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.

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    $\begingroup$ You are looking at $1/0$ which is famously undefined. It does not evaluate to $0$ nor to $\infty $. $\endgroup$ – Ittay Weiss Nov 7 '14 at 0:50
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    $\begingroup$ They're not really equal in the strictest mathematical sense. $\endgroup$ – Arnie Bebita-Dris Nov 7 '14 at 0:51
  • $\begingroup$ Its good to note that $\infty = \lim_{x \rightarrow 0} 1/x $ is not the same as $0^{-1}$ $\endgroup$ – aram Nov 7 '14 at 1:33

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

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    $\begingroup$ could someone please explain what those math symbols mean? $\endgroup$ – Thor Sep 15 '18 at 9:19
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    $\begingroup$ It means, there does not exist a real number $x$ such that $1=0\times x$. $\endgroup$ – k170 Feb 7 at 6:22

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


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