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Thread: Prove sqrt(q) is irrational

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

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

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

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

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

    Please help, thanks a lot.
    Last edited by armeros; Jun 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 $\displaystyle \geq$ 2 be such that $\displaystyle \forall a \in \mathbb{Z} \forall b \in \mathbb{Z}$, q|ab $\displaystyle \longrightarrow $ q|a or q|b. Show that $\displaystyle \sqrt q$ is irrational.

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

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

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

    Please help, thanks a lot.[/QUOTE]
    Last edited by HallsofIvy; Jun 12th 2009 at 01:01 PM.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    Hello!

    First notice that if

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

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

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