Exercise 1.2

We will assume without proof that $\mathbb{Z}$ has unique prime factorizations. (This follows from $\mathbb{Z}$ being a primary ideal domain, and is covered in an abstract algebra course.) This implies that $p$ is a prime iff $p\mid \mathit{ab}\Rightarrow p\mid a$ or $p\mid b$.

Assume there are $m,n$ coprime such that ${(\frac{m}{n})}^2=12$. This implies that $12n^2=m^2$. Consequently, $3\mid m^2$, and since $3$ is prime, $3\mid m$. This means that there is a $p$ such that $m=3p$, so that $12n^2=9p^2$, and consequently $4n^2=3p^2$. Using the primality of $3$, we can conclude that since $3\nmid 4$, $3\mid n^2$, and therefore $3\mid n$. This contradicts the coprimality of $m$ and $n$. Therefore there cannot be any such $m,n$.

(This is the same without assuming any algebra)
If $r$ is a rational number whose square is 12, then ${(r∕2)}^2=3$, so this is equivalent to showing there is no rational number whose square is 3. So suppose ${(m∕n)}^2=3$ where $m,n$ are integers with no common factors. Then $m^2=3n^2$. Either $m=3p$, or $m=3p\pm 1$, and since ${(3p\pm 1)}^2=3(3p^2\pm 2p)+1$, $m$ must be a multiple of 3. But then we have ${(3p)}^2=3n^2$, or $3p^2=n^2$. Hence $n$ is also a multiple of 3, contradicting the assumption that $m$ and $n$ have no common factors.