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venkatcustoms
21-02-2009, 10:17 PM
Dear Friends,
Here you find another theorem which proves 1 = 2

Suppose A =B

Then A – B = A – A (Because A = B)

Add (A – B) both sides;

A – B +(A – B ) = A – A +(A – B )

i.e 2A-2B = A-B

2(A – B) = 1(A – B)

CANCELLING (A-B) BOTH SIDES,

YOU GET 2 =1

If nobody comes with the illogical logic used hereabove( I know Badriji will definitely find it out), I will post the same later

narayanan
21-02-2009, 10:56 PM
Dear friends,

The conclusion reached by Shri VC is:

2 = 1

Multiply both sides by A-B

Then: 2(A-B) = 1 (A-B)

ie. 2x 0 = 1 x 0 [ since it is given that A=B, A-B=0]

ie. 0 = 0

I feel now there is no problem.

01-03-2009, 10:29 PM
Dear friends,

The conclusion reached by Shri VC is:

2 = 1

Multiply both sides by A-B

Then: 2(A-B) = 1 (A-B)

ie. 2x 0 = 1 x 0 [ since it is given that A=B, A-B=0]

ie. 0 = 0

I feel now there is no problem.

Dear friends,
The fallacy reflected in VC's so called theorem, and Shri Naryanan's "Responsive(!) theorem is that Zero can not be used as an operator. That is when Zero occupies numerator as a factor, then zero can not operate as a factor in the Denominator portion.
This can be explained in simple way;-
You can not produce 1 by dividing zero by zero.
This can further be explained as follows;-
25 persons and there are 100 Apples means, Each person gets 4 apples
25 persons, 0 apples means 0/25 each person gets 0 apple.
0 persons, 100 apples means 100/0 this leads to infinity (Apples are there
being shared).
0person, 0 apples leads to 0/0, which can not produce one apple.
That is why we say, Zero can not be an operator in the examples given by VC, and Shri Narayanan.

With regards,