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I'm reading Spivak's Calculus:

2 What's wrong with the following "proof"? Let $x=y$. Then

$$x^2=xy\tag{1}$$

$$x^2-y^2=xy-y^2\tag{2}$$

$$(x+y)(x-y)=y(x-y)\tag{3}$$

$$x+y=y\tag{4}$$

$$2y=y\tag{5}$$

$$2=1\tag{6}$$

I guess the problem is in $(3)$, it seems he tried to divide both sides by $(x-y)$. The operation would be acceptable in an example such as:

$$12x=12\tag{1}$$

$$\frac{12x}{12}=\frac{12}{12}\tag{2}$$

$$x=1\tag{3}$$

I'm lost at what should be causing this, my naive exploration in the nature of both examples came to the following: In the case of $12x=12$, we have an imbalance: We have $x$ in only one side then operations and dividing both sides by $12$ make sense.

Also, In $\color{red}{12}\color{green}{x}=12$ we have a $\color{red}{coefficient}$ and a $\color{green}{variable}$, the nature of those seems to differ from the nature of

$$\color{green}{(x+y)}\color{red}{(x-y)}=y(x-y)$$

It's like: It's okay to do the thing in $12x=12$, but for doing it on $(x+y)(x-y)=y(x-y)$ we need first to simplify $(x+y)(x-y)$ to $x^2-y^2$.

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Red Banana
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    I'm at a loss why there is a close vote for NARQ. This is a perfectly valid question, since the OP did not purport that $1 = 2$ but merely sought the flaw in the "proof". – Lord_Farin Jun 11 '13 at 11:26
  • @Lord_Farin Well, considering that there are illegal reasons for voting to close/downvoting, one of the most chosen is because a question is stupid. I've even created a name for these guys. I've spent a lot of time looking at $(3)$ - I was certain that the problem was there when it was actually on $(2)$. – Red Banana Jun 11 '13 at 11:35
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    @Lord_Farin I even dare to speculate that those guys didn't earn the analytical badge. – Red Banana Jun 11 '13 at 11:44
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    http://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html and http://en.wikipedia.org/wiki/Mathematical_fallacy#Division_by_zero – lab bhattacharjee Jun 11 '13 at 11:59
  • I didn't vote to close, but I would hypothesise that because there was a different proof that $1=2$ on the front page today someone thought that one was enough! I would tend to agree...there are plenty of places on the internet where one can find the solutions to these problems (there are plenty of questions here which do too...), and I do not think it brings much to the site if everyone starts posting their proofs that $1=2$. One question is okay, two is pushing it, and certainly I will vote to close question number three... – user1729 Jun 11 '13 at 12:10
  • @user1729 Yes. But voting to close as a dupe is different of voting to close as not a real question. – Red Banana Jun 11 '13 at 12:24
  • @BandeiraGustavo I didn't say it isn't a real question. Rather, I do not think a flood of such questions is good for the site. If I wanted to try to justify this opinion by trying to fit it into some category that people have come up with, then I would say that it shows a lack of effort from the OP because the solutions are readilly available on the internet. For example, second result. – user1729 Jun 11 '13 at 12:35

2 Answers2

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We have $x = y$, so $x - y = 0$.

EDIT: I think I should say more. I'll go through each step:

$x = y \tag{0}$

This is our premise that $x$ and $y$ are equal.

$$x^2=xy\tag{1}$$

Note that $x^2 = xx = xy$ by $(0)$. So completely valid.

$$x^2-y^2=xy-y^2\tag{2}$$

Now we're adding $-y^2$ to both sides of $(1$) so completely valid and we can see that it's another way of expressing $0 = 0$ as $x=y$, but nothing wrong here yet.

$$(x+y)(x-y)=y(x-y)\tag{3}$$

$$x+y=y\tag{4}$$

Step $(3)$ is just basic factoring, and it is around here where things begin to go wrong. For $(4)$ to be a valid consequence of $(3)$, I would need $x - y \neq 0$ as otherwise, we would be dividing by $0$. However, this is in fact what we've done as $x=y$ implies that $x - y =0$. So $(3)-(4)$ is where things go wrong.

$$2y=y\tag{5}$$

$$2=1\tag{6}$$

As a consequence of not being careful, we end up with gibberish.

Hope this clarifies more!

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We have $x=y$ firstly and from $3$ to $4$ we assume $x\ne y$ simultaneously.

Mikasa
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    We assume $x\neq y$? I didn't get it. With Bryan's answer, I've figured that $x^2-y^2=0$ and $xy-y^2=0$ and it would end as $0=0$. – Red Banana Jun 11 '13 at 11:29
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    @BandeiraGustavo: I mean when you are going from $3$ to $4$, you naturally assume that $x\neq y$. – Mikasa Jun 11 '13 at 11:32