Q.1: 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
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.
Step-by-step explanation:
Let a be any positive integer.
Let q be the quotient and r be remainder.
Then a = bq + r where q and r are also positive integers and 0 ≤ r < b
Taking b = 3, we get
a = 3q + r; where 0 ≤ r < 3
When, r = 0 = ⇒ a = 3q
When, r = 1 = ⇒ a = 3q + 1
When, r = 2 = ⇒ a = 3q + 2
Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m.
⇒ Squares of 3q = (3q)2
= 9q^2
= 3(3q)^2
= 3 m where m is some integer.
Square of 3q + 1 = (3q + 1)^2
= 9q2 + 6q + 1 = 3(3q2 + 2q) + 1
= 3m +1, where m is some integer
Square of 3q + 2 = (3q + 2)^2
= (3q + 2)2
= 9q^2 + 12q + 4
= 9q^2 + 12q + 3 + 1
= 3(3q^2 + 4q + 1) + 1
= 3m + 1 for some integer m.
∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
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