Given: $\{a,b\}\subset \mathbb Q$, $a\ne b$, not perfect squares, $s=\sqrt{a}+\sqrt{b}$, $q=\sqrt{a}-\sqrt{b}$, are $s$ and $q$ both irrationals?

Question

Given: $\{a,b\}\subset \mathbb Q$, $a\ne b$, not perfect squares, $s=\sqrt{a}+\sqrt{b}$ and $q=\sqrt{a}-\sqrt{b}$.

Proof or disproof: $s$ and $q$ are both irrationals.

This is a math contest question. There are good articles in SE related to the sum and difference of irrationals such as [this], for instance. It is easy to find examples of the sum of 2 different irrationals leading to a rational, such as $\sqrt{2}+(1-\sqrt{2})=1$, or to an irrational such as in $\sqrt{2}-(1-\sqrt{2})=2\sqrt{2}-1$. But I'm not sure what happens in the case above, when $a\ne b$.

Full proofs or disproofs are appreciated.

Answer 1

Consider the polynomial $p(x)=(x-s)(x-q)=x^2 -2\sqrt a x +a-b$. It is clear that the the roots of $p(x)$ are $s$ and $q$. Now suppose, for contradiction, $p(x)$ has a rational solution, say $p(r)=0$. That is, $$r^2 -2\sqrt a r +a-b =0$$ and rearranging gives, $$ \sqrt a = \frac{r^2 + a-b}{2r}.$$ But then $\sqrt a$ is rational, contradicting the assumption that $a$ is not a perfect square. Therefore both $s$ and $q$ are irrational.

Answer 2

Let me try to answer my own question. Comments and suggestions are welcome.

Consider the following lemmas, stated without proof:

Lemma 1. (L1) The sum and the ratio of 2 rational numbers are rationals.

Lemma 2. (L2) If $x\ne 0$ is rational and $y$ is irrational, then $xy$ is irrational.

Proposition: If $(a,b)\subset \mathbb Q$, $a\ne b$, not perfect squares, $$s=\sqrt{a}+\sqrt{b}\ \ \text{and}\ \ \ q=\sqrt{a}-\sqrt{b},$$ then $s$ and $q$ are both irrationals.

Proof:

Case 1. $a=0$ and $a\ne b$ or $b=0$ and $a\ne b$.

Trivial, both $s$ and $q$ are irrationals as $s=\sqrt{b}$ and $q=-\sqrt{b}$ in the first situation, and $s=\sqrt{a}$ and $q=\sqrt{a}$ in the second.

Case 2. $a\ne b\ne 0$.

First, notice that (1) $s+q=2\sqrt{a}$, (2) $s-q=2\sqrt{b}$, and (3) $sq=a-b$.

Developing (3), we get $s=\frac{a-b}{q}$. Assuming that $q$ is rational, by L1 we conclude that $s$ is rational. But this is a contradiction, as $s+q=2\sqrt{a}$, an irrational number, and L1 tells us the opposite. Therefore $q$ is irrational.

Now consider again (3) $sq=a-b$. Notice that $a-b$ is rational and $q$ is irrational, as we've just concluded. Assuming that $s$ is rational leads to a contradiction, as by L2, the product of a non-zero rational and an irrational is always an irrational. Therefore, $s$ is also an irrational, and the proof is complete.

Answer 3

$sq=a-b$ so if one is rational, so is the other. If they are both rational, then so is $s+q$.