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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

#61

Post by gizmo686 »

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.

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.

Re: How Craig Barton wishes he’d taught maths

#62

Post by ajross »

thaumasiotes wrote: post #60
ajross wrote: post #21

Earlier quoted context omitted.

> 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.

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.

Re: How Craig Barton wishes he’d taught maths

#63

Post by gizmo686 »

ajross wrote: post #62
thaumasiotes wrote: post #60

Earlier quoted context omitted.

> 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.

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.

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?

> ...

Re: How Craig Barton wishes he’d taught maths

#64

Post by kiba »

mncharity wrote: post #52
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 memor…

My understanding of learning is that it both takes a longer time to actually learn something useful and that we're inefficient in learning or retaining anything because we rushed from one topic to another in our courses.

Re: How Craig Barton wishes he’d taught maths

#65

Post by zozbot123 »

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…

> which is to stress the rule you can do the same thing to both sides of an equation (worrying about things like squaring both sides or multiplying by zero later).

It's a pity that they're being so ambiguous here, because explaining why and when "you can do the same thing" to both sides of an equation is not actually hard! You can apply an injective function that's always defined over the appropriate domain to both sides of an arbitrary equation, and this will preserve the equation entirely because (a = b) is equivalent to (f(a) = f(b)) when f has this property. You can apply a non-injective function with no restriction on its domain, and this may introduce extraneous solutions but will not "miss" any, because (a = b) implies (f(a) = f(b)) if f is always defined. You can apply an injective function, perhaps defined over a more limited domain than the original equality, and this will not introduce extraneous solutions but may "miss" some, because (f(a) = f(b)) implies (a = b) if f is injective, but the converse is not true given any restriction on f's domain. Of course, if these functions are defined in terms of x, then you get to worry about whether the function is injective or well-defined given some value of x. For instance, multiplication by x is not injective if (x = 0) but it is otherwise.

Re: How Craig Barton wishes he’d taught maths

#66

Post by taleinat »

dorchadas wrote: post #47
a_c wrote: post #29

Earlier quoted context omitted.

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…

I'm a high school math and science teacher, and I completely agree this is a problem. Several of my students struggled greatly with simple arithmetic until I set down and gave them reasons for why a negative times a negative was a positive (which meant I also had to explain to them why multiplication works like it does). I personally think part of the problem is how elementary education is structured. At least in my…

Hi dorchadas, I really like your comment!

I'm looking for great high-school Math teachers for a project. If you wouldn't mind sparing a few moments, please get in touch with me via email (gmail with my username) so I can give you some more details.

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