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this is a puzzle involving some trick!! plz help me solve it...

if A=B,
then prove that 1=2

plz help me. i really need the answer!!
of course 1 cannot be equal to 2, but the proof wud b logically correct...

2007-03-17 17:26:13 · 6 answers · asked by zari p 2 in Science & Mathematics Mathematics

6 answers

A = B
A^2 = AB
A^2 + A^2 = A^2 + AB
2A^2 = A^2 + AB
2A^2 - 2AB = A^2 + AB - 2AB
2A^2 - 2AB = A^2 - AB
2(A^2 - AB) = 1(A^2 - AB)
2 = 1

This is one way, but you end up dividing by zero in the last step. A no-no.

2007-03-17 17:34:42 · answer #1 · answered by tryzub91 3 · 1 0

who ever told you that the proof for this non-sense is logically correct was obviously fooled. It is quite obvious that the thing you are trying to prove is wrong so don't use the word "LOGICALLY".

i'll give you what you are looking for ang show tell you where the "illogical" statement is.

if a = b
multiply by b
ab = b^2
subtract a^2
ab - a^2 = b^2 - a^2
factor
a(b-a) = (a-b)(a+b)
* note that (a-b) is zero since a=b.
so this next step is not logical. any number divided by zero is infinite. thus no proper value.

a = a+b.
but a= b so
a = a+a
a = 2a
if a=1
1= 2(1)
1= 2........not true! //answer

2007-03-18 00:49:02 · answer #2 · answered by 13angus13 3 · 1 0

The classic "proof" involves a division by (A - B) at some point. But if A = B, the division is by 0 and the value is not defined. And, it is the fact that it leads to absurdities such as this that explains WHY it is not defined.

2007-03-18 00:44:44 · answer #3 · answered by Anonymous · 0 0

1 and 2 are just symbols. if you say that 1 = dog and 2 = dog therefore 1=2 by the law of syllogism

2007-03-18 00:31:19 · answer #4 · answered by max l 1 · 0 1

ur question is incomplete... or maybe u can bring "A" and "B" on the same side n then whole square them and using identity of "(a+b)" whole sq.. solve..

2007-03-18 00:30:56 · answer #5 · answered by SSSting 2 · 0 1

A = B
A+A = A+B
2A = A+B
2A-2B = A+B-2B
2(A-B) = A+B-2B
2(A-B) = A-B
2 = 1

2007-03-18 00:42:36 · answer #6 · answered by Charley 2 · 1 0

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