```
Let a, b ∈ Z. Let S = {x ∈ Z : x | a} and let T = {y ∈ Z : y | b}. Prove that S ⊆ T if and
only if a | b.
```

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Answer from Sia

Posted about 1 month ago

Solution by Steps

step 1

We start with the definitions of the sets $S$ and $T$: $S = \{ x \in \mathbb{Z} : x | a \}$ and $T = \{ y \in \mathbb{Z} : y | b \}$

step 2

To prove that $S \subseteq T$ if and only if $a | b$, we need to show both directions of the implication

step 3

Assume $S \subseteq T$. This means every divisor of $a$ is also a divisor of $b$. Therefore, if $d | a$, then $d | b$. In particular, $a$ itself must divide $b$, hence $a | b$

step 4

Now assume $a | b$. This means there exists some integer $k$ such that $b = ak$. Any divisor $d$ of $a$ will also divide $b$ since $b = ak$ implies $d | b$. Thus, $S \subseteq T$

Answer

The statement is proven: $S \subseteq T$ if and only if $a | b$.

Key Concept

The relationship between divisors and subsets in integer sets.

Explanation

The proof shows that the inclusion of the set of divisors of $a$ in the set of divisors of $b$ is equivalent to $a$ dividing $b$.

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