Latest IBM Aptitude Question SOLUTION: prove this (62-63)=1

given 62-63=1
we can write it as 62=1+63
now 62=64.........(1)equation
by substituting (1)eq in the give eq 64-63=1 its simple

(62-63)=1
take individually
62=6-2=4
63=6-3=3
(62-63) -->(4-3)=1(hence proved)

With the help of our general conception of number system and the operations like + , - ,* , / we at once can infer that the above statement can't hold . But since it is proposed to prove then there may be some exciting mathematical fallacy which we have to adopt to be able to prove the given problem.

First consider the following proof:

Let a = b ;[] a,b are natural numbers with none of them = 0]
=> (a^2)=(b^2)
=> {(a^2)-(b^2)} = {(a^2)-ab}
=> (a+b)(a-b) = a*(a-b)
=> (a+b) = a ;[cancelling (a-b) from both the sides ]
=> 2*a = a ;[ since a=b]
=> 2=1 ; [cancelling a from both the sides & since ais not equal to 0]
=> 2*2= 2*1 [multiplying both sides by 2]
=> 4 = 2
=> (4-2)=0
=> 2 = 0
=> 1 + 1 = 0
=> 1 + (63-62)=0; [since obviously 63-62 =1]
=> 1 = -(63-62)
=> 1 = 62 - 63
So, 62 - 63 = 1 ....[Proved]

N.B.: This is impossible to hold.Now, of course there is a fallacy. Many of you have already pointed out the fallacy, I know.But who are still wondering that how it has become possible, I'm revealing the flawfull trick I've played to prove it.
Just refer to the 3rd and 4th lines of the above proof.You can notice that I've cancelled (a-b) from both sides,i.e,dividiing both sides by (a-b). But Initially I've mentioned that (a=b). So, it readily follows that (a-b)=0, and all of us surely aware of the fact that division by 0 ,is undefined.So, it is not allowed in no circumstances to perform the division operation by [(a-b)=0].So, from the 4th line on wards we start getting some unexpected and inexplicable results which contradict our general notion or number system and operations.But this tiny and illegal trick helps us to reach the desired solution.

Please share your feedback ,views ,complaints and ofcourse some other kind of proofs in this context. All of us eagerly awaiting them

Thanks to all..:-):-)

(6-2)-(6-3)
4-3=1

(62-63)=1
|62-63|=1

this can be prove by
=(6/2)-(6/3)
=3-2

62-63=62+1-63-1=1.
Hence proved.
By power star

Shift place of 6:
i.e. 62-63=1 becomes 2^6-63=1
64-63=1

squaring both side.

then you got 1=1