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Why can't you divide a number by zero? It is possible to say $\sqrt{-1}$ is an imaginary number $i$, but why can't you say $\frac{1}{0}$ is also an imaginary number $z$ (for example)?

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marked as duplicate by ShreevatsaR, Amzoti, Davide Giraudo, Martin Argerami, Douglas S. Stones Apr 19 '13 at 17:04

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    $\begingroup$ You seem to be under the impression that an 'imaginary number' is a convenience for a seemingly-impossible mathematical operation. The number $i$ satisfying $i^2=-1$ is supremely important. Indeed, the term 'imaginary' is unconstructive and unenlightening; 'complex' number speaks more to its utility. $\endgroup$ – Ian Coley Apr 19 '13 at 16:00
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    $\begingroup$ You can, but it would be useless. (Also, you'd have to use some other name since "imaginary number" is already taken.) MJD wrote a great answer investigating what would happen if you did, here. $\endgroup$ – ShreevatsaR Apr 19 '13 at 16:33
  • $\begingroup$ Also, the IEEE Standard for Floating-Point Arithmetic (IEEE 754), which is what most computers use, does define division by 0: it says that the result of dividing any number by 0 is the special number "NaN" (which stands for "Not a Number"). The discussion by MJD essentially tells you what happens. $\endgroup$ – ShreevatsaR Apr 19 '13 at 16:42
  • $\begingroup$ @Frank, I at least consider "complex" to be as undeserved a name as "imaginary", but I suppose we're stuck with centuries of tradition at this point... if only we had better names for something so useful! $\endgroup$ – J. M. is a poor mathematician Apr 21 '13 at 11:21
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Suppose $x=\frac{1}{0}$. Then we should have $x\cdot0=1$, this is not possible since $0$ times any number will give $0$.

In contrast, the idea to let $\sqrt{-1}=i$ is extending the number system which could satisfy all the calculation rules by itself.


Maybe it will be easier to understand this by taking an example: Let $x=\frac{1}{0}$. Then $$1=x\cdot0=x(1-1)=x\cdot1-x\cdot1=x-x=0$$LHS$\neq$RHS, a contradiction.

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    $\begingroup$ @exploringnet, no number would satisfy this since $0$ times any number should be $0$ by considering the operation rules of field. $\endgroup$ – Easy Apr 19 '13 at 16:02
  • $\begingroup$ if you were ever to label $1/0$ as a quantity, I think you would be forced to formally define it to br $\infty$. For instance, evaluating a function $f$ on the Riemann Sphere at $\infty$ amounts to evaluating $f(1/z)$ at $z=0$. There are plenty of other instances where we "invert zero", but you must use a great deal of care when doing this. $\endgroup$ – John Martin Apr 19 '13 at 16:06
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    $\begingroup$ In addition to John Martin's comment, even you define it to be $\infty$, there are numerous different directions of $\infty$, in which case we treat them as infinity points in affine geometry. But surely this is off topic. $\endgroup$ – Easy Apr 19 '13 at 16:12
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    $\begingroup$ More carefully, the point is that when you add $i$ to the number system (field), it plays nicely with all the rules, but if you add $1/0$ to the system, you have to remove some rules, or make special exceptions to them. $\endgroup$ – Sharkos Apr 19 '13 at 16:13
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    $\begingroup$ @exploringnet To put Easy's comment another way, if you define $Z\cdot 0=1$, then you're not just putting conditions on $Z$, you're changing the definition of $0$. $\endgroup$ – mdp Apr 19 '13 at 16:26
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Another way to see the problem is by looking at limits: $$\displaystyle \lim_{x->0^+} \frac{1}{x} = \infty$$ But: $$\displaystyle \lim_{x->0^-} \frac{1}{x} = -\infty$$

And since there's no preference to any one-sided limit, the limit should not exist and there isn't a specific value to the function $\frac1x$ at 0.

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