People have asked the staff for LaTeX support on several occasions. It's just very low on their priority list because such a small fraction of users would benefit, basically this math group and a few others.
As for the proof, I'm familiar with it so I won't spoil it for those who are seeing it for the first time.
a+b cannot equal b and b+b cannot equal b
assuming a and b are 1 due to a=b
1+1=1>>>2=1 a+b=b
1+1=1>>>2=1 b+b=b
1=1 a=b
Conquistador, of course I know 2 is not equal to 1. I'm asking you to find the flaw, not to state the obvious.
Is this supposed to be a group over the integers? It seems like a and b are supposed to be the identity element, but with addition over the integers the identity would be 0 not one, so a and b would have to be zero.
a+b=b is not the same as b+b=b?
@ nefariousorator: I'm just talking about the real numbers here with the common definition of addition and multiplication. By the way, if you want to get smart and be so formal, you're wrong when you speak of a "group" when there are two binary operations involved.
@ chessman_calum: The first line says a = b, so then a+b = b+b is certainly true.
Hint to the new ppl: What is a-b?
You cannot divide by a-b if a=b
(a+b)(a-b) = b(a-b)
a+b = b
and later you assume that a=b and later
a+b = b
b+b = b
so first you say that a is differnt from b and later you say a=b 
. Contradiction 
!
@ BigFuzzyOne : "To divide a-b you must assume a different from b otherwise you cannot go on next expression.
(a+b)(a-b) = b(a-b) if a different from b
a+b = b
a=b doesn't follow on the next expression a+b = b because you divide by 0 .
a ≠ b a+b
a=b (a+b)(a-b) = b(a-b)
I hope that i was clear this time !
The wrong is :
a ≠b so a+b =b
now a+b=b
now we asume a=b to get b+b=b and 2=1 ????
First we assume a ≠b and later a=b which lead in contradiction .
Or let's say in elegant words :
a-b≠0 so
(a+b)(a-b) = b(a-b)
a+b = b
now a-b=0 so
a+b = b
b+b = b
in the reasoning we assume 2 assertions that are in contradiction with each other.
Actually i gotten the same problem . Maybe another bug from the site.
I see BigFuzzyOne's point: FT-physicist is 90% but not 100% correct each time. The equations just say that a = b, and then you divide by a-b which must be equal to 0. You can't say "ok, so now a+b = b only holds if a is not equal b" since then you're assuming a contradiction, which only makes things more complicated. You need to GET to a contradiction (dividing by 0), not ASSUME one. Else you could just assume the contradiction 1 = 2 and start deriving crazy stuff from this.
(x+1)^2 = x^2+2x+1
(x+1)^2-(2x+1) = x^2
(x+1)^2-(2x+1)-x(2x+1) = x^2 -x(2x+1)
(x+1)^2 -(x+1)(2x+1) = x^2-x(2x-1)
(x+1)^2 -(x+1)(2x+1)+.25(2x+1)^2 = x^2-x(2x+1)+.25(2x+1)^2
((x+1)-.5(2x+1))^2 = (x-.5(2x+1))^2
sqrt(((x+1)-.5(2x+1))^2)) = sqrt((x-.5(2x+1))^2)
(x+1) - .5(2x+1) = x - .5(2x+1)
(x+1) = x
Therefore, 2=1
An old one, but for those who haven't seen it before: What's wrong with the following reasoning?
a = b
aa = ba
aa - bb = ba - bb
(a+b)(a-b) = b(a-b)
a+b = b
b+b = b
1+1 = 1
2 = 1
By the way, someone should convince erik to implement LaTeX support in these forums :/
Ditto... I had to copy my 'proof' from a Latex document!
The solution to the lack of latex support is clearly to use image files instead. The question is what is the easiest way of achieving this. Doing a manual edit of a screengrab would be easy enough, from any app able to write equations. I recall at least one public domain app that converts latex to gifs, as well. It would be most convenient to do the conversion for the whole post, or at least all the maths in one go.
Well, why should a+b= b if a = b?? I guess a/b would have to be zero, but even then it would be 0 = 0, not 2 = 1.
An old one, but for those who haven't seen it before: What's wrong with the following reasoning?
a = b
aa = ba
aa - bb = ba - bb
(a+b)(a-b) = b(a-b)
a+b = b
b+b = b
1+1 = 1
2 = 1
By the way, someone should convince erik to implement LaTeX support in these forums :/