Pigeonhole

Let's first prove that given three irrational numbers, there are two whose sum is irrational. Let $a, b$ and $c$ be irrational numbers. If the sum of any two of these were rational then $a+b$ and $b+c$ would be rational and, so their difference $a-c$ would be rational.On the other hand, so the sum of this with $a+c$ , which is $2a$ , would have to be rational. But it can't be because $a$ is irrational. Therefore, there are always two numbers, in three irrational numbers, whose sum is irrational. Let's now consider six irrational numbers. Let's call one of those numbers $x$ and divide the other five in two sets, $A$ made up by those whose sum with $x$ is irrational and $B$ made up by those whose sum with $x$ is rational. By the Pigeonhole Principle, one of the sets has at least three elements. Suppose it's $A$ . Then there are two numbers in $A$ whose sum is irrational, so the three numbers we are looking for would be these two and $x$ . Suppose, now, it's B. Considering the set formed by any two elements of $B$ and $x$ we conclude that any two elements of $B$ sum to an irrational number, so we need only choose three elements of $B$ .