Solutions to Assignment 1, Math 220
1 Conjecture a formula for
n
X
k=1
1
1
1
1
=
+
+···+
k(k + 1)
1·2 2·3
n · (n + 1)
from the value of this sum for small integers n. Prove your conjecture
using mathematical induction.
Solution Let Sn =
Pn
1
k=1 k(k+1) .
Then
1
1 1
1 1
2
1
3
S 1 = , S2 = + = , S3 = + +
= ,...
2
2 6
3
2 6 12
4
n
and we conjecture that Sn = n+1 . We prove this via induction. Note
that we have already shown this to be true for n = 1. Suppose Sm =
m
for m ∈ N. Then
m+1
Sm+1 =
m+1
X
k=1
X
1
1
1
=
+
k(k + 1)
k(k + 1) (m + 1)(m + 2)
m
k=1
and so
Sm+1 = Sm +
1
.
(m + 1)(m + 2)
It follows from our induction hypothesis that
Sm+1 =
1
m(m + 2) + 1
m+1
m
+
=
=
.
m + 1 (m + 1)(m + 2)
(m + 1)(m + 2)
m+2
This completes the proof.
2 Use mathematical induction to prove Bernoulli’s inequality : If
1 + x > 0, then
(1 + x)n ≥ 1 + nx,
for all n ∈ N.
Solution Note that if n = 1, (1 + x)n = 1 + nx. Suppose, then, that
(1 + x)m ≥ 1 + mx, for some m ∈ N. Then
(1 +x)m+1 = (1 +x)m · (1 +x) ≥ (1 +mx) · (1 +x) = 1 + (m+ 1)x+ mx2 .
Since mx2 ≥ 0, this is ≥ 1 + (m + 1)x and so we reach the desired
conclusion.
1
2
3 Show that any amount of postage that is an integer number greater
than 11 cents can be formed using just 4-cent and 5-cent stamps.
Solution We proceed by induction. Note that 12 = 3·4+0·5. Suppose,
then, that we can write m = a · 4 + b · 5 for a and b nonnegative
integers and m ≥ 12. If a ≥ 1, then m + 1 = (a − 1) · 4 + (b + 1) · 5
is a representation of m + 1 with nonnegative coefficients a − 1 and
b + 1. If a = 0, since m ≥ 12, we have b ≥ 3 and so we can write
m + 1 = 4 · 4 + (b − 3) · 5. This completes the induction proof.
4 Using our axioms for R, prove the the following assertions :
a: If x ∈ R, then −(−x) = x.
b: If x and y are real numbers of the same sign (x and y are either
both positive or both negative), and x < y, then 1/y < 1/x.
c: If x is a real number with x 6= 0 then x2 > 0.
Solution
Part (a). Note that Axiom A5 says that there is a unique number
−(−x) such that −x + (−(−x)) = 0. Since, combining A2 and A5,
−x + x = x + (−x) = 0,
it follows that this unique number must in fact be x and so −(−x) = x.
Part (b). Let us suppose first that x > 0 and y > 0. If y1 = x1 , then
applying M5 twice (together with M2), we have
1=
a contradiction. If
1
y
1
· y → x = y,
x
> x1 , then M5 and O4 imply that
1>
1
· y.
x
From M2, this gives 1 > y · x1 and, applying M5 and O4 again yields
x > y, another contradiction.
If x < 0 and y < 0, we argue as above, only replacing Axiom O4
with Theorem 11.1 (f).
Part (c). If x < 0, then from Theorem 11.1 (f) (taking y = 0 and
z = x), we have
x2 = x · x > x · 0 = 0
(where we have used Theorem 11.1 (b)). If x > 0, then from O4 and
Theorem 11.1 (b),
x2 = x · x > x · 0 = 0.
3
5 Prove that there is no n ∈ N such that 0 < n < 1 (Use the WellOrdering Property of N).
Solution The existence of an integer in (0, 1) contradicts the fact that
N has a least element, namely 1.