Consider this problem.
Let $V$ be the set of all vectors which are positive rational numbers on which addition of vectors $v,w$ is defined as: $$v+w=vw$$ and the scalar multiplication is the usual scalar multiplication with scalars over the field $F$. Verify the distributive axiom over scalar addition.
My try:
We know that the axiom of distributive property over scalar addition as: $$(\lambda+\mu).\mathbf{v}=\lambda.\mathbf{v}+\mu.\mathbf{v}, \lambda, \mu \in \mathbb{R}, \mathbf{v} \in V$$
Let us do RHS first. We have by addition of vectors definition $$\lambda.v+\mu.v=\lambda \mu v^2$$
Well coming to LHS, I am confused in finding $\lambda+\mu$. But what I think is $\lambda, \mu \in \mathcal{F}$, so the addition of these is normal addition, but I am not really sure. For example
$\lambda=2, \mu=3, v=5, \:\lambda,\mu \in F, v \in V$
So the LHS is $$(2+3)5=5 \times 5=25$$
or is it $$(2+3)5=6(5)=30$$
Please clarify.
Assuming the scalar field is $\Bbb R$ or $\Bbb Q$ or something else typical with the usual scalar addition and multiplication amongst themselves...
For clarification, I will write addition of vectors as $\oplus$.
Addition of scalars is still the usual addition of scalars and will continue to be written as $+$.
Further, for clarification, I will write scalars in black and vectors in blue.
You ask about $(2+3)\color{blue}{5}$. This does equal $5\cdot \color{blue}{5}=\color{blue}{25}$
You ask whether $(2+3)\color{blue}{5}$ should have evaluated as $6\cdot \color{blue}{5}$ but it should not have since this is $2+3$ here, not $\color{blue}{2\oplus 3}$.
However, it is worth pointing out what the exercise is trying to get you to recognize... and that is what happens if we were to try to apply distributivity here.
$(2+3)\color{blue}{5} = (2\cdot \color{blue}{5})\oplus (3\cdot \color{blue}{5}) = \color{blue}{10\oplus 15} = \color{blue}{150}$
and here since $25\neq 150$ we have that distributivity failed for this example.