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    • Thread Starter
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    I will algebraically prove to you that 1=2.

    1. Let a = b
    2. Then a² = ab
    3. a² + a² = a² + ab
    4. 2a² = a² + ab
    5. 2a² - 2ab = a² +ab - 2ab
    6. Hence, 2a² - 2ab = a² - ab
    This can be written as 2(a² - ab) = 1(a² -ab)
    and cancelling the (a² - ab) from both sides gives 1=2.

    Can figure out in which step the fallacy lies? First one to give me the correct answer, along with an explanation of why that step is invalid, gets a personal rating and a follow on TSR
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    a^2 -ab=0

    Dividing by 0 gives an undefined result.

    So you've said  2\times 0 = 1\times 0 so  0=0, not 1=2, you can't cancel 0.
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    a^2-ab = 0, and you can't divide by 0.
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    The mistake lies in the last step. As a squared equals ab, a squared minus ab is equal to 0, so in the last step you are actually dividing by zero, which isn't defined.
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    Yeh what they said ^^
 
 
 
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