# Why does 2 equal 1? [duplicate]

A friend showed me this proof:

Proof: 2 = 1

$$Let \space x= y$$

Multiply both sides by x:

$$x^2= xy$$

Subtract $y^2$ from both sides:

$$x^2-y^2= xy-y^2$$

Factor:

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

Cancel out $(x-y)$ from both sides:

$$(x+y) = y$$

Simplify (Because $x=y$):

$$y+y=y$$

$$2y = y$$

$$2 = 1$$

Where does the logic break down? Everything is done to both sides.

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## marked as duplicate by Git Gud, Thomas Andrews, N. S., Aang, Maisam HedyellooJul 17 '13 at 19:24

Don't divide by $0$. Just...don't do it. –  David Mitra Jul 17 '13 at 19:11
$0*1=0*99999 \Rightarrow 1=99999$ for sure –  igf Jul 17 '13 at 19:12
You can't cancel $0$ in the equality $0\cdot1=0\cdot2$ –  Sami Ben Romdhane Jul 17 '13 at 19:13
After the factor is done, $(x-y)=0$ thus both sides become zero and this exercise is finished. Dupe of these questions: math.stackexchange.com/questions/417324/… and math.stackexchange.com/questions/117998/… –  JB King Jul 17 '13 at 19:18
You cannot cancel out the $(x-y)$. You defined $x=y$, so you end up dividing by $0$.