Inner product - change of axioms Suppose we retain the first three axioms for a real inner product 
$1.~\langle x,y\rangle=\langle y,x\rangle$
$2.~\langle x,y+z\rangle=\langle x,y\rangle+\langle x,z\rangle$
$3.~c\langle x,y\rangle=\langle cx,y\rangle$
but replace the fourth axiom by a new axiom:
$4'.~\langle x,x\rangle=0~$ iff $~x=0$
Prove that either 
$\langle x,x\rangle>0,~\forall x\neq0$
or
$\langle x,x\rangle<0,~\forall x\neq0$
Now there is a hint which suggest that I should assume that $\langle x,x\rangle>0$ for some $x\neq0$ and $\langle y,y\rangle<0$ for some $y\neq0$. Then I need to find and element $z\neq0$ in a space spanned by $\{x,y\}$ with $\langle z,z\rangle=0$.
So my idea is as follows: assume $0$ belongs to the space spanned by $\{x,y\}$ (otherwise - result follows trivially). If it does, it means that there exists $c_1$ and $c_2$ not both equal to zero such that 
$$c_1x+c_2y=0$$
As a result there also exists inner product in this space equal to zero, where $z\neq0$ but $\langle z,z\rangle=0$. But then we have a contradiction to the $4'$ axiom.
Could this work as a proof? If not, what do I need to turn it into proof or how should I approach this problem?  Thanks!
 A: I'm not sure what you mean by "assume $0$ belongs to the space spanned by $\{x,y\}$": the span of any set of vectors is a vector subspace, and thus necessarily contains $0$. The rest of your argument also looks muddled to me (unless I misunderstood it).
Here is what I suggest: you are assuming that you have vectors $v$ and $w$ with 
$\langle v, v \rangle = c_1 > 0$, 
$\langle w,w \rangle = c_2 < 0$.
Now try to evaluate $\langle av + bw, av +bw \rangle$ for arbitrary $a,b \in \mathbb{R}$.  Your answer will come out in terms of $a,b,c_1,c_2$ and $\langle v, w \rangle$.  Then show that you can choose $a$ and $b$, not both zero, so as to get $\langle av+bw,av+bw \rangle = 0$.
Aside: This can be viewed as a question on elementary quadratic form theory: show that a real quadratic form is either positive definite, negative definite, or isotropic.  Since every nondegenerate real quadratic form can be diagonalized with $\pm 1$ coefficients, it comes down to observing that if we have $x^2-y^2$ as a subform then we must be isotropic.  
