Rational number and irrational number

If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x in Ssetminus {r}$ such that $r + x = p in mathbb{Q}$, then $x = p-r in mathbb{Q}$, and if $x in Ssetminus {r}$ is such that $rcdot x = p in mathbb{Q}$, then $x = frac{p}{r} in mathbb{Q}$. In that case, it is clear that all squares are rational.

So we can assume that all elements of $S$ are irrational. For each $alpha in S$, there are at most two elements of $S$ whose sum with $alpha$ is rational. For if there were three or more, say $alpha + beta = r in mathbb{Q}$, $alpha + gamma = s in mathbb{Q}$ and $alpha + delta = t in mathbb{Q}$, then $beta + gamma = r+s - 2alpha notin mathbb{Q}$, $beta + delta = r+t-2alpha notin mathbb{Q}$ and $gamma + delta = s+t - 2alpha notin mathbb{Q}$, so by the assumption we would have $betadelta in mathbb{Q}$ and $gammadelta in mathbb{Q}$, and consequently $$(beta - gamma)delta = (r-s)delta in mathbb{Q},,$$ which by the irrationality of $delta$ implies $r = s$ and thus $beta = gamma$, contradicting our assumption that $beta,gamma,delta$ are distinct.

Thus, if $S$ contains at least $7$ elements, then for every $alpha in S$ there are $beta,gammain S$ such that $alphabeta, alphagamma, betagamma$ are all rational. (Picking $alpha$ excludes at most two numbers, and then choosing $beta$ excludes at most two more.) But then $beta = frac{r}{alpha}$ for some $r in mathbb{Q}$, and $gamma = frac{s}{beta} = frac{s}{r}alpha$ for some $sin mathbb{Q}$, whence $$alpha^2 = frac{r}{s}alphagamma in mathbb{Q},.$$

Actually, we can lower the cardinality required in this argument.

For $alpha in S$, let $E(alpha) = {x in S : xalpha notin mathbb{Q}}$. This is the set excluded by $alpha$ in the above argument, and before that we saw that $E(alpha)$ contains at most two elements, since $E(alpha) subseteq { x in S : x + alpha in mathbb{Q}}$. This yields the trivial bound $operatorname{card}bigl(E(alpha) cup E(beta)bigr) leqslant 4$ if $beta in Ssetminus E(alpha)$ used above. But in fact, we have the bound $operatorname{card}bigl(E(alpha) cup E(beta)bigr) leqslant 3$. For if $alphabeta in mathbb{Q}$ and $alpha + x = r in mathbb{Q}$, then $xbeta = (r-alpha)beta = rbeta - alphabeta$ is irrational unless $r -alpha= 0$, i.e. $x = -alpha$. So if $E(alpha)$ contains two elements, at least one of them also belongs to $E(beta)$. And $E(alpha) cup E(beta)$ can only contain three elements if $-alpha in E(alpha)$ and $-beta in E(beta)$. So we can unconditionally lower the required cardinality to $6$, and we can lower it to $5$ under the condition that $S$ contains at most one pair of negatives.

We cannot lower the required cardinality further, because for any irrational $x$ whose square is also irrational, the set $$S = {x, -x, x^{-1}, -x^{-1}}$$ has the property that for each pair of distinct elements either the sum or the product is rational, but none of the squares of the elements is rational.

Let $a$ be one of these numbers. Then $a^2 in mathbb Q$ or $2a in mathbb Q$.

In the first case we are done. In the second case we have $4a^2 in mathbb Q$, hence $a^2 in mathbb Q$.