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Discrete Structures - Homework 3 Solutions | CS 173, Assignments of Discrete Structures and Graph Theory

University of Illinois - Urbana-ChampaignDiscrete Structures and Graph Theory

Material Type: Assignment; Class: Discrete Structures; Subject: Computer Science; University: University of Illinois - Urbana-Champaign; Term: Spring 2007;

Typology: Assignments

Pre 2010

Uploaded on 03/16/2009

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CS173 Discrete Mathematical Structures
Spring 2007
Homework #3 Solution
Due Sun 02/04/07, 8AM.
(40 points)
1) PROOF TECHNIQUES (15 points, 3 points each)
Prove each of the following problems using the indicated technique:
a) Indirect Proof: n is odd if
3
6
n
+
is odd (n is an integer).
Proof:
We prove that its contrapositive is true. Assume n is even. n can be expressed as 2k.
Substitute it into
3
6
n
+
, we get
3 3
(2 ) 6 2(4 3)
k k
+ = +
. This indicates that
3
6
n
+
is an even number.
Therefore, its contrapositive is true.
b) Proof by Contradiction
: The product of an irrational number and a nonzero rational number is
irrational.
Before proving it by contradiction, here is the outline of a proof by contradiction:
Original problem: we want to show that
p q
is true
.
We construct ( ( ))
p q F
¬ → →
.
Since the whole implication ( ( ))
p q F
¬ → →
is true,
( )
p q
must be false. Therefore,
p q
is true.
( )
p q p q
¬ → ⇔ ∧ ¬
.
In summary, what we need to do is first assume
p q
∧ ¬
,
then derive some contradiction/fallacy
from
p q
∧ ¬
.
Proof:
Assume the product of an irrational number and a nonzero rational number is rational
(
p q
∧ ¬
). Denote the product by
c
d
, the irrational number
k
, and the nonzero rational number
a
b
(
a, b, c,
and
d
are integers and
0
b
≠
). We have
c a cb
k k
d b ad
= ⋅ ⇒ =
. Since a, b, c, and d are
integers, cb and ad are integers too. Therefore, the irrational number k can be expressed as a
fraction, which means it is a rational number. Apparently, this is a contradiction.
( ( )
p q F
∧ ¬ →
). Therefore, the original statement is true (
p q
is true).
c) Vacuous Proof:
All positive real numbers x that satisfy
2
7 12 0
x x
+ + <
are also solutions to the
transcendental equation
sin 0
x x
+ =
.
Proof: The solution for
2
7 12 0
x x
+ + <
is
4 3
x
− < < −
, which is not positive. Therefore, it is
vacuously true.
d) Trivial Proof:
If
0
a
>
and
2
b
≤ −
,
2 2
0
a ab b
+ + ≥
Proof:
2 2 2 2
1 3
( ) 0
2 4
a ab b a b b
+ + = + + ≥
We see that it is always true for ,
a b R
∈
. Therefore,
the statement is trivially true.
pf3

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CS173 Discrete Mathematical Structures

Spring 2007

Homework #3 Solution

Due Sun 02/04/07, 8AM.

(40 points)

  1. PROOF TECHNIQUES (15 points, 3 points each) Prove each of the following problems using the indicated technique:

a) Indirect Proof: n is odd if n^3 + 6 is odd (n is an integer). Proof: We prove that its contrapositive is true. Assume n is even. n can be expressed as 2k. Substitute it into n^3 + 6 , we get (2 )k 3 + 6 = 2(4 k^3 + 3). This indicates that n^3 + 6 is an even number. Therefore, its contrapositive is true. b) Proof by Contradiction: The product of an irrational number and a nonzero rational number is irrational. Before proving it by contradiction, here is the outline of a proof by contradiction: Original problem: we want to show that p  q is true. We construct ( ¬( p → q ))→ F. Since the whole implication ( ¬( p → q)) → Fis true, ¬( p →q) must be false. Therefore, p  q is true. ¬( p → q )⇔ p ∧ ¬q. In summary, what we need to do is first assume p ∧ ¬q , then derive some contradiction/fallacy from p ∧ ¬q. Proof: Assume the product of an irrational number and a nonzero rational number is rational ( p ∧ ¬q ). Denote the product by c d

, the irrational number k, and the nonzero rational number a b (a, b, c, and d are integers and b ≠ 0 ). We have c^ k a^ k cb d b ad

= ⋅ ⇒ =. Since a, b, c, and d are integers, cb and ad are integers too. Therefore, the irrational number k can be expressed as a fraction, which means it is a rational number. Apparently, this is a contradiction. ( ( p ∧ ¬q ) → F). Therefore, the original statement is true (p  q is true).

c) Vacuous Proof: All positive real numbers x that satisfy x 2 + 7 x+ 12 < 0 are also solutions to the transcendental equation x + sin x= 0. Proof: The solution for x 2 + 7 x+ 12 < 0 is − 4 < x< − 3 , which is not positive. Therefore, it is vacuously true. d) Trivial Proof: If a > 0 and b ≤ − 2 , a 2 + ab + b^2 ≥ 0 Proof: 2 2 ( 1 ) 2 3 2 0 2 4

a + ab + b = a + b + b ≥ We see that it is always true for a b, ∈ R. Therefore, the statement is trivially true.

e) Proof by Cases: Prove that the square of an integer not divisible by 5 leaves a remainder of 1 or 4 when divided by 5. Proof: All integers not divisible by 5 can be expressed as 5k+1, 5k+2, 5k+3, or 5k+4. Case 1: (5 k + 1)^2 = 25 k 2 + 10 k + 1 = 5(5 k 2 + 2 )k + 1 Case 2: (5 k + 2)^2 = 25 k 2 + 20 k + 4 = 5(5 k 2 + 4 )k + 4 Case 3: (5 k + 3) 2 = 25 k 2 + 30 k + 9 = 5(5 k 2 + 6 k+ 1) + 4 Case 4: (5 k + 4)^2 = 25 k 2 + 40 k + 16 = 5(5 k 2 + 2 k+ 3) + 1 In all cases, the remainder is either 1 or 4.

  1. SET IDENTITIES (15 points) Let A and B be sets. Prove that ( A I B ) U ( A I B )=A using three different techniques. a) Prove it using a membership table. (4 points, 1 point for each row) A B (^) A I B A I B ( A I B ) U ( A IB) 1 1 1 0 1 1 0 0 1 1 0 1 0 0 0 0 0 0 0 0 Truth values in the first and last columns are the same. Therefore the two sets are identical. b) Prove it by showing one of the sets is a subset of the other and vice versa (To prove that X = Yis to prove that X ⊆ YandY ⊆ X). (6 points, 3 points for each part) Proof: (part 1) ( A I B ) U ( A IB )⊆A We need to show x ∈ ( A I B) U ( A IB )→ x ∈A ( ) ( ) definition of union ( ) ( ) definition of intersection ((

x A B A B x A B x A B x A x B x A x B x A x

I U I

I I

) ) (( ) ) distributive law ... absorption law

B x A x A x B x B x A x A

→ ∈ simplification

(part 2) A ⊆ ( A I B) U ( A IB) We need to show that x ∈ A → x ∈ ( A I B) U ( A IB) The set A can be split into two parts. One part is the intersection A I B(it might be empty), and the other part is A − B. A − Bis actually A I B. These two parts are nonoverlapping, and their union is A itself (using a Venn Diagram may help you understand it). Therefore, if x is an element of A, then it belongs to either A I Bor A I B.