Problem. Let there be six irrational numbers. Prove that there exists three irrational numbers among them such that the sum of any two of those irrational numbers is also irrational.
I have tried to prove it in the following way, but I am not sure whether it is watertight or not as I have just started learning graph theory.
Let there be a graph with 6 vertices. We assign a weight equal to those six irrational numbers to each of the vertices. We join all the vertices with edges and color the edges in the following way:
Edge is colored red if the sum of the weights of its end points is irrational.
Edge is colored blue if the sum of the weights of its end points is rational.
We know that when we color a 6-vertex graph with 2 colors then there must be a monochromatic triangle.
If the triangle is red then we are done.
If it is blue, then let the irrational numbers be a, b and c. Therefore a+b, b+c and c+a are all rational. Which implies 2(a+b+c) and a+b+c is rational. As a+b is rational and hence c is also rational. But this is a contradiction.
Hence, our original statement is proved.
Your proof is OK.
But more easily we can prove more strong and general claim. Assume we have a collection of n irrational numbers. We shall call numbers a and b equivalent if the difference a−b is rational. So we can partition our collection into equivalence classes. We shall call classes C and C′ complementary if c+c′ is rational for any c∈C and c′∈C′. From our partition we can choose such classes which contain in total at least n/2 elements and no two complementary classes are chosen. It remains to remark that a sum of any two chosen elements is irrational. In particular, among 5 irrational numbers we can choose 3 with all mutual sums are irrational. From the other hand, a collection consisting of n/2 numbers √2 and n/2 numbers 2−√2 witnesses that the bound n/2 is strict.