Question

Show that any positive odd integer is of the form (6q + 1) or (6q + 3) or (6q + 5), where q is some integer.

Answer 1

Let a be any odd positive integer. We have to prove that a is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
Since a is an integer, consider b = 6 as another integer.
By 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, it cannot take the values 6q, 6q + 2 and 6q + 4 (since these are divisible by 2)
Also, 6q + 1 = 2 × 3q + 1 = 2k₁ + 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.
Hence, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.