Question

show that any positive odd integer is of the form (4 m + 1)or (4 m + 3) when m is some integer​

Answer 1

Let a be any odd positive integer and b = 4.

By division lemma there exist integer q and r such that

a = 4 q + r, where 0 ≤ r ≤ 4

so a = 4q or, a = 4q + 1 or, a = 4q + 2 or, a = 4q + 3

4q + 1 → 4 is divisible by 2 but 1 is not divisible by 2,

so it is an odd number.

4q + 2 → 4 is divisible by 2 and 2 is also divisible by 2,

so it is an even number.

4q + 3 → 4 is divisible by 2 but 3 is not divisible by 2,

so it is an odd number.

4q + 4 → 4 is divisible by 2 and 4 is also divisible by 2,

so it is an even number.

∴ any odd integer is of the form 4q + 1 or, 4q + 3.

Answer 2

Step-by-step explanation:

4m+1

m=1, 4*1+1=5

m=2, 4*2+1=9

4m+3

m=1, 4*1+3=7

m=2, 4*2+3=11