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.
"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.
For students it might not be immediately obvious why that's a problem for vector spaces, but yes it does mean scalar multiplication won't be closed in the vector space. And more practically speaking, if you tried to solve a system of equations without invertible linear combinations, you'd have no linearity whatsoever. Elementary row operations likewise cease to be invertible, so matrix reduction isn't possible...the whole thing breaks down really.
The point was that you don't need to know the jargon of "field" and the full set of implications. It's enough to know that multiplying integers by non-integer scalars can give non-integers, which means that "scalar multiplication" can produce a thing that is not a "triple of integers". So it's not a well defined vector space operation.
No need for "field" or "closure" or any other jargon not in the question as posed.
Not quite: you also need to know that multiplying by integer scalars instead isn't an option.
The question as posed asked as to use the "obvious" choice of scalar multiplication, and to a student who hasn't yet taken the "field" part on board, it might seem obvious to achieve closure by using the integers for scalars.
In fact, closure under scalar multiplication is there. Pick d in Z and d(a,b,c) = (da,db,dc) is fine.
The real problem is, I need an inverse. So if that exists, we have : e(da,db,dc) = (a,b,c) and e must exist in the set for (a,b,c) != (0,0,0).
Now you are trying to find e that behaves like 1/d , but you've left the set - no good.
