Finding the error in this proof that 1=2 I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:
If $x = y$, then
$$
\begin{eqnarray}
x^2 &=& xy \nonumber \\
x^2 - y^2 &=& xy - y^2 \nonumber \\
(x + y)(x - y) &=& y(x-y) \nonumber \\
x + y &=& y \nonumber \\
2y &=& y \nonumber \\
2 &=& 1
\end{eqnarray}
$$

My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then 
$$
\begin{eqnarray}
x + y &=& y \\
x &=& y - y \\
x &=& y = 0
\end{eqnarray}
$$
Did I find the error? If not, am I close?
 A: That certainly is an error, although there is an error that precedes it.
HINT: Look at all the places you have $(x-y)$ in your proof. What is $x-y$? What are you doing with $x-y$ each time it shows up?
A: In third line you have written:

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

Since $x=y$, we can't cancel $(x-y)$, as that equals 0.
Cancellation law in any Integral domain is the following:

Left cancellation law:  If $a\neq 0$ then $ab= ac$ implies $b=c$.
Right cancellation law:  If $a\neq 0$ then $ba=bc$ implies $b=c$.

A: Hint $ $ When debugging proofs on abstract objects, the error may become simpler to spot after specializing to more concrete objects. In your proof the symbols $\rm\:x,y\:$ denote abstract numbers, so let's specialize them to concrete numbers, e.g. $\rm\:x = y = 3.\:$ This yields the following "proof"
$$\begin{eqnarray}
3^2 &=& 3\cdot3  \\
3^2 - 3^2 &=& 3\cdot 3 - 3^2  \\
(\color{c00}{3 + 3})\:(\color{c00}{3 - 3}) &=& \color{c00}3\: (\color{c00}{3-3}) \\
\color{#c00}{3 + 3} &=&\color{#c00} 3\ \ {\rm via\ cancel}\ \ \color{c00}{3-3}  \\
2\cdot 3 &=& 3  \\
2 &\:=\:& 1
\end{eqnarray}$$
Now we can find the first false inference by finding the first $\rm\color{#c00}{false\ equation}$ above;  if it is equation number $\rm\: n\!+\!1,\:$ then the inference from equation $\rm\:n\:$ to $\rm\:n\!+\!1\:$ is incorrect (above: "via cancel $0$")
Analogous methods prove helpful generally: when studying abstract objects and something is not clear, look at concrete specializations to gain further insight on the general case.
A: Since you already declared:

$x = y$

∴ $x - y = 0$
On step 3, dividing by $x-y$ ($= 0$) is a mathematical error, since it is mathematically invalid to divide anything by $0$.
