# Show that all non integral solutions lie on exactly two lines

• May 20th 2009, 09:08 PM
fardeen_gen
Show that all non integral solutions lie on exactly two lines
Consider the equation $\displaystyle \lfloor x \rfloor \lfloor y \rfloor = x + y$.

Show that all non-integral solutions of this equation lie on exactly two lines.
• May 21st 2009, 04:17 AM
HallsofIvy
Quote:

Originally Posted by fardeen_gen
Consider the equation $\displaystyle \lfloor x \rfloor \lfloor y \rfloor = x + y$.

Show that all non-integral solutions of this equation lie on exactly two lines.

The left hand side is an integer whether x and y are or not. If x and y are not integer, then the "fraction parts" must cancel. Either x= a+ r, y= b- r or x= a- r, y= b+ r, where a and b are integers, 0< r< 1. Those give the two lines. (And, of course, ab= a+b.)
• May 21st 2009, 05:47 AM
TheAbstractionist
Quote:

Originally Posted by HallsofIvy
Either x= a+ r, y= b- r or x= a- r, y= b+ r, where a and b are integers, 0< r< 1.

Hi HallsofIvy.

What about $\displaystyle x=a+r,\ y=b+s$ where $\displaystyle 0<r,s<1$ and $\displaystyle r+s=1?$ (Mmm)
• May 21st 2009, 05:52 AM
pankaj
$\displaystyle x+y=0$ and $\displaystyle x+y=6$
• May 21st 2009, 06:01 AM
Isomorphism
Quote:

Originally Posted by TheAbstractionist
Hi HallsofIvy.

What about $\displaystyle x=a+r,\ y=b+s$ where $\displaystyle 0<r,s<1$ and $\displaystyle r+s=1?$ (Mmm)

Isnt that the same thing as HallsofIvy's suggestion? $\displaystyle r+s = 1 \implies y = b+(1-r) = (b+1)-r$, where b+1 is just another integer.
• May 21st 2009, 11:31 AM
pankaj
[x][y]=x+y

[x][y]=[x]+[y]+{x}+{y} ({x} and {y} denote the fractional parts of x and y respectively)

0$\displaystyle \leq${x}+{y}<2

0$\displaystyle \leq$([x][y]-[x]-[y])<2

1$\displaystyle \leq$(1+[x][y]-[x]-[y])<3

1$\displaystyle \leq$([x]-1)([y]-1)<3

([x]-1)([y]-1)=1,2

Case 1

([x]-1)([y]-1)=1

[x]-1=1 and [y]-1=1

[x]=[y]=2

{x}+{y}=(2)(2)-2-2=0

{x}={y}=0,but then $\displaystyle x$ and $\displaystyle y$ will be integers

Case 2

$\displaystyle ([x]-1)([y]-1)=2$

$\displaystyle [x]-1=-1 and [y]-1=-2$

$\displaystyle [x]=0 and [y]=-1$

{x}+{y}=0-0-(-1)=1 and thus $\displaystyle x+y=0$

Case 3

$\displaystyle ([x]-1)([y]-1)=1$

$\displaystyle [x]-1=-1 and [y]-1=-1$

$\displaystyle [x]=0 and [y]=0$

{x}+{y}=0-0-0=0

{x}={y}=0,but then $\displaystyle x$ and $\displaystyle y$ are integers

Case 4

$\displaystyle ([x]-1)([y]-1)=2$

$\displaystyle [x]-1=1 and [y]-1=2$

$\displaystyle [x]=2 and [y]=3$

{x}+{y}=6-2-3=1

Therefore,$\displaystyle x+y=6$