Short-Cut 2: A power raised to another is multiplied.

${\left({a}^{m}\right)}^{n}={a}^{m×n}$

Be careful here, though:

$a{\left({b}^{3}{c}^{2}\right)}^{5}$ = $a{b}^{15}{c}^{10}$

Practice Problems

 1.       ${x}^{4}\cdot {x}^{2}$ 8. ${\left(5xy\right)}^{3}$ 2.       ${y}^{9}\cdot y$ 9. ${\left(8{m}^{4}\right)}^{2}\cdot {m}^{3}$ 3.       ${z}^{2}\cdot z\cdot {z}^{3}$ 10. ${\left(3{x}^{5}\right)}^{3}{\left({3}^{2}{x}^{7}\right)}^{2}$ 4.       ${\left({x}^{5}\right)}^{2}$ 11. $7{\left({7}^{2}{x}^{4}\right)}^{5}\cdot {7}^{3}{x}^{5}$ 5.       ${\left({y}^{4}\right)}^{6}$ 12. ${5}^{3}+{5}^{3}+{5}^{3}+{5}^{3}+{5}^{3}$ 6.       ${x}^{3}+{x}^{3}+{x}^{3}+{x}^{8}+{x}^{8}$ 13. ${3}^{2}\cdot 9$ 7.       ${4}^{x}+{4}^{x}+{4}^{x}$ 14. ${4}^{x}\cdot {4}^{x}\cdot {4}^{x}$

## The Purpose of Homework and My Response

The purpose of homework is to promote learning.  That’s it.  It’s not a way to earn a grade or something to keep kids busy.  It’s also not something that just must be completed in order to stay out of trouble.  Homework is a chance to try things independently, make mistakes and explore the nature of those mistakes in order to better learn the material at hand.

If students are not learning from the homework, it is a waste of time and effort.  There are a few things that could cause students not to learn from the homework.  Even if the assignments are of high quality, without the reflection and correction piece, students will not learn much from homework.

Reflection and correction go together.  It’s not about getting right answers, but thinking about what caused mistakes, identifying misconceptions or procedural inefficiencies and replacing those.  To reflect a student should NOT erase their incorrect working but instead should write on their homework, in pen, what went wrong and what would have been better.

It is quite possible more can be learned when reviewing homework than any other time.  It is certainly a powerful experience.

Textbooks and videos, tutors and peer help offer little appropriate support to help make homework, or practice, meaningful.  Textbooks only provide correct answers, YouTube videos usually do similar treatment to topics as textbooks offer.

I wish to help students learn and believe that reviewing work that has been done is too powerful of an opportunity to pass.  The trick is, how can I provide reflection and insight when to someone I am not sitting with and talking to?  I think I can help provide this reflection piece by doing all of the practice problems myself on a document camera and discussing pitfalls and mistakes, as well as sharing my thinking about the problems as I tackle them.  Further, I can share typical mistakes I see from students as they are learning topics.

So as I develop the Algebra 1 content I will be working on adding videos and short written responses to the assignments to help students think about what they’ve done, its appropriateness, correctness and their level of understanding.

## Promoting Mathematical Literacy through Non-Procedural Questions

Math was okay until they threw the alphabet in it.

I have heard, Math was okay until they threw the alphabet in it, so many times from so many adults, who I wish were trying to be witty, but are in fact completely serious.  And these are reasonable, intelligent people.  Why does the abstraction of a variable or an unknown become so confusing that it creates a huge disconnect?

I’d like to share with you an assignment that I believe will help students transition from the concrete application and properties of Real Numbers to the abstractions we deal with in Algebra and mathematics beyond.

The desired outcomes of this assignment are:

• Improve mathematical literacy by encouraging students to read the mathematical meanings created by the spatial arrangement of numbers and symbols
• Improve their understanding of the Order of Operations and to help student realize that the Order of Operations is not its own topic, compartmentalized, but rather an over-arching understanding of how math is performed.
• Promote abstract thinking about numbers and their properties
• Introduce some concepts that will come into play later in Algebra, like finding the x-intercepts of a quadratic once it is factored

How to Introduce the Assignment

Students must be aware that they will be dealing with abstract ideas and that there are sometimes more than one right answer.  Also, a student can be right, but not completely right, they could also be wrong, but not always.  By fostering a healthy discussion about these problems you can introduce the idea that in order for something to be mathematically true, it must always be true.  If a single circumstance is untrue, then the statement is untrue.

Consider the problem: Given that a and b are real numbers, and the following is true, what do you know about the numbers a and b?   a×b=0

A student might say, In this case a and b are both zero.

That is correct in one case, but there are many cases where that is not true.  It is true that one of them must be zero.

The First Prompt

Given that a and b are real numbers, and the following statement is true, what can you conclude at the numbers a and b?

a – b = 0

At first have them think and write on their own.  Make sure they’re all working, not avoiding this uncomfortable notion.

After a given amount of time (short, maybe one minute), instruct them to talk with two different people.  Be clear that the expectation is that they take turns, one person shares, the other listens and responds.

After the time is up, have a whole-class discussion, but avoid being the authority until the discussion is winding down.  Only be the authority on the subject to help summarize.

The Second Prompt

With the same properties of the numbers and the statement being true, provide them with the equation:

(a + 5)(b –7) = 0

Conduct conversations as you did with the first, maybe allowing more time for them to talk together as this is a more complicated situation.

The Third Act

Switching gears from properties of variables to applying some properties of real numbers will promote their understanding of when the associative and commutative properties as well as challenge their understanding of how math is written.  We are really trying to promote their ability to read and write math and their fact that the spatial arrangement has meaning in math.

Have the students try and add parenthesis, as many as they like, to the following equation, so that it will be true.  Have them try it on their own first, then provide a short amount of time for peer discussion.

3•2 – 72 + 5 = 80

When you conduct a whole-class discussion, make sure it’s student lead, your role is as a mediator, not a disseminator of facts.

Fourth Act

Instruct students to create a similar problem by making a statement they know is true and removing the parenthesis.   For example, they might make up:

8(5 – 3) + 11 = 27

But would only write

8•5 – 3 + 11 = 27

When they’re completed their problem, have them show you.  Once everybody has a problem, hand out 3×5 cards.  On the front of the card the student will write their problem, without the parenthesis.  On the back, they’ll write their name.  Have them pass the cards forward and you can distribute them at to another class the following day.

Last Thing for the Lesson:

The idea here is the same as with the previous activity, but we are accessing their knowledge from a different angle.  They will be given an expression with parenthesis and be asked if the parenthesis can be removed without changing the value of the expression.

For example:  (5 + 4) – 2, or 11 – (4 – 9).

Introduce these problems in the same fashion, with quiet thinking first, then small group discussions, then whole-class discussion.

The Homework:

The homework is critical here because it will challenge students to think and examine how the way in which math is written changes the meaning.  It will also force them to think about numbers in a general fashion.

Timing

The last thing I’d like to mention is that this could easily be done over two separate days depending on the aptitude of the class you’re teaching.

I hope this is helpful and food for thought.

## Why Teaching Properties of Real Numbers is Important

If you are going to do a fraction review, the lesson here might be of some help.  I believe things are best reviewed in context, but this is a decent set of information that also introduces the real numbers and some other basics of math.

The PDF icon to the left has a lesson outline you can feel free to use with the PowerPoints of in any way you see fit.

The structure is all there in the lessons, but they're not over scripted.  Remember, I believe the majority of a lesson should be spontaneous.  It should be anticipated and prepared for, but how the lesson really unfolds depends on the audience.

Below you will find an overview of how and why I teach real numbers as well as two PowerPoint icons you can download and use as your own.  I only ask that you share where you found them.

Anything you purchase from Amazon.com through the banner below goes to producing more materials, and at no cost to you.

### What Good Is It?

The Real Number Line has always been one of the dullest lessons I have to teach.

Natural Numbers are the set of numbers you can count on your fingers, beginning with one.  The Whole Numbers are the Natural Numbers and Zero...Integers are ...

Blah Blah Blah

I have to teach it because it's in the curriculum.  And I always wonder, what use is it if a student knows the difference between a whole number and a natural number?

It is hypocritical of me to complain in such a fashion because I laud the virtues of education being greater than a set of skills or a body of knowledge.  Education is about learning to think, uncovering something previously unknown that ignites excitement and interest.  Education should change how you see yourself, how you think about the world.  It should enrich our lives.

Teaching the Real Number Line can be a huge first step in that direction, if done properly.

### Math is About Ideas, Not Just Computation

There are some rich, yet entirely approachable, mathematical ideas that can be introduced with the Real Number Line (RNL).  For example, a series of questions to be posed to students could be:

1.  The Natural Numbers are infinite, meaning, they cannot be counted entirely.  How do we know that?
2. The Integers are also infinite.  How do we know that?
3.  Is infinity a number?
4. Which are there more of, Natural Numbers of Integers?  How can you know, if they're both infinite?

The idea of an axiom can be introduced.  Most likely, students assume math is true, or entirely made up, but correct or incorrect, because it is written in a book and claimed to be such by a teacher.  The idea of how we know what we know and if math is an invention or a discovery can be introduced by talking about axioms.  For example:

1. Is it true that 5 + 4 = 4 + 5 ?
2. If a and are Real Numbers, would it always be true that b = b?  (What if they were negative?)
3. Is it also true that b = b - a?  How do we know that?
4. Is the following also true:  If a = b, and b = c, then a = c?  How do we know?

The idea here is not to teach students the difference between the Associative Property and the Commutative Property, but to use these properties to introduce students to math as a topic that can be discussed, and that it is not about answer getting, but instead about ideas.

For more on this topic and a few other related items, visit this page.

### Why Are Some Rational Numbers Non-Terminating Decimals?

If you had a particularly smart group of students, you could pose this question.  I mean, after all, 1/3 = 0.3333333333333333...  And yet, we are told rational numbers include decimals that can be written as a fraction (the ratio of two integers).

How it works is sometimes very clear and clean.  For example, 0.7 is said, "Seven tenths." And "Seven tenths," can also be written as the ratio of seven and ten.  And the number seven tenths is of course equal to itself, regardless of how it is written.  The number 0.27 is said, "twenty seven hundredths," which can easily be written as the ratio of twenty seven, for the numerator, and one hundred, as the denominator.  And this can continue so long as the decimal terminates.  But try the same thing with the a repeating decimal and you do not end up with things that are equal.

The algorithm to convert a repeating, but non-terminator decimal into a fraction is pretty straight forward.

But that does not address why a rational number would be a non-terminating decimal.

Click the PPT Icon to the left to download a lesson on converting repeating decimals into fractions for honors students.  It includes a proof of why the square root of two is irrational.

The simple reason that some rational numbers cannot be expressed as terminating decimals has to do with our numbering system.  We use base 10 numbers.  Our decimal system provides us an easy way to write fractions with denominators that are powers of 10.

That means that we have ten numbers, including zero, that fill up one column, like a car’s odometer.  When you travel 9 miles the odometer will read 000009.  When you travel the tenth mile the odometer will read 000010.

Not all of our number systems are base ten. While the metric system is base 10, or at least translates into powers of ten, the Imperial system is base 12 from inches to feet, but 5,280 feet to the next unit of miles, and so on.

Time is another great example of bases other than ten.  Seconds and minutes are base sixty.  You need sixty seconds before you have an hour, not ten.  But hours are base 24 because 24 hours are needed to make one of the next category, which is days.

In time, 25 minutes of an hour is the ratio:

But in base ten this is 0.4166666666666666... Our decimal system does math in base ten, not base sixty.  This is not 41 minutes!  A typical mistake would be two say 25 minutes is 0.25 of an hour.

Back to our original example of 1/3.  Not all numbers can be cleanly divided into groups of ten, like 3.  If we had a base 3 numbering system, where after the third number we moved, then 1/3 would just be 0.1.  But in our numbering system, 0.1 is one tenth.

Other numbers, like four, translate into ten more easily.  Consider the following:

The only issue remaining is that 2.5/10 is not a rational number because 2.5 is not an integer and rational numbers are ratios of two integers.  This can be resolved as follows:

Let's try the same process with 1/3.

As you can see, we will keep getting ten divided by three, forever.

This is a great example of how exploring a question can uncover many topics within the scope of the course being taught.

I hope this has caused you to pause and think of how exploring questions, relationships and properties in mathematics can lead to greater understanding than just teaching process and answer getting.