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 the positive integer-a
b-3
If b - 3 so
0</= R < b
R =0, 1,2
q= some integer
By using division algorithm
let r =0
a=bq+r
a =3q + 0
a=3q
squaring both side
a^ =(3q) ^
a^= 9q^
a^=3(3q^)
let P= 3q^
a^ = 3P ____________1
Let r = 1
a=bq+ r
a= 3q + 1
squaring both side
a^= (3q+1) ^
a^=(3q)^ + (1) ^ + 2 ×3q×1
a^ =9q^ +1+6q
a^= 9q^+6q+1
a^=3(3q^+2q) +1
let 3q^+2q = P
a^=3p+1________________2
From eq 1 and 2 we can say that
the square of any positive integer is in the form of 3q and 3q+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.