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Math Help - Prove sqrt(q) is irrational

  1. #1
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    [SOLVED]Prove sqrt(q) is irrational

    Let q \geq 2 be such that \forall a \in \mathbb{Z}  \forall b \in \mathbb{Z}, q|ab \longrightarrow q|a or q|b. Show that \sqrt q is irrational.

    The hint of the book said to imitate the proof that \sqrt 2 is irrational. I did it by letting x^2=q and assuming x = \sqrt q is rational. So x = \frac{m}{n} and then x^2= q = \frac{m^2}{n^2} . Thus, qn^2=m^2.

    From here, I must consider if m is even or odd. If m is odd, it must follow that both q and n^2 is odd.

    It does not seem I could reach any contradiction from the assumption \forall a \in \mathbb{Z} \forall b \in \mathbb{Z}, q|ab \longrightarrow q|a or q|b.

    Please help, thanks a lot.
    Last edited by armeros; June 12th 2009 at 08:17 AM.
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  2. #2
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    Quote Originally Posted by armeros View Post
    Let q \geq 2 be such that \forall a \in \mathbb{Z}  \forall b \in \mathbb{Z}, q|ab \longrightarrow q|a or q|b. Show that \sqrt q is irrational.

    The hint of the book said to imitate the proof that \sqrt 2 is irrational. I did it by letting x^2=q and assuming x = \sqrt q is rational. So x = \frac{m}{n}
    x= \frac{m}{n}, reduced to lowest terms. That means that m and n have no common factors, therefore m^2 and n^2 have no common factors.

    and then x^2= q = \frac{m^2}{n^2} . Thus, qn^2=m^2.
    And since m^2 and n^2 have no common factors, q must divide m^2= (m)(m) so, by your condition, q must divide m. m= qp for some integer p and so m^2= q^2p^2 and then qn^2= q^2p^2, n^2= qp^2. Since p^2, and thus p, is a factor of m^2 and so not a factor of n^2, we must have q a factor of n^2, again, by your condition on q, a factor of n, contradicting the fact that m and n have no common factors.

    From here, I must consider if m is even or odd. If m is odd, it must follow that both q and n^2 is odd.

    It does not seem I could reach any contradiction from the assumption \forall a \in \mathbb{Z} \forall b \in \mathbb{Z}, q|ab \longrightarrow q|a or q|b.

    Please help, thanks a lot.[/QUOTE]
    Last edited by HallsofIvy; June 12th 2009 at 01:01 PM.
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    MHF Contributor Bruno J.'s Avatar
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    Hello!

    First notice that if

     q | ab implies q | a or q | b for all a,b \in \Bbb Z,

    then q must be a prime. Indeed if we had q=mn with neither m nor n a unit, then q | mn but q would certainly divide neither m or n.

    Now to show \sqrt{p} is irrational, suppose \sqrt{p}=\frac{a}{b}. Then pb^2=a^2. The right-hand side is a square, hence all prime powers in the factorization of pb^2 should have even exponent; but the exponent of p is odd. Hence there exist no such a,b.
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