use edl to show that the square of any positive integer either of the form 3m or 3m+1 for some integer m.
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Hii friend,
Let n be an arbitrary positive integer.
On dividing n by 3 , let Q be the Quotient and r be the Remainder.
Then, by Euclid's division lemma , we have
n = 3q + r , where r is equal to 0 but less than 3.
Therefore,
n² = (3q+r)²
n² = 9q² + r² + 2 × 3q × r
n² = 9q²+r²+6qr.,.....(1)
In case (1) . When r = 0
Putting the Value of r in (1), we get:
n² = 3q² = 3(3q)² = 3m , where m = 3q² is an integer.
Case (2) , when r = 1.
Putting r = 1 in (1) we get:
n² = (9q²+1+6q) = (9q²+6q) +1 = 3(3q²+2q) +1 = 3m +1 , where m = (3q²+2q) is an integer.
Case (3) when r = 2
n² = (9q²+4+12q) =3(3q²+4q+1)+1 = 3m +1 , where m = (3q²+4q+1) is an integer.
Hence,
The square of any positive integer is in the form of 3m or (3m+1) for some integer n.
HOPE IT WILL HELP YOU.... :-)
Let n be an arbitrary positive integer.
On dividing n by 3 , let Q be the Quotient and r be the Remainder.
Then, by Euclid's division lemma , we have
n = 3q + r , where r is equal to 0 but less than 3.
Therefore,
n² = (3q+r)²
n² = 9q² + r² + 2 × 3q × r
n² = 9q²+r²+6qr.,.....(1)
In case (1) . When r = 0
Putting the Value of r in (1), we get:
n² = 3q² = 3(3q)² = 3m , where m = 3q² is an integer.
Case (2) , when r = 1.
Putting r = 1 in (1) we get:
n² = (9q²+1+6q) = (9q²+6q) +1 = 3(3q²+2q) +1 = 3m +1 , where m = (3q²+2q) is an integer.
Case (3) when r = 2
n² = (9q²+4+12q) =3(3q²+4q+1)+1 = 3m +1 , where m = (3q²+4q+1) is an integer.
Hence,
The square of any positive integer is in the form of 3m or (3m+1) for some integer n.
HOPE IT WILL HELP YOU.... :-)
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