Use division algorithm to show that the square of any positive integer is of the form 3p or 3p + 1
Answers
Step-by-step explanation:
Let a=3q {where q is an integer)
a=bq+r where r =(0,1,2)
a=3q +0
◆squarw both side
a^2=(3q)^2
◆a^2=9q^2
◆a^2=3(3q^2)
◆Where p=3q^2
◆r=1
◆a=3q+1
◆square both side
◆a^2=(3q+1)^2
◆a^2=9q^2+6q+1 ( By identity)
◆a^2= 3(3q^2+2q)+1
◆Where p= 3q^2+2q
------> So square of any positive interger is in form of 3p , 3p+1
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.