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Proving 3 + 2 = 0

  1. Aug 4, 2009 #1
    Assume A + B = C, and assume A = 3 and B = 2.

    Multiply both sides of the equation A + B = C by (A + B).

    We obtain A² + 2AB + B² = C(A + B)

    Rearranging the terms we have

    A² + AB - AC = - AB - B² + BC

    Factoring out (A + B - C), we have

    A(A + B - C) = - B(A + B - C)

    Dividing both sides by (A + B - C), that is, dividing by zero, we get A = - B, or A + B = 0, which is evidently absurd.
     
  2. jcsd
  3. Aug 4, 2009 #2
    Can you elaborate on this step ? What do you mean by "factor out" ?
     
  4. Aug 4, 2009 #3

    Pengwuino

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    Gold Member

    Yes, it is. You can't divide by 0.

    It's like saying : 3 * 0 = 4 * 0 and then attempt to divide by 0 to say that 3 = 4.
     
  5. Aug 4, 2009 #4
    You can not divide by zero, you must do the following:

    [tex]\begin{equation}
    \begin{align}
    A(A+B-C)+B(A+B-C)=&0\\
    (A+B-C)(A+B)=&0
    \end{align}
    \end{equation}[/tex]

    which implies that either:

    [tex]
    A+B-C=0\\
    \text{ or }\\
    A+B=0
    [/tex]

    and of cource the second option is rejected since it contradicts with the first one, unless [tex]C=0[/tex], but if you state at the begining that [tex]A\neq 0, B\neq 0, \text{ and } C\neq 0[/tex], then the second choise definetly must be rejected.

    Regards
     
  6. Aug 4, 2009 #5

    jgens

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    Gold Member

    This really is what Pengwuino said already, but to elaborate . . . In the set of real numbers (R), every non-zero number a has a multiplicative inverse element a-1 such that a * a-1 = 1. However, the set of real numbers does not contain a multiplicative inverse element for zero, that is, there is no real number x such that 0 * x = 1 (a fact that is relatively easy to prove).
     
  7. Aug 4, 2009 #6
    I don't know if I sound like I am robbing straight from your face this problem with "Hey then whatz up?"
    Who says 3+2=0 ?
     
  8. Aug 4, 2009 #7
    Alright, in the first line of your argument, you said "assume A+B=C". And your whole argument stems from cancelling out the factor of A+B-C which means dividing by A+B-C, and from your definition of A+B=C implies A+B-C=C-C or A+B-C=0. Therefore you're dividing by zero.
     
  9. Aug 5, 2009 #8
    You divided by zero?

    OH SHI-
     
  10. Aug 5, 2009 #9
    I feel like the original poster, DARTZ, is just posting an interesting demonstration of what kind of mischief dividing by zero leads to and knows perfectly well he can't do so. He even states that he is dividing by 0 and admits the absurdity. DARTZ said he was dividing by zero, and then nearly every post goes and says "hey, you're dividing by zero". I could be wrong though.
     
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