where is your God now ?

math.problem.mkdn

initial condition

a = b

* a

a² = ab

+ a² - 2ab

a² + a² - 2ab = ab + a² - 2ab

grouping

2(a² - ab) = a² - ab

: (a² - ab)

2 = 1

Where is your God now ?

Am I missing something?

a = b
a2 = ab = c

a2 + a2 - 2(ab) = ab + a2 - 2(ab)
> c + c - 2c = c + c - 2c
> 0 = 0
> 0

2(a2 - ab) = a2 - ab
> 2(c - c) = c - c
> 0 = 0
> 0

Seems to hold?

Yep, it holds well.
Doesn't the above/original hold too though (2 = 1) ?
All the operations seem fine.. :P
(I can give you the mistake in the logic, it's not that hard actually)

a = b = 0?

a = b

a (*a) = b (*a)
> a2 = ab

a2 + a2 - 2ab = ab + a2 - 2ab
> a(a + a - 2b) = a(b + a - 2b)
> a(2a - 2b) = a(a - b)
> 2a(a - b) = a(a - b)
> 2a = a
> a = 0

That'd make your last operation division by 0 and thus and invalid result.

Otherwise, you'll have to tell me, I haven't done crap like this since college :)

haha! :P

well, the fourth pass is : (a2 - ab)
which means divide by that term
where that term is a2 - ab = a2 - a2 = 0
and a and bare not some unrelated variables,
we know that a = b.

so it's actually zero , no matter what a and b are.
so that's not applicable
thus you end up with 2 = 1

which in turn is too bad, cause 0 = 0 looks like two cookies to me!
suddenly, cookies(!) everywhere(!!)

Pfft, I said the magic phrase divide by zero, so I call that a cookie for me.

Yeah, brisbin got it. I remember seeing this in high school. Old trick