Fruits of procrastination

Here’s a slightly badly written proof to a competitive math problem. I guess I could expand it slightly if readers find it unreadable.

The following is a question from the International Mathematics Competition, 1994.

Prove that in any set $S$ of $2n-1$ different irrational numbers, there exist $n$ irrational numbers $\{s_1\dots,s_n\}$ such that for any $\{a_1,\dots,a_n\}\in \Bbb{Q}$ such that

(1) $a_i\geq 0,\forall i\in\{1,\dots,n\}$ and (2) $\sum\limits_k a_k>0$ ,
$\sum\limits_{i=1}^n a_is_i$ is also irrational.

Proof: We will prove this by induction. This is obviously true for $n=1$ . Now let us assume that this is also true for $n$ . We shall now prove the case for $n+1$ : that amongst any $2(n+1)-1$ different irrational numbers, there exist $n+1$ numbers such that the given condition is satisfied. We will prove this by contradiction. Let us call this $2(n+1)-1$ element set $A_{n+1}$ .
Remove any two elements from $A_{n+1}$ . We get $2n-1$ elements. By the inductive hypothesis, there exist $n$ elements $\{s_1,\dots,s_n\}$ such that for any $\{a_1,\dots,a_n\}$ non-negative rational numbers amongst which at least one is strictly positive, $\sum\limits_{i=1}^n a_is_i$ is irrational. Let us call this set of elements $S_n$ . Now one by one, add each element of $A_{n+1}\setminus S_n$ , to $S_n$ to form $n+1$ element sets. If each of these $n+1$ element sets can be put into some linear combination to give us a rational number, then we can subtract the right rational linear combination of $A_{n+1}\setminus S_n$ (which by our assumption should also be rational) to give us $S_n$ back, which should again be a rational number. But that is a contradiction. Hence, there has to exist an $n+1$ tuple of elements in $A_{n+1}$ too such that no linear combination with non-negative rational coefficients, such that at least one coefficient is positive, can give us a rational number.