1. Sep 19, 2008

### sutupidmath

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. Sep 19, 2008

### uart

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.

Last edited: Sep 19, 2008
3. Sep 19, 2008

### dodo

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.

4. Sep 19, 2008

### JG89

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

5. Sep 19, 2008

### sutupidmath

2+(-1)+(-1)(2)=2-1-2=-1?

6. Sep 19, 2008

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 $$a$$

$$a + (-1) + a(-1) = -1$$

7. Sep 19, 2008

### sutupidmath

GOT IT!

I feel dumb now!...lol....