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?

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.

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?

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 $$b=c$$.

• OK let's try this. Because $x=y$, $(x - y) = 0$ which means $(x+y)(x-y) = y(x-y) = 0$? In which case we can just stop there. – jamesbrewr Mar 8 '12 at 19:41
• Yes... you can't proceed after the line $(x+y)(x-y)= y(x-y)$ as $x-y$ is zero. – zapkm Mar 8 '12 at 19:43
• Also, it took me a minute, but I understand the cancellation laws now. Effectively, cancelling a variable is just dividing both sides by that variable. In this case, where $x-y = 0$, this is not allowed because division by 0 is always undefined. Correct? – jamesbrewr Mar 8 '12 at 19:43
• (James - FYI "rings, fields, units and zero divisors" are related objects in a field commonly known as "abstract algebra". For when the time comes... :) ) – The Chaz 2.0 Mar 8 '12 at 19:48
• The Chaz -- I hope the time comes soon. Until very recently I haven't understood why people like the subject so much. I picked up a copy of Spivak's Calculus 3rd Edition and I'm falling in love with it -- and I haven't even gotten to the Calculus yet! I've been a member of this Stack Exchange for a couple of days and the insight you guys have provided is just fantastic. Kudos and keep up the good work! – jamesbrewr Mar 8 '12 at 19:51

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