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Why is a value with a negative exponent equal to the multiplicative inverse but with a positive exponent?

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

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closed as unclear what you're asking by Andrés Caicedo, Macavity, Kf-Sansoo, John Ma, Pedro Tamaroff Jun 4 '14 at 22:01


Is this really homework? – Git Gud Jun 3 '14 at 13:01
Presumably, what you mean is, wny is $$a^{-b}=\frac{1}{a^b}?$$ – Thomas Andrews Jun 3 '14 at 13:02
@ThomasAndrews Or the OP is having a slight existential crisis :) – joshin4colours Jun 3 '14 at 19:23
Is the question about why $x^{-a} = 1/x^{a}$ is mathematically valid or about why so many algebra/precalculus classes insist that students always re-write all the exponents to be positive? – Stan Liou Jun 3 '14 at 21:10
Also, $$\frac{1}{a^{-b}} = a^b$$ – Hurkyl Jun 4 '14 at 5:39

If you know that $$x^y \cdot x^z = x^{y+z} \tag{1}$$ and that $$x^0 = 1 \tag{2}$$ then $$\begin{align} 1 &= 1 \\ a^0 &= 1 \\ a^{b - b} &= 1\\ a^{b} \cdot a^{-b} &= 1 \\ a^{-b} &= \dfrac{1}{a^b}\end{align}$$

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Might be clearer if your last line reads $a^{-b} = \frac{1}{a^b}$. – Arkamis Jun 3 '14 at 13:14
Is there any reason not to remove the second-to-last line ($a^b=1/a^{-b}$)? You could just divide by $a^b$ in the first place, and it should be clearer. – Keen Jun 3 '14 at 14:05
I agree, it looks better with one or the other ending, but both is kinda awkward. – DanielV Jun 3 '14 at 14:17
@DanChristensen Anyone who doesn't define $0^0 = 1$ is deliberately making life hard. There really isn't any debate, just trolling. There is no inconsistency in the definition. Some would argue that because we can't infer $\lim_{x,y \rightarrow 0,0} x^y = 1$ then the definition is unsound, but that's just ridiculous. The existence of limits always has to be checked. – DanielV Jun 3 '14 at 17:31
Applying the formula with $a = 0$ gives the result $0^{-b} = \frac{1}{0^b}$. Maybe there is a way to make sense of this of which I am not aware, but it seems best avoided in high school math. Therefore you should stipulate that $a \ne 0$ (and might as well also say $x \ne 0$), and you don't need to know how to evaluate $0^0$. – David K Jun 4 '14 at 7:21


The reason for this is easy to see: $x^3$ means $x\cdot x \cdot x$, and $x^2$ means $x\cdot x$, so $$\begin{align}x^3 \cdot x^2 &= (x\cdot x \cdot x)\cdot(x\cdot x)\\ &=x\cdot x \cdot x\cdot x \cdot x \\& = x^5 \end{align}$$

We'd like this law to continue to hold when we define $x^\alpha$ for negative $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for negative exponents, then whatever we decide that $x^{-1}$ should mean, it should obey the same law: $$x^{-1}\cdot x^{2} = x^{-1+2} = x^1 = x$$

and so $x^{-1} = \frac1x$ is the only choice.

Similarly, what should $x^{-3}$ mean? If we want the law to continue to hold, we need $$x^{-3}\cdot x^{3} = x^{-3+3} = x^0 = 1$$ and thus the only consistent choice is $x^{-3} = \frac1{x^3}$.



If you put $x=-1$ in this formula, you get $$e^{-1} \stackrel?= 1 -1 +\frac12 - \frac16 + \frac1{24}\cdots$$ and adding up just the first few terms one gets $0.375$, which is already pretty close to $\frac1e \approx 0.368$.


(Much of this is copied from my answer to a similar question earlier.)

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+1 really nice answer – friedo Jun 4 '14 at 3:48

(The OP doesn't seem to understand what a negative exponent means. Assuming this was not actually the question asked for homework. Something like this should have been covered by his teacher.)

Just follow the pattern. For non-zero $x$, we have

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

$x^2=x\cdot x$



Continue dividing by $x$ each time, as above...

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

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

$x^{-3}=\frac{1}{x^2}\div x =\frac{1}{x^3}$

and so on.

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Nice work. The only thing I'd have included is the similarity to the negative ide of the number line, 3,2,1,0,-1,-2, etc. Why does bopping negative not bother OP there but it does for exponents? – JoeTaxpayer Jun 3 '14 at 16:37
This is the way I explain it to people. I think it's the most straightforward way. – daviewales Jun 4 '14 at 17:18

In a vague sense, each integer increment of the exponent in $a^b$ means "multiply by a" where $a$ is the base of the exponent. So $2^2$ means "multiply by two, and multiply by two again" to get "multiply by $2*2 = 4$".

So imagine you have $2^{3-1}=2^2=4$. What does the $-1$ mean in that expression? You originally had "multiply by 2, then multiply by 2, then multiply by 2" or "multiply by two three times", but there was also the $-1$, which somehow undid one of those multiply by 2's. The arithmetic function which is the inverse of multiplication is division. Try it - take a number and multiply it by 2, then divide it by 2. You'll come up with the original answer.

Therefore negative exponents are division by the base, rather than multiplication.

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This is really how we define negative exponents, so the question becomes: why is this a reasonable definition? Let's pretend for the moment that we only knew about positive exponents.

If $n$ is a positive integer, what does $x^n$ mean? It's the product of $x$ with itself $n$ times:

$$x^n = \overbrace{x\cdot x\cdots x}^n$$

So if $m$ is another positive integer, we have

$$x^{m+n} = \overbrace{x\cdot x\cdots x}^{m+n}$$ $$ = (\overbrace{x\cdot x\cdots x}^{m})(\overbrace{x\cdot x\cdots x}^{n})$$ $$ = x^m\cdot x^n$$

Now what should $x^n$ mean when $n$ is zero or negative? It would be nice for the rule $x^{m+n}=x^m\cdot x^n$ to be true even when $m$ or $n$ is not positive, so let's see what that rule tells us.

We would have to have $$x^{0+1} = x^0\cdot x^1$$, and so $$x^1 = x^0\cdot x^1$$ $$x = x^0\cdot x.$$ If $x\neq 0$ then the only way this can be true is if $x^0=1$. So we'll define $x^0=1$ whenever $x\neq 0$, and continue from there.

Now for negative exponents. How can we reasonably define $x^{-n}$ when $n$ is a positive integer (and $x\neq 0$)? Going by our rule $x^{m+n}=x^m\cdot x^n$, we would have to have

$$x^{-n+n} = x^{-n}\cdot x^n$$ $$x^0 = x^{-n}\cdot x^n$$ $$1 = x^{-n}\cdot x^n$$

Divide both sides by $x^n$, and we see that we must have $$\frac{1}{x^n} = x^{-n}.$$

So this is how we must define $x^{-n}$.

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Think of it this way: exponentiation is equivalent to repeated multiplication, in the sense that, for example, $3^4=3\times3\times3\times3$; so a multiplication repeated a negative number of times should use the multiplicative inverse, division. Therefore, a negative exponentiation could be represented as a repeated division, which would be equivalent to $a^{-b}=\frac{1}{a^b}$.

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Play with this a little: $$2\to4\to8\to16\ldots\text{etc}$$or equivalently $$2^1\to2^2\to2^3\to2^4\ldots\text{etc}.$$ So you increase the exponent of $2$ with $1$ each time. Now imagine going in the backward direction (meaning you substract $1$ in the exponent each time) but dont stop at two....

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Hmm, I'm fairly certain I've never seen anyone use the notation $^n\sqrt{a}$ with negative values of $n$, although an "automatic" application of $^n\sqrt{a}=a^{\frac{1}{n}}$ would do that... Our suspicion now is that the question doesn't have anything to do with roots at all: the OP probably meant "base." – rschwieb Jun 3 '14 at 13:19
@rschwieb, +1 for identifying the mistake, I will keep the my wrong notation posted for others to see "How not to write math" – Vikram Jun 3 '14 at 13:47
$\sqrt[n]m$ is typeset as \sqrt[n]m, not as "empty space"^n\sqrt m. – Ruslan Jun 3 '14 at 14:39
@Ruslan, thank you. – Vikram Jun 3 '14 at 15:22
