- #1
Lizwi
- 40
- 0
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?
(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?