Daily Graph 26-07-09

solution set is (0,0)
generalizing
n<=x<n+.5
n<=y<n+.5
2n<=x+y<2n+1
[x+y]=2n
{x}=x-n
{y}=y-n
{x}+{y}=x+y-2n
x+y -2n=2n
x+y=4n (condtion:n<=x<n+.5;n<=y<n+.5)
clearly it is possible only (0,0)

max value of {x}+{y} → 2 when {x}→1 AND {y}→1
min value of {x}+{y} =0 when x,y are integers.

CASE I: integral solutions

Clearly [x+y]=0
this implies that [y]+[x]=0
OR [x]= -[y]

So in integral solutions, the solutions are of the form (n,-n) where n is an integer

CASE II: non-integral solutions

LHS is an integer, So {x}+{y} can be 1 only (when it is zero, case is dealt above and it cant become 2)

So, {y} = 1-{x}

[x+y] = [x]+[y]+1 = {x}+{y}

i.e. [x]+[y] = 0

further as {x}+{y}=1

adding both x+y=1