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

Re: How Craig Barton wishes he’d taught maths

#21

Post by ajross » Sat, Feb 09, 2019, 7:08 PM UTC

gizmo686 wrote: post #15

dxbydt wrote: post #5

Earlier quoted context omitted.

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.

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.

Re: How Craig Barton wishes he’d taught maths

#22

Post by joe_the_user » Sat, Feb 09, 2019, 7:16 PM UTC

More precisely, in order to decide whether it is a good idea, one should assess (i) how difficult it is to give an explanation of why some procedure works and (ii) how difficult it is to learn how to apply the procedure without understanding why it works.

Well, teaching basic math at a commuter college years ago, it felt like the issue of "teaching procedure" to "teaching understanding" was complex. The course I was teaching was close to the end of the math requirements for a significant percentage of the students. I was very attracted to teaching ideas but this group of students essentially had the attitude that they wanted a procedure to memorize rather than an explanation, not matter how complex the procedure. It had a certain logic - mathematical explanation would have touched a world they were happy to and committed to leaving forever soon after this. They'd suffered through this world up this point and thinking about it was more painful than simply acting.

Which is to say, I don't think there any easy answer for how to teach math. The failure of American "new math" years ago is something of a lesson in the push-pull of concepts versus concreteness as they can become ideologies in society at large.

Re: How Craig Barton wishes he’d taught maths

#23

Post by gizmo686 » Sat, Feb 09, 2019, 7:24 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.

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.

Re: How Craig Barton wishes he’d taught maths

#24

Post by skh » Sat, Feb 09, 2019, 7:24 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.

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.

Re: How Craig Barton wishes he’d taught maths

#25

Post by jp57 » Sat, Feb 09, 2019, 7:28 PM UTC

Soon someone from HN is going to come and remove the word ‘how’ from the headline. Why? Because clickbait!

Except, of course, the book really is about how he feels he should’ve taught math.

Re: How Craig Barton wishes he’d taught maths

#26

Post by zazen » Sat, Feb 09, 2019, 7:30 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…

> 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.

Re: How Craig Barton wishes he’d taught maths

#27

Post by theontheone » Sat, Feb 09, 2019, 7:48 PM UTC

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."

Re: How Craig Barton wishes he’d taught maths

#28

Post by dxbydt » Sat, Feb 09, 2019, 7:57 PM UTC

zazen wrote: post #26

ajross wrote: post #20

Earlier quoted context omitted.

> 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…

> 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 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.

Re: How Craig Barton wishes he’d taught maths

#29

Post by a_c » Sat, Feb 09, 2019, 8:08 PM UTC

throwawaymath wrote: post #3

Wow, that vector space question is a great example. It’s the kind of thing that should be straightforward for anyone who has taken a linear algebra course, but I can also totally see students getting it wrong. This is especially the case because it’s actually very easy fundamentally (the set of all integers does not comprise a field, and so a vector space cannot be defined over it). But to my recollection, most of th…

The math education I experienced focused heavily on "how". "How" such and such operation arrive to its conclusion and "how" such and such operation fulfil some "rules". Seldom does it touch on "why". Why certain notion, like linear algebra, heck, maybe even negative number, exists in first place. Procedures like negative times negative gives positive number. Yeah sure, but why? What does that mean really. I think the "why" of everything is what makes one understand anything

Re: How Craig Barton wishes he’d taught maths

#30

Post by throwawaymath » Sat, Feb 09, 2019, 8:16 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…

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.