Show that any positive odd integer is of the form (4m + 1) or (4m + 3)

1.	Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.
Answer» Let s be any positive integer. On dividing s by 4, let m be the quotient and r be the remainder. By Euclid’s division lemma, s = 4m + r, where 0 ≤ r ˂ 4 So we have, s = 4m or s = 4m + 1 or s = 4m + 2 or s = 4m + 3. Here, 4m, 4m + 2 are multiples of 2, which revert even values to s. Again, s = 4m + 1 or s = 4m + 3 are odd values of s. Thus, any positive odd integer is of the form (4m + 1) or (4m + 3) where s is any odd integer.

Show that any positive odd integer is of the form (4m + 1) or (4m + 3), where m is some integer.

Answer»

Let s be any positive integer.

On dividing s by 4, let m be the quotient and r be the remainder.

By Euclid’s division lemma,

s = 4m + r, where 0 ≤ r ˂ 4

So we have, s = 4m or s = 4m + 1 or s = 4m + 2 or s = 4m + 3.

Here, 4m, 4m + 2 are multiples of 2, which revert even values to s.

Again, s = 4m + 1 or s = 4m + 3 are odd values of s.

Thus, any positive odd integer is of the form (4m + 1) or (4m + 3) where s is any odd integer.