Day: April 4, 2013

This exercise proves the irrationality of $\sqrt{12}$. First we make a note about rational number expressed as $p/q$ where $p/q$ is in lowest terms and then use Rudin’s recipe for the irrationality of $\sqrt{2}$.

If $n=p/q$ and $m$ is an integer, $m \ne 1$ that divides both $p$ and $q$ then $p’=p/m$ and $q’=q/m$ are both integers and $n=p’/q’$. Therefore we can always reduce $p/q$ so that there is no integer $m$ (prime or otherwise) such that $m \ne 1$, $m\,|\,p$ and $m\,|\,q$. When we have such $p$ and $q$ we then say that the representation of $n$ is in lowest terms.

Following Rudin’s argument for $\sqrt{2}$, note that the word even is just another way of saying “divisible by 2” and odd is another way of saying “not divisible by 2.”

Assume we have integers $p$ and $q$ such that $\left(\frac{p}{q}\right)^2= 12$. Without loss of generality we can assume that these integers are in lowest terms. (That is to say there is no other integer $\ne 1$ that divides both $p$ and $q$. Then
$$p^2=12q^2 = 4 \cdot 3 \cdot q^2$$

Therefore 3 divides $p^2$ (and thus 3 divides $p$).
If 3 divides $p$ then $p = 3m$ for some integer $m$. Thus 9 divides $p^2$ ($p^2=(3m)^2=9m^2$ )so
$$9 \cdot m^2 = 4 \cdot 3 \cdot q^2$$
and thus
$$3 \cdot m^2  = 4 \cdot q^2$$

Since 3 does not divide 4, 3 must then divide $q^2$ and obviously then $q$. But this is contrary to our assumption that $p/q$ is in lowest terms therefore $p/q$ must not exist, i.e. there is no rational $n = p/q$ such that $n^2=12$.