Question

prove that any positive odd integer is form of 4m or 4m+1​

Answer 1

Let, m be 2

So, 4m = 4×(2) = 8 which is a even number

4m+1 = 4×(2)+1 = 9 which is a odd number..

I hope this will help you...

Plz mark me as brainlist...✌

Answer 2

Step-by-step explanation:

Let a be the positive integer.

And, b = 4 .

Then by Euclid's division lemma,

We can write a = 4q + r ,for some integer q and 0 ≤ r < 4 .

°•° Then, possible values of r is 0, 1, 2, 3 .

Taking r = 0 .

a = 4q .

Taking r = 1 .

a = 4q + 1 .

Taking r = 2

a = 4q + 2 .

Taking r = 3 .

a = 4q + 3 .

But a is an odd positive integer, so a can't be 4q , or 4q + 2 [ As these are even ] .

•°• a can be of the form 4q + 1 or 4q + 3 for some integer q .

Hence , it is solved .