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Homework Help: Sets: Is A ⊆ B? A={x ∈ ℤ | x ≡ 7 (mod 8)} B={x ∈ ℤ | x ≡ 3 (mod 4)}

  1. Mar 2, 2010 #1
    A={x ∈ ℤ | x ≡ 7 (mod 8)}
    B={x ∈ ℤ | x ≡ 3 (mod 4)}

    Is A ⊆ B? Yes

    Since x ∈ A, then xa = 7 + 8a = 8a + 7 = 2(4a + 3) +1. And since the ∈ B are of the form xb = 3 + 4b = 4b + 3 = 2(2b + 1) + 1, both ∈ A,B are odd. A ⊆ B since the ∈ of both sets are of 2p + 1. Q.E.D.

    Is this correct?
     
  2. jcsd
  3. Mar 2, 2010 #2
    No. You want to show that every element of A is an element of B. All you showed is that neither contains an even integer.

    Try writing 7 + 8a in a form that shows it is in B.
     
  4. Mar 2, 2010 #3
    I don't know how to go about doing that.
     
  5. Mar 2, 2010 #4

    Mark44

    Staff: Mentor

    What is this character - ℤ ?

    7 + 8a = 3 + 4 + 4(2a) = 3 + 4(1 + 2a)
     
  6. Mar 2, 2010 #5
    Integers
     
  7. Mar 2, 2010 #6
    So do you see how the conclusion follows? It is pretty straight forward now.
     
  8. Mar 2, 2010 #7
    Since x ∈ A, then x = 7 + 8a = 8a + 7 = 4(2a) + 4 + 3 = 4(2a + 1) + 3. And since the ∈ B are of the form x = 3 + 4b = 4b + 3, all x = 4p + 3 and A ⊆ B. Q.E.D.

    Correct?
     
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