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So I devised this proof that $1=0$. Of course it is false, but I don't know why. Why?

$$\begin{align*} x+1&=y\\ \frac{x+1}{y}&=1\\ \frac{x+1}{y}-1&=0\\ \frac{x+1}{y}-\frac{y}{y}&=0\\ \frac{x-y+1}{y}&=0\\ x-y+1&=0\\ x-y+1&=\frac{x-y+1}{y}\\ y(x-y+1)&=x-y+1\\ y&=1\\ x+1&=1\\ x&=0\qquad * * * *\\ y-1&=x\\ \frac{y-1}{x}&=1\\ \frac{y-1}{x}-1&=0\\ \frac{y-1}{x}-\frac{x}{x}&=0\\ \frac{y-x-1}{x}&=0\\ y-x-1&=0\\ y-x-1&=\frac{y-x-1}{x}\\ x(y-x-1)&=y-x-1\\ x&=1\qquad * * * *\\ 1&=0\\ \end{align*}$$

  • 9
    It might help you find the division by zero error if you use LaTeX to write this out more neatly.2012-06-02
  • 1
    Is the error is division by 0? I already figured out that if, in theory, 1=0, then any given numbers are equal. Therefore, any given number is equal to 0. Therefore, you cannot divide. Is this the problem?2012-06-02
  • 3
    $x=0$ and $(y-1)/x=1$? There's your problem.2012-06-02
  • 2
    No, you are literally dividing by zero going from the 8th to 9th lines.2012-06-02
  • 0
    But x doesn't even have to be 0. You cannot divide at all as I explained in previous comment. For example: 1=0, multiple by 2 to get 2=0, add 7 to get 7=9 out of 1=0. It does not necessarily need to be 0 your dividing by cause all numbers are 0. I was just wondering if there was another reason this was invalid.2012-06-02
  • 4
    Actually, there are many steps where you divide by 0. $x+1=y$ implies $y-x-1=0$ and $x+1-y=0$2012-06-02
  • 0
    But any division is by 0 if conclusion is true.2012-06-02
  • 4
    You can't assume the conclusion.2012-06-02
  • 1
    Besides the facts mentioned by everyone else, I think another problem is at the start, and this might help you with other proofs: What are $x$ and $y$? If $x$ and $y$ are variables representing any real number, then your first 8 lines can be removed and just say, if $y=1$, then $x=0$.2012-06-03
  • 1
    I don't understand the reason for the downvotes to this question; OP has a genuine mathematical question and has shown his work. Moreover, there are some good answers to which indicates that it can't be that bad in any event.2012-06-04

4 Answers 4