Question

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

Answer 1

Answer!

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

Answer:

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.