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 June 20th, 2010, 01:09 AM #1 Member   Joined: Feb 2009 Posts: 76 Thanks: 0 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 June 20th, 2010, 01:35 AM #2 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 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 m1,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. 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"? June 20th, 2010, 10:45 PM #4 Senior Member   Joined: Oct 2007 From: Chicago Posts: 1,701 Thanks: 3 Re: Equivalence relations and Construction of Q Use the definition of equivalence mod 11 Tags construction, equivalence, relations ### equivalence relations

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