Which is wrong in the proof?

Question

Consider the following steps: \begin{align} a &= x \\ a+a &= a+x && \text{[add }a\text{ to both sides]}\\ 2a &= a+x && \text{[}a+a = 2a\text{]}\\ 2a-2x &= a+x-2x && \text{[subtract }2x\text{ from both sides]}\\ 2(a-x) &= a+x-2x && \text{[}2a-2x = 2(a-x)\text{]}\\ 2(a-x) &= a-x && \text{[}x-2x = -x\text{]}\\ 2 &= 1 && \text{[divide both sides by }a-x\text{]} \end{align}

Which step is wrong in this proof?

http://en.wikipedia.org/wiki/Mathematical_fallacy#Division_by_zero — lab bhattacharjee, Dec 26 '13 at 13:49
"Divide both side by a-x". 01=02 , so do you think 1 and 2 are equal?. — hrkrshnn, Dec 26 '13 at 13:50
Vishnu, the downvote(s) here, I think, are because you presented the problem without any indication you had tried to solve it yourself. In the future, please include a description of what you tried and/or where you got stuck. — Barry Cipra, Dec 26 '13 at 14:40
You are assuming $a=x$ i n the beginning and then you divide by zero!! — Mhenni Benghorbal, Dec 26 '13 at 15:13

score 3 · Accepted Answer · answered Dec 26 '13 at 13:57

$$\begin{align} a & = x \\ a + a & = a + x \\ 2a & = a+x \\ 2a - 2x & = a + x - 2x\\ 2 (a -x) &= a-x \end{align}$$

All is OK so far but in the next step you are trying to divide $0$ by $0$ which is not allowed $\frac{0}{0}$ is undefined.

Note: $a = x$ so $a - x = 0$

Which is wrong in the proof?

1 Answers1