Finding the error in this proof that 1=2

Question

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

$$ \begin{eqnarray} x^2 &=& xy \nonumber \\ x^2 - y^2 &=& xy - y^2 \nonumber \\ (x + y)(x - y) &=& y(x-y) \nonumber \\ x + y &=& y \nonumber \\ 2y &=& y \nonumber \\ 2 &=& 1 \end{eqnarray} $$

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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?

Answer 1

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"

$$\begin{eqnarray} 3^2 &=& 3\cdot3 \\ 3^2 - 3^2 &=& 3\cdot 3 - 3^2 \\ (\color{c00}{3 + 3})\:(\color{c00}{\color{#0a0}{3 - 3}}) &=& \color{c00}3\: (\color{#0a0}{3-3}) \\ \color{#c00}{3 + 3} &=&\color{#c00} 3\ \ {\rm via\ cancel}\ \ \color{#0a0}{3-3} \\ 2\cdot 3 &=& 3 \\ 2 &\:=\:& 1 \end{eqnarray}$$

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 $\rm\:n\!+\!1\:$ must be incorrect. Doing this above we infer that the culprit is "${\rm via\ cancel}\ \ \color{#0a0}{3-3}$", $ $ i.e. $\,x\cdot \color{#0a0}0 = y\cdot \color{#0a0}0\not\Rightarrow x = y,\,$ i.e. it divided by $\color{#0a0}0$.

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.

Remark $ $ It is however valid to cancel nonzero polynomials and then deduce consequences, e.g. if $\,f(x,y)\,$ and $\,g(x,y)\,$ are (formal) polynomials then

$$\begin{align}f(x,y)(x-y) &= g(x,y)(x-y)\\[.2em] \Rightarrow\ \ f(x,y) &= g(x,y)\end{align}\qquad$$

therefore $\,f(a,b) = g(a,b)$ for all $a,b\,$ (even if $\,a=b\,$ so $\,x-y = a-b = 0).\,$ For example see here where the polynomial $f$ is a determinant, and we prove Sylvester's Determinant Identity $\rm\ det\:\! (I+AB)=det\:\!(I+BA)\, $ by cancelling $\rm\,det\,A\, $ from $\rm\,det\,$ of $\rm\ (1+AB)A = A(1+B A)$. Beware that this slick proof is often wrongly deemed incorrect even by some professors - see the discussion there.

Answer 2

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?

Answer 3

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 $b=c$.