How Craig Barton wishes he’d taught maths
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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
#2"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.
Re: How Craig Barton wishes he’d taught maths
#3But 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.
Re: How Craig Barton wishes he’d taught maths
#4Wow, 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…
Re: How Craig Barton wishes he’d taught maths
#5Wow, 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…
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?
Re: How Craig Barton wishes he’d taught maths
#6Wow, 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…
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?
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.
Re: How Craig Barton wishes he’d taught maths
#7Wow, 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…
edit: in fact i learned linear algebra from hoffman kunze
Re: How Craig Barton wishes he’d taught maths
#8Reading 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.
Re: How Craig Barton wishes he’d taught maths
#9Wow, 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…
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?
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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A. The set of all complex numbers.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.
'
B. The set of all functions from (0,1) to \mathbb R that are twice differentiable.Two twice differentiable functions can be added together to yield a twice differentiable function
(f(x)+g(x))''=f''(x)+g''(x)
and scaled by a scalar (cf(x))''=c(f''(x))
'
C. The set of all polynomials in x with real coefficients that have x^2+x+1 as a factor.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)
So P(x)+Q(x) = (f(x)+g(x))(x^2+x+1)
And of course c*P(x) = cf(x)(x^2+x+1)
Which is also a polynomial with real coefficients. '
D. The set of all triples (a,b,c) of integers.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 multiply(List list, Real c) {
return list.map((x) => x * c);
}
You'd get a compiler exception since the return type wouldn't match (or you'd get some warning about shortening the precision). '
E. The set of all sequences (x_1,\dots,x_n)\in\mathbb R^n such that x_1+\dots+x_n=0 and x_1+2x_2+\dots+nx_n=0.Adding two sequences
(x_1,...x_n) + (y_1...,y_n)
gets you (x_1+y_1...,x_n+y_n)
which plugging into the above two equations, and rearranging, will show that you'd get (x1+y1) + (x2+y2)... + (xn+yn) = (x1+x2+..xn) + (y1+...+yn) = (0) + (0)
and the same for the second equation.Also, scalar scaling works fine.
'
So only D has any issues.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)
Wouldn't have any solutions in the set of integer triplets. This is a linear equation, and the goal of linear algebra is to provide solutions, so it's better to have the framework yield the answer of x = (.5,0,0) and then say "Oh, the answer lies outside of the original set, so it wasn't a vector space to start with." Well, that's kinda the extrinsic view of it, the intrinsic view would throw an exception :).Re: How Craig Barton wishes he’d taught maths
#10Earlier 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.