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# What if 1=2?

I am one.

The pope is one.

We are two.

But 2=1.

Therefore I am the pope.
just like 2+2=5 for sufficiently large values of 2.
Oh. I thought this was going to be one of these type puzzles:

If you have "A" and "B" both equal to 1
So...

A = B
A - B = B - B (subtract "B" from both sides)
A - B = 0 (since "B - B" = 0)
(A - B)*(A + B) = 0 * (A + B) (multiply both sides by "A+B")
(A - B)*(A + B) = 0 (since anything times zero is zero)
(A - B)*(A + B) / (A - B) = 0 / (A - B) (divide both sides by "A - B")
(A + B) = 0 / (A - B) (the "A - B" components of the left side cancel out)
(A + B) = 0 (since anything divided into zero is zero)

And since A = 1 and B = 1 (the original assertion), this implies that 1 + 1 = 0.

Math geeks unite.
Divide by zero error.
If A=B=1 then
(A - B)*(A + B) / (A - B) = 0 / (A - B) (divide both sides by "A - B")
is undefined.

n/0 is undefined. 0/0 is undefined. You lose on both counts.
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