SAT1000 - P884

Algebra Level pending

If set S S is a non-empty subset of integer set Z \mathbb Z , if a , b S , a b S \forall a,b \in S, ab \in S , then S S is closed under multiplication .

Given that for set T , U T,U : T Z , V Z , T V = , T V = Z T \subseteq \mathbb Z, V \subseteq \mathbb Z, T \cap V = \emptyset , T \cup V = \mathbb Z , and a , b , c T , a b c T \forall a,b,c \in T, abc \in T , x , y , z V , x y z V \forall x,y,z \in V, xyz \in V , then which of the choices is true?

A . At least one of T,V is closed under multiplication. A.\ \textup{At least one of T,V is closed under multiplication.}

B . At most one of T,V is closed under multiplication. B.\ \textup{At most one of T,V is closed under multiplication.}

C . Only one of T,V is closed under multiplication. C.\ \textup{Only one of T,V is closed under multiplication.}

D . Both T,V are closed under multiplication. D.\ \textup{Both T,V are closed under multiplication.}


Have a look at my problem set: SAT 1000 problems

B B D D C C A A

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1 solution

David Vreken
Jun 25, 2020

Consider two examples that fit the given requirements.

Example 1: T T is all the odd numbers, V V is all the even numbers. Both are closed under multiplication.

Example 2: T T is all the odd numbers except ± 3 \pm 3 , V V is all the even numbers and ± 3 \pm 3 . ( V V still meets the requirement x , y , z V , x y z V \forall x, y, z \in V, xyz \in V because even if 3 -3 and 3 3 are chosen as two of the numbers, the third number will make the product even.) T T is closed under multiplication (without 3 3 , no two odds can multiply to 3 3 ), but V V is not closed under multiplication because 3 3 = 9 -3 \cdot 3 = -9 , and 9 T -9 \notin T .

In the first example both are closed, but the second example only one is closed, so out of the given choices we can conclude that A) at least one of T T , V V is closed under multiplication .

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