Possible Duplicate:
Is it possible to construct a quasi-vectorial space without an identity element?
In Apostol Multivariable Calculus, $1.5$ exercise $30 b$, he asks the reader to prove that Axiom $10$ is independent of the other axioms for vector spaces. Axiom $10$ says that $$1 * x=x$$
To do this, it is equivalent to prove that $(-1)*x\not = -x$, since $(-1)*x=-x$ implies that $1*x=x$ (we can prove it using the other axioms. My approach is to find $(V,+,*)$ that satisfies the first $9$ axioms and contradicts the axiom $10$.
My attempts so far:
First, I tried setting $V=\mathbb{R}$, redefining $a*x$ to be variations on multiplication. For example, I tried $a*x=|ax|$ but ran into problems with $|a(x+y)|\not=|ax|+|ay|$ in general.
Then, I tried setting $V=\mathbb{R^2}$ and $a*x=ax^T$ if $a$ is positive and $a*x=ax$ if $a$ is not positive. However, I ran into problems with $(a+b)*x\not=a*x+b*x$ for $a=-1$,$b=1$,$x=(1,0)$.
