Is division by zero a logical error in this proof?

[Lizwi]

That's false!, since this logic a + b = t
(a + b)(a - b) = t(a - b)
a^2 - b^2 = ta - tb
a^2 - ta = b^2 - tb
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4
(a - t/2)^2 = (b - t/2)^2
a - t/2 = b - t/2
a = b

gives false results there must be an error in it because we know that two things will never equal 3 things that's impossible. 2 is not 3, you know this. A person who did this proof should have doubted his logic because it produce the obviously false results.


What do you say?

 

[Michael C]

The mistake is near the end:

(a - t/2)^2 = (b - t/2)^2
is correct, but this does not necessarily mean that
a - t/2 = b - t/2

Since a square root can be positive or negative, (a - t/2)^2 = (b - t/2)^2 implies that:
either
(1) a - t/2 = b - t/2
or
(2) a - t/2 = - (b - t/2)

From (2):
a - t/2 = t/2 - b
a + b = t/2 + t/2
a + b = t
We're back where we started!

 

[mathman]

Since t/2 = (a+b)/2, |a - t/2| = |b - t/2|. Assume a < b, then a - t/2 < 0 and b - t/2 > 0.

 

[cng99]

If a=b, multiplying (a-b) both sides doesn't make any sense because a-b=0. Because that way any equation could be proven true. Just multiply 0 both sides and say 0=0.

 

[NeuroFuzzy]

If a=b, multiplying (a-b) both sides doesn't make any sense because a-b=0. Because that way any equation could be proven true. Just multiply 0 both sides and say 0=0.

No... If you say that x=Sqrt(b)+c-d, then yes this equation implies that 0*x=0*(Sqrt(b)+c-d) which means that 0=0. The opposite, going from 0=0 to x=Sqrt(b)+c-d involves division by zero, which doesn't make sense.

I think Michael C hit the nail on the head.

 

[cng99]

No... If you say that x=Sqrt(b)+c-d, then yes this equation implies that 0*x=0*(Sqrt(b)+c-d) which means that 0=0. The opposite, going from 0=0 to x=Sqrt(b)+c-d involves division by zero, which doesn't make sense.

Well we both basically mean the same thing.