2=1 ?

First Way to prove it.
(n+1)^2 = n^2 + 2n + 1
or, (n+1)^2 - (2n+1) = n^2
subtracting n(2n+1),
or, (n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)
or, (n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2
or, (n - 1/2)^2 = (n + 1/2)^2
or, (n - 1/2) = (n + 1/2)
or, -1/2 = +1/2
or, 1 = 0
or, 2 = 1 !!


Second Way

4*4 = 4 + 4 + 4 + 4

:. 4*4= 4 + 4 +… 4 times.

:. x*x = x + x + … x times.

Differentiating,

2*x = 1 + 1 + 1 … x times.

:. 2*x = x

Hence 2 = 1!!!

Third Way

Every1 here knows the pretty little Binomial thorem going as :

(a+b)n = an + n*b*an-1 + ... + n*bn-1*a + bn
Observing we find that leaving the first and last term, all other terms are a multiple of n.
For some wierd results, let us put n = 0 ;
LHS = (a+b)0 = 1
RHS = a0 + 0 ... + 0 + b0
= 1 + 0 ... 0 + 1
= 2
:. since LHS = RHS
2 = 1

Fourth Way
- 1 = - 1
or, sqrt(-1) = sqrt(-1)
or, sqrt(-1/1) = sqrt(1/-1)
or, sqrt(-1)/sqrt(1) = sqrt(1)/sqrt(-1)
or, i/1 = 1/i
or, i2 = 1
or, -1 = 1
or, 2 = 0
or, 1 = 0
or, 2 =1

Fifth Way

The very interesting infinitie series :

0 = 0 + 0 + 0 + 0 ...

For those who don't find it so interesting, follow me.

0 = ( 1 - 1 )
:. 0 = ( 1-1 ) + ( 1-1 ) + ( 1-1 ) ...

Now all of you would have paid attention in primary school when we were taught the associative law of addition.
Interestingly applying that here,

0 = ( 1-1 ) + ( 1-1 ) ...
or, 0 = 1 + (-1+1) + (-1+1) ...
or, o = 1 + 0 + 0 + 0 + 0 + ...
or, 0 = 1
or, 1 = 2

Sixth Way

- 20 = - 20

or, 25 - 45 = 16 - 36

or, 5² - 5*9 = 4² - 4*9

or, 5² - 5*9 + 81/4 = 4² - 4*9 + 89/4

or, (5 - 9/2)² = (4 - 9/2)²

or, (5 - 9/2) = (4 - 9/2)

or, 5 = 4

or, 2 = 1

The question is, Can you find another way of proving it?