Solutions to Homework 9
Mathematics 503
Foundations of Mathematics
Spring 2014
§5.2:1. (b)
Proposition. For each natural number n ≥ 6, 2n > (n + 1)2
Proof. We will proceed by induction on n. Let P (n) refer to the predicate defined in
the proposition. P (6) is true, since 64 > 49. Now assume that for k ≥ 6, P (k) is true.
We will show that P (k + 1) is also true.
Now,
2k+1 = 2(2k ) > 2(k + 1)2 > 2k 2 + 4k + 2 = (k 2 − 2) + (k 2 + 4k + 4).
Note that k 2 + 4k + 4 = (k + 2)2 and for k ≥ 6, k 2 − 2 > 0; so
2k+1 > (k + 2)2 .
6. (a) We have dy/dx = 1/x, d2 y/dx2 = −1/x2 , d3 y/dx3 = 2/x3 , and d4 y/dx4 = −6/x4 .
(b)
Proposition. For every natural number n,
(−1)n−1 (n − 1)!
dn y
=
.
dxn
xn
Proof. We have already checked in (a) that the above formula is true for n = 1.
Now assume that the formula is true for k ∈ N. We must verify it for k + 1. We
have
dk+1 y
d dk y
=
dxk+1
dx dxk
d (−1)k−1 (k − 1)!
=
by the induction hypothesis
dx
xk
d
= (−1)k−1 (k − 1)! (x−k )
dx
k−1
= (−1) (k − 1)!(−kx−k−1 )
(−1)k k!
=
xk+1
Thus the formula is true when n = k + 1. It follows from the Principle of Mathematical Induction that the formula holds for all n ∈ N.
7.
Proposition. For n = 4, 5, 8, 9, 10 and all natural numbers n ≥ 12, there exist nonnegative integers x and y such that
n = 4x + 5y.
Proof. Note that
4 = 4 · 1 + 5 · 0, 5 = 4 · 0 + 5 · 1, 8 = 4 · 2 + 5 · 0, 9 = 4 · 1 + 5 · 1, 10 = 4 · 0 + 5 · 2.
It is clear that 1, 2, 3, 6, 7, and 11 cannot be written in the form 4x + 5y for any
nonnegative integers x and y.
We will now prove that all integers n ≥ 12 can be written in the above form by strong
induction. First note that
12 = 4 · 3 + 5 · 0, 13 = 4 · 2 + 5 · 1, 14 = 4 · 1 + 5 · 2, 15 = 4 · 0 + 5 · 3.
Now let k be a natural numbers such that k ≥ 15, and assume that all natural numbers
n in the range
12, . . . , k
can be expressed in the specified form. We will show that the same is true for the
integer k + 1. We have
k + 1 = (k − 3) + 4.
Since 12 ≤ k − 3 ≤ k, by assumption, there must exist nonnegative integers x and y
such that k − 3 = 4x + 5y. It follows that
k + 1 = (k − 3) + 4 = 4x + 5y + 4 = 4(x + 1) + 5y.
Since x + 1 and y are both positive integers, we have shown that k + 1 can be expressed
in the desired form. Therefore, by induction all integers n ≥ 12 can be written in this
form.
10.
Proposition. Let x be a real number with x > 0. Then for each natural number n
with n ≥ 2, (1 + x)n > 1 + nx.
Proof. We will prove this statement by induction. First note that
(1 + x)2 = 1 + 2x + x2 > 1 + 2x
since x2 > 0. Thus the result is true for n = 2.
Now let k be an integer with k > 2 and suppose that (1 + x)k > 1 + kx. We will show
that (1 + x)k+1 > 1 + (k + 1)x. We have
(1 + x)k+1 =
>
=
=
(1 + x)(1 + x)k
(1 + x)(1 + kx)
by the induction hypothesis (since 1 + x > 0)
2
1 + (k + 1)x + kx
1 + (k + 1)x
since kx2 > 0.
Thus the statement is true when n = k + 1. By induction, it follows that (1 + x)n >
1 + nx for all natural numbers n ≥ 2.
13. (a)
Proposition. For each n ∈ N,
1
1
1
n
+
+ ··· +
=
.
1·2 2·3
n(n + 1)
n+1
Proof. We will prove this statement by induction. First note that
1
1
=
.
1 · (1 + 1)
1+1
so the statement is true when n = 1.
Now let k ∈ N and suppose that the statement is true for n = k, i.e.,
1
1
1
k
+
+ ··· +
=
.
1·2 2·3
k(k + 1)
k+1
We will show that it is true for n = k + 1, i.e,
1
1
k+1
1
+
+ ··· +
=
.
1·2 2·3
(k + 1)(k + 2)
k+2
We have
1
1
1
+
+ ··· +
1·2 2·3
(k + 1)(k + 2)
1
1
1
1
=
+
+ ··· +
+
1·2 2·3
k(k + 1)
(k + 1)(k + 2)
1
k
+
by the induction hypothesis
=
k + 1 (k + 1)(k + 2)
k(k + 2)
1
=
+
(k + 1)(k + 2) (k + 1)(k + 2)
k 2 + 2k + 1
=
(k + 1)(k + 2)
(k + 1)2
=
(k + 1)(k + 2)
k+1
=
.
k+2
Thus the statement is true when n = k + 1. By induction, it follows that the formula
holds for all natural numbers n.
16. (a)
Proposition. If n ∈ N, then there exist an odd natural number m and a nonnegative integer k such that n = 2k m.
Proof. We will prove this statement using strong induction. First note that 1 =
20 · 1, so the statement is true in the case n = 1, since 0 is a nonnegative integer
and 1 is odd.
Now let l be a natural number and suppose that each natural number n such that
1 ≤ n ≤ l can be expressed in the above form. We will show that l + 1 can be
written in this form too. If l + 1 is odd, then l + 1 = 20 (l + 1) so l + 1 can be
expressed in the desired form. On the other hand, if l + 1 is even, then
l+1=2·
l+1
.
2
Since l+1 is even (l+1)/2 is a natural number. Since, in addition, (l+1)/2 < l+1,
the induction hypothesis implies that (l + 1)/2 = 2k · m for some odd natural
number m and some nonnegative integer k. Thus
l+1=2·
l+1
= 2 · 2k m = 2k+1 m.
2
Therefore, the statement is true for n = l + 1, since k + 1 is a nonnegative integer
and m is odd. By strong induction, it follows that the statement holds for all
natural numbers n.
(b) Suppose n = 2k m = 2q p, where m and p are odd natural numbers, and k and q
are nonnegative integers. Without loss of generality, we may assume that k ≤ q.
Dividing our equation by 2k , we obtain
m = 2q p.
Since m is odd, so is 2q−k p. The only way this is possible is if q − k = 0, i.e.,
q = k. Thus m = p. It follows that the expression of n in the form 2k m is unique.
17. (b) This proof is fine except for two issues. One is that it doesn’t state what the variables x
and y stand for when it references the statement P (k−4). (They should be nonnegative
integers.) Second, a careful reading of the argument shows that it supplies a proof of
P (n) for every n except 10: the cases n = 6, 7, 8, 9 are checked by hand, and then a
proof of P (k + 1) is given for k ≥ 10. To fix this, all that is required is to note that
10 = 0 · 2 + 2 · 5.