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Homework Help: Define the relation problem

  1. Apr 23, 2010 #1
    Define the relation ∼ on ℤ as follows: For a,b ∈ ℤ, a∼b iff. 2a + 3b ≡ 0 (mod 5). The relation ∼ is an equivalence relation on ℤ. Determine all the distinct equivalence classes for this equivalence relation.
    Reflexive if a∼a.
    2a + 3a ⇒ 5a ≡ 0 (mod 5); therefore, the relation is reflexive.
    Symmetric if a∼b, then b∼a.
    2a + 3b ≡ 4(2a + 3b) ≡ 8a + 12b ≡ 3a + 2b ≡ 0 (mod 5); therefore, the relation is symmetric.
    Transitive if a∼b and b∼c, then a∼c.
    a∼b ⇒ 2a + 3b ≡ 0 (mod 5)
    b∼c ⇒ 2b + 3c ≡ 0 (mod 5) By adding the two, we obtain ⇒ 2a + 5b + 3c ≡ 2a + 3c ≡ 0 (mod 5); therefore, the relation is transitive.
    2a + 3b ≡ 0 (mod 5) ⇒ 5 | (2a + 3b) ⇒ 5m = 2a + 3b
    [0] = {a ∈ ℤ | a∼0} = {a ∈ ℤ | 5m = 2a} = {a ∈ ℤ | 2a = 5m} = {..., 5, 10, 15, ...}
    [1] = {a ∈ ℤ | a∼1} = {a ∈ ℤ | 5n = 2a + 3} = {a ∈ ℤ | 2a = 5n - 3} = {..., 1, 6, 11, ...}
    [2] = {a ∈ ℤ | a∼2} = {a ∈ ℤ | 5p = 2a + 6} = {a ∈ ℤ | 2a = 5p - 6} = {..., 2, 7, 12, ...}
    [3] = {a ∈ ℤ | a∼3} = {a ∈ ℤ | 5r = 2a + 9} = {a ∈ ℤ | 2a = 5r - 9} = {..., -2, 3, 8, ...}
    [4] = {a ∈ ℤ | a∼4} = {a ∈ ℤ | 5t = 2a + 12} = {a ∈ ℤ | 2a = 5t - 12} = {..., -1, 4, 9, ...}

    Are these correct?
     
    Last edited: Apr 23, 2010
  2. jcsd
  3. Apr 23, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    Re: Relations

    Yes, I think they are correct. Nice job.
     
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