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.

Even more simple but not elementary proof can be given as follows. The set $\mathbb R$ of reals is a linear space over the field $\Bbb Q$ of rationals. There exists a homomorphism $h:\Bbb R\to \Bbb R$ such that $\ker h=\Bbb Q$ (the image $h(x)$ of the real number $x$ can be easily constructed from a decomposition of $x$ via basis of $\Bbb R$ over $\Bbb Q$ containing $1$). Now,  let $K$ be any collection of  irrational numbers. Since $h(x)\ne 0$ for any number $x\in K$, there exist a subcollection $K’$ of $K$ of size at least half of size $K$, such that all elements $h(x)$ for $x\in K’$ have the same sign. Now let $x$  an $y$ be any elements of $K’$. Then $h(x+y)=h(x)+h(y)\ne 0$. Thus $x+y\not\in\ker h=\Bbb Q$. Similarly we can show that any sum of elements of $K'$ is irrational.  
