1961 AHSME Problems/Problem 4
Problem
Let the set consisting of the squares of the positive integers be called ; thus is the set . If a certain operation on one or more members of the set always yields a member of the set, we say that the set is closed under that operation. Then is closed under:
Solution
Consider each option, case by case.
For option A, note that , so the set is not closed under addition because is not a perfect square.
For option B, note that and . Letting one member of the set be and another member be (where and are positive integers), the product of the two members is . Since is an integer and is a perfect square, the set is closed under multiplication.
For option C, note that , so the set is not closed under division because is not an integer.
For option D, note that , so the set is not closed under extraction of a positive integral root because is not a perfect square.
Thus, the answer is .
See Also
1961 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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