## How Math Fixed Music

### Rational Exponents Sound GREAT

Before we dive in, music is primarily defined by what we hear, not by the analysis and insight provided by math.  For example, an octave is a note whose frequency is double that of its parent note.  The mathematical relationship was discovered after the fact.  The following is an exploration of how math is used in music, but I don’t want to put the cart before the horse here.  The math supports the music, makes it work.  But the math is really fine-tuning what we hear.

Pythagoras developed a musical system that over the years evolved into what we have today. (At least Pythagoras is often credited for it.)  Not until “recently,” however, has one of the major problems with music been resolved (see what I did there with resolve?).

The problem the ancients had is that their octaves didn’t line up.  An octave, as I mentioned early, is a note that has twice (or half) the frequency of another note.  Octaves, in modern western music, share the same names, too.  The note A, at 440 Hz, has an octave at 880 Hz, and also 220 Hz.   (There are infinitely many octaves, in theory, though our ears have a limited range of things we can hear.)  The ancients, however, had a problem because after a few octaves, well, they were no longer octaves.

In western music we have 12 semi-tones, A, A# (or B-flat), B, C, C#, D, D#, E, F, F#, G, G# and then A again.  It’s cyclic, repeated infinitely both higher and lower. Each semi-tone in the next series of 12 notes is an octave of our first series of notes.  And the relationship between notes is what makes them, well, musical, not just sounds.

The problem is defining that relationship.  You see, because each note is slightly higher (has a higher frequency), and each note’s octave is double that frequency, what happens is the notes get further and further apart (the differences in their frequencies increases).

Let’s take a look at the frequencies:

 Note Frequency A 220.00 A# 233.08 B 246.94 C 261.63 C# 277.18 D 293.66 D# 311.13 E 329.63 F 349.23 F# 369.99 G 392.00 G# 415.30 A 440.00 A# 466.16 B 493.88 C 523.25 C# 554.37 D 587.33 D# 622.25 E 659.25 F 698.46 F# 739.99 G 783.99 G# 830.61 A 880.00

As you can see, the differences between consecutive notes is increasing, at an increasing rate!  This is not a linear relationship.  Because of this, the ancients had a very hard time defining what was an A and what was a D, especially when you started moving around between octaves.  Things got jumbled, and out of tune.

It is tricky to find the proportion and rate of change between consecutive notes, any two consecutive notes that is.  That’s where math comes in to save the day.  Let’s build the rate of change, shall we.

First, note that the rate is increasing, at an increasing rate, so we cannot add.  I show that in the video below.  We have to multiply.  When we repeatedly multiply, we can use exponents.  Since we need a note and it’s octave to be doubles, our base number is 2.

Since there are twelve notes between a note and its octave, we need to break the multiple of two into twelve equal, multiplicative parts.  That’s a rational exponent, 1/12.

The number we need to multiply each note by is 21/12.  Each note is one-twelfth of the way to the octave.  It is pretty cool indeed.

For a great read on this topic, consider the book Harmonograph.

## Rational Exponents and Logarithmic Counting …

rational exponents

Rational Exponents

In the last section we looked at some expressions like, “What is the third root of twenty-seven, squared?” The math is kind of ugly looking.

$\sqrt[3]{{27}^{2}}$

The procedures are clunky and it is very easy to lose sight of the objective. What this expression is asking is what number cubed is twenty-seven squared. You could always square the 27, to arrive at 729 and see if that is a perfect cube.

There is a much more elegant way to go about this type of calculation. Turns out if we rewrite this expression with a rational exponent, life gets easier.

$\sqrt[3]{{27}^{2}}={27}^{2/3}$

These two statements are the same. They ask the same question, what number cubed is twenty-seven squared?

By now you should be familiar with perfect cubes and squares. Hopefully you’re also familiar with higher powers of 2 and 3, as well as a few others. For example, you should recognize that 625 is ${5}^{4}.$ If you don’t know that yet, a cheat sheet might be helpful.

Let’s look at our expression again. If you notice that 27 is a perfect cube, then you can rewrite it like this:

${27}^{2/3}\to {\left({3}^{3}\right)}^{2/3}$

Maybe you see what’s going to happen next, but if not, we have a power raised to another here, we can multiply those exponents. Three times two-thirds is two. This becomes three squared.

${\left({3}^{3}\right)}^{2/3}\to {3}^{2}=9$

Not too bad! We factor, writing the base of twenty-seven as an exponent with a power that matches the denominator of the other exponent, multiply, reduce, done!

Let’s look at another.

Simplify:

${625}^{3/4}$

We mentioned earlier that 625 was a power of 5, the fourth power of five. That’s the key to making these simple. Let’s rewrite 625 as a power of five.

${\left({5}^{4}\right)}^{3/4}$

We can multiply those exponents, giving us five-cubed, or 125. Much cleaner than finding the fourth root of six hundred and twenty-five cubed.

What about something that doesn’t work out so, well, pretty? Something where the base cannot be rewritten as an exponent that matches the denominator?

${32}^{3/4}$

This is where proficiency and familiarity with powers of two comes to play. Thirty-two is a power of two, just not the fourth power, but the fifth.

${\left({2}^{5}\right)}^{3/4}$

If we multiplied these exponents together we end up with something that isn’t so pretty, ${2}^{15/4}.$ We could rewrite this by simplifying the exponent, but there’s a better way. Consider the following, and note that we broke the five twos into a group of four and another group of one.

${\left({2}^{5}\right)}^{3/4}={\left({2}^{1}\cdot {2}^{4}\right)}^{3/4}$

Now we’d have to multiply the exponents inside the parenthesis by $¾$, and will arrive at:

${2}^{3/4}\cdot {2}^{3}$

Notice that ${2}^{3/4}$ is irrational, so not much we can do with it, but two cubed is eight. Let’s write the rational number first, and rewrite that irrational number as a radical expression:

$8\sqrt[4]{{2}^{3}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\sqrt[4]{8}$.

There’s an even easier way to think about these rational exponents. I'd like to introduce something called Logarithmic Counting.  For those who don't know what logarithms are, that might sound scary.

Do you remember learning how to multiply by 5s...how you'd skip count?  (5, 10, 15, 20, ...)  Logarithmic counting is the same way, except with exponents.  For example, by 2:  2, 4, 8, 16, 32, ... Well, what’s the fourth step of 2 when logarithmically counting? It’s 16, right? 

Let’s look at ${16}^{3/4}$. See the denominator of four? That means we’re looking for a fourth root, a number times itself four times that equals 16. The three, in the numerator, it says, what number is three of the four steps on the way to sixteen?

2 4 8 16

Above is how we get to sixteen by multiplying a number by itself four times. Do you see the third step is eight?

Let’s see how our procedure looks:

Procedure 1:

${16}^{3/4}={\left({2}^{4}\right)}^{3/4}$

${\left({2}^{4}\right)}^{3/4}={2}^{\frac{4}{1}×\frac{3}{4}}={2}^{3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\text{\hspace{0.17em}}8.$

Procedure 2:

${16}^{3/4}=\sqrt[4]{{16}^{3}}$

$\sqrt[4]{{16}^{3}}=\sqrt[4]{{\left({2}^{4}\right)}^{3}}$

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

$\sqrt[4]{{2}^{4}}×\sqrt[4]{{2}^{4}}×\sqrt[4]{{2}^{4}}=2×2×2,\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8.$

The most elegant way is to realize the 16 is the fourth power of 2, and the fraction $¾$ is asking us for the third entry. What is 3/4s of the way to 16 when multiplying (exponents)?

Let’s look at ${625}^{2/3}.$ Let’s do this three ways, first with radical notation, then by evaluating the base and simplifying the exponents, and then by thinking about what is two thirds of the way to 625.

Now this is going to be a tricky problem because 625 is NOT a perfect cube. It is the fourth power of 5, though, which means that 125 (which is five-cubed) times five is 625.

${625}^{2/3}=\sqrt[3]{{625}^{2}}$

$\sqrt[3]{{625}^{2}}=\sqrt[3]{{\left({5}^{4}\right)}^{2}}\to \sqrt[3]{{5}^{8}}$

$\sqrt[3]{{5}^{8}}=\sqrt[3]{{5}^{3}×{5}^{3}×{5}^{2}}$

$\sqrt[3]{{5}^{3}×{5}^{3}×{5}^{2}}=5×5\sqrt[3]{25},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}25\sqrt[3]{25}$

Pretty ugly!

Exponential Notation:

${625}^{2/3}={\left({5}^{4}\right)}^{2/3}$

${\left({5}^{4}\right)}^{2/3}={\left({5}^{3}×{5}^{1}\right)}^{2/3}$

${\left({5}^{3}×{5}^{1}\right)}^{2/3}={5}^{2}×{5}^{2/3}$

${5}^{2}×{5}^{2/3}=25×{5}^{2/3},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}or\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}25\sqrt[3]{25}$

A little better, but still a few sticky points.

Now our third method.

${625}^{2/3}$ asks, “What is two thirds of the way to 625, for a cubed number?”

This 625 isn’t cubed, but a factor of it is.

${625}^{2/3}={\left(125×5\right)}^{2/3}$
This could also be written as:

${125}^{2/3}×{5}^{2/3}$

I am certain that 5 to the two-thirds power is irrational because, well, five is a prime number. Let’s deal with the other portion.

The steps to 125 are: 5 25 125

The second step is 25.

${125}^{2/3}×{5}^{2/3}=25×{5}^{2/3}$

To summarize the denominator of the rational exponent is the index of a radical expression. The numerator is an exponent for the base. How you tackle the expressions is entirely up to you, but I would suggest proficiency in multiple methods as sometimes the math lends itself nicely to one method but not another.

Practice Problems:



## Cube Roots and Higher Order Roots

other roots

Cube Roots
and

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:

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

## Addition and Subtraction of Square Roots

Mathematical Operations and Square Roots

Part 1

In this section we will see why we can add things like $5\sqrt{2}+3\sqrt{2}$ but cannot add things like $2\sqrt{5}+2\sqrt{3}$. Later we will see how multiplication and division work when radicals (square roots and such) are involved.

Addition and Subtraction: Addition is just repeated counting. The expression $5\sqrt{2}$ means $\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}+\sqrt{2}$, and the expression So if we add those two expressions, $5\sqrt{2}+3\sqrt{2},$ we get $8\sqrt{2}$ . Subtraction works the same way.

Consider the expression $2\sqrt{5}+2\sqrt{3}$. This means $\sqrt{5}+\sqrt{5}+\sqrt{3}+\sqrt{3}.$ The square root of five and the square root of three are different things, so the simplest we can write that sum is $2\sqrt{5}+2\sqrt{3}$.

A common way to describe when square roots can or cannot be added (or subtracted) is, “If the radicands are the same you add/subtract the number in front.” This is not a bad rule of thumb, but it treats square roots as something other than numbers.

$5×3+4×3=9×3$

The above statement is true. Five groups of three and four groups of three is nine groups of three.

$5\sqrt{3}+4\sqrt{3}=9\sqrt{3}$

The above statement is also true because five groups of the numbers squared that is three, plus four more groups of the same number would be nine groups of that number.

However, the following cannot be combined in such a fashion.

$3×8+5×2$

While this can be calculated, we cannot add the two terms together because the first portion is three $–$ eights and the second is five $–$ twos.

$3\sqrt{8}+5\sqrt{2}$

The same situation is happening here.

Common Mistake: The following is obviously wrong. A student learning this level of math would be highly unlikely to make such a mistake.

$7×2+9×2=16×4$

Seven $–$ twos and nine $–$ twos makes a total of sixteen $–$ twos, not sixteen $–$ fours. You’re adding the number of twos you have together, not the twos themselves. And yet, this is a common thing done with square roots.

$7\sqrt{2}+9\sqrt{2}=16\sqrt{4}$

This is incorrect for the same reason. The thing you are counting does not change by counting it.

Explanation: Why can you add $5\sqrt{2}+3\sqrt{2}$? Is that a violation of the order of operations (PEMDAS)? Clearly, the five and square root of two are multiplying, as are the three and the square root of two. Why does this work?

Multiplication is a short-cut for repeated addition of one particular number. Since both terms are repeatedly adding the same thing, we can combine them.

But if the things we are repeatedly adding are not the same, we cannot add them together before multiplying.

What About Something Like This: $3\sqrt{40}-9\sqrt{90}$?

Before claiming that this expression cannot be simplified you must make sure the square roots are fully simplified. It turns out that both of these can be simplified.

$3\sqrt{40}-9\sqrt{90}$

$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$

The dot symbol for multiplication is written here to remind us that all of these numbers are being multiplied.

$3\cdot \sqrt{4}\cdot \sqrt{10}-9\cdot \sqrt{9}\cdot \sqrt{10}$

$3\cdot 2\cdot \sqrt{10}-9\cdot 3\cdot \sqrt{10}$

$6\sqrt{10}-27\sqrt{10}$

$-21\sqrt{10}$

What About Something Like This: $\sqrt{7+7}$ versus $\sqrt{7}+\sqrt{7}.$

Notice that in the first expression there is a group, the radical symbol groups the sevens together. Since the operation is adding, this becomes:

$\sqrt{7+7}=\sqrt{14}$.

Since the square root of fourteen cannot be simplified, we are done.

The other expression becomes:

$\sqrt{7}+\sqrt{7}=2\sqrt{7}.$

Summary: If the radicals are the same number, the number in front just describes how many of them there are. You can combine (add/subtract) them if they are the same number. You are finished when you have combined all of the like terms together and all square roots are simplified.

Practice Problems: Perform the indicated operation.

$\begin{array}{l}1.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{25}-5\sqrt{5}+5\\ \\ \\ 2.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{48}+3\sqrt{3}\\ \\ \\ 3.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\sqrt{75}+8\sqrt{24}+\sqrt{75}\\ \\ \\ 4.\text{\hspace{0.17em}}\text{\hspace{0.17em}}\sqrt{200}+8\sqrt{8}-2\sqrt{32}\\ \\ \\ 5.\text{\hspace{0.17em}}\text{\hspace{0.17em}}-2\sqrt{98}+16\sqrt{2}\end{array}$

## 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!

## 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.

## What Do Grades Really 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 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.

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!