Show that every positive even integer is of the form 4m and that every

1.	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» 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.

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»

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.