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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

Update:

Why can't (a - b) = 0 ? So what's wrong if both sides are zero? Nothing wrong with that step. Sorry.

Update 2:

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

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  • Dan A
    Lv 6
    10 years ago
    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.

  • mark p
    Lv 5
    10 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 7
    10 years ago

    << (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

  • a=0

    b=0

    t=0

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