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Equiv. Class Problem

  1. Jul 10, 2006 #1
    In Q[t], define the equivalence relation ~ as f(t) ~ g(t) precisely when f(t) - g(t) is a multiple of t^2 - 5. We define the addition and multiplication of equivalence classes as [f(t)] + [g(t)] = [f(t) + g(t)] and [f(t)] * [g(t)] = [f(t) * g(t)]
    (Assume: ~ is an equivalence relation, Addition/Mutliplication of equivalence classes is well-defined, and every equivalence class contains exactly one element of the form a + bt, where a, b in Q)

    a) Find a, b in Q such that [3t^3 - 5t^2 + 8t - 9] = [a + bt]
    b) Find a, b in Q such that [2t + 7] * [7t + 11] = [a + bt]
    c) Find two equivalence classes whose square is equal to [5]
    d) Find a, b in Q such that [a + bt]^2 + [-2][a + bt] = [19] (Two possible answers)
    e) Find a, b in Q such that [2 - t] * [a + bt] = [1]
    f) Which equivalence classes are zero divisors?
    g) Which equivalence classes have multiplicative inverses?
    h) How many equivalence classes are there whose square is equal to [6] ?

    Note: * means multiplication

    Now, I think I found the following solutions. Can someone verifty these and help solve the remaining parts?

    a) [35t - 54]
    b) [71t + 147]
    c) ??
    d) ??
    e) ??
    f) [0] = [t^2 - 5] (are these the only zero divisor classes??)
    g) ??
    h) ??
  2. jcsd
  3. Jul 10, 2006 #2


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    What does [t] behave like? This should give you a better feel for this field (whose elements are the equivalence classes).
  4. Jul 10, 2006 #3
    I'm lost with this problem and I have a handful of these to do. If I had an example of how to solve one of these I'm confident I can figure out the rest (I learn by example and since there aren't any with the materials I have (not even any odd solutions in the back!) I'm really desparate). Please, is there any way you could post the solutions to this one with some intermediate explanations.
  5. Jul 10, 2006 #4


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    If you think about what I said for a few minutes, you'll get all the answers very easily, along with a deeper understanding. No one is going to do the problems for you, and you should be grateful for that.
  6. Jul 11, 2006 #5

    matt grime

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    How did you get the answers to the parts you have done? If you explain that then we'll have a better idea of what you understand.
  7. Jul 11, 2006 #6


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    Maybe I should have been more clear. [t^2-5]=[t]^2-[5]=[0], so in a sense, [t] behaves just like sqrt(5). See how far you can push this analogy.
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