Question

show that any positive odd integer is of the form 3m or 3m+1 or 3m+2 where ​

Answer 1

Step-by-step explanation:

Proof:

We know that from Euclid’s division lemma for b= 3

Let us assume that any positive integer ‘n’ be of the form 3q or, 3q+1 or 3q+2.

If n= 3q,

On squaring we get,

⇒ n2= (3q)2 = 9q2

⇒ n2= 3(3q2)

⇒ n2= 3m, where m is some integer [m = 3q2]

If n= 3q+1,

On squaring we get,

⇒ n2= (3q+1)2 = 9q2 + 6q + 1 { Solved using the identity (a+b) 2 = a2 + b2 + 2ab}

⇒ n2= 3(3q2 +2q) + 1

⇒ n2= 3m + 1, where m is some integer [m = 3q2 +2q]

If n= 3q+2,

On squaring we get,

⇒ n2= (3q+2)2 = 9q2 + 12q + 4 { Solved using the identity (a+b) 2 = a2 + b2 + 2ab}

⇒ n2= 3(3q2 + 4q + 1) + 1

⇒ n2= 3m, where m is some integer [m = 3q2 + 4q + 1]

Therefore, the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.

Hence proved

Answer 2

Step-by-step explanation:

this is write down the answer ok

best of luck