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Proving that cube root 7 is irrational

  1. Sep 29, 2007 #1
    Hi guys,

    How would you prove that [tex]\sqrt[3]{7}[/tex] is irrational without using the unique factorization thrm? I tried proving that [tex]\sqrt[3]{7}[/tex] is rational but it didn't seem to get me anywhere...


    EDIT: Looks like I posted this in the wrong forum.
    Last edited: Sep 29, 2007
  2. jcsd
  3. Sep 29, 2007 #2


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    I would be inclined to mimic the classic Euclid proof that [itex]\sqrt{2}[/itex] is irrational.

    Assume that [itex]\sqrt{7}[/itex] is rational. That is, assume [itex]\sqrt{7}= \frac{m}{n}[/itex] for integers m and n, reduced to lowest terms. Then, cubing both sides, [itex]7= \frac{m^3}{n^3}[/itex] so 7n3= m3. That tells us that m3 is a multiple of 7. Can you use that to prove that m itself must be a multiple of 7? Remember that in proving that if m2 is even, the m must be even, we have to show that the square of an odd number is always odd. Here, you will have to look at numbers that are not multiples of 7. Is it possible for the third power of a number that is not a multiple of 7 to be a multiple of 7? You will have to look at 6 different cases.
    Last edited by a moderator: Oct 1, 2007
  4. Sep 30, 2007 #3
    Yes I can prove that m[tex]^{3}[/tex] mod 7 = 0 implies mod 7 = 0. I assume that the next step would be set m=7k. Then plug it in to the previous formula which would yield something like

    [tex]7n^{3} = 7 * 49k^{3}[/tex] using this I can prove that n % 7 = 0. So m/n has a common factor, which contracts one of the premises.

    And that would be the proof, correct?
  5. Oct 1, 2007 #4


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    Yes, that was exactly what I meant.
  6. Oct 2, 2007 #5
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