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Possible Duplicate:
Division by $0$

What is the exact value of $1\over0$? In $y=$ $1\over{x}$, it shows it is $\infty$ and $-\infty$ (infinite and minus infinite).

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Division by zero cannot be defined in a meaningful way.

If you interpret $1/0$ as a one-sided limit of $1/x$ (and accept improper limits), then

$$\lim_{x\to 0^+} \frac1x = +\infty \qquad\text{and}\qquad \lim_{x\to0^-} \frac1x = -\infty.$$

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Let $x \in \mathbb{R}^*$. By definition, $1/x$ is the unique real $y$ such that $xy=1$. So you see that you can't extend this definition to include $1/0$, because for all real $y$, $0y=0$.

Another possibility would be to extend $x \mapsto 1/x$ at $0$ by continuity, but there is no such extension as $\lim\limits_{x \to 0^{\pm}} \frac{1}{x} = \pm \infty$.

In fact, there is no really natural way to define $1/0$.

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    $\begingroup$ ... unless you identify $+\infty$ with $-\infty$. Don't. $\endgroup$ Nov 10, 2012 at 10:11
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    $\begingroup$ @HagenvonEitzen: Sometimes, it is quite natural to identify $+\infty$ with $-\infty$. But only do it when necessary, and if you know what you are doing. The result of doing so is known as the projective line. $\endgroup$ Nov 10, 2012 at 10:28
  • $\begingroup$ @HaraldHanche-Olsen: Yes, I know. My "Don't" was therefore intended for the level where the current question arose (I as in a kind of typing hurry and hope that your comment made things clearer) $\endgroup$ Nov 10, 2012 at 15:20
  • $\begingroup$ @HagenvonEitzen: I expected as much, and my comment wasn't really directed at you – I only named you to let you know I commented. $\endgroup$ Nov 10, 2012 at 15:29
  • $\begingroup$ @HaraldHanche-Olsen No problem and thanks again $\endgroup$ Nov 10, 2012 at 15:31

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