Graph theoretic proof: For six irrational numbers, there are three among them such that the sum of any two of them is irrational. 
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:

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*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.

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*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.

It's actually possible to demonstrate that this is NOT a graph-theoretic problem.
The graph-theoretic  condition equivalent to having a finite collection of irrational numbers as vertices, and recording (with edges) which pairs have rational sums is 

Graph that is a disjoint union of complete bipartite graphs $K_{m,n}$.

The graph has that structure for any finite collection of irrational numbers, and every such finite graph can be realized by some irrational numbers. 
All of the binary relation structure of the graph is an encoding of a simpler unary structure, the partition of the vertex set into pairs of subsets (the mod $\mathbb{Q}$ equivalence classes of the numbers, and their negatives).  To answer any question about the graph one looks at the partition, not the edges.
A maximum edge-free subset (independent set) in such a graph is a union of the larger half of each partition-pair.  The cardinality is $\sum \max(m,n)$ which is always at least $\lceil V/2 \rceil$ if the graph has $V$ vertices.
So the first answer by @AlexRavsky, that used the partition directly without introducing a graph, seems to be the optimal argument.
