Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
James, you might have noticed that I added the tag "fake-proofs" to the question. If you click on that tag, you will find many such proofs. A few of them involve division by zero (as does yours). –  The Chaz 2.0 Mar 8 '12 at 19:46
1  
Thanks, @TheChaz, I didn't know the tag existed. –  jamesbrewr Mar 8 '12 at 19:49
add comment

3 Answers 3

up vote 16 down vote accepted

Hint $\ $ When debugging proofs on abstract objects, the problem 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"

$$\begin{eqnarray} 3^2 &=& 3\cdot3 \\ 3^2 - 3^2 &=& 3\cdot 3 - 3^2 \\ (3 + 3)\:(3 - 3) &=& 3\: (3-3) \\ 3 + 3 &=& 3 \\ 2\cdot 3 &=& 3 \\ 2 &\:=\:& 1 \end{eqnarray}$$

Now you can determine which inference is incorrect by determining the first false equation above; if it is equation number $\rm\: n\!+\!1,\:$ then the inference from equation $\rm\:n\:$ to $\rm\:n\!+\!1\:$ is incorrect.

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.

share|improve this answer
1  
That's a good way to think about it. I'll give it a try next time I run across a similar problem. –  jamesbrewr Mar 8 '12 at 20:02
5  
I wish I could upvote this more than once. –  Henning Makholm Mar 8 '12 at 20:14
add comment

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?

share|improve this answer
1  
Unfortunately I read the other reply before yours which blatantly points out what you're getting at, so I didn't get the chance to find it myself. –  jamesbrewr Mar 8 '12 at 19:42
add comment

In third line you have written $$(x+y)(x-y)= y(x-y)$$ as $x=y$, we can't cancel $(x-y)$.

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 $b=c$.

share|improve this answer
    
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 8 '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 8 '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 8 '12 at 19:43
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 8 '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 8 '12 at 19:51
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.