Irrational 2.0

Let the six numbers be six points, called: $A, B, C, D, E$ and $F$ . If two numebr's amount is rational, then we link them with a red line segment, if the amount is irrational, then we link with a blue line segmnent. Since there should be a triangle which sides are colored the same ( Ramsey-theorem ), we will show that there will be a blue triangle. Suppose there is a red triangle, let the numbers be $x$ , $y$ and $z$ . Then $x+y, x+z, y+z$ would be rational numbers, so the sum of them is rational too: $2x+2y+2z$ . From that $x+y+z$ would has to be rational, but since $y+z$ is rational, $x+y+z-(y+z)=x$ would have to be rational, but x is irrational. So there can't be a red triangle, and there must be a blue triangle.

Let the six irrational numbers be $a_i$ for $i=1,\ldots,6$ . First, observe that if $a_i+a_j$ and $a_i+a_k$ are both rational, then their sum $2a_i+a_j+a_k$ is rational as well; hence, $a_j+a_k$ is irrational (for otherwise $2a_i$ and thus $a_i$ would be rational too). Consequently, there do not exist triples all of whose pairwise sums are rational. We call this property $T$ .

Consider the five sums $a_1+a_i$ for $i=2,\ldots,6$ . Either at least three of these are rational or at least three are irrational. If three are rational, then those three $a_i$ have pairwise irrational sums (otherwise $T$ would be violated). If three are irrational, then because of $T$ not all pairwise sums of the three $a_i$ can be rational. Suppose $a_i+a_j$ is irrational. Then $a_1, a_i, a_j$ have pairwise irrational sums.

Let assume that the statement is $false$ .

So, there always exist three such that the sum of any two of them is rational.

Let the six irrational numbers are $A, B, C, D, E$ and $F$ .

Here the three numbers are $A, B,C$ , the sum of any two of these numbers is rational.

So,

A+B=p \tag{1}

B+C=q \tag2

C+A=r \tag{3}

Here , $p,q,r$ are rational.

$(1)-(2)+(3)\rightarrow p-q+r=2a$

$2a$ is irrational but $p-q+r$ is rational .

So, $p-q+r\neq2a$ Therefore , it is proved that , the statement is true.