Laws of indices

Question

Well, given any real number $x$ and any positive integer $n$, the number $x^{n}$ is defined to be the product $x.x.x. ... x.x $ ($n$ times). But, how do we define $x^{r}$ when $r$ is a negative integer, a positive rational number and a negative rational number. Plus, why was it necessary to define $a^{0} = 1$ and not any other number, for all real numbers $a$? I'm in real number arithmetic and not in complex numbers.

Answer 1

When you look at the positive integer definition, you discover that $x^n * x^k = x^{n+k}$. That rule is so attractive that it makes sense to try to define $x^r$, for other $r$, so that it, too, will have this nice property.

Note that the rule above also implies that for integer exponents, $(x^p)^q = x^{pq}$.

If we want the rule to hold for all exponents, let's see what that means for $0$ and $n$:T $$ x^0 * x^n = x^n, $$ so $x^0$ has to be $1$.

And then $x^{-n} * x^n = x^{-n + n} = x^0 = 1$, so $x^{-n}$ has to be $1/x^n$. And in fact if we manage to define $x^r$ for every $r$, the same argument shows that $x^{-r} = 1/x^r$.

And now let's use that corollary of the rule to see about fractional exponents.

For integer $q$, we need (by the corollary) $$(x^{\frac{1}{q}})^q = x^{q \cdot \frac{1}{q}} = x^1 = x.$$ So $x^{1/q}$ is the $q$th root of $x$. And thus more generally, $x^{p/q}$ is the $q$th root of $x$, raised to the $p$th power.

What about $x^a$ for arbitrary real-numbers $a$?

Well, right about there you run into the need to deal with continuity and limits. It's pretty reasonable to say "For every $x$, I'd like to have $r \mapsto x^r$ be a continuous function of $r$. Once you do so, you can then prove (with a LOT of work) that $r \mapsto x^r$ is a continuous function on the rationals. Every such function has a unique continuous extension to the reals (that, too, requires some proof, but it's not too hard). So the definition of $x^a$ for an irrational number $a$ turns out to be this:

"Let $r_1, r_2, \ldots$ be a sequence of rationals whose limit is $a$; define $x^a$ to be $lim_{n \rightarrow \infty} x^{r_n}$."

Not very satisfying the first time you see it, but not so bad after a while.

Answer 2

Let's look at the patterns established by the definition of $x^n$:

$x^n = x \cdot x \cdot ... \cdot x$ ($n$ times)

$x^3 = x \cdot x \cdot x$

$x^2 = x \cdot x$

$x^1 = x$

Every time we subtract one from the exponent, we divide by $x$. And anything divided by itself (except 0 of course) is 1.

$x^0 = 1$

Continuing the pattern...

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

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

So,

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

For rational exponents, start with the rule that $x^{a + b} = x^ax^b$. $a$ values of $x$ multiplied by $b$ values of $x$ yields $a + b$ values of $x$ multiplied together.

What if $a = b = 1/2$?

$x^{1/2 + 1/2} = x^{1/2}x^{1/2}$

$x^1 = x = x^{1/2}x^{1/2}$

That's the definition of the square root. The square root of a number $x$ is a number, $\sqrt{x}$, which when multiplied by itself, yields the original number $x$. Therefore,

$x^{1/2} = \sqrt{x}$.

The cube root can be determined with a rational exponent similarly.

$x^{1/3 + 1/3 + 1/3} = x^{1/3}x^{1/3}x^{1/3}$

$x^1 = x = x^{1/3}x^{1/3}x^{1/3}$

$x^{1/3} = \sqrt[3]{x}$

This is extensible for any natural number $n$:

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

And raising any such number to a power can control the numerator of the fractional exponent.

$(x^{\frac{1}{n}})^m = x^{\frac{m}{n}}$

So, $x^{\frac{m}{n}} = (\sqrt[n]{x})^m$

That covers positive rational exponents.

For negative rational exponents, apply the above:

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

to arrive at:

$x^{-\frac{m}{n}} = x^{(-1)\cdot\frac{m}{n}} = \frac{1}{x^{\frac{m}{n}}}$