Question

Show that every positive even integer is of the form 4m and that every positive odd integer is of the form 4m + 1 for some integer m.

Answer 1

Let n be any arbitrary positive odd integer.
On dividing n by 4, let m be the quotient and r be the remainder. So, by Euclid’s division lemma, we have
n = 4m + r, where 0 ≤ r ˂ 4.

As 0 ≤ r ˂ 4 and r is an integer, r can take values 0, 1, 2, 3.
⇒ n = 4m or n = 4m + 1 or n = 4m + 2 or n = 4m + 3
But n ≠ 4m or n ≠ 4m + 2 ( ∵ 4m, 4m + 2 are multiples of 2, so an even integer whereas n is an odd integer)
⇒ n = 4m + 1 or n = 4m + 3
Thus, any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.