Numbers 8 - Solutions for Even and Odd Integers, Divisibility, Primes and Composites

Numbers 1
COMPSCI 120, S2, 2019
Name:
Course:
Tutor:
College:
Date:
1. Even and odd integers

Numbers 2
a)
Solution
Suppose x is the even integer and y is the odd integer
The definition of odd and even integers is such that:
x=2 a∧ y=2 b+1
Now, y−x =2b +1−2 a=2b−2 a+1=2 ( b−a ) + 1
Let b−a=s
Therefore, y−x =2 s+1 which is odd
b)
Solution
Suppose x is an even integer and y is an odd integer.
The definition of odd and even integers is such that:
x=2 a∧ y=2 b+1
Now, xy= ( 2 a ) ( 2 b+1 ) =4 ab+2 a=2(2 ab+a)
Let 2 ab+ a=s
Therefore, xy=2 s which is even
c)
Solution
To show that a2 +b2 +1
2

Numbers 3
The necessary conditions are:
For a2 +b2 +1
2 to be an integer, the numerator must be an even integer.
For the numerator to be an even integer, a2+ b2must be an odd integer.
Now, for a2+ b2 to be an odd integer, both a and b cannot be even since this
would give an even integer since the product of any two even integers is even.
With either a or b being odd a2+ b2 will be odd since this will be the sum of an odd
and an even integer which is odd.
Therefore, a2 +b2 +1
2 is an integer for a being an even integer and b being an odd
integer.
2. Divisibility
a)
Solution
If a and b are two integers with a∨b∧b∨a
By definition, a∨b if there exists an integer m such that b=ma
And, b∨a if there exists an integer m such that a=mb
This is only true if ma=mb, ¿ which a=b
b)
Solution