a2 = a*a=ab
a2- b2 = ab- b2.
=>(a+b)(a-b)=(a-b)b
a+ b = b //cancelling (a-b) on both sides
As a = b => 2 b = b.
i.e., 2=1 //cancelling b on both sides
subtract 1 from both sides,
=>1=0
Were you amazed on seeing this proof for the first time. I was wondered with this proof for the first time. So when i was thinking about it, I could see a misunderstanding in the concept.
You could see a comment like "//cancelling (a-b) on both sides". Actually what does this cancelling on both sides mean ??? Its nothing but dividing on both sides. If both LHS and RHS contains some common terms, when on division by the same term on both sides yeilds the removal of those terms (which sometime we call as cancelling). So when you consider here, (a-b) is common. (a-b) equals to zero as a=b. So we cannot "divide anything by zero" to yeild a value apart from infinity. So dis proof is ABSURD !!!
Here is another proof to prove 1=0 through integration.
∫ (1/x) dx
u = 1/x , dv = dx
du = -1/x2 dx , v = x
= 1 + ∫ (1/x) dx
I'm damn sure that this proof is also wrong. Just figuring out whats wrong in it !!! Can you help me out if u could spot the mistake ???
And the last one is 0!(zero factorial). I was thought in the school days as "performing no multiplication is equivalent to multiplying by one ". But now i could interpret as
n! = (n-1)! x n
Rearranging:
(n-1)! = n!/n
when n=1:
0! = (1-1)! = 1!/1 = 1/1 = 1
Lets start to dig(g) the word of mathematics !!!!
