16
0. PRELIMINARIES
Example.
Every integer greater than 27 can be written as a · 5 + b · 8 for a, b > 0.
Proof.
Let Z0 = {n 2 Z|n
28} and S = {n 2 Z0|n = a · 5 + b · 8 for a, b
0}.
Clearly, 28 2 S since 28 = 4 · 5 + 1 · 8.
[Two proofs.]
[Using MI] Suppose n 2 S, n = a · 5 + b · 8 with a, b
Then a
If a
If b
3 or b
0.
3.
3,
n + 1 = (a · 5 + b · 8) + ( 3 · 5 + 2 · 8) = (a
3,
n + 1 = (a · 5 + b · 8) + (5 · 5
3) · 5 + (b + 2) · 8.
3 · 8) = (a + 5) · 5 + (b
In either case, n + 1 2 S, so S = Z0 by MI.
3) · 8.
[Using SI] Note 29 = 1 · 5 + 3 · 8, 30 = 6 · 5 + 0 · 8, 31 = 3 · 5 + 2 · 8, and
32 = 0 · 5 + 4 · 8 are in S.
For n > 32, assume {k 2 Z0|28  k < n} ✓ S.
[Show n 2 S.] Since n
5 2 S, 9 a, b
03 n
n = (a + 1) · 5 = b · 8 =) n 2 S, so S = Z0 by SI.
5 = a · 5 = b · 8 =)
⇤