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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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Veljko Nikolic
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8 Answers8

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

Omran Kouba
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DanielV
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    Might be clearer if your last line reads $a^{-b} = \frac{1}{a^b}$. – Emily Jun 03 '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 03 '14 at 14:05
  • I agree, it looks better with one or the other ending, but both is kinda awkward. – DanielV Jun 03 '14 at 14:17
  • So the upshot is, regarding the OP's question, that this makes sense because it is consistent with the basic laws of exponents. +1, simple but effective explanation. – MPW Jun 03 '14 at 14:33
  • @MPW Well I think this explanation is sufficient for a beginner, but it doesn't establish consistency, all it really establishes is "if $a^{-b}$ is anything, then it is ${1 \over a^b}$". What if there is another proof that $a^{-b} = {2 \over a^b}$ ? Consistency requires a much more global approach. Thanks for the vote btw~ – DanielV Jun 03 '14 at 14:48
  • @DanielV Your assumption that $x^0=1$ is somewhat controversial, to say the least. (The old $0^0$ debate again.) – Dan Christensen Jun 03 '14 at 15:24
  • @DanChristensen: I think we can safely assume that $x$ is meant to be a positive real number. – Nick Matteo Jun 03 '14 at 15:25
  • @Kundor That should be made explicit, e.g. $x^0=1$ for $x\neq 0$ – Dan Christensen Jun 03 '14 at 15:28
  • Why number your equations only to not refer to them? – OJFord Jun 03 '14 at 15:55
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    @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 03 '14 at 17:31
  • @OllieFord Good observation, I was going to use them, but then couldn't find a way to include them without making the answer look cluttered. I left them in just in case the poster had any questions, it makes them easy to refer to in comments. I prefer that equations are named even if the names aren't used just because it makes discussion easier. – DanielV Jun 03 '14 at 17:33
  • Fair enough - I can see that that would be cluttered. I partly wondered if it was consequence of some community guideline on writing 'good answers'. – OJFord Jun 03 '14 at 18:01
  • @DanielV Last I heard, they are still teaching that $0^0$ is undefined in high schools. I don't want to start another flame war, but I am partial to this notion myself. Interested readers should see my "Oh, the Ambiguity!" blog posting (dated Oct. 9, 2013) at http://www.dcproof.wordpress.com – Dan Christensen Jun 03 '14 at 18:40
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    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 04 '14 at 07:21
  • @DanielV It seems only logical that it's not necessarily 1... $0^{0}$ is simply indeterminable. How can you say that $0^{3}$ is $0$. $0^{2}$ is $0$. $0^{0.5}$ is $0$, $0^{0.00001}$ is $0$ but $0^{0}$ MUST be $1$? – Cruncher Jun 04 '14 at 15:31
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    For the real numbers $0$, and $0$, $0^0$ is undefined. For the natural numbers $0$ and $0$, we must surely have $0^0 = 1$, since there is one function from the empty set to itself. – Steven Gubkin Jun 04 '14 at 16:10
  • @StevenGubkin Are you suggesting that we don't assume $\mathbb{N} \subset \mathbb{R}$? That sounds troublesome. – DanielV Jun 04 '14 at 17:29
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    @DanielV I only mean to draw attention to the fact thinking about exponentiation in $\mathbb{R}$ we must, by definition, use limits, and so it is appropriate to leave $0^0$ undefined. OTOH, when thinking of exponentiation defined on the natural numbers, this has a natural combinatorial definition $x^y = |X^Y|$ where $X$ is a set with $x$ elements, $Y$ is a set with $y$ elements, and $X^Y$ is the set of functions from $Y$ to $X$. – Steven Gubkin Jun 04 '14 at 18:49
  • BTW, technically the natural numbers are not a subset of the reals, at least if you believe in the usual development of these things in set theory. There is a natural injective map from the naturals into the reals however. – Steven Gubkin Jun 04 '14 at 18:50
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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.)

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MJD
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  • I really liked the extra context of the rule applying in the definition of e. Could you say in what other situations mathematicians made a bad notational choice they had to backtrack? – honestSalami Apr 16 '20 at 04:04
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    Have you seen https://math.stackexchange.com/questions/537383/why-is-x-frac12-the-same-as-sqrt-x/#comment1224789_537391 ? – MJD Apr 16 '20 at 09:58
  • For a larger example, consider the discovery of complex numbers. – MJD Apr 16 '20 at 10:12
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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.

Foo Barrigno
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(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

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

$x^2=x\cdot x$

$x^1=x$

$x^0=1$

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.

Dan Christensen
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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? – JTP - Apologise to Monica Jun 03 '14 at 16:37
  • This is the way I explain it to people. I think it's the most straightforward way. – daviewales Jun 04 '14 at 17:18
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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 $x^{-n}$.

bradhd
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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....

gebruiker
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$\sqrt[-b]{a}=a^{(\frac{-1}{b})}=\frac{1}{a^{\frac{1}{b}}}=\frac{a^0}{a^{\frac{1}{b}}}=a^{(0-\frac{1}{b})}$

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Vikram
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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 03 '14 at 13:19
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    @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 03 '14 at 13:47
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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 $a^{-b}=\frac{1}{a^b}$.

Aidan F. Pierce
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