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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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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. 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.

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  • $\begingroup$ Is there any proof in math that equates one number with another [different] number? $\endgroup$
    – user855113
    Nov 29, 2020 at 16:56
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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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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 $b=c$.

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  • $\begingroup$ 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. $\endgroup$
    – jamesbrewr
    Mar 8, 2012 at 19:41
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    $\begingroup$ Yes... you can't proceed after the line $(x+y)(x-y)= y(x-y)$ as $x-y$ is zero. $\endgroup$
    – zapkm
    Mar 8, 2012 at 19:43
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    $\begingroup$ 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? $\endgroup$
    – jamesbrewr
    Mar 8, 2012 at 19:43
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    $\begingroup$ (James - FYI "rings, fields, units and zero divisors" are related objects in a field commonly known as "abstract algebra". For when the time comes... :) ) $\endgroup$ Mar 8, 2012 at 19:48
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    $\begingroup$ 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! $\endgroup$
    – jamesbrewr
    Mar 8, 2012 at 19:51
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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 $0$.

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