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Unformatted text preview: e).
Proof. Let n ∈ N and a, b, c ∈ Z. Now for (i) we simply note that n0 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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 Spring '09
 W.Alabama

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