How Craig Barton wishes he’d taught maths(gowers.wordpress.com) |
How Craig Barton wishes he’d taught maths(gowers.wordpress.com) |
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.
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.
But to my recollection, most of the popular linear algebra textbooks[1] don’t spend time showing why the integers cannot form a vector space because it’s “easy.” Instead they spend time tediously walking through examples of bizarre sets defined over R and C to show which axioms are fulfilled and which are not.
In a similar vein to the way students might overthink the elementary probability question, I could see university students trained to disprove each of A), B), C) and D) - perhaps making a mistake along the way - instead of quickly scanning the options and picking out the one which simply isn’t defined over a field.
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1. I’m thinking of Friedberg et al, Hoffman & Kunze, Axler, Strang, etc.
edit: in fact i learned linear algebra from hoffman kunze
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 they admit linear relations and are thus solvable as linear systems.
More precisely, a vector space V is any set S defined over a field F which is closed under both vector addition and scalar multiplication, where the elements of S are called vectors and the elements of the underlying field F are called scalars. The term "closed" means that for every pair of vectors x, y in V there exists a vector x + y in V, and for every scalar c in F and vector x in V there exists a vector cx in V. There are eight axioms in total, for things like associativity and commutativity, but those aren't germane to this particular example. What's important is that vector spaces are what allow you to form linear combinations of things, which is the scaffolding you need to prove things like linear dependence and independence; whether or not a system of linear equations has no solutions, one solution or infinitely many solutions, etc.
Fields are the algebraic structures which formalize the elementary arithmetic you're already familiar with over sets like the the complex numbers, the real numbers, the rationals, etc. They are sets which are closed under "regular" addition and multiplication. Notably, integers do not comprise a field because integers do not have multiplicative inverses. Multiplicative inverses are the axiomatic way of establishing that in any field, division must be possible. So concretely, there is no multiplicative inverse 1/n for any integer n. There is in the set of rationals, but not the set of integers. Therefore integers are not closed under multiplication, and they cannot comprise a field.
Since the integers do not comprise a field, you cannot define a vector space over the integers, because the scalars used to define scalar multiplication in vector spaces are just elements of the underlying field. 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.
So circling back to the specific question: it's asking which of the given sets comprises a vector space. You can make all kinds of abstract vector spaces (e.g. the set of all polynomials over a field, the set of all polynomials with degree at most n over a field, the set of all continuous functions, etc). But if you stick with the definition of a vector space, you don't need to tediously test each of the given sets for the eight axioms. You just have to remember the integers don't comprise a field, so the set of all triples of integers can't be a vector space either.
Hopefully that's clear, let me know if you'd like me to clarify anything.
The precise definition of what we mean by that is covered by the vector space axioms.
http://mathworld.wolfram.com/VectorSpace.html
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You can add two complex numbers, and multiply them by a real (or a complex number in this case) and still have a complex number.
So it's a vector space.
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Two twice differentiable functions can be added together to yield a twice differentiable function
(f(x)+g(x))''=f''(x)+g''(x)
(cf(x))''=c(f''(x))
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Given two polynomials P,Q, which have x^2+x+1 as a factor they can be expressed as
P(x) = f(x)(x^2+x+1)
Q(x) = g(x)(x^2+x+1)
P(x)+Q(x) = (f(x)+g(x))(x^2+x+1)
c*P(x) = cf(x)(x^2+x+1)
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Almost works, except multiplying by a scalar would yield a tuple of real numbers. Kinda silly. In programming, if you attempted to write a function
function List<Int> multiply(List<Int> list, Real c) {
return list.map((x) => x * c);
}
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Adding two sequences
(x_1,...x_n) + (y_1...,y_n)
(x_1+y_1...,x_n+y_n)
(x1+y1) + (x2+y2)... + (xn+yn) = (x1+x2+..xn) + (y1+...+yn) = (0) + (0)
Also, scalar scaling works fine.
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The short explanation as to why we insist on scalars being real or complex is that a major goal of linear algebra is to provide ways of solving equations. And you really want to perform division to solve equations, which real and complex numbers let you do. Integers aren't closed over division, so they aren't good for solving linear equations.
For example, if your vector space is integer triplets then
2 * x = (1,0,0)
I personally think part of the problem is how elementary education is structured. At least in my state, elementary school teachers are expected to be extreme generalists, and only have to take a math class or two -- and then nothing above simple college algebra (which is fine). But they always complain about how difficult it is, and they don't understand math, often taking the state's required exam multiple times because they can't pass math. Yet these are the people we allow to teach kids math; it's a huge issue when they're being taught math by people who don't understand why it works, only the algorithms they've memorized.
Coincidentally, this is also why the 'new math' was so lambasted -- these teachers (and often, parents) don't understand how numbers work, thus they think it's useless to teach kids to subtract 20 and add 2 instead of subtracting 18. Despite the fact that one is much easier to do mentally, and allows you to get a good sense of how subtraction and addition interact.
I'm not a fan of charter schools, but if I had money, I'd start an elementary charter school where the subjects were taught by people who understood that and not generalists. And students would get like multiple hours of recess a day, especially those first few years. Just pure, unstructured play time. But that's a rant for another day.
"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 16 and over in a further education college. They were constantly tripping over really quite basic little skill issues and that prevented them from seeing how to tackle the longer and more complex problem solving questions.
"I would go for something roughly equivalent [in the solving of equations such as 4x - 8 = 2x + 2], but not quite the same, 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). Then the problem of solving linear equations would be reduced to a kind of puzzle: what can we do to both sides of this equation to make the whole thing look simpler?"
The idea of just playing with the notation is one I fully intend to try but getting people to think in that abstract way is hard work.
As a computer guy who hates state machines and was always obsessed with math, I feel that just about everything about maths is taught wrong from the get go.
Just the other day I learned about something inductive function got me curious about: linear ordering of structures as proof of termination. Turns out it's been studied in math for long: it's called a well-order. Fine.. thing is we're taught about linear recursion in HS .. but we have no pragmatic notion of induction except ~~ P n-1 => P n ~~ It's so cryptically compressed that I suspect no student beside aspies and other prodigies can have the slightest clue about that. Yet it's so important (and so obvious when shown).
Reading that article I now realise it was that I lacked fluency. I didn't instinctively "know" how to do simultaneous equations because unlike my peers I hadn't spent two years doing them, so I had to remember how to solve them every single time.
All I can say now is thankfully there is the Khan Academy which rapidly improved my mathematical understanding when I needed it.
Wikipedia says "the scalars can be taken from any field, including the rational, algebraic, real, and complex numbers, as well as finite fields."
Note the "I think [...] almost everyone would get this question right (though I’d love to do the experiment)". This is a familiar state. Widespread. Call it, teachers who have not yet had their "oh shit!" moment.
One of the blog comments points at Eric Mazur's (Harvard, physics) oft-repeated talk "Confessions of a Converted Lecturer". Who describes the first time he gave students a Force Concept Inventory. Worried about wasting their time with such easy questions. :) Unaware physics education research was about to become a focus of his career.
Many have been surprised by "Minds of Our Own" (1997) https://www.learner.org/resources/series26.html The short (3 min) introductory video shows MIT and Harvard students struggling to light a bulb with a battery and a wire. Full episodes are below (by clicking on "VoD" buttons).
Harvard Center for Astrophysics has both first-tier astronomy and astronomy education programs. When meeting a new CfA graduate student, I've a little drill, prompting for the color of the Sun, and then of sunlight. They almost always get the first wrong, and then get a conflict, often with a nice "oh, wait, that doesn't make sense does it" moment. The collision of two bits of non-integrated and flawed understanding. Of the few who get it right, halfish (but small N) learned it from CfA instruction on common misconceptions in astronomy education, rather than from their own astronomy education.
But perhaps mathematics is doing better at robust integrated understanding than are astronomy, physics, chemistry, biology and medical school. It seems possible at least.
It's not just people who have had, or not had, their "oh shit!" moment. Professions too. Medicine realizing that medical errors were a major cause of mortality. Realizing even cheap easy universally-approved interventions (aspirin for ER chest pain) weren't consistently being executed. Realizing other industries had decades of experience on how to pursue quality, to which medicine had been oblivious. When the New York Times babbles about "Truth" and "The Journalism You Deserve", I shake my head and think, there's a field that has no clue how badly it's doing, how much work on process quality it's unaware of; a field that has not yet had its "oh shit!" moment.
Except, of course, the book really is about how he feels he should’ve taught math.
I also agree fully. A little while back I did some support tutoring for A-level maths students. The number of students who turned up who mysteriously "had problems with longer questions"... I wish I'd known the example of calculating the perimeter of the rectangle with fractions. That would have really helped explain why the problem wasn't really the length of the question, it was the fact that the student had never properly learned the component skills separately.
Unfortunately, the problem of building impressive-looking edifices on shaky foundations is absolutely endemic in British high-school maths teaching. Thousands of students who never quite understood fractions are "learning" calculus through being taught recipes, and the easier exam questions are formulaic enough that they get through with Cs at least, without any mathematical understanding.
The A-level statistics modules, in particular, have very impressive _sounding_ syllabuses. Students learn T-tests, Chi-squared tests, all this sophisticated statistical machinery. If all these students really understood this stuff, Britain would have a vast army of highly trained statisticians. But nothing of the sort is true, of course: students are just learning a recipe for processing numbers. I can't imagine the carnage if a statistics exam asked the students to write an essay explaining the principle by which a T-test works.
Pardon my rant, this has been on my mind for a while.
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.wordpress.com/2012/11/20/what-maths-a-level-d...
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.
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.
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.
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.
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...)
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 about. It was the surest, saddest sign that the powers that be had given up on genuinely educating the average A-level maths student.
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.
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)?
For instance the real numbers are a vector space over the rationals. They are a different vector space over the reals. They are not a vector space over the complex numbers and are not a vector space over the integers. But they are a module over the integers. But not a module over the complex numbers.
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.
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.
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.
"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.
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.
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.
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.
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.
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.
The whole purpose of this exercise is to see if there was a way to come up with a straightforward, reasonably informal, multiple choice question that would expose a fundamental understanding in basic university math concepts like "vector space" in the same way we see in primary math.
And instead all you people want to do is natter over the ways in which someone could cleverly make the "wrong" answer right. It's... beyond missing the point, it's actively working against the whole goal of the exercise.
By itself, this is a minor complaint (you cannot include every example in you choices, although I do think that an example which could not be viewed as an R-vector space would be good to include). However, when you explain why Z3 is not a vector space, your explanation must be correct. An explanation which also excludes Q3 is incorrect.
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.
> 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?
> ...
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.
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.
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.
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.
I'm afraid you have badly misremembered this stuff.