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Homework Help: Proof of the irrationality of the square root of primes ?

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data

    If p is prime [tex]\sqrt{p}[/tex] is prime.

    Are there flaws in my proof ?

    2. Relevant equations

    3. The attempt at a solution

    Assume [tex] \sqrt{p}=\frac{m}{n} [/tex] and [tex]gcd(m,n)=1[/tex]

    [tex] p = \frac{m^{2}}{n^{2}}[/tex]

    Since p is an integer [tex] n^{2}|m^{2}[/tex] but [tex]gcd(m,n)=1[/tex]

    [tex]n^{2} =1[/tex] ( Should I justify this step ? I don't deem it necessary.)

    [tex] m^{2} = p\Rightarrow m^{2}|p \Rightarrow m|p[/tex] but p is prime and only has factors of 1 and p. So since [tex]m, m^{2}[/tex] divide p [tex]m=1[/tex] this leads to a contradiction since [tex]p =/=1[/tex] .

    Are there flaws ?
  2. jcsd
  3. Sep 13, 2010 #2


    Staff: Mentor

    This isn't a comment about your work, but the statement you're trying to prove seems flawed to me. If p is a prime, then [itex]\sqrt{p}[/itex] will be irrational; hence not an integer. When we talk about numbers being prime or composite, the tacit assumption is that we're talking about integers.
  4. Sep 13, 2010 #3


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    Well, there is an obvious flaw here! You mean to say "If p is prime then [tex]\sqrt{p}[/tex] is irrational."

    Other than that your proof looks good.

    No, saying that gcd(m,n)= 1 and that [itex]n^2[/itex] divides [itex]m^2[/itex] is sufficient.

  5. Sep 13, 2010 #4
    Hahaha. Yes it is flawed that was an honest mistake.
  6. Sep 13, 2010 #5


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    Does this step necessarily rely on the fact that every natural number's prime factorization is unique? The only way that I can think of proving this statement uses it, but I want to know if I'm over looking somethig.
  7. Sep 13, 2010 #6
    Which statement ? There are a couple there.
  8. Sep 13, 2010 #7


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    The fact that n2 = 1.
  9. Sep 13, 2010 #8
    The way I see it , it is a consequence of the fact that [tex] gcd(m,n) =1[/tex] and that [tex] n^{2}|m^{2}[/tex].

    [tex] n|n^{2}[/tex]but n does not divide [tex] m^{2}[/tex] since it implies that [tex]n|m[/tex]. For all this to be true [tex]n^{2}=1[/tex].

    Perhaps a complete proof of this fact may use unique factorization but I think I may be able to prove it by contradiction.
  10. Sep 13, 2010 #9


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    Well, here's why I think that it does assume the uniqueness of prime factorizations. In this specific case, we have that


    where [itex]p[/itex] is a prime number. Now, this clearly means that


    Assuming the Fundamental Theorem of Arithmetic holds, this is an obvious contradiction if [itex]n^2 \neq 1[/itex] since the left side has an even number of prime factors while the right side has an odd number of prime factors. However, if we don't assume that this theorem holds, then we wouldn't know if it was possible to find another prime factorization of [itex]n^2p[/itex] with the same prime factorization as [itex]m^2[/itex] and vice versa.

    I know that the Fundamental Theorem of Arithmetic is true, so there's absolutely nothing wrong with your proof. I was just wondering if it implicitly assumed that the FTA holds.
  11. Sep 13, 2010 #10
    I guess I see your point.
    I believe that I arrived at my conclusion from simple arguments of divisibility.
    To be honest I wasn't thinking about FTA when I wrote this prove.

    I don't think I needed FTA for this.

    I may be wrong, considering I am not very knowledgable in mathematics ( engineering student here).

    But what I did seems like a natural step to me and I believe the fact that [tex] n^{2} =1[/tex] follows for the divisibility statements I stated previously. FTA is a corrolary of divisibility, is it not ?
    Last edited: Sep 13, 2010
  12. Sep 13, 2010 #11


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    Your arguments are fine and I was never disputing their validty.

    If your proof assumes the FTA, then this isn't strictly true; whether or not you stated it explicitly, you used some consequence of it.

    It is a natural step. Again, I just think that it relies on the FTA (I'm not sure that it does though, which is why I asked in the first place).
  13. Sep 13, 2010 #12
    [tex]n^{2}|m^{2} \Rightarrow n|m^{2}[/tex] from here is obvious to me ( without FTA) that n=1 since [tex] gcd(m,n)=1[/tex].

    [tex] n|ab \Rightarrow n|a[/tex] or [tex] n|b[/tex] taking [tex] a=b= m[/tex].

    Did I use FTA or any consequence of it ? I honestly do not know!
  14. Sep 13, 2010 #13


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    You didn't use FTA because you didn't prove anything. You just said it was 'obvious'. That's not a proof. The rest of your proof is much more obvious. n|ab -> n|a or n|b is generally only true if n is prime. I think most people would say that the most important part of the proof was in the part you omitted. I also think you should try and prove it.
  15. Sep 13, 2010 #14


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    No disputes here.

    Why do you think that it's obvious without the FTA?

    You've stated the theorem incorrectly, it should read: If n is prime and n|ab, then n|a or n|b. While it's certainly true that if n|m2, then n|m, you can't use the above theorem unless we can guarantee that n is a prime.

    Moreover, taking this particular case into consideration, since m2/n = np, this seems to be nothing more than a restatement of the fact that m2 = n2p. But then, my original reasoning* concerning the use of the FTA still applies, so I'm not sure that this is any better.

  16. Sep 13, 2010 #15
    [tex] n^{2}|m^{2}\Rightarrow (n| m^{2}\Rightarrow n|m) [/tex]

    I assume the last part in the bracket is where you guys say FTA is used. Right ?

    I suppose that FTA is needed and I am wrong.

    Doesn't this hold without n been prime ?

    Are you saying it is possible n|a*b but neither divides a or b ? Can you give an example ?

  17. Sep 13, 2010 #16


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    6|2*3 but 6 doesn't divide 2 or 3.
  18. Sep 13, 2010 #17

    Yes, I was defintely wrong. I would like to blame this on the fact that is way past my bed time but this actually ignorance. Thankfully I am not a math or physic major so this error is not as horrendous; still unpleasent though!

    I used FTA in my proof without knowing. To go from [tex] n|m^{2}[/tex] to [tex]n|m[/tex] requires the unique factorization theorem.

    Thanks for pointing out my stupidity and the gaps in my reseasoning.

    I guess my professor was right when we said there is always something to gain from a proof.

    Perhaps, I should pay more attention to the so called "facts" or "trivialities" I use in my proof. I often take things for granted which is a bad approach to mathematics.
  19. Sep 13, 2010 #18


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    In terms of your original proof, there were no gaps in your reasoning. The FTA certainly holds and so it's a reasonable thing to assume. Moreover, I don't think that you've said anything particularly stupid. You may have stated a theorem a little incorrectly, but that's not a huge deal, especially if you learn something from it. I think that you're being too hard on yourself.
  20. Sep 14, 2010 #19
    i think to go from [tex] n|m^{2}[/tex] to [tex]n|m[/tex] does not require unique factorization theorem, but gcd(n,m)=1

    unique factorization theorem, used for [tex] n^2|m^{2}[/tex] to [tex]p|m[/tex] for some prime p that divides n
  21. Sep 14, 2010 #20
    Well how can we argue that [tex]n|m[/tex] from [tex]n|m^{2}[/tex] which taking about common factors ? Trying to show that [tex]m^{2}[/tex] has the same factors as m requires unique factorization. Without it I would have a hard time proving the later.

    The idea is to express m as a product of primes and then show that only the exponents only differ by a multiple of 2. In which case we can say for certain that [tex]m^{2}[/tex] does not have factor non-common with m.

    Can you figure out a way without FTA ? I would be interested in seeing it.
    Last edited: Sep 14, 2010
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