Answer / prasant kumar

ya it is possible..



a^2-b^2 =(a+b)(a-b)
so a^2-a^2=(a+a)(a-a)..............eqn 1

and a^2-a^2=a(a-a).......eqn 2

hence after comparing both we will get the result that

a=(a+a)
=>1=2

Answer / sandeep

1^0=1^1
if a^m=a^n,m=n
so,0=1
adding 1 on both sides
0+1=1+1
1=2

Answer / guest

Actually it is not true,but some manipulation can do.

let a=b
multiply by b
ab=b^2
ab-a^2=b^2-a^2
a(b-a)=(b+a)(b-a) [actually b-a=0; 0 can't be
cancelled]
a=b+a
a=a+a [a=b]
a=2a
1=2

Answer / manoj kumar paikray

Here question is that 1=2,
so we have to proved that R.H.S=L.H.S
as L.H.S=1
=1*X
=1*0(let X=0)
=0
then R.H.S=2
=2*X
=2*0(as X=0)
=0=L.H.S
hence L.H.S=R.H.S
SO 1=2

Answer / jayasree

yes, it is possible.
A2-b2 = (a+b) (a-b) as equn 1
Lets take b2=a, substitute in equn 1,
A2-a =(a+b) (a-b)
A2-a = a2-ab+ab-b2
-a = -b2
A=b2
Take log on both sides,
Log a = 2 log b
Assume a=b=0, and subsitute,
Log 0 = 2 log 0
1 = 2 (1) (i.e, log 0 = 1)
1=2

Answer / mms zubeir

Definitely, it is not possible but it is possible with
magics ( remember a magic is nothing but hiding the truth).

Answer / bavas ahamed

a=b
a^2=b^2
a^2=ab
(a^2-b^2)=ab-b^2
(a-b)*(a+b)=b(a-b)
a+b=b
if b=2,a=1
therefore 1=2

Answer / anitha venkatesan

yes the above answer by Manoj Kumar is correct.

Answer / kumaresan

one=two
one=3 letter word
two=3 letter word
3=3 so 1=2

Answer / kumaresan

1=2
one=two
3 letter word = 3 letter word
3=3
so 1=2

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