Question 1110263: Consider the sets A = {x ∈ Z | x = 6s + 1 for some s ∈ Z} and B = {x ∈ Z | x = 3t + 1 for some t ∈ Z}. Prove the following:
a: A ⊆ B
b: ¬(B ⊆ A)
So sorry I haven't tried anything, but I have no idea where to even begin.
Answer by greenestamps(13214)   (Show Source):
You can put this solution on YOUR website!
Do you understand the definitions of the two sets?
Set A contains all integers that are 1 more than a multiple (positive or negative) of 6: A = {..., -11, -5, 1, 7, 13, ...}
Set B contains all integers that are 1 more than a multiple (positive or negative0 of 3: B = {..., -11, -8, -5, -2, 1, 4, 7, 10, 13, ...}
It is easy to see that every element in A is also in B, but not the other way around; that proves that A is a subset of B but B is not a subset of A.
I'm not going to try to do a formal proof.... You can try it if that's what you need.
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