I don't remember the trick to 1 = 0?

Question

There's a trick that starts out assuming a=b, and ends up showing that 1 = 0. I don't remember the steps inbetween, just that it's invalid because somewhere along the way you divide by zero (a-b). Anybody remember?

Answer 1

assume a=b
=> a * a = a * b (Multiply both sides by a)
=> a^2 = ab
=> a^2 - b^2 = ab - b^2 (substract b^2 from both sides)
=> (a - b) (a + b) = b (a - b)
=> a + b = b (canceling a-b from both sides)
=> b + b = b
=> 2b = b
=> 2 = 1

NOTE: canceling (a - b ) from both sides is actually dividing by zero.

Answer 2

1 is equivalent to 0 mod 1. =]

Answer 3

yes :
1 = 1

x*1 = x*1

x = x
x - x = 0

(x-x) / (x-x) = 0/(x-x)
1 = 0/(x-x)

1=0 ( since 0 divided by something is always 0 )

Answer 4

There argument goes something like this:

1 + infinity = infinity
1 + infinity - infinity = infinity - infinity
Therefore,
1 = 0

The problem with that argument is that infinity minus infinity is not zero. It is undefined.

Answer 5

WELL, ALL I KNOW IS THAT, 1 IS NOT EQUAL TO 0. WHICH MAKES THAT/YOUR STATEMENT FALSE.