epdf.pub_the-nuts-and-bolts-of-proofs-third-edition-an-intr

Special Kinds of Theorems 67
for all real numbers y. This implies that:
/(%)) =f(g(y)y
Because/is a one-to-one function, we obtain:
Ky) = giy)
for all real numbers y. Therefore, h = g.
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2. Division theorem. Let a and b be two integer numbers such that a>0
and b > 0. Then, there exist two unique integers q and r, where ^ > 0
and 0 < r < b such that a = bq + r.
(We have already proved the existence of numbers with the required
properties earlier in the book; see Example 5 in the section on Basic
Techniques. Here, we want to include a different proof that uses
mathematical induction.)
Proof: Consider different cases and use mathematical induction.
1. If b=l, consider q = a and r = 0.
2. If a = 0, consider ^ = 0 and r = 0.
3. The statement is clearly true for all numbers a < b. Consider ^ = 0
and r = a.
4. Assume that a>b.
If a = b, the statement is trivially true.
Assume that the statement is true for a generic number n> b. Then
consider a = n-\-l.
a = n+l = (bqi + ri) + 1
= bqi + (ri + 1).
Because 0 < ri < fc, then 1 < ri + 1 < fe.
If ri + 1 < b, then just set qi=q and r\-\-l = r.
If ri + 1 = b, we can write
a = bqi+b = bq-\-r
where q = qi + l and r = 0.
We are now going to prove the uniqueness part of the statement. Let
us assume that there are two pairs of integers, q and r, q' and / such
that
a = bq-\-r = bq^ -\-r'