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June 20th, 2010, 01:09 AM   #1
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Equivalence relations and Construction of Q

I dont understand this problem
Show the ring of equivalence classes of modulo n, namely , every non-zero element is invertible iff
Find the inverse or reciprocal of the class determined by 5 in .

This is what I just have:



Can someone please help me work my way through this problem. Thanks in advance
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June 20th, 2010, 01:35 AM   #2
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Re: Equivalence relations and Construction of Q

The title seems to have nothing to do with the problem.
Anyway, the ideas used in the proof are

*If n is prime, then gcd(m,n)=1 for m<n.
*If n is not prime, there is some m>1,k>1 for which mk = n.

Basically, if n is prime, then (Z/(n)-{0}) forms a group under multiplication mod n. So every element is invertible
If n is not prime, then there is some pair (m,k) of non-zero elements for which mk=0=m0, so there is not a unique solution to the equation mj=l. (because if j is a solution, them j+k is also a solution)

For the second part (finding the inverse of 5), you want to find some k in {0,1,...,10} for which 5k = 1 (mod 11). That is, 5k = 11n+1 for some n.

Let me know if you need something more specific for the first part, but do try to collect those ideas and put it all together.

Edit: Oh, and this fits better in the abstract algebra section, so I'm moving it there.
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June 20th, 2010, 07:54 PM   #3
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Re: Equivalence relations and Construction of Q

Quote:
Originally Posted by cknapp
The title seems to have nothing to do with the problem.
For the second part (finding the inverse of 5), you want to find some k in {0,1,...,10} for which 5k = 1 (mod 11). That is, 5k = 11n+1 for some n.
Thanks. So,
k = 9 and n = 4
so 5k = 11n +1 =5(9)= 11(44)+1 = 45
Now what?
How did you form "5k = 11n +1"?
remeday86 is offline  
June 20th, 2010, 10:45 PM   #4
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Re: Equivalence relations and Construction of Q

Use the definition of equivalence mod 11
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