Show that there is no vector $e$ such that $e × ~x = ~x$ for all $x$? (I apologize for not putting an arrow over my vectors, as I couldn't figure out how to type them)
Basically I'm trying to show that vectors do not have a multiplicative identity. But I can't find a procedure that would help me do this. Would it be enough to say that $e$ would have to be equal to $1$ for this to work, which isn't possible because vectors are not constants?
 A: Given $e \in \mathbb{R}^3$, we have
$$ e \times e =0 \neq e, $$
so no matter what $e$ is, it can't be an identity for $\times$.
A: Assume:
$$
\vec{e}\times\vec{x} = \vec{x}
$$
$$
(\vec{e}\times\vec{x}).\vec{x} = \vec{x}.\vec{x} = 0 = ||x||^2
$$
The only vector that satisfies this is $\vec{x} = \vec{0}$
A: There is more than one kind of vector product.
(1). In an inner-product space such as $R^n$,the inner product (or "dot product") is a scalar, and so for $n>1$, the product $x\cdot y \not \in R^n$.
(2).In $R^3$ the outer product, written $x\times y,$ does belong to $R^3$ , but $x\times x=(0,0,0)$ for any $x\in R^3.$ 
(3). Many vector spaces are also rings, and many of them have a multiplicative identity.
Example 1: Let $P[0,1]$ be the set of polynomials $p:[0,1]\to R$, with (i)...$(a\cdot p)(x)=a p(x)$ for $a\in R$ and $p\in P[0,1],$...(ii) with $(p+q)(x)=p(x)+q(x)$... (iii) with $(p \cdot q)(x)=p(x)q(x).$ Then $P[0,1]$ is a vector space over $R.$ The polynomial $I(x)=1$ for all $x\in [0,1]$ is the unique multiplicative identity of $P[0,1].$
Example 2: Any field $F$ is  a $1$-dimensional vector space over itself.
