8

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?

Aryan
  • 442
jamesbrewr
  • 575

4 Answers4

26

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.

Bill Dubuque
  • 272,048
5

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?

KReiser
  • 65,137
4

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 $b=c$.

Aryan
  • 442
zapkm
  • 1,896
  • 13
  • 20
  • 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 08 '12 at 19:41
  • 1
    Yes... you can't proceed after the line $(x+y)(x-y)= y(x-y)$ as $x-y$ is zero. – zapkm Mar 08 '12 at 19:43
  • 4
    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 08 '12 at 19:43
  • yes exactly.. you can think like this... – zapkm Mar 08 '12 at 19:45
  • "Cancellation law in any field [..]"; also in any ring: if $a$ is a unit :) –  Mar 08 '12 at 19:45
  • @J.D.. yes in any ring... thanks. – zapkm Mar 08 '12 at 19:46
  • 1
    (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 08 '12 at 19:48
  • 2
    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 08 '12 at 19:51
  • @PradipMishra Since you didn't add "if $a$ is a unit", you should either add it, or replace "Ring" by integral domain. – M Turgeon Mar 08 '12 at 19:51
  • @MTurgeon, yes my mistake.. edited. – zapkm Mar 08 '12 at 19:56
1

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 $0$.

Aryan
  • 442
pxxdx
  • 11