$\underline{\text { Agakhanov N.K. }}$.

We will call a non-empty (finite or infinite) set $A$, consisting of real numbers, complete if for any real numbers $a$ and $b$ (not necessarily distinct and not necessarily in $A$), if $a+b$ is in $A$, then $ab$ is also in $A$. Find all complete sets of real numbers.
Let $A$ be a complete set. Since it is not empty, we can choose an element $a \in A$. Then $a+0=a \in A$, so $a \cdot 0=0 \in A$. Since

$(-x)+x=0 \in A$, we now get that $(-x) \cdot x=-x^{2} \in A$ for all real $x$. By the arbitrariness of the choice of $x$, it follows that the set $A$ contains all negative numbers.

Finally, for any $b>0$, from the fact that the number $(-b)+(-b)=-2 b$ lies in $A$, we get that $b^{2}=(-b)(-b) \in A$. Thus, $A$ contains all positive numbers as well.

## Answer

The set of all real numbers $\mathbf{R}$ only.

Send a comment