Something to think about....

You doubt that 2=1?  Consider the following:

     a = x            [true for some a's and x's]
   a+a = a+x          [add a to both sides]
    2a = a+x          [a+a = 2a]
 2a-2x = a+x-2x       [subtract 2x from both sides]
2(a-x) = a+x-2x       [2a-2x = 2(a-x)]
2(a-x) = a-x          [x-2x = -x]
     2 = 1            [divide both sides by a-x]

So then, 2=1, right?  =D

Impossible!  So where is the mistake? Think about it.
You may not like the first step (a=x). But we do this kind of thing all the time in Algebra. It's true for plenty of a's and x's. Assume that a is the number of ears on my head, and x is the number of ears on your head.  In that case a=x.  (if a is not equal to x, forgive me for mentioning it)
Anyway, all of the steps are perfectly legal except for the last one, dividing both sides by a-x. What is a-x? Well, a=x (step 1), so a-x=0. In the last step, we divided by zero. That's not allowed. And this puzzle is a good example of why it is not allowed.  =)