Let a=1 and b=1. Now, we have

a=b

or, \(a^{2}\) = ab

or, \(a^{2}\) - \(b^{2}\)= ab-\(b^{2}\)

or, (a+b)(a-b) = b(a-b)

or, (a+b)=b

Thus on putting the values we get,

1+1=1

or, 2=1

Let a=1 and b=1. Now, we have

a=b

or, \(a^{2}\) = ab

or, \(a^{2}\) - \(b^{2}\)= ab-\(b^{2}\)

or, (a+b)(a-b) = b(a-b)

or, (a+b)=b

Thus on putting the values we get,

1+1=1

or, 2=1

No vote yet

3 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestSigh. This post has been posted way too much...just stop.

Log in to reply

well obviously, 2=2 so your theory kinda goes down the drain.

Log in to reply

Mathametic fallacy derp....you cant cancel out zeroes in an eqn..

Log in to reply

a-b=0 , old joke

Log in to reply

1/0 is not equal to 2/0

(sean S. is Right )

Log in to reply

It's called mathematical fallacy. To go from (a+b)(a-b)=b(a-b) to a+b=b you must divide by (a-b) but because a=1 and b=1, a-b=0 and dividing by 0 is indeterminate. If we were able to divide by 0 then I could just as easily say 1/0=2/0 so 1=2

Log in to reply

nice

Log in to reply

STOP DIVIDING BY ZERO. IT IS ILLEGAL IN MATH FOR A REASON.

Log in to reply

in your proof, \( (a-b)=(1-1)=0 \) and you can't divide by 0

Log in to reply