Given our irrational real numbers $a$, let $a+b = p, ab' = q$ be two rational numbers we hope exist. Then $b = p-a, b' = \frac{q}{a}$ and we want to show that there exist $p, q$ such that $a+\frac{q}{a}, a(p-a)$ Are irrational. If $a+\frac{q}{a}= x$ some irrational then $a$ satisfies the (irrational) quadratic $a^{2}-xa+q = 0$. If $a$ is not the root of a rational quadratic then any rational $q$ produces an irrational $x$. Otherwise $a$ is the root of some rational quadratic with some rational constant term, so pick $q$ not equal to that constant term and $x$ cannot then be rational. Similarly, if $a(p-a) = y$ some rational then $a$ satisfies the (irrational) quadratic $a^{2}-pa+y = 0$. Again, if $a$ is not the root of a rational quadratic then any rational $p$ produces an irrational $y$. Otherwise $a$ is the root of some rational quadratic with some rational $a$ -coefficient, so pick $p$ not equal to that constant term and $y$ cannot be rational.

Definition: Let the minimal polynomial $m(x)$ over $F[x]$ (in practice, over $\mathbb{Q}[x]$) of some $a$ be the unique monic polynomial of smallest degree such that $m(a) = 0$. If $a$ is rational then $m(x) = x-a$. Otherwise $m(x)$ is of some higher degree. Lemma: Given a polynomial $p(x)$ over $F[x]$ such that $p(a) = 0$, the minimal polynomial $m(x) | p(x)$. Proof: Division algorithm. The remainder $r(x)$ when $m(x)$ divides into $p(x)$ must either be $0$ or be a polynomial of degree lower than $m(x)$ such that $r(a) = 0$ - contradiction. Corollary (conjugate root theorem): If $p+\sqrt{q}$ is a root of a rational quadratic, where $p, q$ are rationals and $q$ is not the square of a rational, then $p-\sqrt{q}$ is also a root.

If the monic polynomial of $a$ over the rationals is quadratic, then $a$ cannot be the root of any other monic rational quadratic (or else the difference between the two quadratics, which is both linear and rational, must have $a$ as a root, contradicting $a$ irrational). Hence if we fix one of the coefficients to be a rational not equal to the corresponding coefficient in the minimal polynomial of $a$ then the other coefficient cannot be rational. I am basically just using the idea of the uniqueness of the minimal polynomial, but the other results following from that idea are interesting too.