Proving axiom 1 of inner product

Question

I am doing one of the problems in my book, but I am uncertain if my approach is correct.

Problem: $(u,v)=-u_1u_2u_3$

Let $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$.

\begin{align} (u,v)&= u_3u_2u_1 \\ &= (v,u) \end{align}

Axiom 1 proved. Is this correct?

Answer 1

It's totally incorrect.

It's a function with two arguments that is defined here: $(u, v):=-u_1u_2u_3$.
Putting in $x,y$, we get $(x,y)=-x_1x_2x_3$.
Putting in $v,u$, we get $(v,u)=-v_1v_2v_3$.
Now we can easily show up two specific vectors $u, v$ such that $(u, v) \ne (v, u)$.