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?
uart
Sep19-08, 11:41 AM
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 f(-1,y) = -1[/tex] for all y, and similarly that [itex]f(x,-1) = -1 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 x=x_0. We get:
f(x_0,y) = x_0 + (x_0 + 1) y,
a simple linear function of y with non zero gradient (as x_0 \neq -1)
Since f=-1 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.
Dodo
Sep19-08, 01:22 PM
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.
JG89
Sep19-08, 01:41 PM
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
sutupidmath
Sep19-08, 02:06 PM
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
What about a=2, b=-1, so
2+(-1)+(-1)(2)=2-1-2=-1?
statdad
Sep19-08, 02:26 PM
if the requirement is that neither the numbers , a, b can equal -1 , then
a + b + ab = -1
does not have any solutions, as the factorization of a + b + ab + 1 shows.
However, if either a or b can be -1 , you have infinitely many solutions. (If we choose b = -1 , then for any a
a + (-1) + a(-1) = -1
sutupidmath
Sep19-08, 02:46 PM
if the requirement is that neither the numbers , a, b can equal -1 , then
a + b + ab = -1
does not have any solutions, as the factorization of a + b + ab + 1 shows.
However, if either a or b can be -1 , you have infinitely many solutions. (If we choose b = -1 , then for any a