Why is $v/\|v\|$ not a unit vector?

Solution 1:

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

Solution 2:

$\vec{v} = 0$, then: $$\frac{1}{\|v\|}\vec{v} = ?$$

The norm, $\| \|$, is normally $>0$ except only if $v = 0$.

Solution 3:

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.

Solution 4:

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.