By this logic, should R not be a vector space, as it is not closed under scalar multiplication by i, or Q not be a vector space, as it is not closed under scalar multiplication by sqrt(2)?
For instance the real numbers are a vector space over the rationals. They are a different vector space over the reals. They are not a vector space over the complex numbers and are not a vector space over the integers. But they are a module over the integers. But not a module over the complex numbers.
But the point being spoken to here is that the explanation is backwards: you can't choose 1/n from Z. Therefore you can't use it as a scalar, so you'd never even break closure in the vector space. The hypothesis doesn't work before you can engage that contradiction.
Because it's a definitional thing. A "scalar" is routinely defined as a real number, not an integer.
And you're absolutely right that it makes no sense, which is the whole point of the multiple-choice question. Four of those answers are plausible, the other requires you to make assumptions (like a redefinition of scalar) not in the question as posed.
If you've seen someone define a scalar as a real number, that's really only because they're informally stating their underlying field is R.
The whole purpose of this exercise is to see if there was a way to come up with a straightforward, reasonably informal, multiple choice question that would expose a fundamental understanding in basic university math concepts like "vector space" in the same way we see in primary math.
And instead all you people want to do is natter over the ways in which someone could cleverly make the "wrong" answer right. It's... beyond missing the point, it's actively working against the whole goal of the exercise.
By itself, this is a minor complaint (you cannot include every example in you choices, although I do think that an example which could not be viewed as an R-vector space would be good to include). However, when you explain why Z3 is not a vector space, your explanation must be correct. An explanation which also excludes Q3 is incorrect.
"Or perhaps they wouldn’t like A because the scalar field [the complex numbers] is the same as the set of vectors (unless, that is, they thought that the obvious scalars were the real numbers)."
In this case, while there is an an acknowledgement that you could take the reals as your scalars, it is regarded as the secondary of the "natural" choices.
Or, in my example, example, there is no way to view Q as a vector space over R, but it is clearly a vector space. There is an entire field of algebra (field theory), that relies on the fact that, for example, Q(sqrt(2)) is a 2 dimensional vectorspace over Q.
Well, this is totally untrue. A scalar is defined as a non-vector quantity, a single element as opposed to a multidimensional list of them.
I give up on this thread. It's a bunch of people not just willfully misunderstanding the linked article, but actively campaigning against the whole idea of math education in an attempt to prove how much smarter than each other they are. This is... awful, folks.
> Could one devise a university-level question that would catch a significant proportion of people out in a similar way? I’m not sure, but here’s an attempt.
> Which of the following is not a vector space with the obvious notions of addition and scalar multiplication?
> ...
