Introduction to Mathematical Thinking

Prompted I think by Downes, I decided to look into Keith Devlin’s MOOC on Introduction to Mathematical Thinking.

It was nice to see some acknowledgement of the origin of the ‘MOOC’ term with Downes and Siemens, but not so much to see that brief acknowledgement followed by a claim that a concept only attains respectability when adopted by the likes of Harvard, MIT, and Stanford.

I will be interested in seeing how the peer evaluation system works out and in the effectiveness of the discussion forum software. Unfortunately I joined the course too late to really check out the latter although it does seem to have facilitated a number of effective collaborative learning groups.

I do have some comments with regard to the actual course content:

In the section on Number Theory in Lecture 9B a definition of b|a is given which does not require either the uniqueness of the quotient or the condition that b is nonzero. This definition is then reinforced by requiring students to endorse it in the first Quiz (at around 2:13), but in a subsequent Quiz (at around 4:45) the student is expected to deny the truth of 0|0. This actually led to some discussion in the fora which does not as yet seem to have generated any instructor response.

In the same lecture the Unique Factorization Theorem (Fundamental Theorem of Arithmetic) is proved by using Euclid’s Lemma to establish uniqueness but the proof of that Lemma is omitted as “outside the scope of this course”(around 17:20). I think this is unfortunate because it is against the spirit of the subject to unnecessarily accept things without proof – especially when the proof is not that hard[1], but more importantly because early emphasis on unique factorization makes the student think of it as fundamental (not surprising given the popular name of the theorem) and so to end up using it to prove the Lemma. Several students did try this in the discussion groups but so far without any correction from the instructors (and although there is some value in letting them work things out themselves there is a danger in this type of course that some will “escape” with the wrong idea without ever having had the chance to be corrected).

And in the Analysis section Lecture 10 the fact that the rationals are densely ordered is given (at about 4:00 in Lecture 10A) and reinforced (in a Quiz at about 7:15 and again at about 1:30 in Lecture 10B) without distinguishing it from the more commonly referenced property of topological density in the reals – (which is usually just called density without the modifier). This may lead students into error in subsequent courses since they may be tempted to use the proof of densely ordered as a proof of topological density – which it is not since eg (-1,-1/2)U(1/2,1) is densely ordered but not dense in R.



[1] Here’s a fairly compact proof of Euclid’s Lemma (with a more compact but still essentially complete version in bold):

We’ll first show how to find the greatest common factor of two whole numbers as a difference of multiples and then use the fact that the gcf is 1 if either of the numbers is a prime which does not divide the other to show that if prime p divides ab then it must also divide at least one of a or b.

Any common factor of two whole numbers is also a factor in any sum or difference of multiples of them.
[In symbols, if a=rd and b=sd then na+mb=nrd+msd=(nr+ms)d and it works similarly for differences. (For convenience we’ll work with integer arithmetic from now on so that we don’t have to list all the different sign combinations.)]

Any division of such sums or differences of multiples gives a remainder which is also a sum or difference of multiples.
[If na+mb=q(ja+kb)+r, then r=na+mb-q(ja+kb)=(n-qj)a+(m-qk)b ]

A sum or difference of whole number multiples of two numbers is sometimes called an integer linear combination but I’ll just call it a “combination” for the rest of this proof (and so will refrain from using that word “combination” for other types of algebraic combinations).

The smallest such combination is zero [Take each as the multiplier for the other and subtract], so when dividing any of them by the next to smallest, the remainder must be zero (since it must be smaller than the one we were dividing by). Thus the smallest non-zero combination divides exactly into all of them (including a and b as special cases). So the smallest non-zero combination of a and b is a common factor of a and b.

And since it is a multiple of any other common factor it must be greater than or equal to all of them.
So it is the greatest common factor.

If p is prime and does not divide x then gcf(p,x)=1
So if p does not divide either a or b, then we must have integers j,k,m,n such that ja+kp=1 and mb+np=1.
So 1=(ja+kp)(mb+np)=jmab+janp+mbkp++knpp=rab+sp

with r=jm and s=jan+mbk+knp
So 1 is a combination of ab and p which means that gcf(p,ab)=1 and so p cannot divide ab (since if it did the gcf would be p).

We have shown that if prime p does not divide at least one of a or b then it cannot divide the product ab, which is the same as saying that if prime p divides ab then it must divide at least one of a or b.

Note: A slightly less symmetric version of the last bit uses less algebra but doesn’t feel quite so compelling for me:
If prime p does not divide a
then there are integers m and n with 1=np+ma
This gives b=(np+ma)b=npb+mab
so if p|ab ie ab=qp for some q, then b=nbp+mab=nbp+mqp=(nb+mq)p
So if prime p does not divide a but does divide ab then it must divide b.

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