Question

1. Show that any positive odd integer is of the form 4k+1 or 4k +3, where k is some integer.​

Answer 1

Answer:

By definition, a = 2q is even. Let q = 2k. Then a = 4k. So, a + 1 = 4k + 1 and a + 3 = 4k + 3.

a + 5 = 4k + 4 + 1 = 4(k + 1) + 1

a + 7 = 4k + 4 + 3 = 4(k + 1) + 3

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Let any positive integer be 'a' and dividing it by 4, we get

a = 4k + r where r = 0,1,2 and 3

1. When r = 0, then, a = 4k + 0 => a = 2(2k) ----------> even
2. When r = 1 ,then, a = 4k + 1 ------------------------> odd
3. When r = 2, then, a = 4k + 2 => a = 2(2k + 1)--------------> even
4. When r = 3 , then , a = 4k + 3 ------------------------------->odd

From above, it is proved.