Prove that the sum of k irrational numbers is not always rational for 1 $\le$ k < n

Question

Let k and n be positive integers satisfying 1 $\le$ k < n. Prove that there do not exist n irrational numbers so that no matter how we choose k of them, their sum is always rational.

First off, I'm a bit confused as to how the question is worded, but my interpretation is this: For a finite set of any n irrational numbers, there is always a subset of size k such that those numbers will add up to be irrational.

Given that interpretation, I attempted to used induction, but was only able to prove the base case where n=2 and k=1. This problem is in my textbook under the chapter about Ramsey Theory, so I've been trying to think of how to represent the problem in terms of a graph, but I'm struggling mightily. Any guidance is greatly appreciated.

Answer 1

Suppose the sum of any two is rational, and pick three numbers $a,b,c$. Then $$a=\frac12((a+b)+(a+c)-(b+c))$$ is rational, contradicting the premise.

Similarly, suppose the sum of any three is rational, and pick four numbers $a,b,c,d$. Then $$a=\frac13((a+b+c)+(a+b+d)+(a+c+d)-2(b+c+d))$$ is rational, again a contradiction.

This pattern continues; for general $k$, we pick $k+1$ numbers $a,\ldots,z$ and then

$$a=\frac1k(\Sigma-(k-1)T)$$ where $\Sigma$ is the sum of the $k-1$ terms that include $a$, and $T=b+\cdots+z$ is the single term that doesn't include $a$. $\Sigma$ and $T$ are rational, so $a$ must be rational.

Answer 2

This doesn't really appear to be related to graph theory, and is only spiritually related to Ramsey theory. The question can be settled just by considering some linear equations.

Consider the case $k=2$, $n=3$ as an example. If you have the equations \begin{align} x_1 + x_2 \phantom{+ x_3} &= q_1 \\ x_1 \phantom{+ x_2} + x_3 &= q_2 \\ \phantom{x_1 +} x_2 + x_3 &= q_3 \end{align} where $q_1, q_2, q_3$ are some unknown rational numbers, can you solve for one of the $x_i$? If you do, is the answer rational?

Can you generalize this to arbitrary $k$, with $n=k+1$? (This implies the arbitrary $k$ and arbitrary $n$ case, since we can always ignore $n-k-1$ of the variables.)

Hover to reveal the spoiler to see how to do this, or don't do that if you want to figure out the answer yourself.

Adding all $k+1$ equations together and dividing by $k$ shows that the total $x_1 + x_2 + \dots + x_{k+1}$ is rational, but we are also given that $x_1 + x_2 + \dots + x_k$ is rational. Therefore their difference $x_{k+1}$ is also rational, contrary to our assumption that we had $k+1$ irrational numbers.