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Need help with proof

  1. Feb 2, 2005 #1
    Hello all,

    Prove that [tex] \sqrt[3]{3} [/tex] is irrational.

    Heres what I have so far:
    [tex] \sqrt[3]{3}=\dfrac{m}{n}[/tex]
    [tex]3=\dfrac{m^3}{n^3}[/tex]
    [tex]m^3=3n^3 [/tex]

    Now, I think I am to assume that there is more than one factor of three on the right, or something, I don't know. Can someone point me in the right direction?

    Thanks.
     
  2. jcsd
  3. Feb 2, 2005 #2

    dextercioby

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  4. Feb 2, 2005 #3

    HallsofIvy

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    Any integer n is either
    1) a multiple of 3: n= 3m in which case n2= 9m2= 3(3m2)

    2) a multiple of 3 plus 1: n= 3m+1 in which case n2= 9m2+ 6m+ 1= 3(3m2+ 2m)+ 1

    3) a multiple of 3 plus 2: n= 3m+2 in which case n2= 9m2+ 12m+ 4= 9m2+ 12m+ 3+ 1= 3(3m2+ 4m+1)+ 1.

    In other words, if n2 is a multiple of 3, then n must be a multiple of 3.
     
  5. Feb 2, 2005 #4

    dextercioby

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    Halls,it's the 3-rd power. :wink: I've given him the link to Hurkyl's general proof in the other thread in HS homework section...

    Daniel.
     
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