Let A={n∈Z ∣ n=5r for some integer r} and B={m∈Z ∣ m=20s for some integer s}.
a. Since n=5=5⋅1 and 1∈Z, we conclude that n=5∈A. On the other hand, n=5=20⋅s for each integer s, and therefore, n=5∈/B. We conclude that it is not true that A⊆B.
b. If m∈B then m=20s for some integer s. It follows that m=5(4s),4s∈Z, and consequently, m∈A. Therefore, B⊆A.
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