show that the square of any positive interger is of the from 3m or 3m+1 for some integer m
Answers
Answer:Use Euclid division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 for some integer m. Let a be any positive integer and b = 3. =) a = 3q + r, r = 0 or 1 or 2.
Explanation:
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,
a²=9q²
a²=3(3q²)
a²=3m
m= 3q²
a = 3q + 1
On squaring both the sides ,
a²= (3q+1)²=9q²+6q+1
a²=3(3q²+2q)+1
a²=3m+1
m=3q²+2q
When, a = 3q + 2
On squaring both the sides,
a²=(3q+2)²=9q²+4+12q
a²=3(3q²+4q+1)+1 (4=3+1)
a²=3m+1
m=3q²+4q+1
Therefore , the square of any positive integer is either of the form 3m or 3m+1
Explanation: