How Craig Barton wishes he’d taught maths • Hacker News Board

Re: How Craig Barton wishes he’d taught maths

#31

Post by throwawaymath » Sat, Feb 09, 2019, 8:19 PM UTC

ajross wrote: post #21

gizmo686 wrote: post #15

Earlier quoted context omitted.

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)?

> 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.

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.

Re: How Craig Barton wishes he’d taught maths

#32

Post by throwawaymath » Sat, Feb 09, 2019, 8:21 PM UTC

skh wrote: post #18

gizmo686 wrote: post #15

Earlier quoted context omitted.

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)?

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 the…

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.

Re: How Craig Barton wishes he’d taught maths

#33

Post by throwawaymath » Sat, Feb 09, 2019, 8:24 PM UTC

ajross wrote: post #20

throwawaymath wrote: post #11

Earlier quoted context omitted.

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 p…

> 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…

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.

Re: How Craig Barton wishes he’d taught maths

#34

Post by gizmo686 » Sat, Feb 09, 2019, 8:24 PM UTC

dxbydt wrote: post #28

zazen wrote: post #26

Earlier quoted context omitted.

> 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.

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 f…

>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.

Re: How Craig Barton wishes he’d taught maths

#35

Post by throwawaymath » Sat, Feb 09, 2019, 8:28 PM UTC

theontheone wrote: post #27

I consider myself pretty strong at math (in university right now) and I was stumped by the vector space question. I never considered, actually, what domain scalars should be drawn from. Wikipedia says "the scalars can be taken from any field, including the rational, algebraic, real, and complex numbers, as well as finite fields."

Does Wikipedia actually say that? That's pretty misleading. For any vector space V defined over a field F, V is only closed under scalar multiplication using the scalars of F. You can't choose scalars from arbitrary fields for any given vector space. The scalars have to be chosen from the underlying field of the particular vector space.

Re: How Craig Barton wishes he’d taught maths

#36

Post by dxbydt » Sat, Feb 09, 2019, 8:29 PM UTC

throwawaymath wrote: post #30

dxbydt wrote: post #28

Earlier quoted context omitted.

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 f…

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.

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.

Re: How Craig Barton wishes he’d taught maths

#37

Post by dxbydt » Sat, Feb 09, 2019, 8:34 PM UTC

gizmo686 wrote: post #34

dxbydt wrote: post #28

Earlier quoted context omitted.

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 f…

>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.

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

Re: How Craig Barton wishes he’d taught maths

#38

Post by burlesona » Sat, Feb 09, 2019, 8:43 PM UTC

throwawaymath wrote: post #6

burlesona wrote: post #4

Earlier quoted context omitted.

As a bumbling idiot who never did university level math, I don’t understand the question or the answer. (I did understand the probability one and knew the right answer to that at least.) I’m trying to catch up. Would you be kind enough to explain it?

Sure. Operations in applied linear algebra - such as matrix multiplication and solving systems of linear equations - are formalized by the theory of vector spaces, much like calculus is formalized through the theory of analysis. Vector spaces are algebraic structures which axiomatize the linearity you need to carry out these operations. If you can establish your equations exist in a vector space, you can prove that t…

Thank you, that was really helpful!

Re: How Craig Barton wishes he’d taught maths

#39

Post by avmich » Sat, Feb 09, 2019, 8:54 PM UTC

keithpeter wrote: post #2

Quotes from OA that struck me as on the button... "A prejudice that was strongly confirmed was the value of mathematical fluency. Barton says, and I agree with him (and suggested something like it in my book Mathematics, A Very Short Introduction) that it is often a good idea to teach fluency first and understanding later." Agree fully with Barton and OA here. Until recently I taught GCSE Maths re-take students aged…

Yes, the first one looks like an important quote :) . When I studied math, I usually had trouble understanding or memorizing a rule unless I had at least a rough idea why it holds. In this case I'd suspect just remembering rules and then using them without understanding would - often - cause inconveniences or later errors, when a rule is remembered incorrectly. So maybe it's subjective - what should be taught first?

Re: How Craig Barton wishes he’d taught maths

#40

Post by zazen » Sat, Feb 09, 2019, 8:55 PM UTC

dxbydt wrote: post #28

zazen wrote: post #26

Earlier quoted context omitted.

> 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.

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 f…

>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.