Why is $a^{-x}$ defined to be equal to $\frac{1}{a^x}$?

Question

I have searched the reason behind this definition in two textbooks and haven't found any. They just state that this is the definition but don't ever give any motivation for why this is truth.

Edit:

I figured out why that relation must be true about 10 seconds after posting the question. Instead of looking for explanations I should have thought for a while. Still, thank you everyone.

It is so that you can extend the rule $a^n\times a^m=a^{n+m}$ to negative values of $m$ and $n$. Assuming you know $a^0=1$ for $a\neq 0$ — Mark Bennet, Dec 26 '18 at 20:54
An alternative duplicate target. Make your pick, there's no shortage. — Jyrki Lahtonen, Dec 26 '18 at 21:03

score 7 · Accepted Answer · edited Dec 26 '18 at 20:55

7

A fundamental property of power is $a^x\cdot a^y=a^{x+y}$ for nonzero $a$.

Then, $a^x\cdot a^{-x}$ should be $a^{x-x}=a^0=1$. Therefore $a^{-x}$ should be $1/a^x$.

edited Dec 26 '18 at 20:55

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

answered Dec 26 '18 at 20:52

ajotatxe

65,084

Dang! Beat me to it by seconds. $+1$ For OP: if these two things did not correspond to the same number, exponent laws would not really make sense (at least not fully). – Cameron Williams Dec 26 '18 at 20:53

score 4 · Answer 2 · answered Dec 26 '18 at 20:53

4

$a^0=1$

$a^0=a^{x-x}=a^x a^{-x}=1$.

Thus : $a^{-x}$ is the multiplicative inverse of $a^x$.

answered Dec 26 '18 at 20:53

Mauro ALLEGRANZA

94,169

score 4 · Answer 3 · edited Dec 26 '18 at 20:57

As you said, this is a matter of definition, but this choice is not random. Here is a short explanation why mathematicians chose this definition.

If you define exponentiation for $a\in \mathbb Z^+$ and $b\in \mathbb Z^+$ using repeated multiplication, or define it recursively as $$a^1=a$$ $$a^{b+1}=a\cdot a^b$$ Then the following fundamental property follows: $$a^x\cdot a^{y}=a^{x+y}$$ In order to define $a^b$ for $b\in\mathbb Z\setminus\{0\}$, we may simply assume that this property holds for all $x,y\in\mathbb R$. If we take this to be true, then we may deduce the value of $a^0$, since $$a^0\cdot a^{b}=a^{b}\implies a^0=1$$ From this it would follow that $$a^x\cdot a^{-x}=a^0=1\implies a^{-x}=\frac{1}{a^x}$$ which is the proposition that you are wondering about.

score 1 · Answer 4 · answered Dec 26 '18 at 20:56

Short answer:

For positive integers $m$ and $n$ the identity $$ a^{m+n} = a^m a^n $$ is obvious when you define exponentiation as repeated multiplication.

That identity is so important that we want to preserve it when we allow other kinds of values for the exponents.

The first consequence is $$ a^0 = 1 . $$ Then it's easy to show that $a^{-n}$ must be $1/a^n$.

The rules for rational exponents follow too - e.g. the $1/2$ power is the square root.

Extending the definition to all real numbers is a little more subtle.

score 1 · Answer 5 · answered Dec 26 '18 at 21:06

We originally establish that for $q>0$ we have:

$$a^q \cdot a = a^{q+1}$$ We rearrange this to:

$$a^q=a^{q+1} \div a$$ Let $q=0$. We then see that: $$a^0=a^1\div a = 1$$ Then $q=-1$ leads to $$a^{-1}=1\div a$$ A continuation of this process, combined with extensive use of $$\frac{(\frac ab)}{c}=\frac{a}{bc}$$ implies that: $$a^{-n}=\frac{1}{a^n}$$ is a suitable notation to extend the first statement I gave to all $q\in\Bbb Z$

Why is $a^{-x}$ defined to be equal to $\frac{1}{a^x}$?

5 Answers5