4. 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.
[Hint: Let x be any positive integer then it is of the form 39, 3q+1 or 3q + 2. Now square
each of these and show that they can be rewritten in the form 3m or 3m +1.]
Answers
Given: Any positive integer
To prove: Any positive integer is in the form of 3q,3q+1,3q+2
Proof:
We know that from Euclid’s division lemma for b= 3
Let us assume that any positive integer ‘n’ be of the form 3q or, 3q+1 or 3q+2.
If n= 3q,
On squaring we get,
⇒ n2= (3q)2 = 9q2
⇒ n2= 3(3q2)
⇒ n2= 3m, where m is some integer [m = 3q2]
If n= 3q+1,
On squaring we get,
⇒ n2= (3q+1)2 = 9q2 + 6q + 1 { Solved using the identity (a+b) 2 = a2 + b2 + 2ab}
⇒ n2= 3(3q2 +2q) + 1
⇒ n2= 3m + 1, where m is some integer [m = 3q2 +2q]
If n= 3q+2,
On squaring we get,
⇒ n2= (3q+2)2 = 9q2 + 12q + 4 { Solved using the identity (a+b) 2 = a2 + b2 + 2ab}
⇒ n2= 3(3q2 + 4q + 1) + 1
⇒ n2= 3m, where m is some integer [m = 3q2 + 4q + 1]
Therefore, the square of any positive integer is of the form 3m or 3m + 1 but not of the form 3m + 2.
Hence proved
Answer:
Let 'a' be any positive integer.
On dividing it by 3 , let 'q' be the quotient and 'r' be the remainder.
such that ,
a = 3q + r , where r = 0 ,1 , 2
When, r = 0
∴ a = 3q
When, r = 1
∴ a = 3q + 1
When, r = 2
∴ a = 3q + 2
When , a = 3q
On squaring both the sides,
When, a = 3q + 1
On squaring both the sides ,
When, a = 3q + 2
On squaring both the sides,
Therefore , the square of any positive integer is either of the form 3m or 3m+1.
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