Question

Let C = {n ∈ Z | n = 6r – 5 for some integer r} and D = {m ∈ Z | m = 3s + 1 for some integer s}.

Answer 1

Its is true that C ⊆ D means Every element of C is present in D

According to he question,

Let C = {n ∈ Z | n = 6r – 5 for some integer r}

D = {m ∈ Z | m = 3s + 1 for some integer s}

We have to prove : C ⊆ D

Proof : Let n ∈ C

Then there exists an integer r such that:

n = 6r - 5

Since -5 = -6 + 1

=> n = 6r - 6 + 1

Using distributive property,

=> n = 3(2r - 2) +1

Since , 2 and r are the integers , their product 2r is also an integer and the difference 2r - 2 is also an integer then

Let s = 2r - 2

Then, m = 3r + 1 with r some integer and thus m ∈ D

Since , every element of C is also an element of D

Hence ,  C ⊆ D proved !

Similarly, you have to prove D ⊆ C

To know more about integers here

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