use division algorithm to show that the square of any positive integer is of the form 3p 3p+1
Answers
Answer:
Let: 'a' be the positive integer
⇒ Euclid's Algorithm: a = bq + r [0<r<b]
a = 3q + r - (1) [0<r<3]
= Let: r = 0
∴ By Squaring on both sides, we get
= (a)² = (3q)²
= (a)² = 9q²
= (a)² = 3 x 3q²
= a² = 3p
= Let: r = 1
= a = 3q + 1
∴ By Squaring on both sides, we get
= (a)² = (3q + 1)²
= (a)² = (3q)² + 1 + 2 x (3q) x 1
= (a)² = 3q² + 6q + 1
= a² = 3p + 1
Let: r = 2
∴ By Squaring on both sides, we get
= (a²) = (3q + 2)²
= (a²) = 9q² + 4 + (2 x 3q x 2)
= (a²) = 9q² + 12q + 3 + 1
= (a²) = 3 (3q² + 4q + 1)
= (a²) = 3p + 1
∴ Square of any positive interger will be of the form of 3p (or) 3p + 1 where 'P' is an Integer (∈) 'I'.
Answer:
Let us consider a positive integer a
Divide the positive integer a by 3, and let r be the reminder and b be the quotient such that
a = 3b + r……………………………(1)
where r = 0,1,2,3…..
Case 1: Consider r = 0
Equation (1) becomes
a = 3b
On squaring both the side
a2 = (3b)2
a2 = 9b2
a2 = 3 × 3b2
a2 = 3m
Where m = 3b2
Case 2: Let r = 1
Equation (1) becomes
a = 3b + 1
Squaring on both the side we get
a2 = (3b + 1)2
a2 = (3b)2 + 1 + 2 × (3b) × 1
a2 = 9b2 + 6b + 1
a2 = 3(3b2 + 2b) + 1
a2 = 3m + 1
Where m = 3b2 + 2b
Case 3: Let r = 2
Equation (1) becomes
a = 3b + 2
Squaring on both the sides we get
a2 = (3b + 2)2
a2 = 9b2 + 4 + (2 × 3b × 2)
a2 = 9b2 + 12b + 3 + 1
a2 = 3(3b2 + 4b + 1) + 1
a2 = 3m + 1
where m = 3b2 + 4b + 1
∴ square of any positive integer is of the form 3m or 3m+1.
Hence proved.