show that any positive integer is of the form 3q or 3q + 1 or 3q + 2
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Answered by
6
Let x be the integer
x=3q
x²=9q²
x²=3(3q²)
x²=3q [let 3q² be q]
x=3q+1
x²=(3q+1)²
x²=9q²+6q+1
x²=3(3q²+2q)+1
x²=(3q+1) [let 3q²+2q be q]
x=3q+2
x²=(3q+2)²
x²=9q²+12q+4
x²=3(3q²+4q+1)+1
x²=(3q+1) [3q²+4q+1 be q]
it proves q² is in the form of 3q and (3q+1) but not in the form of (3q+2)
May this helful
x=3q
x²=9q²
x²=3(3q²)
x²=3q [let 3q² be q]
x=3q+1
x²=(3q+1)²
x²=9q²+6q+1
x²=3(3q²+2q)+1
x²=(3q+1) [let 3q²+2q be q]
x=3q+2
x²=(3q+2)²
x²=9q²+12q+4
x²=3(3q²+4q+1)+1
x²=(3q+1) [3q²+4q+1 be q]
it proves q² is in the form of 3q and (3q+1) but not in the form of (3q+2)
May this helful
himanshu222133patzks:
In my school they showed the same method but I am no able to understand the logic behind this thing....can any one tell me
Answered by
5
let a be any positive integer
then
b= 3
a= bq+r
0≤r<b
0≤r<3
r= 0,1,2
case 1.
r=0
a= bq+r
3q+0
3q
case 2.
r=1
a= 3q+1
3q+1
case3.
r=2
a=3q+2
hence from above it is proved that any positive integer is of the form 3q,3q+1 and 3q+2
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