# What is the name of this principle?

1. Oct 31, 2008

### natski

Hi all,

When solving a set of equations with n unknown parameters, you need at least n equations to do this, but perhaps more. What is the name of this principle?

Cheers,
Natski

2. Oct 31, 2008

### HallsofIvy

Staff Emeritus
The "principle" you state is not true. For example, the equation x2+ y2+ z2= 0, although only one equation in 3 unknown parameters has the unique solution x= y= z= 0 in the real numbers. If that is not what you meant, please explain more.

3. Oct 31, 2008

### natski

Ok, but you can't get all of the solutions with just that one equation. Perhaps a better problem would be to consider n non-linear equations.

4. Oct 31, 2008

### yuiop

They are sometimes known as simultaneous equations.

5. Oct 31, 2008

### natski

Yup, I know that. I want to know the name of principle which dictates how many equations you need in order to find the solutions to n parameters.

6. Oct 31, 2008

### nasu

Who said the solutions must be real?
1,1 and i*sqr(2) is also a solutions. And many other sets.

Last edited: Oct 31, 2008
7. Oct 31, 2008

### nasu

The "principle" refers usually to linear equations. And n equations must be linear independent, if you want to get n unknowns.
I don't think I ever seen this called a principle and even less given some specific name.

8. Oct 31, 2008

### Bill_B

According to PlanetMath, the rank-nullity theorem says that "the number of variables minus the number of independent linear constraints equals the number of linearly independent solutions."

I don't think this is exactly what you're asking for, but it's as close as I could get.

9. Oct 31, 2008

### Office_Shredder

Staff Emeritus
Well hell, who says the solutions must be complex? (1,1,1) is a solution in Z3

10. Oct 31, 2008

### HallsofIvy

Staff Emeritus
I did. And since the original question just referred to 'a system of equations' without any restrictions, I am free to state a problem dealing with any system of equations in whatever number system I choose as a counter-example.

Last edited: Oct 31, 2008
11. Oct 31, 2008

### HallsofIvy

Staff Emeritus
Are you suggesting that there are solutions to x2+ y2+ z2= 0 other than x= y= z= 0? If so, tell me what they are! If not, I just gave you an example of 1 non-linear equation which, by itself, determines 3 solutions.

In fact, in the real number system, given any positive integer n, the system
$$x_1^2+ x_2^2+ \cdot\cdot\cdot+ x_n^2= 0$$
completely determines all n solutions. What you are attempting to assert is true of linear systems of equations, not systems of equations in general.

Last edited: Nov 1, 2008
12. Oct 31, 2008

### CRGreathouse

Halls, what (if anything!) can be said about other systems of equations (perhaps polynomials of degree < d)?

13. Oct 31, 2008

### Office_Shredder

Staff Emeritus
I'm tempted to type an uncountable list of solutions here, but I probably would be at it all night

14. Nov 1, 2008

### HallsofIvy

Staff Emeritus
And I am tempted to call myself an uncountable list of names! Of course, I meant x2+ y2+ z2= 0, as I had initially.

The point is still that the "principle" enuciated in the original post simply does not exist!

Last edited: Nov 1, 2008
15. Nov 2, 2008

### maze

If you are thinking of complex x,y,z, then the example can be easily extended to

$$x \bar{x} + y \bar{y} + z \bar{z} = 0$$

so that there is only 1 solution.

16. Nov 2, 2008

### HallsofIvy

Staff Emeritus
No, he was thinking of my unfortunate typo where I wrote "= 1" rather than "= 0" but thank you for extending my point to the complex numbers.

There is no "name" for the principle initially enunciated because we are not in the practice of giving names to incorrect statements!

17. Nov 2, 2008

### Werg22

I think the OP is referring to systems of linear equations. In that case, it's not a principle but simply a trivial fact easily seen when studying systems of a linear equations in their matrix representations: if the number of variables exceeds the number of equations, there must be free variables and hence an infinite number of solutions. In general, there exists no more than 1 solution when there are n linearly independent equations in F^n, where F is the field from which the entries of the matrix (coefficients of the equations) are coming.

18. Nov 2, 2008

### arildno

Totally incorrect.

For example, the equation:

x+y+z=x+y+z+1 has NO solutions, not infinitely many.

19. Nov 2, 2008

### HallsofIvy

Staff Emeritus
The original poster said specifically, in post 3, "Perhaps a better problem would be to consider n non-linear equations. "

20. Nov 2, 2008

### Werg22

Why do you have to be so obnoxious? Obviously I did not mean this applied to systems with inconsistent equations. There are more civil ways to correct someone.