system of equations

[triclino]

How do we solve the system of the following equations:

 

1) [math] x+y+z=0[/math]

 

2) [math] x^2+y^2+z^2 =0[/math]

 

3)[math] x^3+y^3+z^3 =0[/math]

[shyvera]

If x, y, z are all real, the three equations can be reduced to just one, namely the second one.

[mooeypoo]

Moved to homework help.

 

triclino, please tell us what you've tried and where you got stuck. We would love to help, but we're not in the habit of feeding answers with a spoon.

 

~moo

[triclino]

Moved to homework help.

 

triclino, please tell us what you've tried and where you got stuck. We would love to help, but we're not in the habit of feeding answers with a spoon.

 

~moo

 

I do not know how even to start with this system of equations .

 

Do you .

 

If yes please show me.

 

On the other hand answers like the one given by shyvera is a clever way of saying i do not know

[D H]

If x, y, and z are limited to the reals, what is the only solution to equation (2)?

[triclino]

If x, y, and z are limited to the reals, what is the only solution to equation (2)?

 

D.H the solution x=y=z=0 is the solution to every system of equations that are equal to 0.

 

This i suppose is well known to every one of us .

 

Here we want solutions other than those equal to zero

[D H]

There are no other solutions if x, y, and z are restricted to the reals.

[triclino]

There are no other solutions if x, y, and z are restricted to the reals.

 

How can you say that without a proof??

[chocochoco]

U can solve it with your pocket calculator!

[D H]

How can you say that without a proof??

You don't need a proof for everything, triclino. Suppose a mathematician, after some arduous calculations in a math paper, arrived at

 

[math]x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2 = 0[/math]

 

Nobody would blink an eye if the next sentence in the paper was "and thus the only solution in the reals is the trivial solution."

 

If you need a proof, first prove that

1. The sum of a finite set of non-negative numbers is non-negative.
   
2. The sum of a finite set of non-negative numbers, at least one of which is positive, is positive.

 

Note that I said a finite set. Whether [math]1+1+1+\cdots = \sum_{n=1}^{\infty}1 = -1/2[/math] or [math]1+2+3+\cdots = \sum_{n=1}^{\infty} n = -1/12[/math] are different questions.

[shyvera]

 

1) [math] x+y+z=0[/math]

 

2) [math] x^2+y^2+z^2 =0[/math]

 

3) [math] x^3+y^3+z^3 =0[/math]

If x, y, z are real, then obviously x = y = z = 0 is the only solution. Suppose, to make it a little more challenging, that the variables can be complex as well as real. Then notice that

 

[math]0=(x+y+z)^3=x^3+y^3+z^3+3(x+y+z)(xy+yz+zx)-3xyz=-3xyz[/math]

 

Thus xyz = 0, i.e. one of x, y, z is 0. Letting one of them be 0, substituting into (1) and (2), and squaring the first equation, you should find that the product of the other two is 0 as well – which, on substituting back into the first equation again, should mean all three are equal to 0. Hence the trivial solution is again the only solution.

 

NB: I make it that if x, y, z are, in general, elements of an integral domain whose characteristic is not 2 or 3, then the trivial solution is the only solution.

[triclino]

You don't need a proof for everything, triclino. Suppose a mathematician, after some arduous calculations in a math paper, arrived at

 

[math]x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2 = 0[/math]

 

Nobody would blink an eye if the next sentence in the paper was "and thus the only solution in the reals is the trivial solution."

 

And if he was asked to prove it ,then he would realize that it is an impossible task

 

Can you prove it??

[D H]

Of course. It is trivial. You are the one who needs help here, why don't you try it?

[triclino]

So if somebody is sick and needs help he/she must do not go to the Doctor but try to fix him/her self

[mooeypoo]

So if somebody is sick and needs help he/she must do not go to the Doctor but try to fix him/her self

They should go to the doctor and listen to what the doctor says - or look for a second opinion - rather than call the doctor and idiot the minute the doctor utters his answer.

 

This isn't up for a debate, triclino. You are to be nice, or you aren't to be here at all. Case closed. Move on.

 

~moo

[D H]

triclino, you are never going to learn if you do not try. That is why students are given homework problems. So, a homework problem for you:

 

The reals are a field (i.e., they have operators + and * with the standard axioms for those operators). What makes the reals special is that they are a totally order field. This means there is an operator ≥ whose characteristics include, for a,b,c in R,

1. If a ≥ b and b ≥ c then a ≥ c.
   
2. a ≥ b or b ≥ a.
   
3. If a ≥ b then a + c ≥ b + c.

 

Given the above axioms prove that

1. The sum of two non-negative reals is non-negative.
   Hint: In other words, prove that if a≥0 and b≥0 then a+b≥0.
   
2. The sum of a finite set of non-negative reals is non-negative.
   Hint: Use the above result.