MHB Complete Sets of Real Numbers: Find All

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A set A of real numbers is defined as complete if for all a, b in A, the conditions a + b in A and ab in A hold. The discussion establishes that since A is nonempty, it must contain 0, leading to the conclusion that all negative numbers are included in A. Furthermore, it is shown that all positive numbers can also be derived from elements in A, ultimately leading to the conclusion that A must equal the entire set of real numbers, R. This problem originates from the 2016 Russian mathematical olympiad for junior high school students. The final result confirms that the only complete set of real numbers is A = R.
Evgeny.Makarov
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Call a nonempty (finite or infinite) set $A\subseteq\Bbb R$ complete if for all $a,b\in\Bbb R$ such that $a+b\in A$ it is also the case that $ab\in A$. Find all complete sets.
 
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Evgeny.Makarov said:
Call a nonempty (finite or infinite) set $A\subseteq\Bbb R$ complete if for all $a,b\in\Bbb R$ such that $a+b\in A$ it is also the case that $ab\in A$. Find all complete sets.
[sp]Since $A$ is nonempty it contains some real number $a$. Then $0+a = a\in a$, so the completeness condition implies that $0 = 0a\in A.$

Next, $x + (-x) = 0 \in A$, for every real number $x$. Therefore $-x^2 = x(-x) \in A$. But every negative number is of this form. Therefore $(-\infty,0] \subseteq A.$

Finally, $ \frac12y + \frac12y = y$. So if $y\in A$ then $\frac14y^2 =\bigl(\frac12y\bigr)^2 \in A$. But every positive number is of the form $\frac14y^2$ for some negative number $y$, and therefore belongs to $A$.

Conclusion: $A = \Bbb R.$
[/sp]
 
Correct. This is a problem from the regional round of the 2016 Russian mathematical olympiad for junior high school students.
 
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