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 2 years ago
If \(a_1, a_2,...a_n\) is a complete set of residues modulo \(n\) and \(gcd(a, n)=1\), then prove that \(aa_1, aa_2,...aa_n\) is also a complete set of residues modulo \(n\).
[Hint: It suffices to show that the numbers in question are incongruent modulo n ]
 2 years ago
If \(a_1, a_2,...a_n\) is a complete set of residues modulo \(n\) and \(gcd(a, n)=1\), then prove that \(aa_1, aa_2,...aa_n\) is also a complete set of residues modulo \(n\). [Hint: It suffices to show that the numbers in question are incongruent modulo n ]

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jasonchutko
 2 years ago
Best ResponseYou've already chosen the best response.1We have to show that any pair of them is incongruent. We will try to prove the contrapostive by using the fact that: \[a_i \not\equiv a_j (\mod n) \text{ for } i \neq j\] We suppose \[aa_i \equiv aa_j (\mod n)\] and will show that \[i = j\] Because \[aa_i \equiv aa_j (\mod n)\] that means that \[aa_i  aa_j\] must be a multiple of n, say \[a(a_i  a_j) = nk k\epsilon \mathbb{Z}\] By the definition of divides, \[n  a(a_i  a_j)\] By the hypothesis, we know that \[a \text{ and } n\] are coprime, so we know that \[n  (a_i  a_j)\]Thus we have shown that \[a_i \equiv a_j (\mod n)\] and then \[a_i = a_j \text{ and } i = j \] since we need \[a_i\] to form the complete set for \[n\] We have thus proven that any pair of \[aa_i \text{ and } aa_j\] will be incongruent when \[i \ne j \] and finally that the set of \[aa_i\] will form the complete set of residues modulo n.

2bornot2b
 2 years ago
Best ResponseYou've already chosen the best response.1@jasonchutko, thanks a lot!

jasonchutko
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry my LaTeX is off a little bit :P

2bornot2b
 2 years ago
Best ResponseYou've already chosen the best response.1@jasonchutko How many members of the set S={\(x:1\le x\le 100\)} are of the form 4n1. I guess the answer is 25, but is there a way to prove it or show it?

2bornot2b
 2 years ago
Best ResponseYou've already chosen the best response.1@UnkleRhaukus How many members of the set S={x:1≤x≤100} are of the form 4n1. I guess the answer is 25, but is there a way to prove it or show it?

2bornot2b
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, n is an integer, sorry, I should have mentioned that @UnkleRhaukus If you want you can answer it here http://openstudy.com/updates/4f7fe5e1e4b0505bf0829928
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