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How Craig Barton wishes he’d taught maths

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Re: How Craig Barton wishes he’d taught maths

#51

Post by skh »

throwawaymath wrote: post #32
skh wrote: post #18

Earlier quoted context omitted.

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 vect…

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.

Re: How Craig Barton wishes he’d taught maths

#52

Post by mncharity »

joe_the_user wrote: post #22

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…

This is a widespread tradeoff. I'd a conversation with a first-tier college biology professor, about a way to give a more integrated, transferable understanding of a topic. He liked it, but observed, my students will shortly be taking the MCAT (high-stakes medical school entrance exam), and our time together is limited, and the MCAT doesn't test for understanding of the topic, only for something superficial and memorizable, so, I would be doing my students a disservice if I reallocated time to understanding.

Perhaps early primary school is an opportunity to escape this tension. With weaker test constraints, and more years of payoffs over which to amortize the costs of better understanding.

Re: How Craig Barton wishes he’d taught maths

#53

Post by [deleted] »

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.

[deleted]

Re: How Craig Barton wishes he’d taught maths

#54

Post by MarcusBrutus »

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…

The integers are totally closed under multiplication. They lack an inverse under multiplication that's why they aren't a field.

Re: How Craig Barton wishes he’d taught maths

#55

Post by ska »

dxbydt wrote: post #5
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?

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.

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.

Re: How Craig Barton wishes he’d taught maths

#56

Post by winchling »

avmich wrote: post #39
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?

As philosopher Daniel Dennett puts it: competence comes before comprehension. It's totally possible to do something well, as animals do, without understanding what one is doing. But for comprehension one needs to have something in place to reason about and make connections.

And this isn't the full picture. Motivation comes before competence. One needs reasons to acquire skills: they have to address problems in one's mind if the mind is to fully engage. Which is why coercive education with its curricula, exams, etc, largely fails.

Re: How Craig Barton wishes he’d taught maths

#57

Post by mncharity »

zazen wrote: post #17
keithpeter wrote: post #16

Earlier quoted context omitted.

Going back around the millennium or before when I last taught A level maths at college, we had them in over the summer before term started for a two week intensive algebra and basics course. Seemed to help. The original author (Tim Gowers, a Fields medallist and professor of mathematics at Cambridge) has a totally hilarious blog post about being asked to coach a teenager doing A level maths... https://gowers.wordpres…

Thanks for linking that, it's a great read. I really should read more of Gowers' posts. The phrase "memory works far better when you learn networks of facts" was a happy find - I've never been able to express that idea so concisely. I remember discovering they'd moved "differentiation from first principles" away to a further-maths module, as if it's a peripheral, difficult little oddity for the keen kids to hear abou…

> memory works far better when you learn networks of facts

One challenge with teaching a more rough-quantitative Fermi-question-ish introduction to sciences, is it's more sensitive to integration and correctness of understanding. With a Trivial-Pursuit memorize and regurgitate style of "understanding", damage from misconceptions and fragmentation of knowledge is local. Whereas rough-quantitative reasoning benefits from being able to... slide around the knowledge space. Jagged misconceptions and fragmented knowledge seriously impedes the sliding. I imagine memory is similar. Nice phrase.

Re: How Craig Barton wishes he’d taught maths

#58

Post by akalin »

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…

> If you try to define a vector space over a set without multiplicative closure, the vector space cannot be closed under scalar multiplication. Among other things, linear combinations stop being invertible (or even possible in general), and linear relations don't exist.

Mind clarifying this part? As someone else already pointed out, the integers are multiplicatively closed, but I suspect you're using "multiplicatively closed" to also mean "closed under multiplicative inverses". But I don't see how linear combinations stop being possible, e.g., "3x + 2y" is still a linear combination in a Z-module, or what it means for a linear combination to be invertible.

(Also not sure what exactly you mean by linear relations not existing if you have a module and not a vector space...)

Re: How Craig Barton wishes he’d taught maths

#59

Post by thaumasiotes »

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…

Your heavy emphasis on the idea that fields are closed under addition and multiplication is very counterproductive here, because the integers are closed under addition and multiplication.

The integers violate the field axiom that every element in the field (other than 0, the additive identity) must have a multiplicative inverse in the field. But that has nothing to do with the requirement of being closed under multiplication.

Re: How Craig Barton wishes he’d taught maths

#60

Post by thaumasiotes »

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.

> 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.

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