Basically, I got in an argument with my professor over a test question.
Let V be the set of all positive real numbers with opertaions $\vec X + \vec Y = \vec X\vec Y$,
and $k\vec X = \vec X^k $. Verify the 10 axioms to show that V is a vector space
My solution went as follows:
For the 5th axiom,
For each $\vec u$ in V, there is an object $-\vec u$ in V, called a negative of $\vec u$, such that $\vec u+(-\vec u) = (-\vec u)+\vec u = 0$
I used the fact that I found the zero vector to be 1 in the previous step ($\vec 0 = \vec 1$), and stated that since addition becomes multiplication, and that multiplication by zero vector yields zero vector, then $-\vec X = \vec 1.$
$$ \vec X+ -\vec X= \vec 0 \\
-\vec X\vec X= \vec0\\
\text{Since multiplication by zero vector equals zero vector, $-\vec X = \vec 1$} $$
My professor stated that my answer was incorrect and that the right answer is
$$ \vec x + (-\vec x) = \vec 0\\
\vec x + \frac 1 {\vec x}\\
\frac {\vec x} {\vec x} = \vec 0\\
\vec 1= \vec 0 $$
This left me very confused.
1- why is my answer wrong? is it not true that multiplication by zero vector yields zero vector?
2- is it correct to say that $ \vec X \frac {\vec 1} {\vec X} = \vec 1$ when the rules for vector multiplication were not even mentioned in the problem?
thanks
Best Answer
I think there's been some confusion arising from notation and also some confusion between vector operations and real number operations. It's a bit difficult for me to see exactly where the confusion is taking place in your case, but I'll respond to each of your questions in turn with my own way of thinking about it.
Following @amd's advice, whenever talking about an object in a vector sense, I will render it with the vector arrow on top ($\vec x$), otherwise the operations refer to the standard operations among real numbers. Hence we may write the definition of vector addition and scalar multiplication like so: \begin{align} \vec x + \vec y &:= xy \\ k \vec x &:= x^k \end{align} where $k$ can denote any real number (positive or negative)
For any vector space, multiplication by the scalar zero returns the zero vector. You can see that in our case this way: Let $\vec x$ denote any vector in our space (i.e. a positive real number). Then \begin{align} 0 \vec x := x^0 = 1 = \vec 0 \end{align} And $1$ is the correct vector zero in our case since raising any positive real number to the power $1$ (i.e. adding it in the vector sense) returns the original input: \begin{align} \vec x + \vec 1 &= x^1 = x = \vec x \\ \therefore 1 &= \vec 0 \end{align}
$\vec x \frac{1}{\vec x} = \vec 1$ is really bad notation. But it is "correct" if interpreted properly. I would write the interpretation like so: \begin{align} \vec x + (-\vec x) = x \cdot \frac{1}{x} = 1 = \vec 0 \end{align} Hence $-\vec x = \frac{1}{x}$: i.e. the vector negative is the real reciprocal, and so the notation $x \cdot \frac{1}{x}$ (without the arrows) is well-defined because here we are talking about multiplication among real numbers, not vectors.