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Sun, 19 Jul 2015

Math.SE report 2015-04

[ Notice: I originally published this report at the wrong URL. I moved it so that I could publish the June 2015 report at that URL instead. If you're seeing this for the second time, you might want to read the June article instead. ]

A lot of the stuff I've written in the past couple of years has been on Mathematics StackExchange. Some of it is pretty mundane, but some is interesting. I thought I might have a little meta-discussion in the blog and see how that goes. These are the noteworthy posts I made in April 2015.

  • Languages and their relation : help is pretty mundane, but interesting for one reason: OP was confused about a statement in a textbook, and provided a reference, which OPs don't always do. The text used the symbol !!\subset_\ne!!. OP had interpreted it as meaning !!\not\subseteq!!, but I think what was meant was !!\subsetneq!!.

    I dug up a copy of the text and groveled over it looking for the explanation of !!\subset_\ne!!, which is not standard. There was none that I could find. The book even had a section with a glossary of notation, which didn't mention !!\subset_\ne!!. Math professors can be assholes sometimes.

  • Is there an operation that takes !!a^b!! and !!a^c!!, and returns !!a^{bc}!! is more interesting. First off, why is this even a reasonable question? Why should there be such an operation? But note that there is an operation that takes !!a^b!! and !!a^c!! and returns !!a^{b+c}!!, namely, multiplication, so it's plausible that the operation that OP wants might also exist.

    But it's easy to see that there is no operation that takes !!a^b!! and !!a^c!! and returns !!a^{bc}!!: just observe that although !!4^2=2^4!!, the putative operation (call it !!f!!) should take !!f(2^4, 2^4)!! and yield !!2^{4\cdot4} = 2^{16} = 65536!!, but it should also take !!f(4^2, 4^2)!! and yield !!4^{2\cdot2} = 2^4 = 256!!. So the operation is not well-defined. And you can take this even further: !!2^4!! can be written as !!e^{4\log 2}!!, so !!f!! should also take !!f(e^{2\log 4}, e^{2\log 4})!! and yield !!e^{4(\log 4)^2} \approx 2180.37!!.

    They key point is that the representation of a number, or even an integer, in the form !!a^b!! is not unique. (Jargon: "exponentiation is not injective".) You can raise !!a^b!!, but having done so you cannot look at the result and know what !!a!! and !!b!! were, which is what !!f!! needs to do.

    But if !!f!! can't do it, how can multiplication do it when it multiplies !!a^b!! and !!a^c!! and gets !!a^{b+c}!!? Does it somehow know what !!a!! is? No, it turns out that it doesn't need !!a!! in this case. There is something magical going on there, ultimately related to the fact that if some quantity is increasing by a factor of !!x!! every !!t!! units of time, then there is some !!t_2!! for which it is exactly doubling every !!t_2!! units of time. Because of this there is a marvelous group homomophism $$\log : \langle \Bbb R^+, \times\rangle \to \langle \Bbb R ,+\rangle$$ which can change multiplication into addition without knowing what the base numbers are.

    In that thread I had a brief argument with someone who thinks that operators apply to expressions rather than to numbers. Well, you can say this, but it makes the question trivial: you can certainly have an "operator" that takes expressions !!a^b!! and !!a^c!! and yields the expression !!a^{bc}!!. You just can't expect to apply it to numbers, such as !!16!! and !!16!!, because those numbers are not expressions in the form !!a^b!!. I remembered the argument going on longer than it did; I originally ended this paragraph with a lament that I wasted more than two comments on this guy, but looking at the record, it seems that I didn't. Good work, Mr. Dominus.

  • how 1/0.5 is equal to 2? wants a simple explanation. Very likely OP is a primary school student. The question reminds me of a similar question, asking why the long division algorithm is the way it is. Each of these is a failure of education to explain what division is actually doing. The long division answer is that long division is an optimization for repeated subtraction; to divide !!450\div 3!! you want to know how many shares of three cookies each you can get from !!450!! cookies. Long division is simply a notation for keeping track of removing !!100!! shares, leaving !!150!! cookies, then !!5\cdot 10!! further shares, leaving none.

    In this question there was a similar answer. !!1/0.5!! is !!2!! because if you have one cookie, and want to give each kid a share of !!0.5!! cookies, you can get out two shares. Simple enough.

    I like division examples that involve giving cookies to kids, because cookies are easy to focus on, and because the motivation for equal shares is intuitively understood by everyone who has kids, or who has been one.

    There is a general pedagogical principle that an ounce of examples are worth a pound of theory. My answer here is a good example of that. When you explain the theory, you're telling the student how to understand it. When you give an example, though, if it's the right example, the student can't help but understand it, and when they do they'll understand it in their own way, which is better than if you told them how.

  • How to read a cycle graph? is interesting because hapless OP is asking for an explanation of a particularly strange diagram from Wikipedia. I'm familiar with the eccentric Wikipedian who drew this, and I was glad that I was around to say "The other stuff in this diagram is nonstandard stuff that the somewhat eccentric author made up. Don't worry if it's not clear; this author is notorious for that."

  • In Expected number of die tosses to get something less than 5, OP calculated as follows: The first die roll is a winner !!\frac23!! of the time. The second roll is the first winner !!\frac13\cdot\frac23!! of the time. The third roll is the first winner !!\frac13\cdot\frac13\cdot\frac23!! of the time. Summing the series !!\sum_n \frac23\left(\frac13\right)^nn!! we eventually obtain the answer, !!\frac32!!. The accepted answer does it this way also.

    But there's a much easier way to solve this problem. What we really want to know is: how many rolls before we expect to have seen one good one? And the answer is: the expected number of winners per die roll is !!\frac23!!, expectations are additive, so the expected number of winners per !!n!! die rolls is !!\frac23n!!, and so we need !!n=\frac32!! rolls to expect one winner. Problem solved!

    I first discovered this when I was around fifteen, and wrote about it here a few years ago.

    As I've mentioned before, this is one of the best things about mathematics: not that it works, but that you can do it by whatever method that occurs to you and you get the same answer. This is where mathematics pedagogy goes wrong most often: it proscribes that you must get the answer by method X, rather than that you must get the answer by hook or by crook. If the student uses method Y, and it works (and if it is correct) that should be worth full credit.

    Bad instructors always say "Well, we need to test to see if the student knows method X." No, we should be testing to see if the student can solve problem P. If we are testing for method X, that is a failure of the test or of the curriculum. Because if method X is useful, it is useful because for some problems, it is the only method that works. It is the instructor's job to find one of these problems and put it on the test. If there is no such problem, then X is useless and it is the instructor's job to omit it from the curriculum. If Y always works, but X is faster, it is the instructor's job to explain this, and then to assign a problem for the test where Y would take more time than is available.

    I see now I wrote the same thing in 2006. It bears repeating. I also said it again a couple of years ago on itself in reply to a similar comment by Brian Scott:

    If the goal is to teach students how to write proofs by induction, the instructor should damned well come up with problems for which induction is the best approach. And if even then a student comes up with a different approach, the instructor should be pleased. ... The directions should not begin [with "prove by induction"]. I consider it a failure on the part of the instructor if he or she has to specify a technique in order to give students practice in applying it.

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