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

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


closed as unclear what you're asking by Andrés E. Caicedo, Macavity, DeepSea, user99914, Pedro Tamaroff Jun 4 '14 at 22:01

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  • $\begingroup$ Is this really homework? $\endgroup$ – Git Gud Jun 3 '14 at 13:01
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    $\begingroup$ Presumably, what you mean is, wny is $$a^{-b}=\frac{1}{a^b}?$$ $\endgroup$ – Thomas Andrews Jun 3 '14 at 13:02
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    $\begingroup$ @ThomasAndrews Or the OP is having a slight existential crisis :) $\endgroup$ – joshin4colours Jun 3 '14 at 19:23
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    $\begingroup$ 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? $\endgroup$ – Stan Liou Jun 3 '14 at 21:10
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    $\begingroup$ Also, $$\frac{1}{a^{-b}} = a^b$$ $\endgroup$ – 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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    $\begingroup$ Might be clearer if your last line reads $a^{-b} = \frac{1}{a^b}$. $\endgroup$ – Emily Jun 3 '14 at 13:14
  • $\begingroup$ 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. $\endgroup$ – Keen Jun 3 '14 at 14:05
  • $\begingroup$ I agree, it looks better with one or the other ending, but both is kinda awkward. $\endgroup$ – DanielV Jun 3 '14 at 14:17
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    $\begingroup$ @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. $\endgroup$ – DanielV Jun 3 '14 at 17:31
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    $\begingroup$ 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 $0^0$. $\endgroup$ – David K Jun 4 '14 at 7:21

When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$

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 $x^{-3} = \frac1{x^3}$.

But there is more to it than that. Further mathematical developments, which you have not seen yet, confirm these choices. For example, one shows in analysis that as one adds more and more terms of the infinite sum $$1 + x + \frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \cdots$$ the sum more and more closely approaches the value $e^x$, where $e$ is a certain important constant, approximately $2.71828$. One can easily check numerically that this holds for various integer values of $x$. For example, when $x=1$, and taking only the first five terms, we get $$1 + 1 + \frac12 + \frac16 + \frac1{24}$$

which is already $2.708$, quite close to $e^1$, and the remaining terms make up the difference. One can calculate $e^2$ by this method and also by straightforward multiplication of $2.71828\cdot2.71828$ and get the same answer.

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 $\frac1e \approx 0.368$.

If it didn't work out this way, we would suspect that something was wrong somewhere. And in fact it has often happened that mathematicians have tried defining something one way, and then later developments revealed that the definition was not the right one, and it had to be revised. Here, though, that did not happen.

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


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.


(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.

  • $\begingroup$ 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? $\endgroup$ – JoeTaxpayer Jun 3 '14 at 16:37
  • $\begingroup$ This is the way I explain it to people. I think it's the most straightforward way. $\endgroup$ – daviewales Jun 4 '14 at 17:18

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 $x^{-n}$.


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....


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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    $\begingroup$ 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." $\endgroup$ – rschwieb Jun 3 '14 at 13:19
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    $\begingroup$ @rschwieb, +1 for identifying the mistake, I will keep the my wrong notation posted for others to see "How not to write math" $\endgroup$ – Vikram Jun 3 '14 at 13:47

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