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 \(n^2=12\).