Why is $v/\|v\|$ not a unit vector? I have a homework question that is seriously stumping me. It was a true/false statement.
The statement is 

"If $v$ is any vector in an inner product space $V$, then $v/\|v\|$ is a unit vector" 

and according to my professor, the statement is false.
Isn't dividing a vector by its length the very definition of normalizing a vector?
Edit: Thanks everyone! I didn't even consider the null vector.
 A: It's not true for any vector. There is at least one vector for which it's undefined : the null vector.
Now, it's true that for every non null vector, it's an unit vector
A: $\vec{v} = 0$, then:
$$\frac{1}{\|v\|}\vec{v} = ?$$
The norm, $\| \|$, is normally $>0$ except only if $v = 0$.
A: I consider your professor's question to be badly stated.
The expression

If $v$ is any vector in an inner product space $V$, then $v/\|v\|$ is a unit vector.

can be translated into symbols as
$$\forall v \in V:  \left\| \frac{v}{\|v\|}  \right\| =1. \tag{1}\label{first}$$
If your professor has true/false as possible answers then either \eqref{first} is true or its negation, which is
$$ \exists v \in V:  \left\| \frac{v}{\|v\|}  \right\| \ne 1. \tag{2}\label{second}$$
Now note that \eqref{second} isn't true either since it "breaks" if you try to plug in $0\in V$ (for one can't divide by zero).
So while one can see what was meant by the question and that your professor meant false to be the correct answer, I, being a student, am always somewhat peeved when I see something like this on an exam.
A: Great question.  Yes, the zero vector is the counter-example.
I want to also add that the zero matrix is good to remember when needing counter-examples --  it has been a good one, especially for exams, from my own experience.
