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Equiv. Class Problem |
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| Jul10-06, 07:59 PM | #1 |
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Equiv. Class Problem
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) ?? |
| Jul10-06, 09:24 PM | #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).
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| Jul10-06, 09:52 PM | #3 |
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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.
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| Jul10-06, 10:58 PM | #4 |
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Recognitions:
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Equiv. Class Problem
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.
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| Jul11-06, 04:07 AM | #5 |
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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.
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| Jul11-06, 11:50 AM | #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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