Islamic University of Gaza
Faculty of Engineering
Department of Computer Engineering
Fall 2011
ECOM 2311: Discrete Mathematics
Eng. Ahmed Abumarasa
Discrete Mathematics
Sec 1.6
The Foundations: Logic and Proof, Sets, and Functions
Introduction to proofs
Chapter 1:
The Foundations: Logic and Proof, Sets, and Functions
1.4: Introduction to proofs
Methods of Proving Theorems:
 Direct Prove: A direct proof of a conditional statement
q is constructed
when the first step is the assumption that p is true; subsequent steps are
constructed using rules of inference, with the final step showing that q must
also be true.
 Proof by Contraposition: Proofs by contraposition make use of the fact
that the conditional statement p q is equivalent to its contrapositive,
.

Proofs by Contradiction: Because the statement
is a
contradiction whenever r is a proposition, we can prove that p is true if we can
show that



VACUOUS PROOFS:
is true when we know that p is false.
TRIVIAL PROOFS:
is true when we know that q is true.
PROOFS OF EQUIVALENCE:
Exercise:
1.16.10 Use a direct proof to show that the product of two rational numbers is rational.
Solution:
First number x = a/b, b ≠0
Second number y =c/d, d≠0
X*y = a*c/b*d sense b ≠0 and d≠0 then b*d≠0.
Let a*c = l and b*d = m so x = l/m so x is rational number.
1.16.11 Prove or disprove that the product of two irrational numbers is irrational.
Solution:
To disprove it we need only one case doesn`t match the law.
Now let X = √2, irrational number.
X*X = 2 is rational number. (Product of two irrational numbers is not an irrational number).
1.6.16 Prove that if m and n are integers and m n is even, then m is even or n is even.
Solution:
Let
p: m*n is event.
H: m is event.
L: n is event.
(m ᴠ n)
Using prove by Contraposition, not(m ᴠ
not p
So: m is odd and n is odd
m= 2k +1
n= 2l +1
m*n = (2k +1)*(2l+1) = 4kl + 2k + 2l + 1 = 2(2kl +k + l) +1, let (2kl +k + l) = v
so m*n = 2v +1 so m*n is odd
by Contraposition if m*n is even, then m is even or n is even.
1.6.27 Prove that if n is a positive integer, then n is odd if and only if 5n + 6 is odd.
We must prove two implications.
First, we assume that n is odd and show that 5n + 6 is odd.
By assumption, n = 2k + 1 so 2n +6 = 5(2k +1) =6 = 10k + 11 = 10k +10 +1 = 2(5k +5) +1 = 5V +1
where v = (5k +5), so 5n +6 is odd.
Second, we have to show that if 5n + 6 is odd then n is odd.
Using indirect proves:
Let n is event so n = 2k
5n +6 = 10k +6 = 2(5k +3) so it is event so by contraposition if 5n + 6 is odd then n is odd.
From the two cases we found that n is odd if and only if 5n + 6 is odd.
External question: Prove that if x and y are real numbers, then max(x, y) + min(x, y) = x + y.
Solution:
Like this question we cane analysis it to its cases.
Max(x, y) has two cases if x is the max or y is max
Min(x, y) has two cases if y is the min or x is min
Case 1:
Let x is the max so y is the min so max(x, y) + min(x, y) = x +y.
Case 2:
Let y is the max so x is the min so max(x, y) + min(x, y) = y +x = x +y.
So in all cases max(x, y) + min(x, y) = x +y.