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2=1 ?!?!?!

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    Explain:

    a=b
    a^2=ab
    2a^2=a^2+ab
    2a^2-2ab=a^2-ab
    2(a^2-ab)=(a^2-ab)
    2=1
    ?????


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    a^2 - ab = 0 by a = b.
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    this is OLD AND DEAD!
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    Wow. You broke Maths. You are now the indisputed master of the universe. How does it feel? :rolleyes:
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    I'm amazed people never get tired of this.
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    You cannot divide by 0.

    Using variables, it would work.

    Using real numbers, it will not work.
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    lol i still love that proof even though it isn't really a proper proof it confuses people quite a bit - but it does show why you shouldn't divide by 0
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    TROLL
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    omg 1.999999999999999999999999999999 9999999999 =2 omg
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    (Original post by tamimi)
    omg 1.999999999999999999999999999999 9999999999 =2 omg
    No, it doesn't.
    1.999.... = 2
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    (Original post by Instincts_2012)
    You cannot divide by 0.

    Using variables, it would work.

    Using real numbers, it will not work.
    I don't understand what you mean by 'using variables, it would work'. You can't divide by a variable which potentially takes the value of zero without branching your reasoning into "either <expression> = 0 or <everything from dividing by the expression onwards>".
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    (Original post by Trollin)
    No, it doesn't.
    1.999.... = 2
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    (Original post by tamimi)
    Guilty as charged. But seriously, the pedantry of mathematicians knows no bounds :P
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    (Original post by tamimi)
     1. \dot{9} does equal  2

    But what the OP has written is nonsense!
  15. Offline

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    (Original post by raheem94)
     1. \dot{9} does equal  2

    But what the OP has written is nonsense!
    I know that is widely accepted but I just can't see it that way. Wouldn't it be true saying:

    1.\dot{9} = 2 - \frac{1}{\infty}

    That's just my thought. But then again, the infinity fraction tends to zero... However that doesn't mean it shouldn't be there.
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    (Original post by Micky76)
    I know that is widely accepted but I just can't see it that way. Wouldn't it be true saying:

    1.\dot{9} = 2 - \frac{1}{\infty}

    That's just my thought. But then again, the infinity fraction tends to zero... However that doesn't mean it shouldn't be there.
    I can't say much about it.

    Read this.
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    (Original post by As_Dust_Dances_)
    You've just assumed a equals b and that's like saying c also equals d which isn't the case!
    Lol what?
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    (Original post by Slumpy)
    I'm amazed people never get tired of this.
    because nobody stops for a second and thinks, they just get overwhelmed by the sheer plausibility and why nobody has seen it yet
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    Trifling!
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    this doesn't work with 0 values a good example is 2x=x (/x) 2=1 2x-x=0 x=0.In the above proof a^2-ab=0 because it is equal to aa-aa=0 and 2 times that will equal 1 times that.

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