Prove sqrt(q) is irrational

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June 12th 2009, 05:05 AM
armeros

[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.
June 12th 2009, 06:16 AM
HallsofIvy

Quote:

Originally Posted by armeros

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.

Quote:

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]
June 15th 2009, 05:32 PM
Bruno J.

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 $a,b$ .