4 ms·

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 spa

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 8y ago
> 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.
- zazen 8y ago
  > 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 8y ago
    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.
    - throwawaymath 8y ago
      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.
      - dxbydt 8y ago
        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.
      - gizmo686 8y ago
        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.
    - gizmo686 8y ago
      >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 8y ago
        if you don’t give me basis how’ll i span the space ?
        zazen 8y ago
        You talked about having "a member" and getting to every other member with scalar multiplication only. But even given a 2-member basis for R2, how are you planning on using 2 members, with scalar multiplication only? I'm afraid you have badly misremembered this stuff.
- throwawaymath 8y ago
  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.

ajross 8y ago

> 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.

zazen 8y ago
> 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 8y ago
  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.
  - throwawaymath 8y ago
    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.
    - dxbydt 8y ago
      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.
    - gizmo686 8y ago
      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.
  - gizmo686 8y ago
    >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 8y ago
      if you don’t give me basis how’ll i span the space ?
      - zazen 8y ago
        You talked about having "a member" and getting to every other member with scalar multiplication only. But even given a 2-member basis for R2, how are you planning on using 2 members, with scalar multiplication only? I'm afraid you have badly misremembered this stuff.
throwawaymath 8y ago
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.