[Added by PLC: This question is a followup to this already answered question.]
Keep the axioms for a real inner product (symmetry, linearity, and homogeneity). But make the fourth be $$\langle x,x \rangle = 0 \text{ if and only if } x = 0.$$
I want to prove that either $\langle x,x \rangle > 0$ or $\langle x,x \rangle < 0$ for all $x \neq 0$.
Note: $c_1 = \langle x,x \rangle > 0$ and and $c_2 = \langle y,y \rangle < 0$.
Here's the sketch of the proof I want: Assume $\langle x,x \rangle > 0$ for some $x$ and $\langle y,y \rangle < 0$ for some $y$. I'm trying to find a $z \neq 0$ such that $\langle z,z\rangle = 0$, where $z$ is in the space spanned by $\{x,y\}$. By contradiction, we know that $\langle x,x \rangle < 0$ or $\langle x,x \rangle > 0$ for all $x \neq 0$.
Unfortunately, my proof doesn't work that way, and I don't think it proves what I want it to prove.
I say let $\langle z,z \rangle = \langle ax + by, ax + by\rangle$, for $a,b \in \mathbb{R}$. Then $$\langle z,z \rangle = a^2 \langle x,x \rangle + 2ab \langle x,y \rangle + b^2 \langle y,y \rangle = 0.$$ Let $a = \langle y,y \rangle$ and $b = \langle x,x \rangle$. After plugging $a$ and $b$ in, I get: \begin{align*} & \langle y,y \rangle + 2\langle x,y \rangle + \langle x,x \rangle\langle y,y \rangle = 0 \\ \implies& c_2 + 2 \langle x,y \rangle + c_1 c_2 = 0 \\ \implies& 2\langle x,y \rangle = -(c_1 c_2 + c_2). \end{align*} Thus, \begin{align*} \langle z,z\rangle= &= c_1 c_2 + 2(-c_1 c_2 - c_2) + c_1 c_2 \\ &= -2c_1c_2 + 2(c_1c_2 c_2) \\ &= -c_1c_2+c_1c_2 +c_2 = 0 \end{align*} Then $c_2 = 0$
The only thing I can think to do now is to claim a contradiction: we said $c_2 < 0$. But I don't think this proves what we want to prove that either $\langle x,x \rangle < 0$ OR $\langle x,x \rangle > 0$ for all $x \neq 0$.
I think my issue is that I don't know how to choose $a,b$ to make $\langle z,z \rangle = 0$. Someone please offer some help.
Suppose that $\langle x, x \rangle > 0$ and $\langle y, y \rangle < 0$. I claim that there are $a,b \in \mathbb{R}$ such that $ax+by \neq 0$ and $\langle ax+by, ax+by \rangle = 0$. This shows that if $q(v) = \langle v, v \rangle$ assumes both positive and negative value, then there is some nonzero $v$ with $q(v) = 0$.
We have $\langle ax+by, ax+by \rangle = a^2 \langle x,x \rangle + 2ab \langle x,y \rangle + b^2 \langle y,y \rangle$. If we view this as a quadratic equation in $a$, its discriminant is
$\Delta = 4b^2 \langle x,y \rangle^2 - 4 b^2 \langle x,x \rangle \langle y, y \rangle$.
Because of the assumptions on the sign of $\langle x,x \rangle$ and $\langle y, y \rangle$, $\Delta > 0$ when $b \neq 0$. So choose your favorite nonzero value of $b$; then the quadratic formula shows that the equation $\langle ax+by,ax+by \rangle = 0$ can be solved for $a$.