Exponents

Quick aside

Finally, we're at the starting chapter of "Art of Problem Solving". Do not be discouraged, as I took a long roundabout way of getting here, explaining many topics that usually are not covered before this. This was made in order to build strong foundations, as we're now equipped to deal with a strange new operator.

But before that, a small notation change, since multiplication is going to be very common, it is now going to be made implicit. If you see terms seemingly connected by "no operator", you can simply sneak in a $*$. For example:

$$3(5+6) \Rightarrow 3*(5+6)$$ $$3x+7 \Rightarrow 3*x+7$$ $$(x+1)(x+2) \Rightarrow (x+1)*(x+2)$$ $$x(x+1)(x+2) \Rightarrow x*(x+1)*(x+2)$$

The reason for this is that we're going to start seeing a whole bunch of "x" with different multiplications and operands, and typing the $*$ symbol each time is going to clutter the expression. As an example:

$$3x + 3xy + 3zxy \Rightarrow 3*x+3*x*y + 3*z*x*y$$

It's a lot more concise. Do not worry either, since it's multiplications as usual, we can do our usual method of grouping up, so that the above can be:

$$3x(1+y+yz)$$

Definition

The operator is defined as "^". But usually it's not actually written at all, and instead opted to use "superscript". So "2^3" would be:

$$2^3$$

The way this operator works is very similar to the multiplication. It is thus defined:

$$a^b = \overbrace{a*a*a*\ldots *a}^{b \text{ times}}$$

Our previous example extends to:

$$2^3 = 2*2*2 = 8$$

The "2" is called "base" and the "3" is called exponent. Think of it like the base being our radio channel, and the exponent like a volume knob (a very wild one). Modifying the exponent modifies how many times we multiply our base with itself. In our system of operator priority this has highest priority except for parentheses. So it will attach to the first term it sees.

$$2*3^3 = 2*(3^3)$$

Only the 3 gets the "exponential" treatment.

This has already profound implications, as we can imagine. First of all, since we're simply multiplying "2"s together, what happens if we multiply another "2" three more times?

$$2*2*2*(2*2*2) = 2^3*(2*2*2) = 2^3*2^3 = 2^6$$

The 6 comes from the fact we're indeed multiplying the 2 six times in total. What if instead was just two more times?

$$2*2*2*(2*2) = 2^3*(2*2) = 2^3*2^2 = 2^5$$

And just like in our unary example with addition, it seems we're simply adding the two exponents together. Therefore we can produce the following law:

$$a^b*a^c = a^{b+c}$$

Of course, this does not work if the two bases are different, as our definition would break apart.

$$2^3*3^3 = 2*2*2*3*3*3 = \ldots ?$$

In the above case not all hope is lost though, as we can kind of get by if we use the commutative property:

$$2*2*2*3*3*3 = 2*3*2*3*2*3$$

Let's use the associative property to group "2*3":

$$2*3*2*3*2*3 = (2*3)*(2*3)*(2*3)$$

And from our definition above, we're indeed multiplying $2*3$ three times.

$$(2*3)*(2*3)*(2*3) = (2*3)^3 = 6^3$$

So it seems that if we have the same exponent, we can actually multiply bases together. Which brings to a second law:

$$a^n*b^n=(a*b)^n$$

Let's play with exponents

The zero exponential

Let's think about the implications a moment. Take the first law, and consider "is it truly any number?".

$$a^b*a^c = a^{b+c}$$

What if $c=0$?

$$a^b*a^0 = a^{b+0}$$

Well, I have got no clue what $a^0$ is, so we're going to use our trusty variable "x" as a placeholder.

$$a^b*x = a^{b+0}$$

We do know what $b+0$ is though, so we simplify that...

$$a^b*x = a^b$$

So if we multiply this quantity $a^b$ by $x$ we get back the same quantity... Knowing our laws, we know there's an element that has that function within multiplication! Therefore we know $x=1$! Extending those laws:

$$\frac{a^b*x}{a^b} = \frac{a^b}{a^b}$$ $$x = 1$$

Wait, what was "x" supposed to represent?

$$a^0 = x$$ $$a^0 = 1$$

Huh. Therefore it seems that if we were to multiply a "zero times", we do not actually get zero, but one instead. This also seems to work for any "a" I put at all...? Wait, what would $0^0$ mean? Is it still $1$?

$$0^b*0^c = 0^{b+c}$$

Let's put our $c=0$...

$$0^b*0^0 = 0^{b+0}$$ $$0^b*0^0 = 0^b$$ $$\frac{0^b*0^0}{0^b} = \frac{0^b}{0^b}$$

Wait wait wait, stop right here, you violated the law! Since $0^b$ extends to $0*0*0*\ldots*0$, it means that whole result is zero?

$$\frac{0*0^0}{0} = \frac{0}{0}$$

Uh-oh, we just divided by zero. That means this division technically has all possible results (by our definition of inverse), and we cannot actually keep a single result in there. Therefore, as sad as it is to say, for today $0^0$ will have to remain undefined (we'll see why sometimes this can have a value, in the limits section).

This is generally why we avoid using base zero, as it generates so many complications due to the division being so problematic.

The negative exponential

Okay, back to our maybe actual numbers. What if $c < 0$? Let's assume something like $c = -d$, and try to keep our first law to the test.

$$a^b*a^c = a^{b+c}$$

Let's substitute our c for our new value:

$$a^b*a^{-d} = a^{b+(-d)}$$ $$a^b*a^{-d} = a^{b-d}$$

It seems like a negative value ends up in a subtraction. We do not have a value for this yet, let's try to see how this works in actual numbers? Taking $b=3$ and $d=2$:

$$a^3*a^{-2} = a^{3-2}$$

Let's assume our $x=a^{-2}$ since we do not know what that value even is, and we're keeping it distant from all the others.

$$a^3*x = a^{3-2}$$ $$a^3*x = a^1$$

$a^1$ is "a multiplied just once..." therefore only "a":

$$a^3*x = a$$

Okay, we're getting somewhere... Let's try dividing by $a^3$ both sides?

$$\frac{a^3*x}{a^3} = \frac{a}{a^3}$$

For the left side we know we can simplify easily, but for the right one, let's try to expand it a bit?

$$x = \frac{a}{a*a*a}$$

We know we can simplify that too, but the numerator clearly will still need to be there, so we expand it to just "multiply by one":

$$x = \frac{a*1}{a*a*a}$$

Now we can remove an "a" from both top and bottom...

$$x = \frac{1}{a*a}$$

And we know that $a*a$ is $a^2$. Let's go back to our "x" definition and plug it back in...

$$a^{-2} = \frac{1}{a^2}$$

A-ha! If we were to redo all this with a different number for "d", we would obtain a similar result, therefore, we can now generalize!

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

That's nice. Wait a moment, we know those "a" at the bottom of the fraction could be spread out... Let's take back $a^{-2}$ for a moment.

$$a^{-2} = \frac{1}{a^2}$$

Expand that fraction...

$$a^{-2} = \frac{1}{a*a}$$

But we know from our fraction article, we can separate the denominators as multiplying them by $\frac{1}{a}$, what if we apply it here?

$$a^{-2} = \frac{1}{a}*\frac{1}{a}$$

This seems to also work with any quantity of b (our minus 2). Therefore we get a second generalization, for free!

$$a^{-b} = \frac{1}{a^b} = \overbrace{\frac{1}{a}*\frac{1}{a}*\ldots*\frac{1}{a}}^{b \text{ times}}$$

Wait a moment, we just defined something similar! If we take our $\frac{1}{a}$ as a term...

$$a^{-b} = \frac{1}{a^b} = \left(\frac{1}{a}\right)^b$$

All of those are equal ways to write the same thing! We can swap freely, based on our convenience in manipulating the expressions.

Fractional base

Let's take a small detour, what if our base was fractional instead?

$$\left(\frac{a}{b}\right)^n = \overbrace{\frac{a}{b}*\frac{a}{b}*\ldots*\frac{a}{b}}^{n \text{ times}}$$

We know from our fraction exploration we can simply join all the numerators and denominators together by multiplying them, if the two fractions are multiplied together.

$$\overbrace{\frac{a}{b}*\frac{a}{b}*\ldots*\frac{a}{b}}^{n \text{ times}} = \frac{\overbrace{a*a*\ldots*a}^{n \text{ times}}}{\overbrace{b*b*\ldots*b}^{n \text{ times}}}$$ $$\frac{\overbrace{a*a*\ldots*a}^{n \text{ times}}}{\overbrace{b*b*\ldots*b}^{n \text{ times}}} = \frac{a^n}{b^n}$$

This part is simply a corollary of the above, as we now know a fraction simply keeps the exponent in both numerator and denominator. For the reader:

• Prove the following:

$$\left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}$$

Multiplication of exponents

Let's see what multiplying exponents together does.

We try to see how $a^{b*c}$ expands. First we expand our multiplication definition.

$$a^{b*c} = a^{\overbrace{b+b+\ldots+b}^{c \text{ times}}}$$

We can now see that it's made up of many "+" symbols. As we know from before, a plus in the exponent means a multiplication in the actual number. So we translate:

$$a^{\overbrace{b+b+\ldots+b}^{c \text{ times}}} = \overbrace{a^b * a^b * \ldots * a^b}^{c \text{ times}}$$

That's a repeated multiplication! We apply the definition as we know it...

$$\overbrace{a^b * a^b * \ldots * a^b}^{c \text{ times}} = \left(a^b\right)^c$$

So putting it all together:

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

By the way, $a^{b^c} \neq \left(a^b\right)^c$. See for example:

$$\begin{aligned} 3^{(3^2)} &=_? (3^3)^2 \\ 3^{9} &=_? (27)^2 \\ 19683 &=_? 729 \end{aligned}$$

Therefore the convention is to treat serial exponentiation as the exponentiation of the exponent:

$$a^{\small b^{\small c}} = a^{(\small b^{\small c})}$$

This is because the other way around can be expressed as a simple multiplication of exponents, as we've seen.

Inverse

As we know, every operation we define, usually has an inverse. In case of addition, subtraction. For multiplication, division. In case of of exponentiation, it's the nth-root. Let's first define it mathematically, as we usually do for every other symbol (let's first call it $\square$ so we can immediately tell it's a binary operator):

$$a^b = c \iff c\square b = a$$

The actual symbol and notation is:

$$a^b = c \iff \sqrt[\small b]{c} = a$$

And by corollary:

$$\sqrt[\small b]{a^b} = a$$

There's a few issues with this operator already. First of all, it's a lot more difficult to calculate by hand. We can approximate it by rounding to the nearest integer, but it's not going to be close. Second, we have actual issues considering $b=0$. Since $a^0=1$, the only possible zero-th root would be of 1, but writing

$$\sqrt[\small 0]{1}$$

runs also into issues since we've demonstrated all numbers (except 0) are a possible result. So in general, we assume that $b \neq 0$.

Also, since exponentiation to the power of 2 is common, (as it's called squaring a number), the 2nd-root is just called square root, and it foregoes the root itself.

$$\sqrt{4} = \sqrt{2*2} = \sqrt{(2)^2} = 2$$

In the above expression, the power of 2 and the root cancel each other out, as they are inverse. The issue of course, is that the whole number has to be squared for that to cancel out. An example here:

$$\sqrt{8} = \sqrt{2*2*2} = \sqrt{(2*2)*2} = \sqrt{2^2*2} = ???$$

We cannot easily cancel out the root. To better understand what sorts of manipulations we can do, we are going to delve into an extra type of number we've neglected as exponent.

Fractional exponent

We've explored in an exponent what it means to add, subtract, even multiply. Our next target is division, better represented by fractions.

$$a^{\large \frac{1}{b}} = ???$$

Let's see if we have some law we can use to find out something more.

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

What happens if we take $c=\frac{1}{b}$?

$$a^{\normalsize b*\frac{1}{b}} = \left(a^b\right)^{\normalsize \frac{1}{b}}$$ $$a^{1} = \left(a^b\right)^{\normalsize \frac{1}{b}}$$ $$a = \left(a^b\right)^{\normalsize \frac{1}{b}}$$

So a power of $\frac{1}{b}$ is enough to bring $a^b$ into back to $a$. But wait, didn't we just say that's the inverse operation?

$$\sqrt[\small b]{a^b} = a$$

Therefore we discover that:

$$a^{\normalsize \frac{1}{b}} = \sqrt[\small b]{a}$$

And we can generalize it to (if you want to demonstrate it as an exercise, just remember that $\frac{c}{b} = c*\frac{1}{b}$):

$$a^{\normalsize \frac{c}{b}} = \sqrt[\small b]{a^c}$$

Okay! What about where we got stumped before?

$$\sqrt{8} = \sqrt{2*2*2} = \sqrt{(2*2)*2} = \sqrt{2^2*2} = ???$$

Well, we now know square root is just exponentiation to the power of $\frac{1}{2}$!

$$\sqrt{2^2*2} = \left(2^2*2\right)^{\normalsize \frac{1}{2}}$$

And we know when the base is a multiplication, we can just split it off in two...

$$\left(2^2*2\right)^{\normalsize \frac{1}{2}} = \left(2^2\right)^{\normalsize \frac{1}{2}}*2^{\small \frac{1}{2}}$$ $$\left(2^2\right)^{\normalsize \frac{1}{2}}*2^{\small \frac{1}{2}} = 2^{2*\frac{1}{2}}*\sqrt{2}$$ $$2^{2*\frac{1}{2}}*\sqrt{2} = 2^1*\sqrt{2} = 2*\sqrt{2}$$

Therefore it seems that we can break up a multiplication inside a root , as long as we keep the root on both terms that are then multiplied together.

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

This cannot be done with addition or subtraction. What about division? Let's test.

$$\sqrt{\frac{a}{b}} = \sqrt{a*\frac{1}{b}} = \sqrt{a}*\sqrt{\frac{1}{b}}$$

Wait a moment, can we do a bit better? We know this is exponentiation in disguise so...

$$\sqrt{a}*\sqrt{\frac{1}{b}} = \sqrt{a}*\left(\frac{1}{b}\right)^{\small \frac{1}{2}}$$

Wait, if we reference (49) we can do better!

$$\sqrt{a}*\left(\frac{1}{b}\right)^{\small \frac{1}{2}} = \sqrt{a}*\frac{1}{b^{\frac{1}{2}}}$$ $$\sqrt{a}*\frac{1}{b^{\frac{1}{2}}} = \sqrt{a} * \frac{1}{\sqrt{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$

And we've proven it works for division too!

Spoilers

As a taste of what's to come, here's a bit of a preview of what's upcoming. First of all, the square root is not a straightforward operation. For example:

$$\sqrt{4} = 2$$

because we know $2*2=4$. But if we think about it, there's something weird. I've mentioned before that $(-2)*(-2)=4$. So technically, this root shouldn't be only positive. And exactly, the true result of a square root is both positive and negative:

$$\sqrt{4} = \pm 2$$

The $\pm$ symbol stands for "plus/minus" as it's just the two symbols stacked on top of each other. This allows us to define that there's actually two values in there, which share the same "absolute value".

$$\sqrt{4} = 2 \land \sqrt{4} = -2$$

So our equations start to break down too with the introduction of exponents. This is due to having only shown equations that are "linear", ergo they're to a single power ($x^1$ to be precise). We're going to see equations of the type:

$$x^2 = 4$$

And while a square root is going to be our inverse method for getting "x", it's important to note that it will produce two possible values of $x=2$ and $x=-2$. Our dreams of having at most a single value of "x" are now completely gone.

Arguably, $\sqrt[3]{8}$ does not have the same issue! As we know that $(-2)*(-2)*(-2) = -8$ instead, the value is retained.

It seems like values like $(-2)^a$ are going to be more problematic too, this is going to be depending if "a" is even or odd, as the value determines our total value of "*" operations we do, and whether we'll be able to remember our sign (plus or minus) or not.

A note on absolute value:

$$|a|$$

is always a positive value, no matter whether a was positive or negative before. It represents the amount of "xs" or "-xs" the value represents, without a care if they were positive or negative "xs".

$$|5| = 5$$ $$|-5| = 5$$ $$|0| = 0$$

A little bit of a question: can we ever put a negative number under a square root?

$$\sqrt{-1} = ???$$

Arguably, for what we know as of now, any form of $a*a$ or $(-a)*(-a)$ ended up always giving a positive number. In case of $-a$ due to the "minuses" canceling out $(-2)*(-2)=4$.

So, for the moment, our answer is no. That does not mean every root is to be thrown out, as we're still able to give good answers like

$$\sqrt[\small 3]{-8} = -2$$

because we know

$$(-2)*(-2)*(-2) = (-2)^3 = -8$$

Interestingly enough, we're going to soon explore the implications of what if the answer is yes.

Inequality

Our inequality expressions break a little as we transcend the linear equations. They will require a more thorough study of the variables we put forward, and a singular inequality can now spawn multiple.

As an example:

$$x^2 < 4$$

We know both $2$ and $-2$ are points in which $x^2$ becomes 4. If we take $-3$ we get to $9$, so we're already out, and taking $3$ also does the same. Taking a point in between like $1$ instead gives us $1*1=1$, so it's still less than $4$. Therefore that inequality becomes:

$$-2 < x < 2$$

For more complex equations, like

$$x^2 + 4x +4 > 0$$

We'll need to consider some more tricks in order to reduce the values we consider. In this specific case, we can employ a trick, knowing that $4x=2x+2x$:

$$x^2 + 2x+2x +4 > 0$$

We group up the terms like so:

$$(x^2 + 2x)+(2x+4) > 0$$

And in the first group, we use the distributive property of multiplication to group up an "x":

$$(x(x + 2))+(2x+4) > 0$$

And in the second we group up "2":

$$x(x + 2)+2(x + 2) > 0$$

Now we group up the term $x+2$:

$$(x + 2)(x + 2) > 0$$

And we know that's a repeated multiplication and we can express it as a square.

$$(x + 2)^2 > 0$$

We know that a square always produces positive numbers, except for $0^2=0$, so we just take the base $x+2$ and determine that it has to be different than zero.

$$x + 2 \neq 0$$ $$x \neq -2$$

If we were to substitute $-2$ to the equation, we would get:

$$(-2)^2 + 4*(-2) +4 > 0$$ $$4 - 8 + 4 > 0$$ $$0 > 0$$

Which does not satisfy it, but every other number does.

This was a special case that we could solve algebraically. More complex cases will have to be done through function analysis, as they won't have always easy answers.