MATH 1200 (SECTION E): ANSWERS FOR HOMEWORK 4
FARID ALINIAEIFARD
1. Use Mathematical Induction to prove that for all integers n,
1+
1 1
1
1
+ + ··· + 2 ≤ 2 − .
4 9
n
n
Proof. We proceed the proof by mathematical induction.
Let the statement p(n) be the inequality 1 + 41 + 19 + · · · + n12 ≤ 2 − n1 . Note that P (1)
is true since 1 ≤ 2 − 12 = 1. Suppose that P (k) is true for every positive integer k,
i.e.,
1 1
1
1
1 + + + ··· + 2 ≤ 2 − .
4 9
k
k
Therefore, we have
1 1
1
1
1
1
1 + + + ··· + 2 +
≤
2
−
+
(1)
4 9
k
(k + 1)2
k
(k + 1)2
Consider that
k(k + 1) ≤ (k + 1)2 ⇒
Note that
1
1
≥
.
k(k + 1)
(k + 1)2
1
1
1
= −
.
k(k + 1)
k k+1
Therefore,
1
1
1
−
≥
,
k k+1
(k + 1)2
and so
1
1
1
−
≥
(2)
k (k + 1)2
k+1
We can conclude from (1) and (2) that
1 1
1
1
1
1 + + + ··· + 2 +
≤2−
2
4 9
k
(k + 1)
k+1
Which means that when P (k) is true then P (k + 1) is true. So by Mathematical
induction
1+
1 1
1
1
+ + ··· + 2 ≤ 2 −
4 9
n
n
for all positive integers n.
1
2
AUTHOR’S NAME
2. How many subsets does {1} have? How many subsets does {1, 2} have? Find a
formula for the number of subsets of {1, 2, . . . , n}, and use mathematical induction
to prove your formula.
Proof. The number of subsets of {1} is 2 = 21 . The number of subsets of {1, 2} is
4 = 22 . We claim that the number of subsets of {1, 2, . . . , n} is 2n and we prove it
by mathematical induction. When n = 1, our claim is true. Suppose that for every
positive integer k, {1, 2, . . . , k} has 2k subsets. Let A = {1, 2, . . . , k, k + 1}. For every
subset S of A either k + 1 ∈ S or k + 1 6∈ S. The number of subsets of A that not
containing k +1 is same as the number of subsets of {1, 2, . . . , k}, and so by induction
hypothesis is equal to 2k . Every subset of A that contains k + 1, is a union of the
set {k + 1} and a subset of {1, 2, . . . , k}. The number of such subsets is same as the
number of subsets of {1, 2, . . . , k}, which is equal to 2k by mathematical hypothesis.
Since the number of subsets of {1, 2, . . . , k + 1} is the sum of subsets containing k + 1
and the subsets not containing k + 1, we conclude that the number of subsets of A is
equal to 2k + 2k = 2k+1 .