Swapnil Saxena
Last Activity: 13 Years ago
The non integral solutions can be deduced through the following algorithm
Ttake x= Ix + Fx and y= Iy + Fy (I for Integral part and F for fractional part)
Then [x][y] = x+y
IxIy = Ix + Fx + Iy + Fy
==> IxIy = (Ix + Iy) + (Fx + Fy)
==> Iy( Ix-1 ) = (Ix) + (Fx + Fy)
==> Iy = ((Ix) + (Fx + Fy))/( Ix-1 )
For Iy to be integral, (Fx + Fy) should definitely be integral because (Ix-1) is itself integral
Since the only integral value that (Fx + Fy) can take = 1,except 0 (not possible as the solution are non integral and the fractional parts cannot take negative values or more than 0.9999)
Thus the equaion becomes ==> Iy = (Ix + 1)/( Ix-1 ) ==> Iy = 1+2/( Ix-1 )
Solving this equation using graphs(Hyperbolic Curves) ,
The only integral solution for Ix and Iy are (2,3) and (3,2) , (0,-1) , (-1 ,0) (Note: All these are solution of greatest integer function)
Using this result we can easily deduce that (2.99, 3.01), (2.98,3.02), (3.02, 2.98),(3.99, 2.01) are some of the solution of the equation.
In this way, the equation of the first line can be deduced, = x+y-6=0
Using this result we also deduced that (0.99,-0.99),(-0.99,0.99),(0.35,-0.35) are some of the solution of the equation.
In this way, the equation of the first line can be deduced, = x+y=0