Why does zero raised to the power of negative one equal infinity?

Question

I had the question of $0^{-1}$ on a math test and I naturally assumed that this evaluates to zero, but from what I have seen from various sources it is equal to infinity which I do not quite understand. I would sooner believe that this it is just undefined.

You are looking at $1/0$ which is famously undefined. It does not evaluate to $0$ nor to $\infty $. — Ittay Weiss, Nov 07 '14 at 00:50
They're not really equal in the strictest mathematical sense. — Jose Arnaldo Bebita Dris, Nov 07 '14 at 00:51
Its good to note that $\infty = \lim_{x \rightarrow 0} 1/x $ is not the same as $0^{-1}$ — aram, Nov 07 '14 at 01:33

k170 · Answer 1 · 2014-11-07T01:03:48.487

13

$$ 0^{-1}=\frac{1}{0}=\mbox{undefined} $$ because $$ \lnot\exists x\in\mathbb{R}:1=0\times x $$

edited Nov 07 '14 at 01:03

answered Nov 07 '14 at 00:53

k170

9,045

3

could someone please explain what those math symbols mean? – Thor Sep 15 '18 at 09:19
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It means, there does not exist a real number $x$ such that $1=0\times x$. – k170 Feb 07 '21 at 06:22
¬ = not; ∃ = exists; ∈ℝ = any x belonging to the Real numbers set; : = such that; × = times. Please note that × are two different x signs, even different from the x letter. – AMDP Oct 10 '23 at 16:49

score 12 · Answer 2 · edited Nov 11 '21 at 18:20

Strictly speaking we say $0^{-1}$ is undefined.

$x^{-1}$ is the multiplicative inverse of the number $x$. By definition of the multiplicative inverse, the product of a number and its multiplicative inverse equals one. So we would have $0\times 0^{-1} = 1$.

However, by definition of zero and multiplication, the product of zero and any number equals zero. So $0\times 0^{-1} = 0$.

So, unless $0=1$, these definitions conflict! Hence the multiplicative inverse of zero is undefined.

Another way.

We can examine the behaviour of the function $f(x)=x^{-1}$ as $x$ approaches zero. Plot the curve $y=1/x$ to visualise what is happening. (It is a hyperbola.)

On the right side, the limit of $x^{-1}$ tends towards positive infinitude as $x$ tends downwards to zero.

$$\lim_{0<x\to 0} \frac 1 x = +\infty$$

On the left side, the limit of $x^{-1}$ tends towards negative infinitude as $x$ tends upwards towards zero.

$$\lim_{0>x\to 0} \frac 1 x = -\infty$$

So there is a discontinuity in the function $f(x)= x^{-1}$ at $x=0$. Thus the quantity of $0^{-1}$ is indefinite.

Why does zero raised to the power of negative one equal infinity?

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