Question

From the proof of (ii) . Explain/Show why -n+ 1Sm-kn-1 is true by construction. . Explain/Show why 0 is the only number divisible by n in the range -n+1 ton-1 Proposition 6.24. Fix a modulus nEN. (i) is an equivalence relation on Z. (ii) The equivalence relation-has exactly n distinct equivalence classes, namely (ii) We need to prove that every integer falls into one of the equivalence classes [0], [1],..., [n -1], and that they are all distinct. For each m E Z, we can, by the Division Algorithm (Theorem 6.13), find integers q,r with 0 <r<n-1 such that m-qn+r. In particular, m- r is divisible by n. But then mEr, and by Proposition 6.4, [m]-[r], which is one of the equivalence classes Now assume 0 < m, k < n-1 and imi [k]. Our goal is to show that m = k. Again by Proposition 6.4, [m] [k] is equivalent to m k, i.e., n divides m-k. But by construction, and the only number divisible by n in this range is 0, that is, m-k.

Answer 1

$-n+1\le m-k\le n-1$ because
?Tm
where the remainder after division by can range from $0\le r\le n-1$

Thus, $0\le m\le n-1$ and $0\le k\le n-1$

So the least/greatest values is $-n+1\le m-k\le n-1$

As the least value is obtained when $m=0,\,k=n-1\Rightarrow m-k=-n+1$ (m is least possible, k is largest possible) and vice versa

Note that the consecutive multiples of are
2n, -m. 0, n, 2n

Noting that $<-n<-n+1\le 0\le n-1<n<$

So the only multiple of between
-n 1 and n-1
is so we must have $m=k$