Download Chapter 1: The Foundations: Logic and Proofs Section 1.7

Discreet Mathematics
Chapter 1:
The Foundations: Logic and Proofs
Section 1.7:
Introduction to Proofs
Q.2.
Use a direct proof to show that the sum of two even integers is even.
Solution:
Assume n, m even numbers.
So we can conclude there are a, b where:
n=(2×a) 1
m=(2×b)  2
from 1 and 2 n+m=2×a + 2×b=2(a+b) which is even number.
Q.18.
Prove that if n is an integer and 3n + 2 is even, then n is even using
a) a proof by contraposition.
b) a proof by contradiction.
Solution:
1. Proof by contraposition:
We need to prove that if n is odd, then 3n + 2 is odd.
Assume that n is odd.
Then we can write n = 2k + 1 for some integer k.
Then 3n + 2 = 3(2k + 1) + 2 = 6k + 5 = 2(3k + 2) + 1. – odd number
Thus 3n + 2 is odd.
2. proof by contradiction.
We need to show that if 3n+2 is even and n is odd, then there is a contradiction.
Let n be an arbitrary integer.
Assume by way of contradiction that 3n + 2 is even and n is odd.
Then n = (2k +1) for some integer k
and so 3n+2 = 3(2k +1)+2 = 6k +5 = 2(3k + 2) + 1
So 3n + 2 is odd. So 3n + 2 is odd and even.
Contradiction!
Q.8.
Prove that if n is a perfect square, then n + 2 is not a perfect square.
Solution:
By contradiction.
Assume that both n and n + 2 are perfect square.
This means there exist nonnegative integers a, b such that:
n = a2
n+2 = b2.
Then 2 = (n+2)-n = b2 -a2 = (b-a)(b+a):
Therefore :
b+a = 2 and b-a = 1,
so 2b = 3 and b = 3/2;
contradiction with the fact that n is an integer.