MTH 260 SECTION 3.4 27 DELTA COLLEGE Let S be the subset of ...

MTH 260 SECTION 3.4 27
DELTA COLLEGE
Let S be the subset of ordered pairs of integers defined recursively by
Basis step: (0, 0) ∈ S.
Recursive step: If (a, b) ∈ S, then (a, b + 1) ∈ S, (a + 1, b + 1) ∈ S, and (a + 2, b + 1) ∈ S.
a) List the elements of S produced by the first four applications of the recursive definition.
b) Use strong induction on the number of applications of the recursive step of the definition
to show that a ≤ 2b whenever (a, b) ∈ S.
c) Use structural induction to show that a ≤ 2b whenever (a, b) ∈ S.
Solution
a) After the basis step:
After the first recursive step:
S = {(0, 0)}
S = {(0, 0),
(0, 1), (1, 1), (2, 1)}
After the second recursive step:
S = {(0, 0),
(0, 1), (1, 1), (2, 1),
(0, 2), (1, 2), (2, 2), (3, 2), (4, 2)}
After the third recursive step:
S = {(0, 0),
(0, 1), (1, 1), (2, 1),
(0, 2), (1, 2), (2, 2), (3, 2), (4, 2),
(0, 3), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)}
After the fourth recursive step:
S = {(0, 0),
(0, 1), (1, 1), (2, 1),
(0, 2), (1, 2), (2, 2), (3, 2), (4, 2),
(0, 3), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3),
(0, 4), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (7, 4), (8, 4)}

Think of the points of S being generated on a grid and you will have a good picture of the
points generated after each step.
b) Using strong induction on the number n of applications of the recursive step, where n is
a non-negative integer, let P (n) be the statement:
“After n applications of the recursive step, a ≤ 2b whenever (a, b) ∈ S.”
The statement P (0) is true, since after 0 applications of the recursive step the only member
of S is (0, 0) and 0 ≤ 2(0).
Now assume that the statements P (j) are true for 0 ≤ j ≤ k and show that P (k + 1) is true.
Take a member (a, b) of S that has been generated after k + 1 applications of the recursive
step. If (a, b) occured after j applications of the recursive step, and j ≤ k, then we know that
a ≤ 2b by assumption. If (a, b) occured only after k + 1 applications of the recursive step then
there is a member (α, β) ∈ S, generated after fewer applications of the recursive step, such that
(a, b) is either
(α, β + 1), (α + 1, β + 1), or (α + 2, β + 1)
For each of these three possibilities a ≤ 2b because α ≤ 2β (by assumption):
i) For the first pair: a = α ≤ 2β < 2(β + 1) = 2b.
ii) For the second pair: a = α + 1 ≤ 2β + 1 < 2(β + 1) = 2b.
iii) For the third pair: a = α + 2 ≤ 2β + 2 = 2(β + 1) = 2b.
Therefore P (k + 1) is true and P (n) is true for all non-negative integers n.
c) Now using structural induction we prove the statment:
“a ≤ 2b whenever (a, b) ∈ S.”
In the basis step of the recursive definition (0, 0) is the only member of S and 0 ≤ 2(0).
In the recursive step we generate a new element (a, b) of S from an existing element (α, β) ∈ S
in one of three ways:
(α, β + 1), (α + 1, β + 1), or (α + 2, β + 1)
By assumption α ≤ 2β, and in each of the three cases above, that guarentees that a ≤ 2b
i) For the first pair: a = α ≤ 2β < 2(β + 1) = 2b.
ii) For the second pair: a = α + 1 ≤ 2β + 1 < 2(β + 1) = 2b.
iii) For the third pair: a = α + 2 ≤ 2β + 2 = 2(β + 1) = 2b.
Therefore a ≤ 2b whenever (a, b) ∈ S.
Notice that the two arguments are very similar, but the structural induction argument avoids
the awkward reference to the number of applications of the recursive step.