Question

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

Answer 1

Hint: In this problem, we have used Euclid division lemma, Euclid division lemma says that if $ a $ and $ b $ are $ 2 $ positive integers, then $ a $ is equal to sum of the product of $ b $ and $ q $ and $ r $ .

Complete step-by-step answer:
Given that the number is a positive odd number.
Now, we use the Euclid division lemma which says that one number can be written in terms of another number with some remainder terms in it.
The remainder term will be always less than the other number.
That is take $ a $ and $ b $ are two positive integers then,
 $ a = bq + r $ where $ 0 \le r < b $
Let us take a positive integer to be $ a $ and take $ b $ as four. Hence,
  $ a = 4q + r $
Where, the remainder is in between $ 0 $ to $ 3 $ .
Since, we know that $ r $ is an integer greater than or equal to $ 0 $ and less than $ 4 $ .
Hence, the possibilities of $ r $ is either $ 0 $ , $ 1 $ , $ 2 $ or $ 3 $ .
Suppose if the value of r is $ r = 1 $ .
Then the equation will become,
 $ a = 4q + r $
 $ a = 4q + 1 $
Since, we know that $ 4q $ is always an even number if we add one to that even number then the number becomes odd.
So this implies that $ 4q + 1 $ is always an odd integer.
Suppose if $ r = 3 $ , then the equation will become,
 $ a = 4q + r $
 $ a = 4q + 3 $
Since, we know that $ 4q $ is always an even number if we add three to that even number then the number becomes odd.
So this implies that $ 4q + 1 $ is always an odd integer.
Therefore, any odd integer is of the form $ 4q + 1 $ or $ 4q + 3 $ .

Note: We can prove this theorem using contradiction that it is assumed that $ 4q + 1 $ is even so the number can be written as $ 2k $ where $ k $ is integer. On rearranging we will get $ q $ as rational but we have assumed $ q $ to be integer this implies that $ 4q + 1 $ is odd.