Friday 29 July 2011

Review: Mathematics Education for a new era: video games as a Medium for learning

June 18th, 2011 | by Sol |

How can we rethink the current Math education paradigm to consider the wealth of available technology? Can video games teach basic math skills or are they just a waste of time? Video game helps students to Math proficiency? What are the key elements in a game that promotes the mastery of mathematics? These are some of the "game change" questions, Keith Devlin tackles in his new book "mathematics education of a new era: video games as a Mediumfor learning."

Keith Devlin is well qualified to explore these important issues. He is researcher focused on using different media to teach mathematics. He is the author of 30 books; a number of them to examine how we learn Math. Devlin has published more than 80 research articles and he has won numerous prestigious prizes and awards. And many of by Keith Devlin as "Math guy" on National Public Radio, the man who since 1995 has takes important mathematical ideas from current events and explain them as general audience can understand them.

I am very fascinated by the question of how computers and specific game can help students to learn Mathematics. I must admit that I am somewhat apprehensive about video games the power to change the Math education, but it is, because all I have seen, are mathematical exercises with sweet graphics and sound effects thrown in to try to keep pupils. Devlin does not address these types of games, but. He thinks games where pupils identify deeply with the game. Garden, or "is" an avatar is an important way to help kids identify with games. In a successful game the player gets a role and the game supports him or her to grow and mature in this role. If a game is designed to improve Math proficiency can include avatars that are competent in the subject and give experience to the player "grow into" competence. Devlin makes the seemingly subtle, but very important distinction between "do Math" and "Math." He claims, and i agree that a person who perceives himself as a person who grows as a "mathematical able" person is far more likely to succeed than a person who simply "do the Math." The former is an identity, while the latter is that many, a task.

I lead Math circles in Santa Fe. We collect one night a month. I throw out a math problem, ask many questions and encourage group members, which include adults and children, to investigate the problem, to make and test conjectures, and, most importantly, I invite the group to have fun. People as a group, they learn things, and they have fun. After having read the Devlins book realize I have created an environment for learning, which has most of the ten most important features of the game that Devlin borrows for his work:

failure not hurt risk is a part of the game need for immediate feedback is used to "Star" trial and error is almost always the best plan, there is always an answer, I can do the math competition is always fun and familiar bosses and rules is less important group action and conflict

I Devlins focus on games, assumes the right games. I hope that the educational game designers use his ideas when designing educational opportunities. And in the meantime, teachers (and Math circle leaders) would do well to borrow some of the ideas in what works in the virtual worlds in their classrooms.

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Thursday 28 July 2011

Wild about Math blogs 5/27/11

June 3rd, 2011 | by Sol |

Welcome to wild about Math blogs!

This is the last edition. If you want to put on your own personal Math blog carnival, I recommend you follow the large list of blogs at Mathblogging.org and let us all know about the items you like. I trøde, I had a rather large list of Math blogs in my RSS reader; their list is much larger. You can subscribe to parts or all of their list via RSS and you can even follow the Twitter feeds of a bunch of Math bloggers.

I discovered some really wonderful BBC Math radio shows. See here.

Math teachers play Carnival # 38 is up on mathematics and multimedia.

I have spend time on the website, strong Math Pickle enjoys the simple yet deep and difficult to solve mathematical puzzles and games there since "inspired people" are particularly noteworthy. There are some familiar faces on the page, Martin Gardner, We Hart and James Tanton to name a few. And there are a lot of people I don't know much about whom I have read on. Here is an inspired person from the list:

Leo Moser seems to have been the first person who are in favour of unresolved problems used in K-12 education. He asked many hard problems with child-like eagerness: "what is the area in at least the House that a device worm can live comfortably?" importance which shape can cover a worm regardless of how he curls up?

And also from the site Math Pickle is a video of a fun division games with some deep things happening under the surface.

MAA NumberADay has an interesting bit in number theory:

The product of four primes in a prime quadruplet (with the exception of 5, 7, 11, 13) always ends in 189. Example: 101 x 103 x 107 x 109 = 121, 330, 189.

I was curious about why this should be and I got this tip from Wikipedia:

All primary quadruplets except {5, 7, 11, 13} is of the form {30n + 11, 30n + 13, 30n + 17, 30n + 19} for some integer n. (this structure is necessary to ensure that none of the four prime numbers are divisible by 2, 3 or 5).

You can see why the product should end in 189 (apart from the first set)?

Sue on Math Math writes has a challenging Math problem from "Rediscovering mathematics, by Shai Simonson."

Finally, here is a funny number line cartoon from xkcd.

Hat tip to Ars Mathematica.

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Tuesday 26 July 2011

Topology antics

June 4, 2011 | by Sol |

Here is a very clever topology trick!

More information on Division by zero.

Hat tip to the multiplication of infinity.

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Monday 25 July 2011

Review: Mathematica in action

June 14th, 2011 | by Sol |

[Editor's note: If you are using Mathematica you can appreciate the revision I just published on my player with Mathematica blog.]

"Stan Wagon, and have exchanged a number of e-mails about Mathematica. Some messages in the dialogue I realized that I should review his latest book: Mathematica ® in action: Problem solving through visualization and calculation. Before I immersed myself in the book I knew I it because I enjoy Stan's playful relationship with Mathematica and enjoy receiving the simple and elegant little programs that Stan would send me. "

Read the review.

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Friday 22 July 2011

Three high school courses via MIT OpenCourseWare

June 4, 2011 | by Sol |

If you do not have discovered the MIT courses for high school students, you owe yourself a look. There are courses in mathematics, science and the humanities, all available for free via the Internet, with the course notes, and homework assignments included. These rates are a fraction of the more than 2000 MIT OpenCourseWare classes.

Three of the courses is in mathematics:

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Wednesday 20 July 2011

A fixed point Brouwer's invariance of domain theorems, and Hilbert's fifth problem

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We recall Brouwer’s famous fixed point theorem:

Theorem 1 (Brouwer fixed point theorem) Let {f: B^n \rightarrow B^n} be a continuous function on the unit ball {B^n := \{ x \in {\bf R}^n: \|x\| \leq 1 \}} in a Euclidean space {{\bf R}^n}. Then {f} has at least one fixed point, thus there exists {x \in B^n} with {f(x)=x}.

This theorem has many proofs, most of which revolve (either explicitly or implicitly) around the notion of the degree of a continuous map {f: S^{n-1} \rightarrow S^{n-1}} of the unit sphere {S^{n-1} := \{ x \in {\bf R}^n: \|x\|=1\}} to itself, and more precisely around the stability of degree with respect to homotopy. (Indeed, one can view the Brouwer fixed point theorem as an assertion that some non-trivial degree-like invariant must exist, or more abstractly that the homotopy group {\pi_{n-1}(S^{n-1})} is non-trivial.)

One of the many applications of this result is to prove Brouwer’s invariance of domain theorem:

Theorem 2 (Brouwer invariance of domain theorem) Let {U} be an open subset of {{\bf R}^n}, and let {f: U \rightarrow {\bf R}^n} be a continuous injective map. Then {f(U)} is also open.

This theorem in turn has an important corollary:

Corollary 3 (Topological invariance of dimension) If {n > m}, and {U} is a non-empty open subset of {{\bf R}^n}, then there is no continuous injective mapping from {U} to {{\bf R}^m}. In particular, {{\bf R}^n} and {{\bf R}^m} are not homeomorphic.

This corollary is intuitively obvious, but note that topological intuition is not always rigorous. For instance, it is intuitively plausible that there should be no continuous surjection from {{\bf R}^m} to {{\bf R}^n} for {n>m}, but such surjections always exist, thanks to variants of the Peano curve construction.

Theorem 2 or Corollary 3 can be proven by simple ad hoc means for small values of {n} or {m} (for instance, by noting that removing a point from {{\bf R}^n} will disconnect {{\bf R}^n} when {n=1}, but not for {n>1}), but I do not know of any proof of these results in general dimension that does not require algebraic topology machinery that is at least as sophisticated as the Brouwer fixed point theorem. (Lebesgue, for instance, famously failed to establish the above corollary rigorously, although he did end up discovering the important concept of Lebesgue covering dimension as a result of his efforts.)

Nowadays, the invariance of domain theorem is usually proven using the machinery of singular homology. In this post I would like to record a short proof of Theorem 2 using Theorem 1 that I discovered in a paper of Kulpa, which avoids any use of algebraic topology tools beyond the fixed point theorem, though it is more ad hoc in its approach than the systematic singular homology approach.

Remark 1 A heuristic explanation as to why the Brouwer fixed point theorem is more or less a necessary ingredient in the proof of the invariance of domain theorem is that a counterexample to the former result could conceivably be used to create a counterexample to the latter one. Indeed, if the Brouwer fixed point theorem failed, then (as is well known) one would be able to find a continuous function {F: B^n \rightarrow S^{n-1}} that was the identity on {S^{n-1}} (indeed, one could take {F(x)} to be the first point in which the ray from {f(x)} through {x} hits {S^{n-1}}). If one then considered the function {G: B^n \rightarrow {\bf R}^n} defined by {G(x) := (1+\|x\|) F(x)}, then this would be a continuous function which avoids the interior of {B^n}, but which maps the origin {0} to a point on the sphere {S^{n-1}} (and maps {S^{n-1}} to the dilate {2 \cdot S^{n-1}}). This could conceivably be a counterexample to Theorem 2, except that {G} is not necessarily injective. I do not know if there is a more rigorous way to formulate this connection.

The reason I was looking for a proof of the invariance of domain theorem was that it comes up in the very last stage of the solution to Hilbert’s fifth problem, namely to establish the following fact:

Theorem 4 (Hilbert’s fifth problem) Every locally Euclidean group is isomorphic to a Lie group.

Recall that a locally Euclidean group is a topological group which is locally homeomorphic to an open subset of a Euclidean space {{\bf R}^n}, i.e. it is a continuous manifold. Note in contrast that a Lie group is a topological group which is locally diffeomorphic to an open subset of {{\bf R}^n}, it is a smooth manifold. Thus, Hilbert’s fifth problem is a manifestation of the “rigidity” of algebraic structure (in this case, group structure), which turns weak regularity (continuity) into strong regularity (smoothness).

It is plausible that something like Corollary 3 would need to be invoked in order to solve Hilbert’s fifth problem. After all, if Euclidean spaces {{\bf R}^n}, {{\bf R}^m} of different dimension were homeomorphic to each other, then the property of being locally Euclidean loses a lot of meaning, and would thus not be a particularly powerful hypothesis. Note also that it is clear that two Lie groups can only be isomorphic if they have the same dimension, so in view of Theorem 4, it becomes plausible that two Euclidean spaces can only be homeomorphic if they have the same dimension, although I do not know of a way to rigorously deduce this claim from Theorem 4.

Interestingly, Corollary 3 is the only place where algebraic topology enters into the solution of Hilbert’s fifth problem (although its cousin, point-set topology, is used all over the place). There are results closely related to Theorem 4, such as the Gleason-Yamabe theorem mentioned in a recent post, which do not use the notion of being locally Euclidean, and do not require algebraic topological methods in their proof. Indeed, one can deduce Theorem 4 from the Gleason-Yamabe theorem and invariance of domain; we sketch a proof of this (following Montgomery and Zippin) below the fold.

— 1. Invariance of domain —

Now we prove Theorem 2. By rescaling and translation invariance, it will suffice to show the following claim:

Theorem 5 (Invariance of domain, again) Let {f: B^n \rightarrow {\bf R}^n} be an continuous injective map. Then {f(0)} lies in the interior of {f(B^n)}.

Let {f} be as in Theorem 5. The map {f: B^n \rightarrow f(B^n)} is a continuous bijection between compact Hausdorff spaces and is thus a homeomorphism. In particular, the inverse map {f^{-1}: f(B^n) \rightarrow B^n} is continuous. Using the Tietze extension theorem, we can find a continuous function {G: {\bf R}^n \rightarrow {\bf R}^n} that extends {f^{-1}}.

The function {G} has a zero on {f(B^n)}, namely at {f(0)}. We can use the Brouwer fixed point theorem to show that this zero is stable:

Lemma 6 (Stability of zero) Let {\tilde G: f(B^n) \rightarrow {\bf R}^n} be a continuous function such that {\|G(y)-\tilde G(y)\| \leq 1} for all {x \in f(B^n)}. Then {\tilde G} has at least one zero (i.e. there is an {y \in f(B^n)} such that {\tilde G(y)=0}).

Proof: Apply Theorem 1 to the function

\displaystyle x \mapsto x - \tilde G(f(x)) = G(f(x)) - \tilde G(f(x)).

\Box

Now suppose that Theorem 5 failed, so that {f(0)} is a not an interior point of {f(B^n)}. We will use this to locate a small perturbation of {G} that no longer has a zero on {f(B^n)}, contradicting Lemma 6.

We turn to the details. Let {\epsilon > 0} be a small number to be chosen later. By continuity of {G}, we see (if {\epsilon} is small enough) that we have {\|G(y)\| \leq 0.1} whenever {y \in {\bf R}^n} and {\|y-f(0)\| \leq 2\epsilon}.

On the other hand, since {f(0)} is not an interior point of {f(B^n)}, there exists a point {c \in {\bf R}^n} with {\|c-f(0)\| \leq \epsilon} that lies outside {f(B^n)}. By translating {f} if necessary, we may take {c=0}; thus {f(B^n)} avoids zero, {\|f(0)\| \leq \epsilon}, and we have \displaystyle \|G(y)\| \leq 0.1 \hbox{ whenever } \|y\| \leq \epsilon. \ \ \ \ \ (1)

Let {\Sigma} denote the set {\Sigma := \Sigma_1 \cup \Sigma_2}, where

\displaystyle \Sigma_1 := \{ y \in f(B^n): \|y\| \geq \epsilon \}

and \displaystyle \Sigma_2 := \{ y \in {\bf R}^n: \|y\| = \epsilon \}.

By construction, {\Sigma} is compact but does not contain {f(0)}. It is a perturbation of {f(B^n)}, and crucially there is a continuous map {\Phi: f(B^n) \rightarrow \Sigma} defined by setting \displaystyle \Phi(y) := \max( \frac{\epsilon}{\|y\|}, 1 ) y. \ \ \ \ \ (2)

Note that {\Phi} is continuous and well-defined since {f(B^n)} avoids zero.

By construction, {G} is non-zero on {\Sigma_1}; since {\Sigma_1} is compact, {G} is bounded from below on {\Sigma_1} by some such that ; in particular, {P} does not vanish on {\Sigma_1}. At present, it is possible that {P} vanishes on {\Sigma_2}. But as {P} is smooth and {\Sigma_2} has measure zero, {P(\Sigma_2)} also has measure zero; so by shifting {P} by a small generic constant we may assume without loss of generality that {P} also does not vanish on {\Sigma_2}. (If one wishes, one can use an algebraic geometry argument here instead of a measure-theoretic one, noting that {P(\Sigma_2)} lies in an algebraic hypersurface and can thus be generically avoided by perturbation. A purely topological way to avoid zeroes in {\Sigma_2} is also given in Kulpa’s paper.)

Now consider the function {\tilde G: f(B^n) \rightarrow {\bf R}^n} defined by

\displaystyle \tilde G(y) := P( \Phi( y ) ).

This is a continuous function that is never zero. From (3), (2) we have is such that {\|y\| > \epsilon}. On the other hand, if {\|y\| \leq \epsilon}, then from (2), (1) we have \displaystyle \| G(y) \|, \| G( \Phi(y) ) \| \leq 0.1

and hence by (3) and the triangle inequality \displaystyle \| G(y) - \tilde G(y) \| \leq 0.2 + \delta.

Thus in all cases we have

\displaystyle \| G(y) - \tilde G(y) \| \leq 0.2 + \delta \leq 0.3

for all {y \in f(B^n)}. But this, combined with the non-vanishing nature of {\tilde G}, contradicts Lemma 6.

— 2. Hilbert’s fifth problem —

We now sketch how invariance of domain can be used to establish the solution to Hilbert’s fifth problem (as formulated in Theorem 4). Our main tool is the Gleason-Yamabe theorem, in the form stated in Theorem 4 of this previous post:

Theorem 7 (Gleason-Yamabe theorem) Every locally compact Hausdorff group has an open subgroup that is the projective limit of Lie groups.

A locally Euclidean group is of course locally compact Hausdorff; it is also first countable, and hence (by the Birkhoff-Kakutani theorem, discussed in this post) is also metrisable. Because of this, it is not difficult to show that the open subgroup given by the Gleason-Yamabe theorem in this case can be obtained as a projective limit

\displaystyle G := \lim_{\stackrel{\leftarrow}{n}} G_n

of a countable sequence {(G_n)_{n \in {\bf N}}} of Lie groups, with continuous homomorphisms from {G_{n+1}} to {G_n} for each {n}.

Using Cartan’s theorem to obtain the smoothness of outer automorphisms on Lie groups as in previous posts, we see that to show that a topological group is a Lie group, it suffices to locate an open subgroup that is a Lie group. In view of these reductions, it suffices to show

Proposition 8 Let {G} be a locally Euclidean group that is the inverse limit \displaystyle G := \lim_{\stackrel{\leftarrow}{n}} G_n

of Lie groups. Then {G} is itself isomorphic to a Lie group.

We now prove the proposition. We first observe by shrinking the {G_n} if necessary (and using Cartan’s theorem that locally compact subgroups of a Lie group are still Lie) we may assume that all the projection maps in the inverse limit are surjective; indeed, by local compactness we may take {G_n = G/H_n} where {H_n} are a sequence of compact normal subgroups converging to the identity.

As continuous homomorphisms of Lie groups are automatically smooth (as proven in this previous post), the projection maps from {G_{n+1}} to {G_n} induce an associated projection map from {{\mathfrak g}_{n+1}} to {{\mathfrak g}_n} at the Lie algebra level. As the former maps are surjective, the latter maps are also. In particular, the dimensions of the finite-dimensional Lie algebras {{\mathfrak g}_n} are non-decreasing in {n}.

Another consequence of surjectivity is that every tangent vector in {{\mathfrak g}_n} can be lifted up to {{\mathfrak g}_{n+1}}. Continuing this process and passing to the inverse limit, it is possible to show that every one-parameter subgroup in one of the factor groups {G_n} can be lifted up (locally, at least) to a one-parameter subgroup of {G}. In fact, one can show that a neighbourhood of the origin in {{\mathfrak g}_n} can be lifted up into {G}, to provide a continuous injection of an open subset of that Lie algebra into {G}.

Now we crucially use the invariance of domain (Corollary 3) to conclude that the dimensions of the Lie algebras {{\mathfrak g}_n} must be bounded in {n} (indeed, they cannot exceed the dimension of the locally Euclidean group {G}). As these dimensions are non-decreasing, they must therefore be eventually constant; by discarding finitely many of the factor groups, we may thus assume that all the {G_n} have the same dimension. In particular, the kernels of the projection maps from {G_{n+1}} to {G_n} are zero-dimensional Lie groups and are thus discrete. As {G_n = G/H_n}, this means that the {H_n/H_{n+1}} are also discrete, thus {H_{n+1}} is an open subgroup of {H_n}, and so (by compactness of {H_n}) has finite index. As the {H_n} converge to the trivial group as {n \rightarrow \infty}, we conclude that each of the {H_n} are profinite, and in particular are totally disconnected. Thus, for any {n}, we have a short exact sequence

\displaystyle 0 \rightarrow H_n \rightarrow G \rightarrow G_n \rightarrow 0

describing {G} as the extension of a Lie group {G_n} by a totally disconnected compact group {H_n}. Furthermore, we can lift a small neighbourhood of the identity of {G_n} to {G} in a continuously injective manner. This neighbourhood is connected, and acts on the normal subgroup {H_n} by conjugation. But {H_n} is totally disconnected, and so all orbits of this action must be points; thus, the lift of this neighbourhood commutes with {H_n}. As a consequence, one can show that {G} has the local topological (and group) structure of the direct product {H_n \times G_n}. But {G} is also locally Euclidean; as {H_n} is totally disconnected, these two statements are only compatible if {H_n} is discrete. Thus the Lie group {G_n} is an open subgroup of {G}, and we are done.

Remark 2 The above argument shows that any metrisable group which is “finite-dimensional” in the sense that it does not contain continuous injective images of non-trivial open sets of Euclidean spaces {{\bf R}^n} of arbitrarily large dimension, has the local structure (as a topological space) of the product of a Euclidean space and a totally disconnected space. Of course, direct products of Lie groups and totally disconnected groups provide one such example of this claim. A more non-trivial example is given by the solenoid groups, a simple example of which is the (additive) group \displaystyle G := ({\bf Z}_p \times {\bf R}) / {\bf Z}^\Delta

where {{\bf Z}^\delta := \{ (n,n): n \in {\bf Z} \}} is the diagonally embedded copy of the integers. This is a compact Hausdorff group that has a short exact sequence \displaystyle 0 \rightarrow {\bf Z}_p \rightarrow G \rightarrow {\bf R}/{\bf Z} \rightarrow 0

and has the local structure of {{\bf Z}_p \times {\bf R}/{\bf Z}}, although {{\bf R}/{\bf Z}} does not embed into this group and so this is not a direct or semi-direct product. (Topologically, it can be viewed as the set {{\bf Z}_p \times [0,1]} after identifying {(x,0)} with {(x+1,1)} for every {p}-adic integer {x \in {\bf Z}_p}.) It has topological dimension one, in the sense that it contains continuous injective images of open subsets of {{\bf R}^n} if and only if {n \leq 1}. It is also the projective limit of the Lie groups \displaystyle G_n := (({\bf Z}/p^n{\bf Z}) \times {\bf R})/{\bf Z}^\Delta,

each of which is isomorphic to the circle {{\bf R}/{\bf Z}}, but in a manner which becomes increasingly “twisted” as {n} increases (which helps explain the terminology “solenoid”). It is an instructive exercise to verify that {G} is connected, but not path connected or locally connected. Thus we see that connected locally compact groups can contain some significant non-Lie behaviour, even in the abelian setting.

Remark 3 Of course, it is possible for locally compact groups to be infinite-dimensional; a simple example is the infinite-dimensional torus {({\bf R}/{\bf Z})^{\bf N}}, which is compact, abelian, metrisable, and locally connected, but infinite dimensional. (It will still be an inverse limit of Lie groups, though.)

Remark 4 The above analysis also gives another purely topological characterisation of Lie groups; a topological group is Lie if and only if it is locally compact, Hausdorff, first-countable, locally connected, and finite-dimensional. It is interesting to note that this characterisation barely uses the real numbers {{\bf R}}, which are of course fundamental in defining the smooth structure of a Lie group; the only remaining reference to {{\bf R}} comes through the notion of finite dimensionality. It is also possible, using dimension theory, to obtain alternate characterisations of finite dimensionality (e.g. finite Lebesgue covering dimension) that avoid explicit mention of the real line, thus capturing the concept of a Lie group using only the concepts of point-set topology (and the concept of a group, of course).