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.

Question

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.

Accepted Answer

∻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 $. ⚹◊How can you demonstrate that if $a | b$, then every divisor of $a$ is also a divisor of $b$?⍭ Generate me a similar question◊

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