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This question already has an answer here:

Wasn't quite sure if this was suitable for Puzzling; however it has a problem/puzzle in it, so decided to give it a shot. Please appropriately downvote if you feel you should.

Yesterday morning I came to class as usual, and a student came up to me. He said he had a proof that could revolutionize mathematics.

Being a Grade 11 teacher, I didn't believe him.

So I asked him to tell me about the proof, he said "I've divided it into several steps, and said it seemed to work perfectly well, with zero problems." Then he told me "It's a proof that 1 is equal to 0."

So I asked if he could find any mistakes, he responded "None that I could find, sir, however it wouldn't surprise me if there was. I had thought of it after you taught us factoring. I think you'd like to take a look".

I indeed did, and he showed me, mentioning that it was indeed stating $1=0$.

Obviously this is false, and by looking at the paper I noticed his trivial error almost immediately.

Assume x and y are two nonzero numbers.
$x = y$
$x^2 = x*y $
$x^2 - y^2 = x*y - y^2 $
$x + y = y $
$2*y = y $
$2 = 1 $
$1 = 0$

However, I have a challenge for you, and this is to find the error my student made.

But not only that, you must explain how everything he told me was true.

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marked as duplicate by Deusovi, CodeNewbie, A E, AJL, Engineer Toast Oct 21 '15 at 12:04

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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    $\begingroup$ Similar question here. $\endgroup$ – Len Oct 21 '15 at 3:18
  • $\begingroup$ Not a duplicate. There's an additional word problem about"0 problems" $\endgroup$ – warspyking Oct 21 '15 at 13:38
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The errors lies here:

There is an implicit division by 0 between steps 3 and 4. Step 3 states: $x^2-y^2=xy-y^2$. Factoring this produces $(x+y)(x-y)=y(x-y)$. Going to step 4 requires that you divide through by $(x-y)$, but step 1 implies that $x-y=0$.

He is technically correct in what he said because:

He said the proof has "zero problems". This is correct; there is a problem with the number 0 in his proof.

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    $\begingroup$ Good job :) Did you think this puzzle was appropriate? $\endgroup$ – warspyking Oct 20 '15 at 21:49
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The student divided by zero between the third and fourth lines. He wrote the third line as:

$(x + y)(x-y) = y(x-y)$

And proceeded to cancel the term $(x-y)$. This can be done if and only if $(x-y) \neq 0$, but since the hypothesis is $x=y$ we arrive at a contradiction.

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