[dxbydt]

you need closure on scalar mult. if (2,3,4) is a valid int triple & 1/3 is your scalar then (2/3,1,4/3) throws you out of the group so there goes your closure. unlike op, you don’t really need to know about fields to solve this.

[gizmo686]

If the triple is defined over the integers, why would you allow 1/3 as a scalar in the first place.

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)?

[skh]

Something being sidestepped in the post you responded to is that what is being talked about is a valid algebraic object with lots of structure to it. It’s called a module which you can think of as a sort of vector space. It’s just that the scalars may not have the property that they have multiplicative inverses. (I’m deliberately focusing on rings that are integral domains for the nitpickers.). When talking about these objects you have to include the underlying scalar set.

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.

[throwawaymath]

Everything you've said is true, but circling back to the example given, we still can't choose a scalar 1/n for integral n. Yes the integers are a ring, and yes you can define a module over a ring which generalizes a vector space.

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.

[skh]

You are correct and I wasn't trying to criticize what you wrote. gizmo686's post (the one I responded to) indicated a sense of insight into these issues. I wanted gizmo686 to feel justified in his/her thoughts. Namely, that what we call modules are natural objects and they look at feel like vectors spaces on the surface.

[ajross]

> why would you allow 1/3 as a scalar in the first place.

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.

[skh]

Consider the possibility that it does make sense but that you aren’t aware of why it makes sense. A vector space has much more structure than a module and the distinction is not unimportant. Also, scalars are not defined as a real number. Scalars are elements of the base field. When talking about a vector space one must always specify the base field. This is important and is the point of the problem in question. For instance the real numbers are a vector space over the real numbers and that vector space structure is different than the vector space structure of the real numbers as a vector space over the rational numbers.

[throwawaymath]

Huh? A scalar is very specifically a field element by definition. This is why it's important to specify the field you're working with when you talk about a vector space - a scalar is not going to be a real or a complex if your field isn't R or C.

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.

[ajross]

I keep seeing you people in this thread and wondering, with all respect, what planet you're coming from.

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.

[gizmo686]

Because you are going to have students who mark the answer as correct, and you need to be prepared to explain to them why it is wrong. In addition, you explanation of why it is wrong should be accurate, and should not suggest that other correct answers are also wrong. Returning to the original question, why is it that Z3 is not a vector space, but Q3 is. If you say that neither of these are vector spaces, then you have a misunderstanding about what a vector space is which the question would miss because the author forgot to include Q3 as an option.

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.

[gizmo686]

Except that we have vector spaces with scalars that are not the reals all the time. For instance, consider this excerpt from the article:

"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.

[thaumasiotes]

> Because it's a definitional thing. A "scalar" is routinely defined as a real number, not an integer.

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.

[ajross]

Not to undergraduates in early mathematics courses it's not. This is a term introduced in grade school, for goodness sake.

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.

[gizmo686]

Maybe because the article was talking about undergraduates:

> 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?

> ...

[throwawaymath]

While that's the right idea, I'd push back against not needing to know about fields since the scalars are just field elements. If you try to define a vector space over the integers, it's more accurate to say you can't choose 1/n as a scalar, because 1/n doesn't exist in your underlying field. Your closure ends before you even get to choose the element.

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.

[ajross]

> I'd push back against not needing to know about fields since the scalars are just field elements.

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.

[throwawaymath]

I suppose. All I'm getting at is that since Z doesn't contain 1/n for integral n, you wouldn't be able to use it as a scalar in the first place. So if you extrapolate from there, you have to choose a different route to show that defining the vector space doesn't work because you can't trigger the contradiction that fails scalar multiplicative closure.

[zazen]

> It's enough to know that multiplying integers by non-integer scalars can give non-integers

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.

[dxbydt]

This isn’t quite right. When I personally learnt these things in an undergrad program in math in the US, we learnt monoids. Then we learnt semigroups. Then groups. Then abelian groups. Then vector spaces. Then on the midterm we got questions exactly like the one we are debating here - is this guy a vector space, is that guy a semigroup, is that guy abelian etc. At that point, none of us knew what a ring was, what a field was etc. In the US you learn things like cosets and Lagrange’s theorem way before you even get to fields. That’s why I said you don’t need fields.

If you have (2,3,4) and want to navigate to (5,6,7) who is also in your space and you have scalar mult as your tool of choice then mult with 2 gets you to (4,6,8) but then you are stuck. Soon you realize no matter what you do you can’t navigate that space without fractions.

A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult. Addition is just freebie because you can rephrase it as bunch of scalar mults.

[zazen]

>If you have (2,3,4) and want to navigate to (5,6,7) who is also in your space and you have scalar mult as your tool of choice then mult with 2 gets you to (4,6,8) but then you are stuck. Soon you realize no matter what you do you can’t navigate that space without fractions.

One of us is very confused. It seems to me that I also can't get from (2,3,4) to (5,6,7) by pure scalar multiplication even if fractions are allowed. If I pick a scalar factor of 2.5 to make 2 -> 5 work, then I get (5, 7.5, 10). If I pick anything else, the result won't start with 5.

>>A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult.

Really no. You can only access parallel vectors by scalar multiplication. E.g. if your vector space is R2, given a starting vector and scalar multiplication, you can anything in a line with the direction of that vector, but nothing pointing in a different direction. That's more or less why it's called "scalar" multiplication - it scales the original vector, but doesn't change its direction.

[dxbydt]

you can change the direction, -1 is a scalar. but yeah, you are right about the rest. cheers!

[gizmo686]

>A working definition of a space might be - you have a member in that space, you can get to every other member by just scalar mult.

Isn't this a 1 dimensional space. Eg. Consider the vector space R2 over R.

If you have the vector (1,0), there is no way to arrive at the vector (1,1) through just scalar multiplication.

[dxbydt]

if you don’t give me basis how’ll i span the space ?

[throwawaymath]

You learned cosets and Lagrange's theorem before you learned fields? Did you take a course in abstract algebra before you took analysis? If so that seems a little unconventional to me, but I don't see another explanation since fields are taught in analysis.

[gizmo686]

That would be typicall if you go the algebra route. In an introductory algebra class, you would typically open with group theory. The first deep theorem you cover would be Lagrange, whose proof is normally based on cosets.

Typically, students don't start on an algebra track until after a fair amount of analysis, but there is no real reason for that to be the case. Its a shame too since, as someone who prefers algebra myself, I (totally unfairly) blame analysis for giving math a bad image.

[dxbydt]

yeah i took abstract algebra before real analysis. we did hit rings & fields in algebra but by that time it was finals week & they got minimal coverage. we used Herstein, that’s the order in that book.

[ska]

You are doing this the wrong way around, as your scalars are in Z and so you can't just "pick" 1/n, any more than you can pick pi.

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.

[zazen]

But if you don't know about fields, you could use the integers instead of the reals for your scalars, and imagine you've made a perfectly good closed vector space that way.