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I've always taken for granted that $x^{-1} = \frac{1}{x}$ because it works and everything, but how can I see that this must be true?

I am not looking for answers like "If you have $x^n / x^m$ then this is $x^{n-m}$ and so $x^{0}/x^1 = 1/x = x^{0-1} = x^{-1}$ because then I am stuck wondering whether it probably makes sense that $x^n$ can be defined when $n$ is negative (I hesitate to just assume that because we can plug numbers into something that it will be valid).

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    $\begingroup$ This is the definition of 1/x. $\endgroup$ – Tac-Tics Aug 4 '16 at 3:09
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    $\begingroup$ @turkeyhundt I don't like appealing to "common sense" results like this because it would lead one to believe that negative binomials (for example) aren't a thing due to the factorial function, and yet there is another definition that works (generalized version). I am after something a little more rigorous. $\endgroup$ – Sean Hill Aug 4 '16 at 3:11
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    $\begingroup$ @SeanHill It comes from the general rule $x^{-n}=\frac{1}{x^n}$ The proof that encompasses integers $n$ already requires some more detailed math. In your post it seems like you want proof of certain rules of exponents. For that you need some more tools $\endgroup$ – imranfat Aug 4 '16 at 3:16
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    $\begingroup$ What do you mean by "it provably makes sense"? $\endgroup$ – Eric Wofsey Aug 4 '16 at 3:21
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    $\begingroup$ For positive exponents, we may describe $x^n$ using repeated multiplication. It satisfies the relation $x^{n+m}=x^nx^m$ when $n$ and $m$ are positive. To keep the pattern consistent, we may define $x^0:=1$ and $x^{-n}=1/x^n$ when $-n$ is negative. The reason we choose this definition is because then $x^{n+m}=x^nx^m$ provably holds for all integers $n$ and $m$. In other words, $\{\cdots,x^{-2},x^{-1},x^0,x^1,x^2,\cdots\}$ under multiplication works the same way that $\{\cdots,-2,-1,0,1,2,\cdots\}$ works under addition. $\endgroup$ – arctic tern Aug 4 '16 at 3:43
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So there is a subtle question here, We can verify that $x^n * x^m = x^{n+m}$ when $n,m$ are natural numbers and $x$ is a natural number. Now the "natural generalization" here is to just decide that that formula applies for all numbers. From there we can compute it for negative values. Of course there are other, much less practical versions of the exponent, one could make up where this doesn't hold, and the addition rule holds only when n,m are natural numbers.

What that effectively amounts to is finding functions

$$f(x) | f(n+m) = f(n)f(m)$$ When $n,m$ are natural numbers. Exponentiation, in the traditional sense is one of the smoother solutions to this, but I'm sure you can invent some freakish counterexample which does very very unpredictable things on inputs that aren't natural numbers.

For example consider the functional equation:

$$ f(n+m) = (f(n) + frac(n)+ 0.5 \delta(x) )(f(m) + frac(m) + 0.5\delta(x))$$

On positive integers this reduces to

$$ f(n+m) = f(n)f(m)$$

But elsewhere it does something WILDLY different. So this function if looked at from just the natural numbers would like identical to our exponentiation, but outside of there would be very very strange looking, in principle $f(-1)$ would not be anything like $\frac{1}{x}$

But, the guiding principle here then isn't "WHAT IS MATHEMATICALLY CORRECT", it is, what makes most sense to use. And the most natural tool then is the original exponential the way we are used to.

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    $\begingroup$ To do this we introduce $frac(X)$ which is the fractional part of $x$. Example: if $x = 1.2344333$ then $frac(x) = .2344333$ naturally $frac(x) = x$ if $x$ is an integer. Now the next function I utilize is $\delta(x) = |x|/x$ (graph that at desmos.com to see what it is doing). We can then do the following: observe that $(\delta(x)-x)$ is 0 if x is positive and -2 otherwise. Furthermore $frac(x) = 0$ when x is an integer. Then $\delta(x)-x+frac(x)$ is only 0 over the positive integers - the natural numbers call this expression H(x). Then $$f(n+m) = (f(n)+H(n))(f(m)+H(m))$$ $\endgroup$ – frogeyedpeas Aug 4 '16 at 18:45
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    $\begingroup$ Necessarily reduces to the normal exponentiation equation: $$f(n+m)=f(n)f(m)$$ over the natural numbers, but elsewhere is doing something wildlyyyy different. Giving you an example of "exponentiation" that doesn't obey the $x^{-1} = \frac{1}{x}$ law. i noticed I had a typo in the answer $\endgroup$ – frogeyedpeas Aug 4 '16 at 18:49
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    $\begingroup$ Graphing the thing at Desmos, looks like a line going out to the right at $x=1$, and another line going out to the left at $x=-1$? The sign function? $-1$ if $x$ is negative, positive $1$ if $x$ is positive? $\endgroup$ – Sean Hill Aug 4 '16 at 18:52
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    $\begingroup$ So basically you have made a function that works as intended for the positives, but the mechanics behave in unexpected ways in the negatives because certain pieces that normally "cancel out" no longer do? $\endgroup$ – Sean Hill Aug 4 '16 at 18:53
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    $\begingroup$ Yes, and so to answer your question: there is no "Logical reason" to accept that $x^{-1} = \frac{1}{x}$ that is purely because of a convention $\endgroup$ – frogeyedpeas Aug 4 '16 at 18:55
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Given $a,b,c$ from any set $Y$ with a multiplication defined and $1$ is the corresponding multiplication identity. By definition,

  • $\frac{a}{b}$ is the unique $y \in Y$ (if exists) such that $yb = a = by$.
  • $c^{-1}$ is the unique $y \in Y$ (if exists) such that $yc = 1 = cy$.

Substitute $a$ by $1$ and $b, c$ by $x$, we find both definition of $\frac{1}{x}$ and $x^{-1}$ reduce to the unique $y$ in $Y$ (if exists) such that $yx = 1 = xy$. As a result, $\frac{1}{x} = x^{-1}$ whenever they make sense.

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Its not true always... infact.

Its entirely depends on which set you are working.

One more notable thing is $x^{-1}$ is just a notation to represent a inverse of an element $x$(mostly for multiplication or the second operation we consider in a ring).

Now, we will to come to our case.

Suppose my set is $\mathbb R$, the set of real numbers. Then,

the multiplicative inverse will be easily obtained by $\frac 1 x$.

Hence, $x^{-1}=\frac 1 x$.

But, suppose my set is $\mathbb Z$, the set of integer.

we are fail to say that $x^{-1}=\frac 1 x$ since, the existence of $x^{-1}$ is itself a question mark. That is, what is the inverse of $2$?

Concluding that, $x^{-1}$ is just a representation for the inverse of an element in a given set in the given set with respect to the second operation(or multiplication).

I hope it will help..

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    $\begingroup$ I don't think that the OP meant $x^{-1}$ as a multiplicative inverse. $\endgroup$ – Andres Mejia Aug 4 '16 at 4:33
  • $\begingroup$ @AndresMejia Since the OP didnt quote anything in his/her post. I have answered in one point of view. $\endgroup$ – David Aug 4 '16 at 4:36
  • $\begingroup$ Well, the OP does mention something of the form $a^{-n}$ as well, and given that you've mentioned $\mathbb{R}$ as the field in question, this answer is insufficient. For example, set $n=\pi$. Then $a^{-\pi}$ isn't at all obvious from a discussion of multiplicative inverses. $\endgroup$ – Andres Mejia Aug 4 '16 at 4:39
  • $\begingroup$ @AndresMejia I think I didnt communicate properly. Kindly note that, I have discussed only about $x^{-1}$ not $x^n$. Which means given an element in a set(may be in a group), what is the $x^{-1}$, the inverse of $x$ in that set(or in group). $\endgroup$ – David Aug 4 '16 at 4:46
  • $\begingroup$ I dont think that anything you said is incorrect, but its worth mentioning to the OP that you are restricting the conversation to $x^{-1} $ and tbat this wont work for negative exponentiation in general. Either way, its an otherwise lucid answer :) $\endgroup$ – Andres Mejia Aug 4 '16 at 4:49
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I am assuming you agree $x^n=\frac{x^{n+1}}{x}$ for positive integer $n$ (we need to begin from somewhere anyway since otherwise we should start with the philosophy of numbers!). This will get a recursive formula for calculation of the previous terms, hence an extension to zero and negative powers ($n=0$ and $n<0$).

For $n=0$ we get $$x^0=\frac{x^{0+1}}{x}=\frac{x}{x}=1$$ Similarly, for $n=-1$ it becomes $$x^{-1}=\frac{x^{-1+1}}{x}=\frac{x^{0}}{x}=\frac{1}{x}$$ where we used $x^0=1$ from the previous equation. The same approach is used for $x^{-2}$ etc.

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    $\begingroup$ No, you are using the division rule of exponents, which is inherently related to the multiplication rule of exponents, which is what the OP does not want. Moreover, if you don't prove these rules in the first place, you haven't proven anything. $\endgroup$ – imranfat Aug 4 '16 at 3:43
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I suppose we need to start somewhere.

definition: $$e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$$

We note that $e^{\log(a)x}=(e^{\log(a)})^x=a^x$, where $\log(x)$ is the inverse to $e^x$. Also note that this is well defined since the sum converges by the ratio test.

See here for the equivalent definitions, and you can use the limit definition to prove your desired identity, by taking the definition of $a^x$ given above.

The point is, that exponentiation is the unique function where the rate or growth is proportional to the value of the function itself. We like this property over $\mathbb{N}$ (i.e: $a^{n+1}/a^n=a$) [along with $a^{n+m}=a^n\cdot a^m$], so it is reasonable to extend it to the real numbers, and the $\exp$ function is a way to do this rigorously, although on the natural numbers, the function should coincide with the usual definition of exponentiation...


If your interest was just over $\mathbb {Q} $ note that that $x^{-1} $ is the multiplicative inverse, as has been outlined by other answers.

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  • $\begingroup$ I don't understand, because by definition $b^{\log_b(a)} = a$, so why does $e$ matter? How do we know that's how $e$ is defined via that summation? $\endgroup$ – Sean Hill Aug 4 '16 at 16:58
  • $\begingroup$ Well, if you go to the aforementioned wiki, you can see different and equivalent definitions. You have to start with something. This was the definition I thought might be most cohetent. Maybe the limit definition would have been better. We have to define exponentiation. $\endgroup$ – Andres Mejia Aug 4 '16 at 17:00
  • $\begingroup$ It looks like $e$ is just being defined out of thin air, though. Why is this defined the way it is? $\endgroup$ – Sean Hill Aug 4 '16 at 17:08
  • $\begingroup$ Maybe Im not communicating. I havent defined $e$ anywhere, only the exponential function. Perhaps it would have been clearer if I used exp (x) $\endgroup$ – Andres Mejia Aug 4 '16 at 17:09
  • $\begingroup$ This is a hard question to answer. I'm not sure how to prove your intuition for what $e$ is. You have to accept one of the definitions to make sense of exponentiation in $\mathbb{R}$. You can alternatively define $x^{-1}=\frac{1}{x}$ over $\mathbb{Q}$ and use the fact that all $r \in \mathbb{R}$ have some sequence of $q_{i} \in \mathbb{Q}$ so that $q_{i} \to r$. Then, define $r^{x}=\lim_{i \to \infty} q_{i}^{x}$. You need to start somewhere though. You can start with $\ln(x)=\int_{0}^{x} \frac{1}{t} dt$ and define $e^{x}$ to be the inverse of this function...there are a lot of ways to proceed. $\endgroup$ – Andres Mejia Aug 4 '16 at 17:30
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Things work this way:

From the observation of $x^{n+m}=x^nx^m$, it is natural to try to generalize to negative numbers and write

$$x^{n+(-m)}=x^nx^{-m},$$ then $$x^{-m}=\frac{x^{n+(-m)}}{x^n}=\frac{x^{n-m}}{x^n}.$$

In particular, with $n=1$ this yields

$$x^{-1}=\frac1x.$$

Adopting this convention, one can check that the known properties of exponents apply to the negative ones, like

$$x^{nm}=(x^n)^m.$$

Indeed,

$$x^{n(-m)}=x^{-nm}=\frac1{x^{nm}}=\frac1{(x^n)^m}=(x^n)^{-m}.$$

So the convention makes things consistent, which is what you call "true".

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  • $\begingroup$ But I don't understand what makes it true just because we can write it down. For example I could write one of those proofs where the denominator gets divided by $0$ somewhere and the results end up being nonsense, but otherwise for any other input the math would work out. How do we know this works and that we can apply the frameworks the same way to negative integers? I know we can because "it works," but is that all there is to it? We define it this way because the results happen to work out properly? $\endgroup$ – Sean Hill Aug 4 '16 at 17:06
  • $\begingroup$ @SeanHill: you know it works by checking. Imagine you defined $x^{-m}$ as being $-x^m$. Then you couldn't write $x^{n-m}=x^nx^{-m}$, but you could write other theorems such as $x^{-n-m}=x^{n+m}$. So things aren't true by themselves, they are true because we choose them to be true in a way that pleases us. $\endgroup$ – Yves Daoust Aug 5 '16 at 8:31
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One may simply proceed like this

We know that Number(or Variable) * Its Inverse = 1
Similarly X*(Inverse of X) = 1
Implying Inverse of X = 1/x ;

And by convention we assume Inverse of a number as ${number}^{-1}$ i.e number raised to the power of minus 1;

Another :

More simply as Positive powers means Multiplication and so Negative powers means division simple ;)

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It's a matter of definition. We can define $x^{-1}=$ a purple elephant wearing a tutu if we had a good reason to.

We have defined that if $k$ is an integer and $k \ge 2$ then $x^k$ will be defined as $x^k = \underbrace{x\cdot x \cdot x .... \cdot x}_{k\text{ times}}$. This is well defined and makes sense.

An immediate result is $x^k x^m = (\underbrace{x\cdot x \cdot x .... \cdot x}_{k\text{ times}})(\underbrace{x\cdot x \cdot x .... \cdot x}_{m\text{ times}})=\underbrace{x\cdot x \cdot x .... \cdot x}_{k+m\text{ times}}=x^{k+m}$.

That's a great little fact! Our rule: If $k,m$ are integers so that $k\ge 2, m\ge 2$ then $x^kx^m = x^{m+k}$.

It is useful. But is seems kind of silly to require that $k,m \ge 2$. After all $x*x^2 = x^3$ and surely we can make the rule more useful.

It's be nice if we could allow $k, m$ to equal $1$. But $x^1 = \underbrace{x\cdot x \cdot x .... \cdot x}_{1\text{ time}}$ makes absolutely no sense!

But we can define $x^1 = $ a purple elephant wearing a tutu if we had a good reason. And we do have a good reason to define $x^1 = x$. Now our rule extends to $k,m$ being integers $\ge 1$.

And it is consistent. Because we are just writing down $x$s on paper and counting them, right? $x^1 = x$ is just one $x$.

Well. $1*x^k = x^k*1 = x^k$ and $1$ is a multiplicative identity and multiplying by $1$ is the same as leaving things alone so if $x^k=\underbrace{x\cdot x \cdot x .... \cdot x}_{k\text{ times}}$ means writing down $x$ $k$ times then multiplying by $1$ is the same are writing .... zero $x$s.

So it is useful, and we have a good reason, to extend our definition to $x^0 = 1$. And now our rule is $x^kx^m = x^{k+m}$ if $k$ and $m$ are non-negative integers.

I think you see where I'm going.

If $k > m\ge 0$ we have $x^k(\frac 1x)^m=(\underbrace{x\cdot x \cdot x .... \cdot x}_{k\text{ times}})(\underbrace{\frac 1x\cdot \frac 1x \cdot \frac 1x .... \cdot \frac 1x}_{m\text{ times}}) =\underbrace{x\cdot x \cdot x .... \cdot x}_{m-n\text{ times}}$.

Wouldn't it make sense, and wouldn't it be useful, and most of all wouldn't it be consistent, to define that $x^{-m} = (\frac 1x)^m$?

The answer is yes, it would.

.......

So.... tl;dr.....

Why does $x^{-1} = \frac 1x$?

Because it can.

Because its useful.

Because it works.

And because it's consistent.

And because we say so (and because we can say so).

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Provided we have that $x^nx^m =x^{n+m}$ and $x^n/x^m =x^{n-m}$, we can use this property to prove that $a^{-n}=1/a^n$ by contradiction.

Suppose $$a^{-n}\ne \frac{1}{a^n}$$

$a^{-n}\ne \frac{1}{a^n}$

$a\cdot a^{-n}\ne a\cdot\frac{1}{a^n}$

$a^{1-n}\ne \frac{a}{a^n}$

$a^{1-n}\ne \frac{a^{1}}{a^n}$

$a^{1-n}\ne a^{1-n}$

$\Rightarrow\Leftarrow$: contradiction

$\therefore a^{-n}= \frac{1}{a^n}$

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  • $\begingroup$ What's the role of contradiction here? $\endgroup$ – egreg Feb 7 at 23:27
  • $\begingroup$ @egreg If I understand your question properly, the contradiction serves to exclude the possibility of the desired equation being false, thereby proving the mutually exclusive alternative of the equation being true. $\endgroup$ – SmoovOpr8r Feb 7 at 23:35

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