Math 2112 Solutions
Assignment 4
2.2.19 ∃x ∈ R such that ∀y ∈ R− , x > y.
(a) ∀x ∈ R ∃y ∈ R− such that x > y.
(b) Both the original statment and the new statement are true.
2.2.25 If the square of an integer is even, then the integer is even.
Contrapositive: If an integer is odd, then the square of the integer is odd.
Converse: If an integer is even, then the square of the integer is even.
Inverse: If the square of an integer is odd, then the integer is odd.
4.2.11
1
1·2
+
1
2·3
+ ··· +
1
n(n+1)
=
n
n+1 ,
for all integers n ≥ 1.
Proof: By mathematical induction.
Base Case: Let n = 1. Then LHS =
LHS = RHS.
1
1·2
=
1
2
and RHS =
1
1
Inductive Step: Let k ≥ 1. Assume that 1·2
+ 2·3
+ ··· +
Consider
1
1
1
1
+
+ ··· +
+
1·2 2·3
k(k + 1) (k + 1)(k + 2)
1
2.
Therefore
1
k(k+1)
=
k
k+1 .
By our inductive hypothesis, we know that
1
1
1
1
+
+ ··· +
+
1·2 2·3
k(k + 1) (k + 1)(k + 2)
=
=
=
=
=
k
1
+
k + 1 (k + 1)(k + 2)
k(k + 2) + 1
(k + 1)(k + 2)
k 2 + 2k + 1
(k + 1)(k + 2)
(k + 1)(k + 1)
(k + 1)(k + 2)
k+1
.
k+2
But then P (k + 1) is true.
1
1
1
+ 2·3
+ · · · + n(n+1)
=
Therefore 1·2
induction.
n
n+1 ,
1
for all integers n ≥ 1 by mathematical
4.2.16
Qn
1
i=0 ( 2i+1
·
1
2i+2 )
=
1
(2n+2)! ,
for all integers n ≥ 0.
Proof: Proof by mathematical induction.
Base Case: Let n = 0. Then LHS = 11 · 12 =
LHS = RHS.
1
2
and RHS =
1
2!
= 12 . Therefore
Qk
1
1
1
Inductive Step: let k ≥ 0. Assume that i=0 ( 2i+1
· 2i+2
) = (2k+2)!
. ConQk+1 1
1
sider i=0 ( 2i+1 · 2i+2 ). By the inductive hypothesis, we know that
k+1
Y
(
i=0
1
1
·
)
2i + 1 2i + 2
=
1
1
1
·(
·
)
(2k + 2)! 2k + 3 2k + 4
=
1
.
(2k + 4)!
But then P (k + 1) is true.
Therefore
Qn
1
i=0 ( 2i+1
·
1
2i+2 )
=
1
(2n+2)!
for all n ≥ 0.
4.3.12 For any integer n ≥ 1, 7n − 2n is divisible by 5.
Proof: Proof by mathematical induction.
Base Case: Let n = 1. Then 71 − 21 = 5, which is indeed divisible by 5.
Inductive Step: Let k ≥ 1. Assume that 7k − 2k is divisible by 5. Consider
7k+1 − 2k+1
=
7 · 7k − 2 · 2k
=
=
7k + 7k + 5 · 7k − 2k − 2k
7k − 2k + 7k − 2k + 5 · 7k .
Note that the first two terms are divisible by 5 by our inductive hypothesis,
and the last term is divisible by 5 by the definition of divisibility. Therefore
7k+1 − 2k+1 is divisible by 5.
Therefore, 7n − 2n is divisible by 5 for all integers n ≥ 1.
4.3.30 Use mathematical induction to show that in any round-robin
tournament involving n teams, where n ≥ 2, it is possible to label the
teams T1 , T2 , . . . , Tn so the Ti beats Ti+1 for all i = 1, 2, . . . , n − 1.
Proof: Proof by mathematical induction.
Base Case: Let n = 2. Then there was only one game played. Label the winner
2
as T1 and the loser as T2 .
Inductive Step: Let k ≥ 2. Assume that for all tournaments with k teams,
there is a labelling as described above. Consider a tournament on k + 1 teams.
When we remove one team, say team A, we have a tournament on k teams.
Thus, by our inductive hypothesis, there is a labelling T1 , T2 , . . . , Tk as above.
Either A beats team T1 , A loses to the first m teams (where 1 ≤ m ≤ k − 1)
and beats the (m + 1)st team, ot A loses to all the other teams.
In the first case, A, T1 , T2 , . . . , Tk is a desired ordering, so we relabel our teams
so that A becomes T10 , T1 becomes T20 , and so on.
In the second case, T1 , T2 , . . . , Tm , A, Tm+1 , . . . , Tk is a desired ordering (since
A lost to Tm but beat Tm+1 ), and so we relabel accordingly.
In the third case, T1 , T2 , . . . , Tk , A is a desired ordering (since A lost to everyone, in particular they lost to Tk ), and so we relabel accordingly.
In all cases, we have found the desired labelling. Thus the result holds by
mathematical induction.
4.4.8 Suppose that h0 , h1 , h2 , . . . is a sequence defined as follows:
h0 = 1, h1 = 2, h2 = 3,
hk = hk−1 + hk−2 + hk−3
for all integers k ≥ 3.
a. Prove that hn ≤ 3n for all integers n ≥ 0.
b. Suppose that s is any real number such that s3 ≥ s2 + s + 1. (This
implies that s > 1.83.) Prove that hn ≤ sn for all n ≥ 2.
a. Proof: Proof by strong mathematical induction.
Base Cases: Note that h0 ≤ 30 , h1 ≤ 31 , h2 ≤ 32 .
Inductive Step: Let k > 2. Assume that hi ≤ 3i for all integers i with 0 ≤ i < k.
Consider hk . By our inductive hypothesis, we know that
hk
=
≤
hk−1 + hk−2 + hk−3
3k−1 + 3k−2 + 3k−3
= 32 · 3k−3 + 3 · 3k−3 + 3k−3
= (32 + 3 + 1)3k−3
≤ 33 · 3k−3
=
3k
3
Therefore, hk ≤ 3k .
Thus the result holds by strong mathematical induction.
b. Proof: Proof by strong mathematical induction.
Base Cases: Note that since s > 1.83, h2 ≤ s2 , h3 ≤ s3 , h4 ≤ s4 .
Inductive Step: Let k > 4. Assume that hi ≤ si for all integers i with 2 ≤ i < k.
Consider hk . By our inductive hypothesis, we know that
hk
=
hk−1 + hk−2 + hk−3
≤ sk−1 + sk−2 + sk−3
= s2 · sk−3 + s · sk−3 + sk−3
= (s2 + s + 1)sk−3
≤
=
s3 · sk−3
sk
Therefore, hk ≤ sk .
Thus the result holds by strong mathematical induction.
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