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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$\begingroup$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.
$\endgroup$ 12 $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 1 $\begingroup$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..
$\endgroup$ 6 $\begingroup$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.
$\endgroup$ 5 $\begingroup$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".
$\endgroup$ 2 $\begingroup$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 ;)
$\endgroup$ $\begingroup$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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