. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Answers
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.
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Answer:
Let x be the positive integer, then it is of the form 3q or 3q + 1 or 3q + 2.
Squaring, we get
(3q)² = 9q² = 3 × 3q² = 3m, m = 3q²
(3q + 1)²= 9q² + 6q + 1
= 3(3q² + 2q) +1
= 3m + 1,m = 3q² + 2q
(3q + 2)² = 9q² + 12q + 4
= 9q² + 12q + 3 + 1
= 3(3q² + 4q² + 1) + 1
= 3m + 1,m = 3q² + 4q + 1
››Square of any positive integer is of the form 3m or 3m + 1 for some integer m