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Are all numbers equal? What is mathematically wrong with this proof?
Theorem: All numbers are equal (We are trying to prove that all numbers are equal)
Proof: Choose arbitrary a and b, and let t = a + b. Then
a + b = t
(a + b)(a - b) = t(a - b) ........multiply both sides by (a - b)
a^2 - b^2 = ta - tb ..............Remove brackets
a^2 - ta = b^2 - tb ..............Put terms with b on one side, and those with a.
a^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4 ..............Add (t^2)/4 on both sides.
(a - t/2)^2 = (b - t/2)^2 ....................Factorize both sides
a - t/2 = b - t/2 .............................Find sqrt on both sides
a = b ......................conclusion
So all numbers are the same
Why can't (a - b) = 0 ? So what's wrong if both sides are zero? Nothing wrong with that step. Sorry.
Again, taking the sqrt on both sides can't be wrong! It works on both sides.
If we insist on +/- sqrt, then we have to do it on both sides.
(a - t/2)^2 = (b - t/2)^2 ....................Factorize both sides
a - t/2 = b - t/2 .............................Find sqrt on both sides OR
- (a - t/2) = - (b - t/2)
a = b ......................conclusion
It doesn't change the outcome.....does it"
4 Answers
- Dan ALv 610 years agoFavorite 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.
- mark pLv 510 years ago
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.
- ?Lv 710 years ago
<< (a + b)(a - b) = t(a - b) ........multiply both sides by (a - b) >>
If multiplying with (a-b), than a-b â 0, because you'd get 0 on both sides.
From a-b â 0 => a â 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?
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