SAT1000 - P884

Algebra Level pending

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

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

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

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

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

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

Have a look at my problem set: SAT 1000 problems

B

D

C

A

1 solution

David Vreken
Jun 25, 2020

Consider two examples that fit the given requirements.

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

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

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 $T$ , $V$ is closed under multiplication .

SAT1000 - P884

1 solution

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