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This question already has an answer here:

If we agree that $\textbf{(a) }\dfrac{x}{x}=1$, $\textbf{(b) }\dfrac{0}{x}=0$, and that $\textbf{(c) }\dfrac{x}{0}=\infty^{\large\dagger}$, and let us suppose $z=0$:

$$\begin{align*} z&=0&&\text{given.}\\ \dfrac{z}{z}&=\dfrac{0}{z}&&\text{divide each side by }z.\\ \dfrac{z}{z}&=0&&\text{by }\textbf{(b)}.\\ 1&=0&&\text{by }\textbf{(a)}. \end{align*}$$ Now, from this, we can get say that $\dfrac{0}{0}=1$. What went wrong?


$\dagger$ Or undefined, if you prefer it.

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marked as duplicate by user147263, quid, Michael Grant, JimmyK4542, Ivo Terek Dec 18 '14 at 0:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ If you assume that the inverse of $0$ exists in a ring (with unity), then it's standard proving that $0$ is the only element. $\endgroup$ – egreg Dec 17 '14 at 22:34
  • $\begingroup$ (a) is only true if $x\neq 0$ because other you get your proof to show $0=1$ $\endgroup$ – Squirtle Dec 17 '14 at 22:35
  • $\begingroup$ @egreg Are you referring to Ring Theory? :/ $\endgroup$ – Conor O'Brien Dec 17 '14 at 22:47
  • $\begingroup$ @ConorO'Brien Yes; but since the integers form a ring where $0\ne1$, the inverse of $0$ doesn't exist, unless you prefer to violate the distributive law. $\endgroup$ – egreg Dec 17 '14 at 22:48
  • $\begingroup$ @egreg erm... could you explain to the non-Ring theorist what exactly that means? $\endgroup$ – Conor O'Brien Dec 17 '14 at 22:48
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You cannot divide by zero. So if $z=0$, then you also cannot divide by $z$. Therefore, your second equality $$\frac{z}{z}=\frac0z$$ is ultimately incorrect.

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All answers were given, I'll give one answer offering additional information on division by zero.

Actually it's possible to divide by zero. It's usually not possible to divide by zero, because this operation is meaningless(in some contexts). But sometimes, it's useful to be able to divide by zero and there is a theory devoted to it, called Wheel Theory. Have a look at this article too. Just be careful on how do you use this information, there are contexts in which this is useful. At some of them, it's useless.

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  • $\begingroup$ Hmm... very interesting. Thanks! $\endgroup$ – Conor O'Brien Dec 17 '14 at 23:01
  • $\begingroup$ Very interesting indeed. $\endgroup$ – Miguelgondu Dec 17 '14 at 23:07
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Thou shalt not divide by zero. Your first statement doesn't hold if $x=0$.

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  • $\begingroup$ It's actually possible to divide by zero. Take a look at my answer. $\endgroup$ – Billy Rubina Dec 17 '14 at 23:09
  • $\begingroup$ Very interesting! I'd never heard of such beasts. $\endgroup$ – John Dec 17 '14 at 23:18
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First, there are a couple of comments to make.

(a) is only valid for $x \neq 0$; (b) is only vaild for $x \neq 0$ (c) is not defined.

So you don't have the right to divide by $z$ in your second line.

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We agree about a little more than what you wrote. We agree that

(a) If $x\neq 0$, then $\frac xx=1$

(b) If $x\neq 0$, then $\frac0x = 0$

We do not agree about $(c)$ at all, we think it is undefined.

Using only (a) and (b), you cannot prove that $\frac00=1.$

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You divided by zero on the second line. There are many proofs similar to yours that end up being complete garbage because of division by zero. Although your example doesn't hide it as well.

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  • $\begingroup$ It's actually possible to divide by zero. Take a look at my answer. $\endgroup$ – Billy Rubina Dec 17 '14 at 23:10
  • $\begingroup$ Interesting, +1. $\endgroup$ – MathMajor Dec 17 '14 at 23:11

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