From conditions follows, that each $x \in M$ is non-zero irrational. Let elements of set are placed as graph, and every edge $(x,y)$ is colored red, if $x+y$ is rational, and colored blue if $xy$ is rational 1)In every triangle $xyz$ should be edges of both colors. $x+y,y+z,z+x \in \mathbb{Q} \to x=\frac{1}{2}((x+y)+(z+x)-(y+z)) \in \mathbb{Q}$ $xy,yz,zx \in \mathbb{Q} \to x^2=\frac{xy*zx}{yz} \in \mathbb{Q}$ 2) Can be only one blue edge in every triangle $xyz$ $xy,xz \in \mathbb{Q} \to \frac{y}{z}=r \in \mathbb{Q} \to y+z=z(r+1) \in \mathbb{Q} \to z \in \mathbb{Q}$ Now let there are at least $5$ vertexes: $x,y,z,u,t$ Let $xy,xz$ are red, than $yz$ is blue But then $yu,yt$ are red and so $xu,xt$ are blue, which is contradiction with 2) So can be not more than $4$ vertexes Example for $4$ vertexes: $1+\sqrt{2},-1+\sqrt{2},2-\sqrt{2},-2-\sqrt{2}$