Irrationally irrational irrationals

Claim If $I_1$ , $I_2$ , $I_3$ are three irrational numbers such that $I_1+I_2$ and $I_1+I_3$ are rational, then $I_2+I_3$ cannot be rational.

Proof On the contrary, assume $I_2+I_3$ is rational. Let $I_1+I_2=R_1$ $I_1+I_3=R_2$ $I_2+I_3=R_3$ Adding up all the equations, we get $2(I_1+I_2+I_3)= R_1+R_2+R_3$ $\implies I_1+I_2+I_3= \frac{R_1+R_2+R_3}{2}$ Now note that $I_1= (I_1+I_2+I_3) - (I_2+I_3) = \frac{R_1+R_2+R_3}{2} -R_3= \frac{R_1+R_2-R_3}{2}$ Similarly, $I_2= \frac{R_1-R_2+R_3}{2}$ $I_3= \frac{-R_1+R_2+R_3}{2}$ Since $R_1$ , $R_2$ , and $R_3$ are rational, so should be $I_1$ , $I_2$ , and $I_3$ .This is a contradiction. QED

We now claim that the answer is $N=5$ . First, note that $N=4$ doesn't satisfy the conditions, any counterexample being of the form $\{I_1, R_1-I_1, I_2, R_2-I_2 \}$ , where $I_1$ and $I_2$ are irrational numbers and $R_1$ and $R_2$ are rational numbers.

For $N=5$ , we denote each of the five irrational numbers $\{I_1,I_2,I_3,I_4,I_5 \}$ as $5$ vertices on a graph. Two vertices will be connected by a red line iff their sum is rational, and the line connecting them will be blue otherwise. Note that in such a graph, a monochromatic triangle (triangle with all edges having the same color) cannot appear, because if it does, we would either have a contradiction (when the color of the triangle is red), or we would have a three element set such that the sum of any two of them is irrational (when the color of the triangle is blue). It is known that the only possible graph with $5$ vertices without a monochromatic triangle is as follows ( source: wikipedia ):

Here's the link in case the picture doesn't load: http://s15.postimg.org/vqrzx6k2z/irrational.png

However, note that $I_{1} + I_{4} = (I_{1} + I_{2} ) - (I_{2} + I_{3}) + (I_{3} + I_{4})$ From the picture above, it is clear that $I_{1}+I_{2}$ , $I_{2} + I_{3}$ , and $I_{3} + I_{4}$ are rational, whereas $I_{1} + I_{4}$ isn't. Hence we arrive at a contradiction, so there exists no counterexample for $N=5$ . We thus conclude $\boxed{N=5}$ is the minimum value of $N$ satisfying the conditions.

Firstly, the set $\{a, b, -a, -b\}$ , where $a,b$ are irrational: if we choose 3 elements then by the pigeonhole principle we must choose a number and its negative , which sum to the rational number $0$ , so $N > 4$ .

Consider a set of irrationals $\{a,b,c,d,e\}$ that does not have the property that we can find a subset of 3 such that the sum of any 2 is irrational, i.e., for all subsets of 3, there are 2 which sum to a rational.

Sort it so that $a,b$ do not sum to a rational number. This is possible for any set of size 3 or more, because if we have $A+B = q_1$ , $A+C = q_2$ , $B+C = q_3$ , then combining those equations gives $2A = q_1 + q_2 - q_3$ , contradicting the assumption that $A$ is irrational and $q_i$ rational.

Because we could choose the subset $\{a, b, X\}$ for any $X$ , we must have $c,d,e$ all of the form $q-a$ or $q-b$ , where $q$ is a rational.

But now consider the subset $\{c,d,e\}$ . The sum of any 2 of these must be the sum of two rationals, minus $2a$ , or $2b$ , or $a+b$ . Those are all irrational so the subset $\{c,d,e\}$ contradicts the assumption that $\{a,b,c,d,e\}$ exists, so we have $N=\boxed{5}$ .

If we make a graph where the vertices represent the irrational numbers, and two vertices are connected by an edge iff their sum is rational, then the question can be rephrased as follows: "Find a minimal $N$ so that any subgraph of $N$ vertices have a set of three vertices with no edge connecting any pair of them (some times called an 'anti-triangle')."

This graph cannot have odd cycles. Indeed, if we have irrational numbers $a_1, a_2, \ldots , a_{2n+1}$ such that $a_i + a_{i+1} \in \mathbb Q$ , then $a_1 + a_{2n+1} = (a_1 + a_2) - (a_2 + a_3) + \cdots - (a_{2n} + a_{2n+1}) + 2a_{n+1}$ cannot be rational, since it is the sum of a finite sequence of rational numbers (the parentheses) and one irrational ( $2a_{2n+1}$ ). Especially there cannot be any triangles in the graph.

By the same argument, if we have four vertices $a, b, c, d$ with edges $ab, bc, cd$ , then $ad$ MUST be an edge, for $a + d = (a + b) - (b + c) + (c + d)$ is a sum of three rational numbers. Thus if there are four vertices in a "row", so to speak, then those four vertices actually close to form a square.

Now, let's try to find a maximal subgraph with no anti-triangle. We will start with choosing three points $a, b$ and $c$ . If none of these are connected, then we are done, so we can assume that at least $a$ and $b$ are connected. From here we will split into two cases, depending on whether $c$ is connected to one of the other two or not.

Case 1: $c$ connected to say $b$ .

Introduce a fourth vertex $d$ . If $d$ is connected to neither $a$ nor $c$ (including if $bd$ is an edge), the $acd$ is an anti-triangle and we are done.

So let's say that $cd$ is an edge of the graph. Then $d$ is at the same time connected to $a$ , by the argument above, and the graph must be a square. Any fifth vertex attempted to add to this square will cause an anti-triangle with at least one of the diagonals of the square, since we cannot have any triangles.

Case 2: $c$ isolated.

Pick a fourth vertex $d$ . Now $d$ cannot be connected to both $a$ and $b$ , and therefore, if it is not connected to $c$ , there is an anti-triangle present in the two vertices $c, d$ and one of $a, b$ . So assume $cd$ is an edge.

If $d$ is connected to either $a$ or $b$ , then we again have a square, and that violates the premise of this case, since $c$ is not supposed to be connected to neither $a$ nor $b$ .

So the we're left with $ab$ and $cd$ the only edges. But a fifth vertex $e$ cannot be added to this graph without making an anti-triangle, since it must be unconnected with at least one of $a, b$ and at least one of $b, c$ .

Therefore we conclude that any graph of $5$ or more vertices constructed of irrationals in the way described above will have an anti-triangle. Thus $N = 5$

To have two irrational numbers add up to a rational number, they must be conjugates. In order to have any 2 of a group of 3 irrational numbers to add to be irrational, they must all not be conjugates of each other. Since conjugates come in pairs, we can have at most 4 irrational numbers before there must be three irrational numbers that are not conjugates of each other (because of the Pigeonhole Principle). Therefore, our answer is $\boxed{5}$ .

The only way for 2 irrational numbers to have a rational sum is if they are additive inverses. Assuming the worst case possible, let all of the irrational numbers be additive inverses of each other. In this case, we need to guarantee that there are 3 irrational numbers that have the same sign, so that any 2 of them will have an irrational sum.

To find the smallest value of N, we need to work from the smallest case until we can guarantee that they work. The minimum amount of irrationals would be 3. However, in this case, we could have 2 positive irrationals $( \sqrt{n})$ and 1 negative irrational $(-\sqrt{n})$ which would give us sums of $2\sqrt{n}$ and 0. However, 0 is irrational, so N=3 does not work.

The next case is 4 irrationals. In this case, we could have 2 positive irrationals and 2 negative irrationals, which would give us sums of $2\sqrt{n}, -2\sqrt{n}$ , and 0. Thus, N=4 does not work.

When N=5, there are 6 possible ways to distribute the irrationals.

• 5 positive irrationals
• 4 positive irrationals and 1 negative irrational
• 3 positive irrationals and 2 negative irrationals
• 2 positive irrationals and 3 negative irrationals
• 1 positive irrational and 4 negative irrationals
• 5 negative irrationals

Each of these cases has at least 3 irrationals of the same sign. Therefore, the smallest value of N is $\fbox{5}$

First we must realize that for any 3 irrational numbers in question, at least one of the three sums must also be irrational. Otherwise, if $a+b=r_{1}$ , $b+c=r_{2}$ and $a+c=r_{3}$ , then $r_{1}+r_{2}-r_{3}=2b$ , which must be rational, yet cannot be. So recasting this problem into a graph, you need the lowest value of N such that a graph on N points must contain either 3 points all connected(i.e. a triangle), or 3 points not connected at all. Now in a direct application of Ramsey's Theorem , the answer to this would be 6. However, my earlier statement (that there cannot be a "loop" of 3 numbers, every 2 adding up to a rational) is also true for any odd-tuple of numbers in a loop, or cycle in graph theory terms(notably, 5). So in the counterexample to R(3,3) for N=5, you will notice a cycle of length 5. That could not happen in this example, so the answer is N=5. I know it's not so rigorous, but it can be made so, and I can give specific counterexamples for $N=4$ . $\sqrt{2},10-\sqrt{2},\sqrt{5},10-\sqrt{5}$