Proving R/Q and Q/Z Has No Element of Finite Order

[zcdfhn]

Prove R/Q has no element of finite order other than the identity.

First of all, I have trouble visualizing what R/Q is. But I do know that afterwards you can try to raise an element in R/Q to a power to get to 0, but there will not be a finite number that will be able to do so except zero itself.

So I think I have the jist of the problem figured, but I need help visualizing R/Q. I would appreciate if you could give examples of what numbers belong to what cosets.

I also have a similar question about Q/Z, and I have to prove that every element has a finite order.

Thank you.

 

[matt grime]

You don't try to visualize R/Q (I don't think it's possible to do so, and isn't necessary either), just take the definition: [x] and [y] are the same coset if and only if x-y is a rational number.

Now, the operation is addition of cosets (not multiplication, so no raising things to powers), and you want to show that if [x] is not [0], i.e. x is not rational, then [nx] is not the [0] for any n, i.e. that nx is not rational either for any n. So you see that just writing out what it is that you need to proves makes the question entirely trivial.

 

[zcdfhn]

Thank you so much! You really clarified my view on quotient sets, I finally found a connection to what I learned about the quotient Z/nZ and it makes so much sense now.