Prove: 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 Note by Bodhisatwa Nandi
6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 5 months ago

1/0 is not equal to 2/0

(sean S. is Right )

- 6 years, 5 months ago

a-b=0 , old joke

- 6 years, 5 months ago

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

- 6 years, 5 months ago

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

- 6 years, 4 months ago

nice

- 4 years, 8 months ago

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

- 6 years, 5 months ago