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Math Help - Show that all non integral solutions lie on exactly two lines

  1. #1
    Super Member fardeen_gen's Avatar
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    Show that all non integral solutions lie on exactly two lines

    Consider the equation \lfloor x \rfloor \lfloor y \rfloor = x + y.

    Show that all non-integral solutions of this equation lie on exactly two lines.
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    Consider the equation \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.)
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  3. #3
    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    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 x=a+r,\ y=b+s where 0<r,s<1 and r+s=1?
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  4. #4
    Senior Member pankaj's Avatar
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    x+y=0 and x+y=6
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    Lord of certain Rings
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    Quote Originally Posted by TheAbstractionist View Post
    Hi HallsofIvy.

    What about x=a+r,\ y=b+s where 0<r,s<1 and r+s=1?
    Isnt that the same thing as HallsofIvy's suggestion? r+s = 1 \implies y = b+(1-r) = (b+1)-r, where b+1 is just another integer.
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  6. #6
    Senior Member pankaj's Avatar
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    [x][y]=x+y

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

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

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

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

    1 \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 x and y will be integers

    Case 2

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

     <br />
[x]-1=-1 and [y]-1=-2<br />

    [x]=0 and [y]=-1

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

    Case 3

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

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

    [x]=0 and [y]=0

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

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

    Case 4

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

    [x]-1=1 and [y]-1=2

    [x]=2 and [y]=3

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

    Therefore, x+y=6
    Last edited by pankaj; May 21st 2009 at 06:35 PM.
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