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Thread: Where does this problem fail?

  1. #1
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    Question Where does this problem fail?

    So, someone was showing me a math problem where he said he could get 1 = 2. Here's what he did:

    $\displaystyle x = 1$

    Thus
    $\displaystyle x^2 = x^2$

    So
    $\displaystyle x^2 - x = x^2 - 1$
    Right?

    Then, he factored:
    $\displaystyle x(x -1) = (x - 1)(x + 1)$

    Divided by (x - 1):
    $\displaystyle [x(x-1)]/(x-1) = [(x-1)(x + 1)]/(x - 1)$
    to cancel out (x - 1)

    Resulted in:
    $\displaystyle x = x + 1$
    From where he plugged in x = 1:
    $\displaystyle 1 = 1 + 1$
    Therefore:
    $\displaystyle 1 = 2$

    Please explain to me why this fails, because it seems like it does in so many places, but I want to know why exactly! For instance, dividing by $\displaystyle (x - 1)$ would be the same as dividing by zero, which would fail, but he thinks he's pretty much found where math has a weakness. I need expert verification!
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  2. #2
    Super Member Bacterius's Avatar
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    There is a division per zero while simplifying, which invalidates all the following, sorry. You cannot divide per zero, because it would mean that $\displaystyle 2 = 1$, crumbling the whole theory of mathematics. Look at this :

    $\displaystyle a = b$

    $\displaystyle a^2 = ab$ (multiply by $\displaystyle a$)

    $\displaystyle 2a^2 - 2ab = a^2 - ab$ (add $\displaystyle a^2 - 2ab$)

    $\displaystyle 2(a^2 - ab) = a^2 - ab$ (factorize on left side)

    $\displaystyle 2 = 1$ (simplify by $\displaystyle a^2 - ab$)

    Which is impossible, because $\displaystyle a^2 - ab = 0$ (division per zero).
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  3. #3
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    Ray, you are awesome! Thanks!

    Quick edit: When you say (add $\displaystyle a^2 - 2ab$), do you mean $\displaystyle 2a^2 - 2ab$?
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  4. #4
    Super Member Bacterius's Avatar
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    Quote Originally Posted by audax View Post
    Ray, you are awesome! Thanks!

    Quick edit: When you say (add $\displaystyle a^2 - 2ab$), do you mean $\displaystyle 2a^2 - 2ab$?
    No, I mean $\displaystyle a^2 - 2ab$
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  5. #5
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    can't divide by 0

    note that x-1=1-1=0, so you can't divide by 0, as you have done.
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