Rational Exponents

An exponent that is no longer a count

For a positive integer $n$, the power $a^n$ is repeated multiplication, and from that single meaning you proved the laws you have used ever since:

Each one is a bookkeeping fact about counting factors. Line up $m$ copies of $a$ next to $n$ copies of $a$ and you are holding $m+n$ copies, which is the first law.

Now notice something you may never have questioned. You also write $a^0 = 1$ and $a^{-3} = 1/a^3$, and neither of those is a count. Zero copies and negative-three copies of $a$ are not multiplications anybody can perform. Those values were not discovered by counting; they were selected, and the thing that selected them was the demand that the first law survive. If $a^{m+n} = a^{m}a^{n}$ is to keep holding when one exponent is $0$, then

and dividing by $a^{m}$ (safe, since $a \neq 0$) leaves $a^{0} = 1$. No other value is available. The same pressure fixes negative exponents: $a^{-n} \cdot a^{n} = a^{-n+n} = a^{0} = 1$, so $a^{-n}$ has to be the reciprocal $1/a^{n}$.

That is the move this lesson repeats one more time. Every extension of the exponent notation past “repeated multiplication” has been made by asking a single question: what value would let the laws keep working? Answer it for the exponent $1/n$ and you have rational exponents.

The radical symbol, made precise

You have been writing $\sqrt{\ }$ for a while now. It sat under the quadratic formula, it measured the modulus of a complex number, and it named the irrational roots of polynomials. Nothing about the symbol is new here. What has never been pinned down is which number the symbol names, and that is suddenly about to matter.

Call $r$ an $n$-th root of $a$ when $r^{n} = a$. The trouble is that this description does not single out one number:

• $2$ and $-2$ are both square roots of $4$, since $2^2 = 4$ and $(-2)^2 = 4$.
• $-2$ is the only real cube root of $-8$.
• $-4$ has no real square root at all, because the square of a real number is never negative.

A symbol has to name one number, so we make a choice, and the choice is called the principal root.

For $a \geq 0$ and any integer $n \geq 2$, the symbol $\sqrt[n]{a}$ means the nonnegative real number whose $n$-th power is $a$. For odd $n$, a negative $a$ has exactly one real $n$-th root and it is negative, so $\sqrt[n]{a}$ names that number: $\sqrt[3]{-8} = -2$. For even $n$, a negative $a$ has no real $n$-th root, so $\sqrt[n]{a}$ names nothing real at all.

This is why $\sqrt{9}$ is $3$ and not $-3$, and not $\pm 3$ either. The radical hands you exactly one number. That is also the reason the quadratic formula has to write $\pm\sqrt{b^2 - 4ac}$ with the sign spelled out in front. If you want both roots, you have to ask for both, because the symbol by itself only ever gives you one.

Underneath the convention is a fact worth proving, because the entire rest of the lesson leans on it.

Read that headline once more, because it is the hinge of the whole lesson: a positive number has exactly one positive $n$-th root. A negative number has no positive $n$-th root at all, so the moment the base goes negative there is nothing left for this lemma to pin down. Hold that thought.

What $a^{1/n}$ is forced to be

Here is the central argument. We are free to define $a^{1/n}$ however we like, but we are not going to be free for long.

Notice exactly what each demand bought. The power law forced $x^{n} = a$, and when $n$ is even that still leaves two candidates: for $a = 4$ and $n = 2$, both $2$ and $-2$ square to $4$. Positivity is what picked one of them. Drop either demand and the definition stops being forced, which is a useful reminder that “the definition is natural” is a claim needing an argument, not a feeling.

The general rational exponent

Once $a^{1/n}$ is nailed down for $a > 0$, the rest of the rational exponents come along for free. Keep that hypothesis in view, because it is doing real work: everything in this section is stated for a positive base. The last section of the lesson shows precisely how it all collapses without one.

So let $a$ be positive, let $m$ be an integer, and let $n$ be a positive integer. Apply the power law twice and you get two recipes:

Take the root first and then the power, or take the power first and then the root. Those are different computations, and they had better land on the same number, or the symbol $a^{m/n}$ means nothing.

There is a second worry, and it is easy to miss. The fractions $\tfrac{2}{6}$ and $\tfrac{1}{3}$ are the same rational number. So $a^{2/6}$ and $a^{1/3}$ must be the same value, even though one recipe says “sixth root, then square” and the other says “cube root, then leave alone.” Nothing so far guarantees that.

Both worries are real. For $a > 0$ both are settled, and by the same lemma.

Definition. For $a > 0$, an integer $m$, and a positive integer $n$, $\;a^{m/n} = \left(\sqrt[n]{a}\right)^{m}$.

Computing with rational exponents

In practice, always take the root first. Both routes agree, so you may as well pick the one that keeps the numbers small. For instance, $64^{5/6}$ is $\left(\sqrt[6]{64}\right)^{5} = 2^5 = 32$, whereas the other route asks you to face $\sqrt[6]{64^{5}} = \sqrt[6]{1{,}073{,}741{,}824}$ to reach the same $32$.

The laws survive, with one standing hypothesis

Everything you know about exponents now extends to rational exponents. Throughout this list, $a$ and $b$ are positive and $r$ and $s$ are any rational numbers:

These are not new assumptions. They are consequences of the integer laws plus the definition, and the proof of the first one shows how all of them go. Push everything down to a single positive number and let the integer laws do the work.

Where the laws break: a negative base

That hypothesis $a > 0$ has been standing in front of every result in this lesson. Here is what it is protecting you from.

Take the base $a = -8$ and the exponent $\tfrac{2}{6}$, and compute it three perfectly reasonable looking ways.

Reduce the exponent first. $\tfrac26 = \tfrac13$, and $(-8)^{1/3} = -2$, since $(-2)^{3} = -8$. Reducing a fraction changes nothing about which rational number it names, so this step is honest arithmetic. $-2$ is the real value of $(-8)^{2/6}$.

Power first, then root. $\left((-8)^{2}\right)^{1/6} = 64^{1/6} = 2$, since $2^{6} = 64$. This looks just as legal, but it is not: it pulls the base through the outer power, $\left(a^{m}\right)^{1/n} = a^{m/n}$, and that identity was proved only for $a > 0$. For $a = -8$ it is not licensed, and it hands back the wrong number.

Root first, then power. $\left((-8)^{1/6}\right)^{2}$, and it stops dead: $-8$ has no real sixth root, because an even power of a real number is never negative. This is the tell. If the power-first route were legitimate, this order should reach the same place; instead it cannot even start, which is the sign that the manipulation smuggled something past you.

One route gives the real value, $-2$. The other two are not alternative real values; one is an invalid shortcut and the other exposes why.

Why does the power-first shortcut fail? Go back and reread the proofs. Every one of them, the forcing argument and both halves of well-definedness, spent the same fact: a positive number has exactly one positive $n$-th root. A negative number has no positive $n$-th root whatsoever, so there is nothing for that lemma to pin down, and pulling a negative base through the outer power stops being licensed. The hypothesis $a > 0$ printed above the law list is precisely the hypothesis the proofs consumed.

So what is still safe? Reducing the exponent is always safe: $\tfrac26$ and $\tfrac13$ are the same rational number no matter the base, so $(-8)^{2/6}$ and $(-8)^{1/3}$ must agree, and both equal $-2$. What is not safe is pulling the base through the power law. Both $\left((-8)^{2}\right)^{1/6} = 2$ and $(-8)^{1/3} = -2$ are true statements about real numbers, but only the second one equals $(-8)^{2/6}$. The false step is claiming $\left((-8)^{2}\right)^{1/6}$ equals $(-8)^{2/6}$: that identity, $\left(a^{m}\right)^{1/n} = a^{m/n}$, was proved only for $a > 0$.

This gives a general rule. For $a < 0$ and a rational exponent $\tfrac{m}{n}$ already written in lowest terms, there is no fraction left to reduce, so the only route is root then power: $a^{m/n} = \left(a^{1/n}\right)^{m}$. That route exists exactly when $a^{1/n}$ is a real number, which needs $n$ odd. So $a^{m/n}$ is real when $n$, the reduced denominator, is odd, and has no real value when $n$ is even. Try it on a numerator bigger than $1$:

real, because $\tfrac43$ is already in lowest terms with odd denominator $3$. But $(-16)^{5/2}$ has no real value at all: $\tfrac52$ is already in lowest terms, its denominator $2$ is even, and $-16$ has no real square root to raise to the fifth power.

The same crack runs through the radical form of the laws. The rule $\sqrt{a}\sqrt{b} = \sqrt{ab}$ needs $a$ and $b$ to be nonnegative. Push it to $a = b = -1$, using the convention $\sqrt{-1} = i$ from the complex numbers chapter, and the left side is $i \cdot i = i^{2} = -1$ while the right side is $\sqrt{1} = 1$. The law does not survive the trip.

That last computation is worth being precise about, now that you have complex numbers. “Not a real number” no longer means “meaningless”: $-8$ genuinely has six complex sixth roots, and you could pick one. But nothing in this lesson proved a single exponent law for them, and the $i \cdot i$ calculation shows the laws really do fail out there. Leaving the positive reals means leaving the system in which these rules were proved. Complex exponents are a serious subject, and they are not this one. Use $a > 0$ whenever you use a law, and treat an odd root of a negative number as a value you look up rather than a value you manipulate.

Common mistakes