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Prove or contradict? Help?

  1. Sep 19, 2008 #1
    Let a and b be to real numbers different from -1. Then show that the following is possible by finding values of a and b, or prove that it is impossible?

    a+b+ab=-1

    ?

    I have no clue how to do this one?
     
  2. jcsd
  3. Sep 19, 2008 #2

    uart

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    Science Advisor

    Here's a very ugly proof (that there are no solutions apart from those that contain -1 as one of the coordinates).

    Just consider the LHS as a function of two variables and we're trying to determine points where this function is equal to -1.

    f(x,y) = x+y+xy

    It easy to show that [itex]f(-1,y) = -1[/tex] for all y, and similarly that [itex]f(x,-1) = -1[/itex] for all x. We want to determine if the function is equal to -1 at any points apart from along those two lines.

    Consider a slice of the function at [itex]x=x_0[/itex]. We get:

    [tex]f(x_0,y) = x_0 + (x_0 + 1) y[/tex],

    a simple linear function of y with non zero gradient (as [itex]x_0 \neq -1[/itex])

    Since [itex]f=-1[/itex] when y=-1 and gradient is non-zero then [itex]f(y)[/tex] can not be equal to zero for any other value of y.
     
    Last edited: Sep 19, 2008
  4. Sep 19, 2008 #3
    Given
    a + b + ab + 1 = 0​
    you can factor a, to get
    a(b + 1) + b + 1 = 0​
    and now factoring b+1,
    (a + 1)(b + 1) = 0​
    Thus one of the factors at the left must be zero.
     
  5. Sep 19, 2008 #4
    a + b + ab = -1
    b + ab = -1 - a
    b(1 + a) = -(1 + a)
    b = -1

    a + ab = -1 - b
    a(1 + b) = -(1+b)
    a = -1

    So the only solutions are a = b = -1
     
  6. Sep 19, 2008 #5
    What about a=2, b=-1, so

    2+(-1)+(-1)(2)=2-1-2=-1?
     
  7. Sep 19, 2008 #6

    statdad

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    Homework Helper

    if the requirement is that neither the numbers [tex], a, b [/tex] can equal [tex] -1 [/tex], then

    [tex]
    a + b + ab = -1
    [/tex]

    does not have any solutions, as the factorization of [tex] a + b + ab + 1 [/tex] shows.

    However, if either [tex] a [/tex] or [tex] b [/tex] can be [tex] -1 [/tex], you have infinitely many solutions. (If we choose [tex] b = -1 [/tex], then for any [tex] a [/tex]

    [tex]
    a + (-1) + a(-1) = -1
    [/tex]
     
  8. Sep 19, 2008 #7
    GOT IT!

    I feel dumb now!...lol....
     
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