Thus k n b a so nb a and we have that b amod n

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Unformatted text preview: e). Proof. Let n ∈ N and a, b, c ∈ Z. Now for (i) we simply note that n|0 so that n|(a − a) and a ≡ a(mod n). For (ii), if a ≡ b(mod n), then n|(a − b) so that there is a k ∈ Z such that kn = (a − b). Thus (−k )n = (b − a), so n|(b − a) and we have that b ≡ a(mod n). For (iii), if a ≡ b(mod n) and b ≡ c(mod n), we have that n|(a − b) and n|(b − c) so that n|[(a − b) + (b − c)] which gives n|(a − c). Thus a ≡ c(mod n). Definition 8. If a relation has the properties that it is reflexive, symmetric, and transitive, then we call it an equivalence relation. Thus “≡” to the modulus n is an equivalence relation,...
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