Math.SE report 201504
[ 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 metadiscussion 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 welldefined. 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 math.se
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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