Question

Prove that any positive odd integer is of the form 4n+1 4n+3

Answer 1

Answer :

Step-by-step explanation:

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

Using Euclid Division Lemma

a = bq + r            [ 0 ≤ r < b ]

⇒ a = 4q + r       [ 0 ≤ r < 4 ]

Now, possible value of r :

r = 0, r = 1, r = 2, r = 3,

CASE I :

If we take, r = 0

⇒ a = 4q + 0

⇒ a = 4q

⇒ a = 2 ( 2q )

⇒ a = 2n.      [ n = 2q as an integer ]

So, it is rejected because it is an even number.

CASE II :

If we take, r = 1

⇒ a = 4q + 1

⇒ a = 4n + 1      [ n = q as an integer ]

As it cannot be expressed in 2n form, it is an odd number which is in the form 4n + 1.

CASE III :

If we take, r = 2

⇒ a = 4q + 2

⇒ a = 4n + 2

⇒ a = 2 (2n + 1 )   [ n = q as an integer ]

As it can be expressed in (2n + 1)form, it is an even number.

CASE IV :

If we take, r = 3

⇒ a = 4q + 3

⇒ a = 4n + 3   [ n = q as an integer ]

As it cannot be expressed in 2n form, it is an odd number which is in the form 4n + 3.

Hence, it is proved that any positive odd integer is of the form 4n+1 and 4n+3.

Answer 2

Step-by-step explanation:

Let a be any positive odd integer and b=4

So by applying division lemma to a,b

a=4q+r where 0≤r<4

When r=0

a=4q+0=4q=2(2q)

When r=1

a=4q+1

When r=2

a=4q+2=2(2q+1)

When r=3

a=4q+3

Since 4q,4q+2 are multiple of 2 so they are even numbers

So every positive odd integer is of the form (4q+1) or (4q+3) where q is some integer

mark brainliest