Answer
a. No, R ⊈ T because there are elements in R that are not in T. If R = {2, 4, 6, 8, 10, …} and T = {3, 6, 9, 12, 15, …}, then 2 is one example of an element in R that is not in T.
b. Yes, T⊆R because all elements in T are divisible by 6, all elements in R are divisible by 2, and all elements divisible by 6 are divisible by 2.
The set of elements in T can be described as z = 6a = 2(3a) = 2 * (some integer)
The set of elements in R can be described as y = 2b = 2 * (some integer).
The set of elements in T and R can both be described as 2 * (some integer).
Therefore, T ⊆ R is true.
c. Yes, T⊆S because all elements in T are divisible by 6, all elements in S are divisible by 3, and all elements divisible by 6 are divisible by 3.
The set of elements in T can be described as z = 6a = 3(2a) = 3 * (some integer)
The set of elements in S can be described as x = 3b = 3 * (some integer).
The set of elements in T and R can both be described as 3 * (some integer).
Therefore, T ⊆ S is true.
Work Step by Step
a. No, R ⊈ T because there are elements in R that are not in T. If R = {2, 4, 6, 8, 10, …} and T = {3, 6, 9, 12, 15, …}, then 2 is one example of an element in R that is not in T.
b. Yes, T⊆R because all elements in T are divisible by 6, all elements in R are divisible by 2, and all elements divisible by 6 are divisible by 2.
The set of elements in T can be described as z = 6a = 2(3a) = 2 * (some integer)
The set of elements in R can be described as y = 2b = 2 * (some integer).
The set of elements in T and R can both be described as 2 * (some integer).
Therefore, T ⊆ R is true.
c. Yes, T⊆S because all elements in T are divisible by 6, all elements in S are divisible by 3, and all elements divisible by 6 are divisible by 3.
The set of elements in T can be described as z = 6a = 3(2a) = 3 * (some integer)
The set of elements in S can be described as x = 3b = 3 * (some integer).
The set of elements in T and R can both be described as 3 * (some integer).
Therefore, T ⊆ S is true.