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 of different irrational numbers, there exist irrational numbers such that for any such that
(1) and (2) ,
is also irrational.
Proof: We will prove this by induction. This is obviously true for . Now let us assume that this is also true for . We shall now prove the case for : that amongst any different irrational numbers, there exist numbers such that the given condition is satisfied. We will prove this by contradiction. Let us call this element set .
Remove any two elements from . We get elements. By the inductive hypothesis, there exist elements such that for any non-negative rational numbers amongst which at least one is strictly positive, is irrational. Let us call this set of elements . Now one by one, add each element of , to to form element sets. If each of these 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 (which by our assumption should also be rational) to give us back, which should again be a rational number. But that is a contradiction. Hence, there has to exist an tuple of elements in 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.