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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?

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James, you might have noticed that I added the tag "fake-proofs" to the question. If you click on that tag, you will find many such proofs. A few of them involve division by zero (as does yours). – The Chaz 2.0 Mar 8 '12 at 19:46
Thanks, @TheChaz, I didn't know the tag existed. – jamesbrewr Mar 8 '12 at 19:49
up vote 18 down vote accepted

Hint $\ $ When debugging proofs on abstract objects, the problem 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"

$$\begin{eqnarray} 3^2 &=& 3\cdot3 \\ 3^2 - 3^2 &=& 3\cdot 3 - 3^2 \\ (3 + 3)\:(3 - 3) &=& 3\: (3-3) \\ 3 + 3 &=& 3 \\ 2\cdot 3 &=& 3 \\ 2 &\:=\:& 1 \end{eqnarray}$$

Now you can determine which inference is incorrect by determining the first false equation above; if it is equation number $\rm\: n\!+\!1,\:$ then the inference from equation $\rm\:n\:$ to $\rm\:n\!+\!1\:$ is incorrect.

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.

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That's a good way to think about it. I'll give it a try next time I run across a similar problem. – jamesbrewr Mar 8 '12 at 20:02
I wish I could upvote this more than once. – Henning Makholm Mar 8 '12 at 20:14

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?

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Unfortunately I read the other reply before yours which blatantly points out what you're getting at, so I didn't get the chance to find it myself. – jamesbrewr Mar 8 '12 at 19:42

In third line you have written $$(x+y)(x-y)= y(x-y)$$ as $x=y$, we can't cancel $(x-y)$.

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 $b=c$.

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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

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