Are all numbers equal? What is mathematically wrong with this proof?

Favorite Answer

"
(a - t/2)^2 = (b - t/2)^2 ....................Factorize both sides
a - t/2 = b - t/2 .............................Find sqrt on both sides
"

(a - t/2)^2 = (b - t/2)^2 does not necessarily mean that a - t/2 = b - t/2. The other alternative is that a - t/2 = -(b - t/2)


eg. if a=7, b=3, t=10, then:

(a - t/2)^2 = (b - t/2)^2
(7-5)^2 = (3-5)^2
(2)^2 = (-2)^2
4 = 4

i.e. that step is ok, however:

a - t/2 = b - t/2
7 - 5 = 3 - 5
2 = -2

That step is not ok! This is because you assumed that a - t/2 and b - t/2 were both non-negative, which only occurs when a=b i.e. your "proof" is only valid when a=b to begin with, which is why it leads to that conclusion.

EDIT: x^2 = y^2 does not mean that either x=y or -x=-y, it means that x=y or x=-y. This is the problem with your justification. I do agree with you though that a-b=0 is not a problem. If this is the case, then every step is valid. However, if a-b=0, then a=b, which is what you end up proving anyway....Show more

Your proof is fine until the next-to-last step, where you take the square root.
sqrt(a*a) = +/-a, not just a.

While it is true that (-3)^2 = (3)^2 this does not imply that 3 = -3

Notice that t/2 is basically the average of a and b.
So, (a-t/2) and (b-t/2) have the same magnitude, but opposite signs. The squares are equal, but that doesn't mean the numbers are equal....Show more

<< (a + b)(a - b) = t(a - b) ........multiply both sides by (a - b) >>

If multiplying with (a-b), than a-b ≠ 0, because you'd get 0 on both sides.
From a-b ≠ 0 => a ≠ b , which opposes your final solution.

@Additional Details
Because if (a-b)=0, than every equation is correct:
5 = 3
5(a-b) = 3(a-b); a-b = 0
=> 5*0 = 3*0
=> 0=0
You see?


63n1u5...Show more

a=0
b=0
t=0...Show more