# Prove sqrt(q) is irrational

• Jun 12th 2009, 06:05 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.

• Jun 12th 2009, 07:16 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.

$q | ab$ 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 $a,b$.