Quotient groups Q/Z and R/Q

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Homework Statement


Show that every element of the quotient group [tex]\mathbb{Q}/\mathbb{Z}[/tex] has finite order but that only the identity element of [tex]\mathbb{R}/\mathbb{Q}[/tex] has finite order.

The Attempt at a Solution


The first part of the question I solved. Since each element of [tex]\mathbb{Q}/\mathbb{Z}[/tex] is of the form [tex]\mathbb{Z}+\frac{r}{s}[/tex] if we add this element s times to itself, we get all [tex]\mathbb{Z}[/tex] back, since [tex]s\frac{r}{s}=r[/tex]. But for the second part of the question I have no clue... Can anyone hint me in the right direction?
 

Answers and Replies

  • #2
Dick
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Ok, so a nonidentity element of R/Q has the form Q+r, where r is irrational. What would be wrong with n*(Q+r)=Q?
 
  • #3
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Ok, so a nonidentity element of R/Q has the form Q+r, where r is irrational. What would be wrong with n*(Q+r)=Q?

n*(Q+r)=n*r+Q since Q is normal. has it got something to do with the fact that for p=prime 1/p cannot be written as the sum of two rationals?
 
  • #4
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oh no... 1/7=1/14+1/14
 
  • #5
Dick
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n*(Q+r)=n*r+Q since Q is normal. has it got something to do with the fact that for p=prime 1/p cannot be written as the sum of two rationals?

If x+Q=Q what can you say about x? This has nothing to do with primes.
 
  • #6
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If x+Q=Q what can you say about x? This has nothing to do with primes.

If x is not Q, then this can never be true, since the sum of a non-rational and a rational number is non-rational.

If x is in Q then x+Q = Q since for each element q in Q there exists an element q-x in Q which gives x+q-x=q, and each element in x+Q is in Q.

x must be a rational.
 
  • #7
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Ok, so a nonidentity element of R/Q has the form Q+r, where r is irrational. What would be wrong with n*(Q+r)=Q?

The only thing confusing me a little from the beginning is that if we have a nonidentity element of R/Q it has to have the form Q+r where r is irrational. Does r have to be irrational because Q is the identity of R/Q?
 
  • #8
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I get it now! If x+Q=Q, x must be rational. Therefore, a nonidentity element of R/Q has the form Q+r where r is irrational. Now if n*(Q+r)=Q, n*r+Q=Q and n*r is rational. But this cannot be the case since r is irrational, and we have a contradiction, and a nonidentity element of R/Q does not have finite order.
Thanks for the hints ;)
 

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