Real Numbers — Maths STD 10 — Question

Let 'a' be a given positive integer.
On dividing 'a' by 2, let q be the quotient and r be the remainder.
Then, by Euclid's algorithm, we have
a = 2q + r, where 0 ≤ r < 2
⇒ a = 2q + r, where r = 0, 1
⇒ a = 2q or a = 2q + 1
When a = 2q for some integer q, then clearly 'a' is even.
When a = 2q + 1 for some integer q, then clearly 'a' is odd.
Thus, every positive integer is either even or odd.