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1 = -1

  1. May 29, 2008 #1
    Hey Everyone,

    A while back I found this cool little proof that showed 1 = -1. Now I am fully aware there was a little cheat in there somewhere, but I have lost the little proof. Has anyone come across it, or have anything similar? I just think it's cool, even though in one of the steps there is a mistake. I know it starts with like rooting one, and then putting 1 = (-1)(-1) etc.

    Cheers.

    _Mayday_
     
  2. jcsd
  3. May 29, 2008 #2
    I got another version:

    [tex] \frac{-1}{1}=-1[/tex] and [tex] \frac{1}{-1}=-1[/tex]

    so:

    [tex]\frac{-1}{1}=\frac{1}{-1}[/tex]

    if [tex]\sqrt{-1}=i[/tex]

    then

    [tex]\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}[/tex]

    so:

    [tex]\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}[/tex]

    [tex]\frac{1}{2}(\frac{i}{1})=\frac{1}{2}(\frac{1}{i})[/tex]

    becomes

    [tex]\frac{i}{2}=\frac{1}{2i}[/tex]

    [tex]\frac{i}{2} + \frac{3}{2i} = \frac{1}{2i} + \frac{3}{2i} [/tex]

    [tex]i(\frac{i}{2} + \frac{3}{2i}) = i(\frac{1}{2i} + \frac{3}{2i})[/tex]

    [tex]\frac{-1}{2}+\frac{3}{2}=\frac{1}{2}+\frac{3}{2}[/tex]

    [tex]\frac{2}{2} = \frac{4}{2}[/tex]

    [tex] 1 = 2 [/tex]
     
  4. May 29, 2008 #3
    YOu got a problem right here: [tex]\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}[/tex] Since this gives i=1/i.


    [tex]\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}[/tex]

    Just because a=b doesn't mean that [tex]\sqrt a = \sqrt b. [/tex]
     
    Last edited: May 29, 2008
  5. May 29, 2008 #4
    yes it does (as long as we've agreed on some convention so that [tex]\sqrt{x}[/tex] is a function, which we have)

    And that line is correct. The problem is that in the complex numbers [tex]\sqrt{\frac{a}{b}} = \sqrt{\frac{c}{d}}[/tex] does not imply that [tex]\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{c}}{\sqrt{d}}[/tex]. This is true for the positive real numbers, but not for complex numbers in general.

    But the poster was just asking for "proofs" that 1 = -1. Of course they are all flawed. But to the OP: There are a lot of "proofs" of this, so any more description, if you could remember any part of it, would be useful.
     
    Last edited: May 29, 2008
  6. May 29, 2008 #5
    Hey Mayday,

    You pretty much have a proof in the one that Dirk_mec1 posted.

    Once you get to this step:

    [tex]\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}[/tex]

    you now have

    i/1 = 1/i

    multiply both sides by i

    you have (i^2)/1 = i/i --> -1/1 = 1/1 --> -1 = 1.
     
  7. May 29, 2008 #6
    Thank you! Wait till my class see this stuff!
     
  8. May 30, 2008 #7
    Another one that freaks people out

    a=b
    aa=ab
    aa-bb=ab-bb
    (a+b)(a-b)=b(a-b)
    divide by a-b
    a+b=b
    since a=b then
    2b=b
    2=1
    Naturally this is completely fake, the error in this logic is that when you divide by a-b you are dividing by zero. If you want to can keep repeating this and get like 1=4 and stuff. Kinda freaks people out but make sure you explain it to them in the end =P.
     
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