Rational number and irrational number

Question

There are 197 different non-zero real numbers, and the sum of any two distinct numbers is a rational number or the product is a rational number. Prove that: 197 numbers, each square is a rational number.

The number 197 is not important, so I can assume that there are only 10 numbers to make the question sounds easier:

• when all of the 10 numbers are rational, then their square must be rational.

• Now I assume that there is a irrational number a . Then, the remaining 9 numbers can be either p - a or p / a, while p is a rational number.

However, if there will not be three number like p - a. let b₁ = p₁ - a, b₂ = p₂ - a and b₃ = p - a, then b₁ + b₂ = p₁ + p₂ - 2a, which is not a rational number, so b₁ b₂ = p₁p₂ - a(p₁ + p₂) + a² must be rational number. b₂b₃ and b₁b₃ must also be rational number. Therefore, p₁p₂ - a(p₁ + p₂) + a², p₂p₃ - a(p₂ + p₃)+ a² , p₁p₃ - a(p₁ + p₃)+ a² are rational numbers. p * p is absolutely a rational number, so A₀ = - a(p₁ + p₂) + a², A₁ = - a(p₂ + p₃)+ a² , A₂ = - a(p₁ + p₃)+ a², A is rational number, A₂ -A₁ = a(p₂ - p₁), p₂ - p₁ can only be 0, which contradicts that every number is different.

This shows that there are at most 2 numbers like p - a, and there must be at least 8 numbers like p / a. However I didn't know how to continue the proof, anyone can help me?

Answer 1

If the set - call it $S$ - contains one rational number, say $r$, then all of the numbers are rational, for if $x \in S\setminus \{r\}$ such that $r + x = p \in \mathbb{Q}$, then $x = p-r \in \mathbb{Q}$, and if $x \in S\setminus \{r\}$ is such that $r\cdot x = p \in \mathbb{Q}$, 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 $\beta + \gamma = r+s - 2\alpha \notin \mathbb{Q}$, $\beta + \delta = r+t-2\alpha \notin \mathbb{Q}$ and $\gamma + \delta = s+t - 2\alpha \notin \mathbb{Q}$, so by the assumption we would have $\beta\delta \in \mathbb{Q}$ and $\gamma\delta \in \mathbb{Q}$, 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 $\beta,\gamma\in S$ such that $\alpha\beta, \alpha\gamma, \beta\gamma$ 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 $s\in \mathbb{Q}$, whence $$\alpha^2 = \frac{r}{s}\alpha\gamma \in \mathbb{Q}\,.$$

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

For $\alpha \in S$, let $E(\alpha) = \{x \in S : x\alpha \notin \mathbb{Q}\}$. 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 $\beta \in S\setminus E(\alpha)$ used above. But in fact, we have the bound $\operatorname{card}\bigl(E(\alpha) \cup E(\beta)\bigr) \leqslant 3$. For if $\alpha\beta \in \mathbb{Q}$ and $\alpha + x = r \in \mathbb{Q}$, then $x\beta = (r-\alpha)\beta = r\beta - \alpha\beta$ 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.

Answer 2

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 $a^2 \in \mathbb Q$.