4) Show that cube of any positive integer as of theform 3m or 3m+1 or 3mt8 for some integer m About the author Daisy
Answer: [tex]\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}[/tex] 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. error its square not cube ☆ I HOPE ITS HELP YOU ☆ Reply
Answer:
[tex]\huge\pink{\fbox{\tt{࿐αɴѕωєя࿐}}}[/tex]
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.
error its square not cube
☆ I HOPE ITS HELP YOU ☆
cube of
any positive integer is either of form 9q ,9q+1 or 9q+8 for some integer Q