1961 AHSME Problems/Problem 4

Problem

Let the set consisting of the squares of the positive integers be called $u$; thus $u$ is the set $1, 4, 9, 16 \ldots$. 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 $u$ is closed under:

$\textbf{(A)}\ \text{Addition}\qquad \textbf{(B)}\ \text{Multiplication} \qquad \textbf{(C)}\ \text{Division} \qquad\\ \textbf{(D)}\ \text{Extraction of a positive integral root}\qquad \textbf{(E)}\ \text{None of these}$

Solution

Consider each option, case by case.

For option A, note that $1 + 4 = 5$, so the set is not closed under addition because $5$ is not a perfect square.

For option B, note that $9 \cdot 4 = 36$ and $4 \cdot 25 = 100$. Letting one member of the set be $a^2$ and another member be $b^2$ (where $a$ and $b$ are positive integers), the product of the two members is $(ab)^2$. Since $ab$ is an integer and $(ab)^2$ is a perfect square, the set is closed under multiplication.

For option C, note that $9 \div 4 = 2.25$, so the set is not closed under division because $2.25$ is not an integer.

For option D, note that $\sqrt{4} = 2$, so the set is not closed under extraction of a positive integral root because $2$ is not a perfect square.

Thus, the answer is $\boxed{\textbf{(B)}}$.

See Also

1961 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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