Discrete Mathematics- An Open Introduction - 3rd Edition, 2016a

Selected Solutions 363
F 2k+3 − 1
by recursive def.
Therefore F 0 + F 2 + F 4 + · · · + F 2k+2 F 2k+3 − 1, which is to say P(k + 1)
holds. Therefore by the principle of mathematical induction, P(n) is true
for all n ≥ 0.
2.5.6.
Proof. Let P(n) be the statement 2 n < n!. We will show P(n) is true for
all n ≥ 4. First, we check the base case and see that yes, 2 4 < 4! (as 16 < 24)
so P(4) is true. Now for the inductive case. Assume P(k) is true for an
arbitrary k ≥ 4. That is, 2 k < k!. Now consider P(k + 1): 2 k+1 < (k + 1)!. To
prove this, we start with the left side and work to the right side.
2 k+1 2 · 2 k
< 2 · k! by the inductive hypothesis
< (k + 1) · k! since k + 1 > 2
(k + 1)!
Therefore 2 k+1 < (k + 1)! so we have established P(k + 1). Thus by the
principle of mathematical induction P(n) is true for all n ≥ 4.
2.5.12. The only problem is that we never established the base case. Of
course, when n 0, 0 + 3 0 + 7.
2.5.13.
Proof. Let P(n) be the statement that n + 3 < n + 7. We will prove that
P(n) is true for all n ∈ N. First, note that the base case holds: 0 + 3 < 0 + 7.
Now assume for induction that P(k) is true. That is, k + 3 < k + 7. We must
show that P(k + 1) is true. Now since k + 3 < k + 7, add 1 to both sides.
This gives k + 3 + 1 < k + 7 + 1. Regrouping (k + 1) + 3 < (k + 1) + 7. But
this is simply P(k + 1). Thus by the principle of mathematical induction
P(n) is true for all n ∈ N.
2.5.14. The problem here is that while P(0) is true, and while P(k) →
P(k + 1) for some values of k, there is at least one value of k (namely k 99)
when that implication fails. For a valid proof by induction, P(k) → P(k + 1)
must be true for all values of k greater than or equal to the base case.
2.5.16. We once again failed to establish the base case: when n 0,
n 2 + n 0 which is even, not odd.
2.5.19. The proof will be by strong induction.
Proof. Let P(n) be the statement “n is either a power of 2 or can be written
as the sum of distinct powers of 2.” We will show that P(n) is true for all
n ≥ 1.
Base case: 1 2 0 is a power of 2, so P(1) is true.