Sunday 4 May, 2008

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?

6 comments:

Bring_it_on said...
This comment has been removed by the author.
Sameer Mittal said...

Uh ... I am like the idiot here. But just one contradiction.
In proof 5, when you prove 0=1, that is fine, but then you can't add 1 to both sides, as that proves 1=2, so infact, you have to add 1 to one side, and 2 to the other side, or vice versa. But, I hope you get my point.

You can't add 1 on both sides if it later proves 1=2.

Shubham Goel said...

I too have to say something about proof 5. When you are applying the law of addition, then you take out 1 but the last no. that would be left in the series would be -1 alone and it wouldn't have 1 to go with it. That 1 would be the 1 that is taken out after applying the law so that equals 0. So, according to me it is wrong.

"Sajal" said...

I m BACK

Shubham,

I think you are in class 9 and you aren't equipped with the properties of "infinity"
The proof simply uses the property
"infinity=infinity+1"
That "..." in the first statement meant series going to infinite.

Your reason would have been correct if the series was going till n terms

You'll know more about infinity and its different behavior once you reach class 11.

I hope I made sense

"Sajal" said...

Other moderators, please correct me if I am wrong

Espèra said...

This is a stupid proof, but here goes:
x - x = 0
Or, 1(x - x) = 2(x - x)
Or, 1 = 2

=P