parmenides51 wrote: Let $A$ be an infinite set of real numbers containing at least one irrational number. Prove that for every natural number $n > 1$ there exists a subset $S$ of $A$ with n elements such that the sum of the elements of $S$ is an irrational number. Let $I$ be one irrational number of set $A$. Suppose the contrary of proposed statement : any subset with $n$ elements has a rational sum. Choosing two elements $x,y$ and any subset B of $n-1$ other elements, the sum of elements of $\{x\}\cup B$ and of $\{y\}\cup B$ both are rational, and so is their difference $x-y$ So all elements of $A$ are in the form $I+q_k$ where $q_k$ is rational. And any subset of $n$ elements has a sum of $nI+q_1+q_2+...+q_n$ which is irrational, and so contradiction. Q.E.D.