Another simple math question

[Peterson]

I ran across this 'joke' on the internet this morning. I can't figure out what the 'trick' is. At first glance, I thougt it was the old 'division by zero' trick but it doesn't look like that's it. What am I missing?
Theorem: All numbers are equal.
Proof: Choose arbitrary a and b, and let t = a + b. Then

a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

So all numbers are the same, and math is pointless.

:confused:
I know I'll kick myself when I see what it is.

[Ziggurat]

Originally posted by Peterson

What am I missing?

:confused:
I know I'll kick myself when I see what it is.

Took me a minute, but the mistake is in this step:

(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2

This step is wrong, because it assumes (incorrectly) that both (a-t/2) and (b-t/2) are non-negative. In fact, unless a=b (in which case both terms are zero, and proving a=b is rather a tautology), one of the terms MUST be negative. So you SHOULD have:

a-t/2 = -(b - t/2) = t/2 - b

(which is correct even for a=b), from which we can get

a + b = t.

No surprise now.

[Peterson]

That crossed my mind, that they were assuming that the square root would be positive (instead of + or -). Not as obvious as I thought, so I'm not kicking myself. :)

Thanks!