## Cube Roots and Higher Order Roots

other roots

Cube Roots
and
Other Radicals

Square roots ask what squared is the radicand. A geometric explanation is that given the area of a square, what’s the side length? A geometric explanation of a cube root is given the volume of a cube, what’s the side length. The way you find the volume of a cube is multiply the length by itself three times (cube it).

The way we write cube root is similar to square roots, with one very big difference, the index.

$\begin{array}{l}\sqrt{a}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}square\text{\hspace{0.17em}}\text{\hspace{0.17em}}root\\ \sqrt[3]{a}\to \text{\hspace{0.17em}}\text{\hspace{0.17em}}cube\text{\hspace{0.17em}}\text{\hspace{0.17em}}root\end{array}$

There actually is an index for a square root, but we don’t write the two. It is just assumed to be there.

Warning: When writing cube roots, or other roots, be careful to write the index in the proper place. If not, what you will write will look like multiplication and you can confuse yourself. When writing by hand, this is an easy thing to do.

$3\sqrt{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{8}$

To simplify a square root you factor the radicand and look for the largest perfect square. To simplify a cubed root you factor the radicand and find the largest perfect cube. A perfect cube is a number times itself three times. The first ten are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1,000.

Let’s see an example:

Simplify:

$\sqrt[3]{16}$

Factor the radicand, 16, find the largest perfect cube, which is 8.

$\sqrt[3]{8}×\sqrt[3]{2}$

The cube root of eight is just two.

$2\sqrt[3]{2}$

The following is true,

$\sqrt[3]{16}=2\sqrt[3]{2}$,

only if

${\left(2\sqrt[3]{2}\right)}^{3}=16$

Arithmetic with other radicals, like cube roots, work the same as they do with square roots. We will multiply the rational numbers together, then the irrational numbers together, and then see if simplification can occur.

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

Two cubed is just eight and the cube root of two cubed is the cube root of eight.

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)\left(\sqrt[3]{2}\right)$

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\sqrt[3]{2\cdot 2\cdot 2}$

${2}^{3}×{\left(\sqrt[3]{2}\right)}^{3}=8×\sqrt[3]{8}$

The cube root of eight is just two.

$8×\sqrt[3]{8}=8×2$

${\left(2\sqrt[3]{2}\right)}^{3}=16$

Negatives and cube roots: The square root of a negative number is imagery. There isn’t a real number times itself that is negative because, well a negative squared is positive. Cubed numbers, though, can be negative.

$-3×-3×-3=-27$

So the cube root of a negative number is, well, a negative number.

$\sqrt[3]{-27}=-3$

Other indices (plural of index): The index tells you what power of a base to look for. For example, the 6th root is looking for a perfect 6th number, like 64. Sixty four is two to the sixth power.

$\sqrt[6]{64}=2\text{​}\text{​}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}because\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{2}^{6}=64.$

A few points to make clear.

·         If the index is even and the radicand is negative, the number is irrational.

·         If the radicand does not contain a factor that is a perfect power of the index, the number is irrational

·         All operations, including rationalizing the denominator, work just as they do with square roots.

Rationalizing the Denominator:

Consider the following:

$\frac{9}{\sqrt[3]{3}}$

If we multiply by the cube root of three, we get this:

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{3}}{\sqrt[3]{3}}=\frac{9\sqrt[3]{3}}{\sqrt[3]{9}}$

Since 9 is not a perfect cube, the denominator is still irrational. Instead, we need to multiply by the cube root of nine.

$\frac{9}{\sqrt[3]{3}}\cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}}=\frac{9\sqrt[3]{9}}{\sqrt[3]{27}}$

Since twenty seven is a perfect cube, this can be simplified.

$\frac{9}{\sqrt[3]{3}}=\frac{9\sqrt[3]{9}}{3}$

And always make sure to reduce if possible.

$\frac{9}{\sqrt[3]{3}}=3\sqrt[3]{9}$

This is a bit tricky, to be sure. The way the math is written does not offer us a clear insight into how to manage the situation. However, the topic we will see next, rational exponents, will make this much clearer.

Practice Problems:

Simplify or perform the indicated operations:

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[4]{64}\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{9}+4\sqrt[3]{9}\\ \\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[3]{9}×4\sqrt[3]{9}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt[5]{64}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\sqrt{7}}{\sqrt[3]{7}}\end{array}$

## Multiplying and Dividing Square Roots, Rationalizing the Denominator

1.7.2 square root operations continued

Square Roots
Multiplication and Division

At some point square roots should no longer be considered an operation but rather the most efficient way to express a number. For example, the best way to write one hundred trillion is $1×{10}^{14}$. The best way to express the number times itself that is two is as $\sqrt{2}.$

That provides insight when we consider multiplying a rational number and an irrational number together. It is not confusing for some irrational numbers, like π. Nobody confused 3π because we understand that symbol is the best way to write the number. There’s not a way to rewrite multiples of π other than by writing the multiple in front.

However, $3\sqrt{2}$ is often written as $\sqrt{6}$. There are reasons explained by the order of operations which tell us why this is false, but understanding what the square root of two is perhaps offers the simplest insight.

$\sqrt{2}\approx 1.414$

$3\sqrt{2}=\sqrt{2}+\sqrt{2}+\sqrt{2}$

$3\sqrt{2}\approx 1.414+1.414+1.414$

4.242

The square root of six is approximately 2.449. Not the same thing at all.

The following, however, is true:

$\sqrt{2}×\sqrt{3}=\sqrt{6}$

and

$\sqrt{2×3}=\sqrt{6}$.

The following generalization can be used. Sometimes it is best to write things one way versus another, and it is up to you to decide if rewriting an expression offers insight.

$\sqrt{ab}=\sqrt{a}\cdot \sqrt{b}$

If two numbers are both square roots you can multiply their radicands together. But you cannot multiply the radicand of a square root with rational number like we saw above.

Division is a little more nuanced, but only when your denominator is a fraction.

This generalization is true for division:

$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

For example:

$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{2}.$

This can be calculated two ways.

$\frac{\sqrt{8}}{\sqrt{4}}=\sqrt{\frac{8}{4}}=\sqrt{2}.$

or

$\frac{\sqrt{8}}{\sqrt{4}}=\frac{2\sqrt{2}}{2}=\sqrt{2}.$

But you cannot divide rational numbers into the radicand, or the radicand of a square root into a rational number. Remember, square roots, when simplified, are the most efficient way of writing irrational numbers. If we used k to represent the square root of two, these types of confusing things would not be happening.

Nobody would confuse what is happening with
$\frac{6}{k}$. We simply cannot evaluate that because 6 and k do not have common factors. When k is written as the square root of two, sometimes people just see a 2 and reduce.

The only issue with division of square roots occurs if you end up with a square root in the denominator.

$\frac{5}{\sqrt{2}}$

Denominators must be rational and the square root of two is irrational. However, there’s an easy fix. Remember that $\sqrt{2}×\sqrt{2}=\sqrt{4},$ and $\sqrt{4}=2.$ It is also true that:

$\frac{\sqrt{2}}{\sqrt{2}}=1$ .

To Rationalize the Denominator, which means make the denominator a rational number, we just multiply as follows:

$\frac{5}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{2}$

Sometimes we end up with something like this:

$\frac{5}{3\sqrt{2}}$

Three is a rational number and is perfectly okay in the denominator. If you multiply by the fraction $\frac{3\sqrt{2}}{3\sqrt{2}},$ you can still get the simplified equivalent, but you’ll have extra reducing to do at the end. Instead, just multiply by the irrational portion.

$\frac{5}{3\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{5\sqrt{2}}{6}$.

In summary, to divide or multiply with square roots, you can multiply or divide the radicands. However, if you’re multiplying or dividing rational numbers and square roots, you cannot combine the radicands and the rational numbers.

Practice Problems:

Perform the indicated operations:

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(5\sqrt{7}\right)\left(3\sqrt{14}\right)\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(\sqrt{15}\right)\left(\sqrt{3}\right)\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3\sqrt{2}}{\sqrt{8}}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\sqrt{5}}{\sqrt{3}}\\ \\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{3}{\sqrt{8}}\cdot \frac{\sqrt{2}}{6}\end{array}$

## The Most Important Component of Quality Teaching is …

What do you think the single most important part of effective teaching, in high school, is?

Breaking down classroom management and teaching into a lock and step routine is impossible.  People are too variable.  And, especially in high school, we are talking about the interactions of 150 – plus people a day!

It is because of the nature of how people behave and interact, how our motivations to fit in and get along guide a lot of our decisions that I claim establishing relationships is the single most important aspect of effective teaching, in high school.

I didn’t always feel this way.  I believed that discipline, structure, and content were king.  They’re certainly first tier, but they’re not king: Relationships are.

For me the light first clicked on when I watched an episode of Undercover Boss.  Here's a clip of the episode.

In this episode the corporate offices wanted to see why one location, that was not geographically or demographically different than the other stores, outsold the other stores.  Was it management, something on the retail side?

It turned out this woman, Dolores, had worked there for 18 years and she knew EVERY single customer by name and knew about them.  People just kept coming back because she knew them, took care of their needs because she knew them, and also, because she knew them, they felt welcome.

Do I Really Need a Relationship with the Students?

In high school students don’t have much choice.  They have to come see you daily.  But that alone will not make them respectful, engaged, and willing participants.  Dolores showed me that if you just get to know people, and are warm and welcoming, they’ll be willing and eager to show up.  This translates nicely to high school.

When you have a relationship with students that are far more compliant out of genuine respect.  They’re willing to participate and enjoy being in your class, even if they don’t like your subject (happens to me a lot with math).

By having relationships with students your day is also a lot nicer.  If you’re down, or off, for whatever reason, instead of taking advantage of you, like sharks smelling blood in the water, they’re on their best behavior – if you have a good relationship with them.

How Do I Build a Relationship with So Many Students?

So how can we build relationships with students when you have 35 per class shuffling in and out every 55 minutes or so?  I mean, there’s teaching, testing, checking homework, discipline, interruptions from the office, … the list goes on and on.  How can we develop relationships with students with all of that going on?

The first way is just small talk.  Not everybody is good at that, but it is easy with kids.  Ask them simple things like if they have pets, and then about their pets or if they wish they could have a pet.  Ask them about the nature of their family, how many siblings they have, where they fit in (birth order).

Another way to build this relationship is to have a “Pet Wall” where students can bring pictures of their pets and place them on that part of the wall.  It generates conversation, which is what’s needed to establish these relationships.

Giving sincere compliments is a great way to build relationships.  But, they must be sincere.  There’s almost nothing more insulting than an insincere compliment, there’s certainly nothing more condescending.  When students see you treating others with kindness and generosity it endears you to them.  They gauge a lot of their relationship with you on how you treat others.

How you handle discipline is very important, too.  If you berate a child in an unprofessional manner, you lose a lot of that hard earned relationship with other students.  They may not like the kid who is always a distraction, however, again, they gauge their relationship with you by how they see you treating others.

The last thing I’ll share here is that you can share things about yourself with them.  It can be funny stories or minor conflicts in your life, nothing that crosses a professional boundary, but things to which they can relate.  A story about how your toast fell and landed jelly side up (or down as the case may be), and so on.

It is incredibly difficult to site one thing as most important because no one factor of teaching stands on its own.  If too much focus is placed on one thing, at the expense of others, an imbalance will lead to poor teaching.

All that said, I believe that establishing relationships is the most important thing you can do as a high school teacher.  It will not only make the students more willing, it will also greatly improve the quality of your day!

Let me know your thoughts.  Thanks for reading.

## 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.

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

Let’s answer the square root questions we can answer:

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

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

For a PDF of the following, click here.

For a PowerPoint of this information, click here.

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.

## What Do Grades Really Mean?

What Do Grades Mean

The following is highly contentious.  Many of the situations discussed here should ultimately be considered on an individual basis.  The purpose of this is not to create a rubber-stamp solution to all problems that arise with grade assignment and student ability and or performance, but is to provide a general framework so that those individual decisions can be made in fairness and with respect to what is best for the student.

In a previous post I asked about a student in summer school that obviously knew Algebra 1 (he earned 100% on his quizzes and tests), but failed during the year because he didn’t do his classwork.  The question is, Does he deserve to fail Algebra 1?

When you flip the situation around it is equally interesting.  There are many kids who work hard, but do not really understand or learn the math.  Do they deserve to pass based on the merits of effort?

The real issue with both of these situations is what grades mean, or what should they mean.  When I worked at Cochise Community College I adopted their definition of letter grades which is described below:

A – Mastery

B – Fluency

C – Proficiency

D – Lacking Proficiency

Those are clean and inoffensive definitions of grades.  A student with an A has mastered the material.  To be fluent means you can navigate the materials but not without error.  To be proficient means you can get the job done, but there are some gaps in ability, but the student can demonstrate a measurable level of command of all of the objectives. Students who earn a D are not able to demonstrate proficiency.

A student who struggles with the material does not deserve an A, even if they worked harder than those who earned an A.  This might seem unfair, but unless the objective of the class is to teach the value of hard work, to reward the hardworking, but barely proficient, student with a label of mastery is to cheat the student and cheapen the merit of your class.

Do these definitions mean that a lazy kid that get 95% on the final exam deserves an A, but that a hard working kid that gets a 52% on the same final deserves an F?  I say, with a few qualifications, yes.

Is this really fair to the student who works hard but has not yet realized an appropriate level of mastery to be awarded a passing grade? (I used the phrase, “has not yet,” instead of, “cannot,” to acknowledge the belief that students can learn, and if they are motivated and working, the only question will be the time scale of when they learn the material.)

I would say, for a math class, that the best thing that can happen is they are awarded the appropriate grade, an F.  Consider if this student is given a passing grade and the class is a prerequisite course?  They’re truly set up for failure in the subsequent class.

There is perhaps no worse example of bad teaching that remains within legals bounds than to inappropriately assign grades to students.  If a student deserves a C based on ability, but is given an A based on effort, they will believe they are doing everything right and do not need to improve in order to achieve similar success in subsequent courses.

But to give a student who possesses mastery a failing grade in a class because of lack of work ethic is to teach the student that passing classes is a matter of compliance.  Behave and you’ll be rewarded.  Those kids are taught that grades are not a reflection of knowledge or ability, and that means that education is not about learning.  To me, this is an injustice.

I do not believe in the efficacy of these objective lessons.  That would be, failing a student based on the notion that they do not deserve to pass because they are lazy. I believe that given meaningful and challenging opportunities, most of these highly intelligent, but seemingly lazy, students will show themselves to be hard working with amazing focus and direction and incredible capacity for quality work.

What about percentages.  Is it appropriate that an 80% is a B, if a B means fluency?

When I first began teaching I would have said, absolutely, a student does not deserve an A if they scored an 87% on their test.  Since then I’ve changed my mind.  Some topics require higher than 90% accuracy to be awarded an A, while with other topics, mastery might be far below 90%.

The level of complexity, variability of solutions and length of assessment all must be considered.  This is why sometimes a grading rubric is far superior to assigning grades based on a percentage of correctness or completion.

I teach a curriculum that is designed and tested by Cambridge University, the IGCSE test is what students take.  They have a very different way of assigning and defining grades than we use here in the United States.  Without going into details about how they do the specifics, they assign large portions of credit based on evidence of appropriate thinking.  In other words, if a student demonstrates understanding they will receive passing credit.  But, to achieve a high grade, mastery is truly measured.  And yet, in math at least, the percentages of correctness for mastery are usually in the mid-70’s.  This is because the nature of the questions asked are often non-procedural and the method of solution is not clear, students cannot be trained on how to answer the questions they face on IGCSE exams.

How Do Students Earn Grades

How a student can earn a grade varies, or should, depending on subject and age, and perhaps even minor topic within the subject.  I believe that separating student work into weighted categories is an appropriate method of helping make transparent to the student how their grade will be assigned.  It also by-passes the tricky question of, “What is a point?”  For me, a homework assignment is worth 5 points, they’re assigned daily, except Fridays, for a total of 20 points for the week.  Yet, a quiz might only be worth 12 points, but will be a far more accurate representation of student’s ability on the topic.

By assigning weights to the categories, this can be easily balanced.  This begs the question, how do you weight the categories?

But what about the student who works, performs all assigned tasks, but can only demonstrate a level of understanding best described as “Lacking Proficiency?”  Shouldn’t hard work be rewarded?

And whatever your beliefs on these questions, would your opinion change depending on the age of the student, or perhaps the subject?  Should a Chemistry student be rewarded for effort in the same way they’d be rewarded for effort in a Dance class?

At some point, nobody cares about potential or effort.  If a child’s mother wants his room clean, she knows he has the potential to clean it, but if he fails to do so, the potential matters not.  And if he’s really trying to get it done, but cannot master the discipline to carry through the task, does the effort really matter?

Here is how I set up my grades for high school.  It is nuanced and complicated, but I’ll give the outline.  Note that for college classes I use a different system.

In high school I weigh categories of grades and have changed the percentages and categories over time until I settled on what seems to work best.  These work for my students because it seems to motivate the lazy-smart students and also rewards the hardworking – low aptitude student, because if they remain persistent, they will learn.

Tests – 40%
Quizzes – 25%
Homework – 25%
Other – 10%

I believe extra credit should be awarded for students that perhaps help others, or for extraordinary performance.  However, a student should NOT be allowed to raise their grade through extra credit.  That is, at the end of the term a student is given a pile of work, that if performed, will raise their grade.  This is bad teaching!

The difference between a quiz and a test is similar to the difference between a doctor’s check-up versus an autopsy.  The quiz is a chance to see how things are going and adjust accordingly.  The test is final.  In high school I award credit for homework based on completion, but do not accept late homework.

Rewarding Effort?

While I wish that effort equaled success, it doesn’t always work that way…depending on how you define success.  For example, I can try as hard as possible to paint a world-famous landscape, but will likely fail if my measure of success is producing a world-famous piece of art. That said, I believe there is a reward beyond measure only discovered with true effort.  Our potential, our best, is not static, it changes.  It changes in respect to our current level of effort.  We can never fulfill our potential, you see.  It is always slightly above how hard we are trying.  So, if you’re not really trying, your potential decreases, but if you’re pushing your limits, the limits themselves stretch.  That is the real downfall of those with an inherent talent that never learn to push themselves.  Their potential decreases, dropping down to just higher than their level of effort.

I greatly reward effort, encourage it and makes positive examples of how effort promotes success.  However, I do not assign grades to effort.  How hard someone needs to try in a given subject to be successful varies entirely upon the student’s aptitude.  And suppose you have a truly gifted student, they could be great, if they learn to work hard, right?

Well, perhaps, but there’s more than work ethic involved in greatness.  What role does passion play?  Take a great young musician and over-structure their training and practice, they’ll burn out.  You’ll snuff their passion.

Grades

I asked the boy whose situation started this whole conversation if he felt he deserved to be in summer school.  Before he answered I explained that I didn’t have an expected answer, I didn’t really know if he belonged in summer school or not.  Without hesitation, he said he did deserve summer school, because, he said, he was lazy.

So maybe the kid will learn that if he’s lazy he gets punished.  But he also learns that grades are arbitrary, with respect to ability.

I do not like objective lessons, do not believe them to be effective.  I prefer a punishment that fits the crime, but also one that redirects the offender, allows them to correct their action.

I cannot say in this child’s case specifically, I was not there and I am not judging his teacher, but perhaps a quicker punishment that redirected him could have also taught him that being lazy was unacceptable and at the same time also allowed him to see grades as a reflection of his abilities.

All that said, this is highly contentious and varies incredibly depending on particular situations of students.

Let me know what you think, agree or disagree.  Leave me a comment.

## 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!

For more information, follow this link.

As always, thank you for reading.

## Why Do Some Fractions Repeat as Decimals?

For a PDF of the following, click here.

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.

As always, thanks for reading.

## 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.

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

What about Zero?

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

Reading Math

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

2.      Multiplication is repeated addition

3.      Addition is “skip” counting

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. ${4}^{x}\cdot {4}^{x}\cdot {4}^{x}$