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:

*

*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.
 A: I have tried to prove Alex Ravsky's first proof in the following way. Once again I want to emphasize that I am just learning Graph theory and not very confident about the watertightness of my proof.I will appreciate any improvement. 
Let $T$ be a graph with $V$ vertices. We divide it into $\left\lfloor\frac{n}{2}\right\rfloor$ subsets such that the number of vertices in each subset is equal or differs from each other by atmost 1 (Turan's condition).Let $r$ be minimum number of vertices in the subsets, thus $r+1$ is the maximum number of vertices.
Therefore $(\left\lfloor\frac{n}{2}\right\rfloor)r \leq V < (\left\lfloor\frac{n}{2}\right\rfloor)(r+1) $
By joining the vertices of different subsets in blue and same subsets in red we form a Turan graph (example of such a set of irrational numbers is given by Alex Ravsky for example $\left\lfloor \frac{n}{2} \right\rfloor \sqrt{2}$ s and $\left\lfloor \frac{n}{2} \right\rfloor 2-\sqrt{2}$ s),and we can conclude by the property of Turan graph that there is no $(\left\lfloor \frac{n}{2} \right\rfloor+1)$ clique in $T$.
Now $r \geq 1$ implying V is greater than or equal to $\left\lfloor \frac{n}{2} \right\rfloor$.
If $V=\left\lfloor\frac{n}{2}\right\rfloor$ then it is obvious that there can be no $(\left\lfloor\frac{n}{2}\right\rfloor+1)$ clique. If $r=2$ then $V \geq (n-1)$.
Thus $T$ is the graph with maximum vertices that contains a $\left\lfloor\frac{n}{2}\right\rfloor$ clique but not a $\left\lfloor\frac{n}{2}\right\rfloor +1$ clique. At the same time $r<3$ because otherwise there will be a monochromatic red triangle within the subsets. Thus putting the possible values of $r$ we get that   

$(n-1) \leq V < 3\left\lfloor\frac{n}{2}\right\rfloor  => (n-1) \leq V \leq 3\left\lfloor\frac{n}{2}\right\rfloor -1$  

For all $V$ satisfying above condition has a strict bound of $\left\lfloor\frac{n}{2}\right\rfloor$ for its maximum sized clique.
A: Let we have $n$ vertices instead of six and consider only blue edges in the graph. Similarly to the proof from the question, we can show that the graph contains no (blue) cycles of odd length. Hence it (vertex set) is two-colorable. Thus it has an independent set of vertices of size at least $n/2$, and sum of any two numbers from this set is irrational. 
A: 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.
A: 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 $c\in C$ and $c’\in C’$. 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 $\sqrt{2}$ and $n/2$ numbers $2-\sqrt{2}$ witnesses that the bound $n/2$ is strict. 
A: 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.  
