Show that any positive odd integer is of the form 6q + 1 or, 6q + 3 or, 6q + 5, where q is some integer.
Solution
Let a be any odd positive integer we need to prove that a is of the form 6q + 1, or 6q +3, 6q+ 5, where q is some integer
Since a is an integer consider b = 6 another integer applying Euclid’s division lemma we get
a = 6q + r for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 since 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q + 2 and 6q + 4
(since all these are divisible by 2)
Also, 6q + 1 = 2 × 3q + 1 = 2k 1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2(3q + 2) + 1 = 2k3 + 1, where k3 is an integer
Clearly, 6q +1, 6q +3, 6q + 5 are of the form 2k + 1, where k is an integer
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.
Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, 6q + 5 where q is some integer
Concept insight: In order to solve such problems Euclid’s division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must be of the form 6q + 1, 6q +3, 6q + 5
Basic definition of even (divisible by 2) and odd numbers (not divisible by 2) and the fact that addiction and multiplication of integers is always an integer are applicable here.
