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Every element in R/Q has order 1 or ∞

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Let R and Q be the groups of real and rational numbers under addition, and R/Q be a quotient group (or factor group).

    Part a) If a + Q = b + Q for a,b ∈ R, what can you deduce about a and b?

    Part b) Prove that every element in R/Q has order 1 or ∞.


    2. Relevant equations

    No relevant theorems or anything that I'm aware of. Almost all of the groups we've worked with have been finite. Most of the theorems single out the group in the hypothesis to be a finite group.


    3. The attempt at a solution

    My attempt at part a):
    This question is so vague. It says a lot about them. You almost have to do part b) first to see what it is you need to say. I'm going to say this implies that a ∈ b + H, which is easy to show. Not sure what that will get me.

    My attempt at b):
    Let r + Q ∈ R/Q. Assume the order of r + Q is not ∞. Then (r + Q)^k = kr + Q = e + Q where "e" is the identity of R (that is, 0), for some positive integer k. I will show that k = 1. Suppose k > 1. Then kr + Q = e + Q. Presumably, this is where I would use part a) to claim that kr ∈ e + Q. This means kr = q for some q ∈ Q. I'm not sure how to show a contradiction from there.

    I'm actually pretty sure I'm on the wrong track altogether. I'll be up for a little longer, but maybe it will make more sense if I take a nap.

    P.S. Actually, I just had a thought while writing this. Consider an integer k such that kr + Q = e + Q = Q. This implies kr ∈ Q by part (a). That implies that r itself must be rational, since the product of an integer k and an irrational # would still be irrational. Thus it has order 1, since r + Q = Q for any rational number. If r is irrational, then then kr + Q != e + Q for any integer k, since to do so would imply that kr itself is rational. So the order must be infinite in that case.

    Is that right? I'm so tired and I'm so prone to making accidental conceptual mistakes with quotient groups, I want to make sure I didn't say anything naive/stupid. I would of course write up my solution more rigorously than I have just stated it.

    Edit: And I know multiplication for "kr" is not actually defined. It's actually "r + r + r + ..." a total of k times. I'm abusing the notation for simplicity. The sum of an irrational number k times still won't ever be rational.
     
  2. jcsd
  3. Oct 13, 2009 #2

    Dick

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    There's nothing wrong with what you are saying. I'm not sure what they are looking for in a) either, except that a and b differ by a rational. And I agree with your argument. If r is rational then r+Q=0 in the quotient group R/Q. If r is irrational then it doesn't have finite order in R/Q.
     
  4. Oct 13, 2009 #3
    Ugh. Thank you so much. I'll definitely refine it a little when I put it on paper. As I was writing this question, I swear I had no idea where to go next. So I didn't mean to sort of waste your time by asking it only to answer it in the next paragraph. I'm used to the idea of sets of sets, but I think the notation gets me confused sometimes. I'm so used to compact representations of elements that writing them involving notation for a set and an element which together represents a set that is an element gets me turned around sometimes.

    Oh, and I know that part (a) was intended to help me with part (b). So it was meant to prod me to look at that condition, where a + Q = b + Q. I think my answer there will work since I used it to prove (b). Thanks again.
     
    Last edited: Oct 13, 2009
  5. Oct 14, 2009 #4

    HallsofIvy

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    If a+ Q= b+ Q (in other words, a and b are in the same equivalence class), then a and b differ by a rational number: a- b is rational. That's the simplest thing you can say and probably is what you want.
     
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