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Let $x=5$, $y=7$, $z=6$

$x+y = 2z$

Rearranging, $x-2z = -y$

and $x = -y+2z$

Multiply both sides respectively. $x^2-2xz = y^2-2yz$

$$x^2-2xz+z^2 = y^2-2yz+z^2$$ $$(x-z)^2 = (y-z)^2$$ $$x-z = y-z$$ Hence $x=y$, or $5 = 7$

Well, the conclusion is clearly false, but what went wrong? I think it may be the step in which one square roots both sides because it's taking out one solution?

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    $\begingroup$ Maybe when you took a squre root: $x^2=y^2$ implies $x=y$ or $x=-y$. indeed $x-z=z-y=-1$ $\endgroup$
    – Belgi
    Apr 28, 2012 at 0:42

4 Answers 4

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Hint $\ $ When debugging proofs on abstract objects, the error may become simpler to spot after specializing to more concrete objects. The symbols $\rm\:x,y,z\:$ denote abstract numbers, so let's specialize them to their concrete number values: $\rm\:x = 5,\: y=7,\: z = 6,\:$ yielding this "proof" $$\begin{eqnarray} 5 + 7 &=&\: 2\cdot 6 \\ 5- 2\cdot 6 &=&\: -7 \\ \cdots\ &=&\ \cdots \\ (5-6)^2\! &=&\: (7-6)^2 \\ \color{#c00}{5-6}\ \ \:&=&\:\ \ \color{#c00}{7-6}\: \end{eqnarray}$$ Now we can spot which inference is incorrect by determining the first $\rm\color{#c00}{false\ equation}$ above. If equation number $\rm\: n\!+\!1\:$ is false then the inference from equation $\rm\:n\:$ to $\rm\:n\!+\!1\:$ is incorrect. Doing so we find that last equation being false, which reveals the culprit inference $\,(-1)^2 = 1^2\color{#c00}{\Rightarrow\, -1 = 1}$

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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  • $\begingroup$ I like this answer best so far, since it gives a more general method than just looking at this particular problem. $\endgroup$
    – Doug Spoonwood
    Apr 28, 2012 at 2:06
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What you have is

$$(x-z)^2=(y-z)^2$$

Note that $y-z$ is positive, but $x-z$ is negative. Thus you have to consider the absolut value when taking the square root.

This means

$$|x-z| = y-z$$

But not

$$x-z = y-z$$

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  • $\begingroup$ A few seconds before saving my own answer, I go the usual notice that someone else had answered, and that was this answer. My version is essentially the same thing, but maybe more simply expressed. Here's a confession: I believe in simplicity. $\endgroup$
    – Michael Hardy
    Apr 28, 2012 at 2:31
  • $\begingroup$ ....and Bill Dubuque's got a point too. $\endgroup$
    – Michael Hardy
    Apr 28, 2012 at 2:32
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If $(x-z)^2 = (y-z)^2$ then $x-y = \pm(y-z)$.

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  • $\begingroup$ That is essential what I said. $\endgroup$
    – Pedro ♦
    Apr 28, 2012 at 0:46
  • $\begingroup$ And what you said is essentially what I said. $\endgroup$
    – Michael Hardy
    Apr 28, 2012 at 0:48
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    $\begingroup$ Touchè. I'm just saying there is no need for a duplicate. $\endgroup$
    – Pedro ♦
    Apr 28, 2012 at 0:49
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    $\begingroup$ So, shall we fight a duel to decide which one should be deleted? $\endgroup$
    – Michael Hardy
    Apr 28, 2012 at 2:32
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You have $x=−y+2z$ and $x−2z=−y$ then you multiply both sides. I think you are computing $-xy$ in two ways: for the first $-xy=y^2-2zy$ for the second $-xy=-2zx+x^2$ so you have $y^2-2zy=x^2-2zx$. Adding $z^2$ to both sides gives $(y-z)^2=(x-z)^2$. But now taking square roots gives $|y-z|=|x-z|$.

If you try to remove the absolute value signs then you get $y-z=x-z$ if both are positive or negative but $y-z=-(x-z)$ if these have opposite signs. In the first case this leads to $y=x$, and in the second case $y=-x+2z$. For your example, the second case applies $7=-5+2\cdot6$.

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