Question

Show that any positive odd intiger is of form 4m+1 (or) 4m+3 where m is some integer? ​

Answer 1

Answer:

Step-by-step explanation:

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.

Answer 2

Answer:

4(15)+3

60+3

63

4(9)+1

36+1

37

hence proved