How can this (math) be?
 

How can this be?

Assume that:
a = b
Therefore:
a^2 = ab
a^2 - b^2 = ab - b^2
(a+b)(a-b) = b (a-b)
a+b = b
Since a = b then
2*b = b
2 = 1

The riddle (a mathematical joke) was provided by EE Times. Yes, a simple algebraic flaw exists. But what is it?

This is totally going to bug the crap out of me until I figure it out... But I worked it out and I can't find the mistake... Ah, my math teachers sucked.

a^2 - b^2 = 0. Game over.

Not quite - the problem is that (a-b) is zero, and you are dividing by it in step 3 to 4.

To get from (a+b)(a-b) = b (a-b)
to
a+b = b
you must divide each side by (a-b). That's great, except...

since a=b, then (a-b) = (a-a) = 0

Only Chuck Norris can divide by zero.

If a = b then a^2 - b^2 = ab - b^2 is the same thing as saying 0 = 0. I think you are done at that point.

i contemplated just posting 'pull my finger' here, but instead, my curiosity has me. .....


what does '^' mean in math?

^ means raise to the power of. So a^2 means a squared and a^3 means a*a*a.

thank you. now.....pull my finger

Algebra 1
 

Doesn't anyone remember FOIL? :)

(a+b)(a-b) = b (a-b)
a^2 -ab + ab -b^b
a^2 - b^2
0

Therefore, 2 does not equal 1 :)

RIght, the whole thing hinges on one '0=0' set after another. They sneak in a '2*0=0' and try to equate it with 2*1=1, the mistake is in the very last operation.

You divide by zero to get from step three to four, don't you?

My junior high Algebra teacher showed us this.

Yes, but I think (and I could be wrong, it's been awhile since I was in junior high myself) it's okay, because you're dividing zero by zero, and 0/0 is the one acceptable case where it equals one.

You can't do that either... there's no definite value for it.

No. 0=0 is OK, just like 1=1 is OK in step one. You just can't divide by zero.