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

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

Alan Liang - 8 years ago

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

Jess J - 8 years 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

Sean Sullivan - 8 years ago

1/0 is not equal to 2/0

(sean S. is Right )

Vamsi Krishna Appili - 8 years ago

a-b=0 , old joke

Tan Wei Sheng - 8 years ago

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

Yash Kodesia - 8 years ago

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

Jaidin Medina - 8 years ago

nice

azadali jivani - 6 years, 4 months ago

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

Tan Li Xuan - 8 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$