Equiv. Class Problem


by hopsh
Tags: class, equiv
hopsh
hopsh is offline
#1
Jul10-06, 07:59 PM
P: 2
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) ??
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StatusX
StatusX is offline
#2
Jul10-06, 09:24 PM
HW Helper
P: 2,566
What does [t] behave like? This should give you a better feel for this field (whose elements are the equivalence classes).
hopsh
hopsh is offline
#3
Jul10-06, 09:52 PM
P: 2
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.

StatusX
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#4
Jul10-06, 10:58 PM
HW Helper
P: 2,566

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.
matt grime
matt grime is offline
#5
Jul11-06, 04:07 AM
Sci Advisor
HW Helper
P: 9,398
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.
StatusX
StatusX is offline
#6
Jul11-06, 11:50 AM
HW Helper
P: 2,566
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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