End of Mathematics as we know it (for fun)

What if I told you that I can prove, mathematically, that one plus one equals...one, not two? That, in fact, one equals two? Not philosophically, but literally—mathematically. Wouldn't that shake things up a bit in every field that employs algebra?

Let's consider:

a = b

Then,

a2 = b2. (That's squared; I don't know how to use the proper "raised to" sign here on the computer...)

Then we can subtract b2 from both sides,

a2 - b2 = b2 - b2

but we know that a2 - b2 = (a+b) x (a-b)

So, we have...

(a+b)(a-b) = b2 - b2.

Now, b2 can be written as ab, as a = b. So...

(a+b)(a-b) = ab - b2.

Take b out common from the RHS...

(a+b)(a-b) = b(a-b)

The term (a-b) is common to both sides, and can be eliminated.
Then we have...

a + b = b.

a = b. So, substitute a by b...

b + b = b; (That is...1 + 1 = 2.)

2b = b.

Or...2 = 1.



Note: I don't claim to "invent" this or anything. I remember reading about this somewhere many years ago, and I'm merely presenting it here.

P.S.: The catch...there is always one....is that the term (a-b) cannot be eliminated from both sides. The reason? Well, a-b is a zero-value term because a equals b, and division by zero is not acknowledged. So, alas, technically, this was a fail, but for someone who doesn't quite notice this little thing...it's a very interesting math prank.

Then we can subtract b2 from both sides,

a2 - b2 = b2 - b2

sorry dude, but by this point you've already failed.
you've already stated that a2 = b2, so when you do

a2 - b2 = b2 - b2

you're saying

1 - 1 = 1 - 1 which equals 0 = 0

missed the last line at the bottom explaining it as a prank.....what is the point of this post then?