## Author Archive

### A cool sequence problem…

June 15, 2010

Our oldest son (nearly 10) posed the following challenge:

What comes next in this list?

1, 1, 1, 2, 2, 3, 2, 4, 3, …

Answer and rationale (his and mine) after the jump…

### A billion here, a billion there; soon you’re talking a real portfolio!

May 6, 2010

What is a billion?

• In contemporary USA, it is 109
• In contemporary France, it is 1012
• In the UK prior to the early 1970s, it is 1012
• In the UK after the early 1970s, it is 109
• In most European countries, it is 1012
• In 19th C France, it is 109
• In 14th – 18th C France, it is 1012
• It is a modernized spelling of “bymillion”, a word introduced in 1475 by Jehan Adam for a million2. (He also coined the term “trimillion” for a million3, and similar vocabulary for higher powers, vestiges of which remain in our number systems.)

Apparently, a billion is also a wickedly large number of shares of stock to be trading at one time. For if you accidentally hit the “b” key instead of the “m” key at your computer, and thus execute a trade in billions of shares instead of millions of shares, you might cause the Dow Jones Industrial Average to drop 9 percent in a matter of moments on a Thursday afternoon. Or so I’ve heard.

### A=B implies that 1=1, therefore…

February 2, 2010

I’ve ranted in the past about the fallacy of trying to prove an identity by starting with the equation itself, then manipulating both sides of the equation until you arrive at a valid identity.

While grading some homework this term (involving proofs of trig identities), I found the need to raise the subject again in class.  But my stock example, proving that -3 = 3 by squaring both sides, seemed too transparent.  I wanted something where the fallacy was solely due to proving that False implies True.

I ended up with the following example, which I like a lot, but which I’m certain has been rediscovered by others over the ages.  Still, it’s a good illustration of why we can’t prove identities in this way.

Claim: $\sin x = \cos x$, for all $x$.

Proof: Assume that $\sin x = \cos x$.  If we square both sides, this implies that $\sin^2 x = \cos^2 x$.  Furthermore, since equality is reflexive symmetric, it follows that $\cos^2 x = \sin^2 x$.

Finally, adding these two equations gives us $\sin^2 x + \cos^2 x = \cos^2 x + \sin^2 x$, which reduces to the equation 1=1.  QED.

The careful reader will note that squaring both sides is irrelevant, as is the Pythagorean identity for sine and cosine.  In essence we have a general proof that $A = B$ for any expressions $A$ and $B$:  If $A = B$, then by reflexivity symmetry we know that $B=A$, and thus $A + B = B + A$.  But since addition is commutative, this reduces to the identity $A+B = A+B$.

I prefer the slight obfuscation of the $\sin^2 x + \cos^2 x$ proof over the distilled simplicity of “reflexive symmetric plus commutative”.

### How would Michael Palin say “Quod erat demonstrandum”?

December 17, 2009

The letters QED (or the unabbreviated “quod erat demonstrandum” [“which is what was to be proved”]) are often used to indicate the end of a formal proof.  This longstanding tradition goes back to a style of proof writing, where the culminating sentence of the argument gives a recapitulation of the statement of the theorem; the QED places a stamp of finality on the discussion.

In modern typesetting, the QED has been largely replaced with typographic symbols; typically a solid or hollow rectangle or square is used to demark the end of the proof.  (The cynic in me wonders if these just serve as flags for when the reader should take up reading carefully again.)

How does one indicate the end of a proof in a classroom setting?  Often I will scribble out a square (or whatever symbol our textbook uses); sometimes I’ve written out “Q.E.D.”.  Often I’ll pause, then solicit questions and comments.

But apparently I much more frequently channel Michael Palin.

Today I gave a final exam in Real Analysis II.  This group of students has gotten to work with me on proofs for a full year, so they know my quirks and foibles better than most.

On the last page of the final, most of the students ended their last proof with the phrase:

“And there was much rejoicing.”

This isn’t a phrase I’ve consciously chosen to use in class, but it rings true enough as something I’m sure I *have* said on occasion.  But if it made this much of an impression on the students, I wonder if I use it all the time, not just occasionally.  Hmmm.

Come to think of it, it isn’t such a bad replacement for Quod erat demonstrandum.  Captures the feeling of a good solid proof, well-understood.  And it is much more evocative than a box.

### Memorial Day 2009

May 25, 2009

Last year, we posted a brief discussion of the history of Memorial Day, a US holiday of remembrance of Americans who died in military service for their country.

Following up on last year’s post:   Frank Buckles is now 108, and [naturally] is still the last known surviving American veteran of World War One.  Last year’s NPR interview with him is still available on-line.

Photo taken by TwoPi in January 2008 at Fort Rosencrans National Cemetery, on Point Loma, San Diego,  CA.

### Sunscreen confusion

May 14, 2009

An article in the New York Times describes consumer confusion over the ever-rising SPF numbers (used to rate the efficacy of sunscreen lotions), and their interpretation.

Unfortunately, the NYT adds to the confusion with the following:

The difference in UVB protection between an SPF 100 and SPF 50 is marginal. Far from offering double the blockage, SPF 100 blocks 99 percent of UVB rays, while SPF 50 blocks 98 percent. (SPF 30, that old-timer, holds its own, deflecting 96.7 percent).

Technically they’re right:  doubling the blockage is not the same as halving your radiation exposure.  But in terms of safety, the issue isn’t how much UV exposure you’ve avoided, but rather how much UV actually gets to your skin cells (which would then be a 2% versus 1% comparison).

According to the article, SPF measures how much longer a person wearing sunscreen can be exposed to sunlight before getting a burn, when compared to someone wearing no sunscreen.  Someone wearing SPF 50 can remain in the sun 50 times longer than someone with no sunscreen, and so SPF 100 sunscreen provides the wearer with twice the protection (in terms of time) as SPF 50 sunscreen.

It turns out there is a sense in which SPF100 is not twice as effective as SPF50 in protecting your skin, but it has nothing to do with the 99%/98% comparison.

According to the NY Times, “a multiyear randomized study of about 1,600 residents of Queensland, Australia” found that most users applied at most half of the recommended amount of sunscreen.

“If people are putting on about half, they are receiving half the protection,” said Yohini Appa, the senior director of scientific affairs at Johnson & Johnson, of which Neutrogena is a subsidiary.

But in fact they are receiving far less than half the protection:   a 2007  British Journal of Dermatology study noted that cutting the amount of sunscreen in half did not reduce the effective SPF in half, but rather reduced it geometrically to its square root.

If a person uses half of the recommended amount of an SPF50 sunscreen, they’ll get the protection of an SPF7 (since 7.1 is roughly √50), while similarly underapplying SPF100 sunscreen gets the protection of SPF10.

Apparently, if you’re looking for the protection of an SPF30 product, but like most people tend to under-apply sunscreen, you should be shopping for sunscreen rated as SPF 900.   No word yet on when such products will hit the marketplace.

(One wonders: does this work the other way ’round?  If I apply TWICE the recommended amount of a cheaper SPF8 sunscreen, do I end up with the protection of SPF64 sunscreen?)

### Tetrahedral Pyramids (part 2 of 3)

April 24, 2009

The first post in this series, Pascal’s Pyramid (part 1 of 3), explored number patterns that arise in a 3 dimensional version of Pascal’s Triangle: a pyramid with square cross sections.  Another way to do a 3D version of Pascal’s Triangle would use triangular cross sections.   If you think of building a tetrahedron by stacking oranges in a pile, each orange is in contact with (up to) three oranges that lie in the level above it.

As before, we set the top number in the pyramid to be 1, and assume that at the lower levels, each number will be the sum of the (up to) three adjacent numbers on the previous level.

Also as before, we know to expect the binomial coefficients to appear on each of the three triangular faces of our new pyramid.

In the interior, the numbers generated satisfy the recursion

f(n+1, a+1, b+1) = f(n, a, b) + f(n, a, b+1)+ f(n, a+1, b+1)

where f(0,0,0) = 1, and f(n,a,b) is understood to be zero if any of the parameters go out of range (that is, if a or b is either negative or greater than n, or if b>a).

The resulting pyramid:

And for example, the numbers on the seventh level from the top are:

Examining this triangle, we recognize an odd variation on Pascal’s triangle:  each row is a multiple of a row of the usual Pascal triangle, where the multiplier is the left-most entry in the row.  Since those terms are also terms on an outer face of the pyramid, we know they too are binomial coefficients, and we are led to conjecture that  $f(n, a, b) = { n \choose a } {a \choose b}$.

As before, we can prove this result by induction on n.  The base case, when n=0, works, since ${0 \choose 0} {0 \choose 0} = 1$.

If we assume (for some specific value of n)  that each $f(n, a, b) = {n \choose a} { a \choose b}$, then it follows that

$f(n+1, a+1, b+1) = {n \choose a}{a \choose b} + {n \choose a}{a \choose {b+1}}+ {n \choose {a+1}}{{a+1} \choose {b+1}}$

$= {n \choose a} \bigg( {a \choose b} + {a \choose {b+1}} \bigg) + {n \choose {a+1}}{{a+1}\choose {b+1}}$

$= {n \choose a}{{a+1} \choose {b+1}} + {n \choose {a+1}}{{a+1} \choose {b+1}}$

$= {{n+1} \choose {a+1}}{{a+1}\choose {b+1}}$

And thus we have our identity by induction on n.

This suggests a direction for exploration:  what interesting things do we know about Pascal’s Triangle, that might generalize to three dimensions?

Coming up:  Sierpinski’s Pascal Pyramid (3/3)

### Pascal’s Pyramid (part 1 of 3)

April 20, 2009

In an earlier post, Ξ had mentioned a question from Who Wants To Be A Millionaire that made reference to “a famous “pyramid” of numbers that starts with the number one on top”.  The intended answer was “Pascal’s Triangle” (my emphasis).

That got me wondering what a three dimensional analogue of the Pascal Triangle might look like.

Today we’ll explore a Pascal-like construction of a square pyramid.   The top-most point on the pyramid will be assigned the value 1.  As we move down the pyramid,   let’s assign each point a value  by taking the sum of the 4 terms directly above and adjacent to each point.

If you think about the 4 outer faces, each point along a face only has two points that lie above it.  Thus if we restrict our attention just to one outer face, our construction is identical to the two dimensional Pascal triangle, and for  each face we should get the familiar binomial coefficients.

One can describe  how a given entry depends on those above it:  If we let f(n, a, b) be the entry in the nth layer down from the top, in row a and column b, then the entries that lie above will be in layer n-1, their row will be a or a-1, and their column will be b or b-1.  Thus

f (n, a, b) = f(n-1, a-1,b-1) + f(n-1, a, b-1) + f(n-1, a-1, b) + f(n-1, a, b)

with f(0,0,0)=1, and f understood to be zero if any of the parameters get out of range (e.g. if a or b > n, or is negative).

Numerically, we get the following pattern (for 0 ≤ n ≤ 6):

For example, the entries in the fifth layer are:

Each entry appears to be the product of the first entry in its row, with the first entry in its column.  Since we know the row and column values are binomial coefficients, we conjecture that $f(n,a,b) = {n \choose a} {n \choose b}$.

To prove this, we start by verifying that when n=0, the product of the binomial coefficients is equal to 1.  Proceeding inductively, we assume that the result holds for n-1, and we compute f(n,a,b):

${ {n-1} \choose {a-1}} {{n-1} \choose {b-1}} + {{n-1} \choose {a}} { {n-1} \choose {b-1}} + {{n-1} \choose {a-1}}{{n-1} \choose {b}} + {{n-1} \choose {a}} {{n-1} \choose {b}}$

$= \bigg( {{n-1} \choose {a-1} } + {{n-1} \choose {a}} \bigg) \bigg( {{n-1} \choose {b-1}} + {{n-1} \choose {b}} \bigg)$

$= { {n}\choose {a}} {{n} \choose {b}}$

QED by induction; each entry in Pascal’s Square Pyramid is a product of two binomial coefficients, and is equal to the sum of the four terms immediately above it.

Coming upTetrahedral pyramids (2/3);  Sierpinski’s Pascal Pyramid (3/3)

### Phonetic Phun on a Phriday

April 3, 2009

Last August, Ξ and I were inspired by a post about the International Radiotelephony Spelling Alphabet at our favorite non-math blog, puntabulous, to create a satirical version, one that would be completely confusing and virtually useless.

We posted a preliminary version of this to the comments of that original post.  Since then, the list has evolved, with help from our fellow commentator friends at puntabulous, as well as help from Batman, NP, and several other of Ξ’s coworkers.  What follows is the fruit of that community effort.
Note:  As with the Official version, this is meant to be read aloud.

• A is for aye
• B is for bdellium
• C is for czar
• D is for djinni
• E is for eye
• F is for fyce  (??)
• G is for gnu
• H is for hour
• I is for iajo
• J is for jicama
• K is for knight
• L is for llama
• M is for mnemonic
• N is for night
• O is for one
• P is for philter
• Q is for Quran
• R is for roister    (say it fast)
• S is for Sea
• T is for tsar
• U is for uighur
• V is for vrouw
• W is for why
• X is for Xi
• Y is for you
• Z is for zwieback

### Musing on dictionaries, axioms, and algorithms

April 2, 2009

I have a distinct memory of a specific moment in childhood, sitting in a second grade classroom, when I realized that dictionaries are inherently circular.  I can open the dictionary to see a description of the meaning of the word “proponent”, and read “one who argues in favor of something”.  But then if this is to truly provide meaning for the word “proponent”, I need to also know the meanings of these seven other words.

Each of which I can look up in this same dictionary, and find their respective meanings described in terms of myriad other English words.

Sadly, this process never ceases.  In order for English words to have meaning by this process, one needs to know the meanings of some core set of words, some basis, in terms of which all other English words can be described.

(A parallel example:  If I don’t know any Finnish, then picking up a Finnish dictionary is not going to help me learn Finnish.)

So languages, and dictionaries in particular, are kind of like axiomatic systems.  We start with basic terms or axioms, whose meaning we know, whose truth we assume.  From this ad hoc starting point, we can build structure and meaning, but fundamentally the meaning of any particular thing reduces down to reference to our starting axioms, our undefinable terms.

I imagine creating  a digraph, whose vertices correspond to English words, with edges A→B if the word A appears in the definition of B.  (Such a digraph will depend on the dictionary one chooses, and there are also subtleties driven by the fact that some words take on multiple meanings in different contexts.  But let’s brush over those technicalities for now.)

I have questions.  And I don’t know enough about computational linguistics to even know how to ask them appropriately, or where to look for possible answers.

1. Is this graph weakly connected? (Is everything linked to everything else?)  Or are there subgraphs that are isolated from the rest?  [Almost certainly the use of pronouns and simple verbs will lead to a single connected component; otherwise one might imagine some esoteric field of study all of whose technical vocabulary comprises a single component of the graph.]
2. Is the digraph strongly connected: can I get from every vertex to every other by following the directed links?  If I take the definition of “proponent”, and then examine the definitions of “one”, “who”, “argues”, “in”, “favor”, “of”, “something”, and iterate this process, will I ever find a definition which uses the word “green”, for example? [Presumably in general the answer is no.  There are probably words that are never used in the definition of other words:  highly technical words come to mind, such as “anthrax” or “dyspnea”.]
3. Is there a “basis” for the graph? That is, a minimal set of vertices V containing predecessors for all other vertices in the graph? [Is this called a “rootset” in the context of digraphs?]  {OK, this one I can answer with a Yes, on general principles, since the number of words is finite. One at a time, throw out any superfluous ones.}
4. What is the smallest possible basis for this graph?  How many English words must I know in order to be able to look up and understand the meaning of any other word in the dictionary?
5. Are there reasonably efficient algorithms for generating the smallest possible basis V?  Is this something I could be doing for fun on a PC running Matlab?  Or is this something so unreasonably complex that I’d need something much more powerful to take it on?

Apart from personal curiosity, it seems that the size of such a minimal basis might be used to measure the quality of a dictionary: can the authors describe everything in terms of a relatively small basic vocabulary?

The image Dictionaryindents.jpg was photographed by Thegreenj and posted to Wikipedia under the GNU Free Documentation License (v 1.2 or later).

### What’s a seven letter word for “seven letter word”?

April 1, 2009

Today I was trying out a “Math Jeopardy” game that a colleague had created, and one of the categories was “7 Letter Words”.  An example of the sort of answer/question pair for that category:

ANSWER:  This often is seen when the sun shines following a rain storm.

Question:  What is a “rainbow”?

As I was reading through the questions in this category, my brain started anticipating “A word meaning `seven letter word'”.

Offhand, I didn’t know of a word meaning “seven letter word”.  For that matter, I couldn’t immediately think of any words that meant “a word with n letters”, for any particular value of n.

But if such words existed…  clearly, a word meaning “one letter word” would have more than one letter in it, since we can easily enumerate all the one letter words in English, and check their meanings.   And it seemed pretty likely that a word meaning “one hundred letter word” would have fewer than 100 letters.

AH HA! I thought, for a fleeting moment… if such names start off too long, and eventually are too short, then somewhere in between they must be just right…, until I realized that there was no expectation of continuity, that any putative function for which f(n) = “the number of letters in a word meaning ‘word with n letters'” would map the natural numbers into the natural numbers, and so the intermediate value theorem need not hold.

A bit of thought, a trip to a latin dictionary, and then a forehead slap later, we had a few such words in mind:

• monoliteral     (having one letter)
• centiliteral       (having 100 letters)

Now the root “literal” has seven letters, so we cannot slap a prefix in front of it and get a 7 letter word, much less a 7 letter word meaning “has seven letters”.  But if we can find number prefixes whose length is 7 less than the number they signify, we’d at least be able to create words whose length matched the length they aimed to describe.  And happily, I did manage to create a few examples:

• duodeliteral  (having 12 letters)
• undeliteral  (having 11 letters)
• decliteral  (having 10 letters)

Playing around with this suggests some other fun avenues for exploration:

• In English, the word “four” has 4 letters.  A bit of thought is perhaps enough to convince you that no other english word could use the same number of letters as the word it represents.  What would a proof of that look like?
• What happens in other languages?  Are there languages where more than one word uses “its” number of letters?  Are there languages where there are no such coincidences?

A far more general linguistic/logic topic: adjectives that apply to themselves.  “Short”, or “polysyllabic”, or “English”.  Perhaps “ostentatious”, or “unabbreviated”.  Does “mispelled” count?

But then what of “Nonselfapplicable”?  Does it apply to itself?  Is “nonselfapplicable’ a nonselfapplicable word?

(I see this last paradox is just over 100 years old.  That’s me, always late to the party.)

From now on, I will always associate Goldilocks and the Three Bears with the intermediate value theorem.

### Tidal force, or The Moon and the Mosquito revisited

March 30, 2009

As Ξ noted in an earlier post, the claim that the gravitational pull of a mosquito is stronger than the gravitational pull of the moon is off by a fair bit … roughly five orders of magnitude.

The original author of the claim, George Abell, was an astronomer.  Aren’t astronomers supposed to be good at working with large numbers?  Wondering if he had been misquoted in Scientific American, I set out to find out what I could about this claim.

A google search turned up a fair number of sources that describe Abell’s claims; most of their accounts are similar to that of Lilienfeld and Arkowitz.  One of the more widely read accounts appears in The Skeptics Dictionary (by Robert T. Carroll):

Astronomer George O. Abell claims that a mosquito would exert more gravitational pull on your arm than the moon would (Abell 1979).

The secondary literature is pretty much in agreement: Abell claimed that the mosquito exerts a stronger gravitational pull than the moon.  (And clearly that claim is false.)

A quick stop at the library of a nearby college turned up Abell’s original article.  He had written a piece for the Skeptical Inquirer, a review of a book on the putative effects of the moon on human behavior.  The book’s author had suggested a plausible mechanism for such influence:  the body is largely made up of water, and we all know the moon is a primary cause of tides on the Earth.

Abell’s discussion notes the source of the Moon’s influence on terrestrial tides:  not the gravitational pull of the moon, but rather the difference in that force between the nearest and farthest points on the Earth.  Because of that difference, the Earth is (very slightly) distorted, with its fluid surface in motion attempting to achieve equilibrium.

Abell notes that while the Sun’s gravitational pull on the Earth is more than 100 times stronger than the Moon’s, its tidal force — the difference in the Sun’s pull over the diameter of the Earth — is less than half that of the tidal force of the Moon.

If the Moon’s influence on human behavior were tidal (acting on the fluids in the body), then that tidal effect would  be due to differences in the Moon’s gravitational pull on the fluid in different parts of your body — differences due to the fact that (for a 6 foot tall person) part of their body could be six feet closer to the center of mass of the Moon than other parts of their body.   Abell quotes this difference as being “about one part in $3 \times 10^{13}$ (or 30 trillion) of the weight of that fluid”, and states that the copy of a magazine in the reader’s hand is tens of thousands of times more significant in producing tidal forces.

If we take the weight of the magazine to be roughly 1/2 lb [based on weighing a stack of AMS Notices], or 2.2 Newtons, leading to a mass of 0.2 kg, then the tidal force due to the magazine (over distances ranging from .2 m to .5 m — assuming I hold the book about 20 cm from my torso, and my torso is roughly 30 cm deep) will be

$\frac{G\cdot M \cdot (0.2)}{.2^2} - \frac{G \cdot M \cdot (0.2)}{.5^2 }$

(where G is the gravitational constant, and M is the mass of the fluids in your body), which works out to approximately $2.8 \times 10^{-10} \cdot M$ newtons.  A similar calculation for the lunar tidal force acting on my body when I am standing (taking my height to be 2 meters):

$\frac{G\cdot M \cdot (7.3483 \times 10^{22})}{(3.844\times 10^8 - 2)^2} - \frac{G\cdot M \cdot (7.3483 \times 10^{22})}{(3.844\times 10^8)^2}$

which is roughly $3.5 \times 10^{-13} \cdot M$ newtons.

So indeed, as Abell claimed, it is reasonable to conclude that a small book or magazine in your hand exerts a tidal force on the order of thousands of times stronger than the moon on your body.

But wait!  Where are the mosquitoes?

Abell described a second potential source of lunar influence on the human body.  He writes:

What might matter is the difference between your weight in the presence of the moon’s gravitational effect and what it would be if there were no moon.  At the most, that difference amounts to only 0.01 gram, or about 0.0003 ounces, less than the effect of a mosquito on your shoulder.

In the paragraph leading up to that passage, Abell noted that this isn’t the same as the gravitational pull of the Moon on your body, since the Moon is also pulling on the Earth (so you have a slightly larger acceleration toward the Moon than the center of the Earth does — in effect, this is the tidal force calculation in a different guise).

It isn’t quite clear what Abell means by “the effect of a mosquito on your shoulder”:  is he referring to a gravitational (tidal) effect due to the mosquito, or is he just referring to the weight of the mosquito — the amount by which it increases your weight when it lands?

I’ll leave the analysis of these two interpretations as exercises for our readers.  But the conclusions I found:  if we compute the tidal force of the mosquito on your body, it is far greater than the tidal force of the moon on your body (because of its proximity versus the great distance to the Moon).  If instead we believe that Abell is claiming that the weight (mass) of the mosquito is greater than the 0.01 g that he had computed, then he is wrong:   Ξ had generously estimated the mass of a mosquito at 5 mg; Abell’s estimate of the Moon’s influence on a person’s weight is about twice that.

Abell’s review of Arnold Leiber’s The Lunar Effect: Biological Tides and Human Emotions (1st ed, 1978) appeared in the Skeptical Inquirer 3 (1979) pp 68 – 73.

### Universal sets and the Russell paradox

March 18, 2009

I was working with some students on set theory recently, and we were momentarily puzzled by their textbook’s definition of subset:

Let A and B be two sets contained in some universal set U. […] The set A is a subset of a set B if each element of A is an element of B….  More formally, A is a subset of B provided that for all x ε U, if x ε A then x ε B.

What threw us was the reference to a universal set U.  Why bother with that?  Why wouldn’t we just quantify over all x in A, instead of all x in U?

After a bit of thought, I realized there were a few reasons:

1. This makes defining set equality a bit cleaner:  A=B provided for each x in U, x ε A iff x ε B.  (If we didn’t have a universal set U to refer to, we’d presumably have to do two separate if-then statements, one quantifying over A, the other over B.)
2. It places the formal discussion of set theory in the familiar setting of Venn diagrams.
3. It sets the stage for dealing with set operations (union, intersection, and especially complements).

There is a 4th reason:  a desire to avoid impredicative definitions (according to the Oxford Dictionary of Philosophy, “Term coined by Poincaré ; for a kind of definition in which a member of a set is defined in a way that presupposes the set taken as a whole”;  more loosely, definitions that can potentially be self-referential).

Sidestepping the technicalities,   I wondered if we could easily craft a compelling reason to work within a universal set: what are the consequences of ignoring it?  The obvious place to worry is in computing set complements.  And thinking about self-reference, I wondered what could go wrong in building Øc without reference to a universal set?

If we call that complement M (for “monster”, or “massively huge”, or…), we see a set that contains everything that isn’t in the empty set.   “Cool!”, you’re thinking — that’s a great choice for a universal set U!

Well, hold on a bit.  This M has lots of stuff in it.  17, and {17}, and cool mathy stuff like that.  But it also has my old 82 Impala, and every sock I’ve ever lost, and every subset of the set of all socks that I’ve lost.   Everything you can imagine is an element of M.

For that matter, M even contains itself as an element, since M contains every thing as an element.

Once you see that this monster has the bizarre property that MεM, you probably begin to worry.  Maybe we want a smaller universe, that only has the well-behaved stuff, those things in M that aren’t elements of themselves.

Thus we define $R := \{ x \in M \quad : \quad x \not \in x \}$, the set of reasonable objects.  This set is a better candidate for our universe: it avoids those strange objects that are members of themselves.  (It still has my 82 Impala, though.)

B ut now, is R itself a reasonable object?  Is R ε R ?  If R is a reasonable object, then it should be a member of R, which is a very unreasonable thing for it to do. (That is, if R ε R, then R doesn’t satisfy the membership criteria for R, and shouldn’t be a member of R.)

So apparently then R is unreasonable.  But then we’d expect $R \not \in R$, which is to say that R is a reasonable object, and thus should be a member of R.

In this way, we’re led to a variation on the Russell Paradox.
Apparently building Øc without first pinning down our universe is a bad idea.

The image above is a public-domain harbor map, showing the canning docks in Liverpool England.

Bertrand Russell  first discovered the self-referential paradox of building a set of all sets that are not members of themselves.   He was not (so far as I am aware) related to Baron Russell of Liverpool (Sir Edward Russell), but in any event I shall henceforth look at that image and see “Russell’s Pair of Docks”.

### Thinking mathematically

March 4, 2009

I spent an hour last night in the company of 8-year-old boys.  (My oldest son had a Cub Scout meeting.)  The discussion topic:  safety, including fire safety, and what to do in case of an emergency.

One of the boys in the group got in a “mood”, and every question that was posed to him, he’d twist around into a more extreme predicament:

Scout:  What if your FACE is on FIRE?!?!?

or

Leader:  What if you and your friend are walking across a frozen pond, the ice breaks, and your friend falls in?  What should you do first?

Scout:  What if there’s a giant glacier, and they fall in, but they’re too deep down to reach?  And you’re in the middle of no where, and you can’t go get help?

or

Scout: What if part of the roof falls down, and you’re stuck underneath, and you’re UNCONSCIOUS!  Then what do you do???

My reaction was “Hey, he’s thinking like a mathematician!”  He knows the stock answer that is expected, and he’s asking what happens if we change the hypotheses, considering a related problem where the conclusion doesn’t  follow.   HE’S DOING MATH!!!!!!!!!!!!!!!

And then I realized that no, he isn’t, he’s just being an eight year old at a cub scout meeting.

I’d  love if my students responded to my questions with phrases like “But what if we use fractions instead?”, or “But what if the coefficients are matricies? Can we still complete the square?”  etc….

So the challenge, as we prep for our classes:  find a way to ask questions with obvious answers, that will get students motivated to say “yeah yeah, but what about THIS situation?”, and aim to “lead them” (pushing rope comes to mind) toward the actual course content we want to explore.

If anyone has insights in designing lessons that exploit this inate human cussedness, I’d love to learn more.

### Predicting the Super Bowl

February 1, 2009

It is probably bad form to start off with a disclaimer, but here it goes anyways:  I don’t follow the NFL.  I am aware that the Steelers and the Cardinals are in the Super Bowl, and I am aware that the Cardinals are not from St. Louis, where the Rams play now that they’ve left LA, and the Raiders aren’t in LA anymore either.  (Nor are the Chargers.)  But I’d be hard pressed to name any of the current players for the Steelers or Cardinals, so any opinion I might have to offer about the game will be based on other data.

I am going to a Super Bowl party this afternoon, and there is likely to be discussion (or perhaps -ahem- a “party game” with prizes and such) focused on the final score of the contest.

What strategy would one employ in order to be most likely to predict the final outcome?

After perusing the final scores of 12,595 NFL games (from 1922 to the present), I’ve found a few patterns:

• The most frequently occurring scores for an individual team are:
• 17 points  (7.2% of all scores),
• 14 (6.3%),
• 7 (6.1%),
• 24 (6.0%), and
• 10 (5.9%).
• The least frequently occurring total scores [under 20]  are:
• 1 point (which never can happen under NFL rules),
• 4 points (which happened once:  Racine 10, Chicago 4, on November 25, 1923),
• 5 points (18 times in NFL history),
• 2 points (31 times), and
• 8 points (36 times)
• It is possible for a team to score exactly 1 point under NCAA Football rules, but I haven’t had a chance to search through that ocean of data to see if it has ever happened.  [If a team is attempting a PAT, and a safety occurs, the defending team gets 1 point under the NCAA rulebook [see page FR-108.]
• The most frequently occurring final scores are:
• 20 – 17 (211 times),
• 17-14 (162 times),
• 27-24 (153 times),
• 13-10 (141 times), and
• 24-17 (125 times).
• Modulo 10, the most frequently occurring team scores are:
• 0 (19.3%),
• 7 (19.1%),
• 4 (15.1%), and
• 3 (12.3%);
• The least frequently occurring team score modulo ten is 2 (3.1%).
• Modulo 10, the most frequently occurring pairs of scores are:
• {0,7} (8.1% of all games), followed by
• {4,7} (6.8%),
• {0,3} (5.8%),
• {0,4} (5.2%), and
• {3,7} (4.5%).
• The least likely combinations modulo 10 are:
• {2,2} (0.024% = 3 times in NFL history), followed by
• {9,9} (0.10%),
• {5, 5} (0.14%),
• {5, 9} (0.25%), and
• {2, 8} (0.26%)

My own Super Bowl prediction?  Cool ads and lots of fun chatting with friends.