NTI.30. Let a be an integer and n a natural number. Show that there exists a
unique integer r such that a ≡ r mod n and 0 ≤ r < n.
Proof. Define the set S := {a − kn : k is an integer and a − kn ≥ 0}. The set S only
contains non-negative integers by definition. If we let k := −|a|, then a−kn = a+|a|n ≥ 0
for any natural number n, therefore S contains the element a + |a|n and hence S is nonempty. Since S contains only non-negative integers and is non-empty, it has a least
element by the W.O.P.. Call the least element r.
Since r is in S, it must be of the form a − kn for some integer k; that is r = a − kn.
Rewriting this equation as a − r = nk and using Definition 13, we see that a ≡ r mod n.
Again since r is in S, we know that r ≥ 0 by the definition of S. To prove that r < n
we assume that r ≥ n to arrive at a contradiction. If r ≥ n, then r − n = a − kn − n =
a − (k − 1)n ≥ 0, which means r − n is in S, which is a strictly smaller integer than r.
This contradicts r being the least element. Therefore r is not greater than or equal to n,
hence r < n.
The last part we need to prove is that r is unique. Assume there exists another
integer r0 such that a ≡ r0 mod n and 0 ≤ r0 < n. Using a ≡ r mod n with a ≡ r0 mod n
and Proposition 18, we obtain r ≡ r0 mod n, or n|(r − r0 ) and thus n divides |r − r0 |.
But since 0 ≤ r, r0 < n, it must be that |r − r0 | < n. The possible value for |r − r0 |
are 0, 1, 2, . . . , n − 1, but if |r − r0 | > 0 and n divides |r − r0 |, then this contradicts
Proposition 24. Therefore the only possibility for |r − r0 | is for it to equal 0, which means
that r − r0 = 0, hence r = r0 . 2