# Exponents

After addition, multiplication (serial addition), subtraction (addition in reverse), and division (inverse multiplication), comes exponentiation (serial multiplication). Most of us learned about the basic x² stuff in school — it just means x-times-x. Likewise, x³ just means x-times-x-times-x. Serial multiplication. No problem.

But sometimes it’s x-2 or x½, and it’s hard to see how those work. As it turns out, it can all be understood based on a single axiom:

$\displaystyle{x}^{n}={x_1}\times{x_2}\times\ldots\times{x_n}$

Which is the rule behind the basic understanding you already have. The exponent says how many times to multiply a value times itself. Put in formal math terms, it looks like this:

$\displaystyle\prod^{n}_{i=1}{x}_{i}$

From this axiom we can very easily derive the first theorem:

$\displaystyle{x}^{1}={x}$

Any number to the power of one is just that number because there is only one instance of it in the multiply chain.

We can also derive what is often presented as a second axiom (but is really another derived theorem):

$\displaystyle{x}^{a+b}={x^a}\times{x^b}$

Because:

$\displaystyle{x}^{n}={x}^{a+b}=\left({x_1}\cdot{x_2}\cdot\ldots{x_a}\right)\times\left({x_1}\cdot{x_2}\cdot\ldots{x_b}\right)$

The multiply chain can be broken into two parts consisting of acount instances and bcount instances (where a+b=n). Those two parts are obviously multiplied together, so we derive the second theorem:

$\displaystyle{x}^{a+b}={x^a}\times{x^b}$

Note that it’s possible to treat this equality as the axiom and derive the axiom above (along with everything else). In either case, a single axiom leads to all the theorems.

We derive another key equality by noting that (by addition and the first theorem):

$\displaystyle{x}^{1+0}={x}^{1}={x}$

And therefore (using the second theorem):

$\displaystyle{x}^{1+0}={x}^{1}\times{x}^{0}={x}$

If we divide the last two terms by x¹ (which, per the first theorem, is just x) we end up with the very important third theorem:

$\displaystyle{x}^{0}={1}$

Any number raised to the power of zero is just one.

The theorem for x0 lets us derive a rule that might be surprising. We start by noting that:

$\displaystyle{x}^{(n-n)}={x^0}={1}$

By subtraction and the third theorem. If we re-express this as:

$\displaystyle{x}^{\left(n+(-n)\right)}={x^0}={1}$

Then we then can invoke the second theorem to say:

$\displaystyle{x}^{\left(n+(-n)\right)}={x^n}\times{x^{-n}}={1}$

If we divide the last two terms by xn, we have the fourth theorem:

$\displaystyle{x}^{-n}=\frac{1}{x^n}$

Which says that negative exponents are the inverses of positive ones. A negative exponent just means “one-over” the positive version.

We can also derive what happens with non-integer exponents. We’ll start with a simple example by first noting that:

$\displaystyle{x}^{(\frac{1}{2}+\frac{1}{2})}={x}^{1}={x}$

By addition and the first theorem. As we’ve done above, we can invoke the second theorem to give us:

$\displaystyle{x}^{(\frac{1}{2}+\frac{1}{2})}={x}^{\frac{1}{2}}\times{x}^{\frac{1}{2}}=\left({x}^{\frac{1}{2}}\right)^2={x}$

And if we take the square root of the last two terms, we get a special case of the fifth theorem:

$\displaystyle{x}^{\frac{1}{2}}=\sqrt{x}$

We can make this more general by starting with:

$\displaystyle{x}^{(\frac{1}{n}+\frac{1}{n}+\cdots\frac{1}{n})}={x}^{1}={x}$

Where the fraction 1/n is repeated n times, and then, using the second theorem:

$\displaystyle{x}^{(\frac{1}{n}+\frac{1}{n}+\cdots\frac{1}{n})}=\left({x}^{\frac{1}{n}}\right)^n={x}$

Using the same logic as above (taking the nth root of last two terms) we get the general case of the fifth theorem:

$\displaystyle{x}^{\frac{1}{n}}=\sqrt[n]{x}$

So fractional exponents (with a numerator of one) give us roots.

Finally, recursive exponents such as:

$\displaystyle\left({x}^{a}\right)^{b}$

Note that, by the initial axiom:

$\displaystyle\left({x}^{a}\right)^{b}=(x^a)_{1}\times(x^a)_{2}\times\ldots(x^a)_{b}$

And each xa expands to:

$\displaystyle{x}^{a}={x}_{1}\times{x}_{2}\ldots{x}_{a}$

So, we have x times itself a times, times itself b times, which is just a times b in terms of total instances of x. This gives us the sixth theorem:

$\displaystyle\left({x}^{a}\right)^{b}={x}^{({a}\times{b})}$

Recursive exponents just multiply. This works nicely in reverse. For instance:

$\displaystyle{x}^{\frac{3}{5}}={x}^{(\frac{1}{5}\times{3})}=\left({x}^{\frac{1}{5}}\right)^{3}=\left(\sqrt[5]{x}\right)^3$

Which can be very helpful with fractional exponents that have numerators other than zero.

Amazin’ the theorems we can derive from that single axiom, eh?