Easter Monday Post

Table of Contents

1. Introduction & Rapoport Problem of the Day
2. Rapoport Geometry Problem #2
3. Rapoport Geometry Problem #3
4. Rapoport Geometry Problem #4
5. Opposite Pi Day
6. Remembering John Horton Conway
7. John Conway & the Trapezoid
8. m as the Slope & theta as the Angle
9. The Conway Doomsday Algorithm
10. Conway & the Surreal Numbers
11. Conclusion & Cosmos Episode 11: "Shadows of Forgotten Ancestors"

Introduction & Rapoport Problem of the Day

Today on her Daily Epsilon of Math 2020, Rebecca Rapoport writes:

A 25 ft. ladder is placed against a vertical wall, with the base 7 ft. from the wall. If the top of the ladder slips 9 feet, how far will the foot slide?

This is clearly a Pythagorean Theorem problem. The length of the ladder is the hypotenuse, and the legs are the altitude and base of the ladder:

a^2 + b^2 = c^2
a^2 + 7^2 = 25^2
a^2 + 49 = 625
a^2 = 576
a = 24

So the original altitude of the ladder is 24 feet. When we say that the top slips nine feet, it means that the new altitude is nine feet less than the old altitude:

a = A - 9 = 24 - 9 = 15 feet

We'll let A, B, and C be the new sides of the triangle. Actually, C doesn't change because the length of the ladder is still 25 feet. But A and B certainly change:

A^2 + B^2 = C^2
15^2 + B^2 = 625
225 + B^2 = 625
B^2 = 400
B = 20

Thus the new base of the ladder is 20 feet. When we ask for how far the foot slides, it means we want the difference between the new base and the old base:

B - b = 20 - 7 = 13 feet

Therefore the desired distance is 13 feet -- and of course, today's date is the thirteenth.

Today is the Monday after Easter -- also known as Easter Monday. Many schools that aren't closed the rest of Easter Week are nonetheless closed on Easter Monday -- including my old district. Indeed, on the original pre-coronavirus calendar, this is the end of a long four-day weekend.

As I mentioned in my last post, before last week I've only posted once before on Good Friday, and that was four years ago. But I've never posted before on Easter Monday until today. Of course, in most years I follow the calendar of my old district, where there is never school on Good Friday.

Twice before this year, I followed the calendars of other districts, but in neither year did I make an Easter Monday post. Four years ago, the district I followed took the week after Easter (that is, Easter Week or Bright Week) off. Thus I posted on Good Friday, but not Easter Monday. And the following year I had just left the old charter school, but I continued to follow that calendar anyway. The school took Holy Week off (like the rest of LAUSD), but on a calendar I'd seen before I left, the charter took an extra day off on Easter Monday. I don't know why (professional development???) -- all I know was that there was no school the day after Easter. And so I didn't post that day.

And so this is a first for the blog -- an Easter Monday post. If the schools were still open, I wouldn't have posted either last Friday or today.

But instead, the schools remain closed. This is my sixth spring break/coronavirus break post.

As I mentioned in my last post, this week is full of Geometry questions. I mentioned how many of the traditionalists don't believe that "visual learners" exist or that visual diagrams are needed. Ironically, today's problem is the only Geometry problem this week that's not accompanied by a diagram.

I'll post the other questions in a random order, rather than in chronological order. This time, I hope that you'll work out the questions with me rather than simply guess the date.

Since all of these problems have diagrams -- and since those diagrams are mostly unlabeled -- know in advance that I'm adding the labels myself. So let's dive in.

Rapoport Geometry Problem #2

In Circle A, BC and DE are diameters, F is on the circle, and BF, CD intersect in the circle at P. In addition, Angle BAP is a right angle, AB = 40, AP = 30, FP = x.

For this problem, we see that Triangle ABP is a right triangle with legs 30 and 40. The Pythagorean Theorem thus gives us that length of the hypotenuse BP:

AP^2 + AB^2 = BP^2
30^2 + 40^2 = BP^2
900 + 1600 = BP^2
2500 = BP^2
BP = 50

Now this becomes a Power of a Point problem. We already know BP, and DP, EP are found easily:

AP + DP = AC
30 + DP = 40 (radius of circle)
DP = 10

AP + AE = EP
30 + 40 = EP
EP = 70

We never quite reached Power of a Point in the U of Chicago text, where it is Lesson 15-7. But we can use it today:

BP * FP = DP * EP
50x = 10 * 70
50x = 700
x = 14

Therefore the desired length is 14 -- and of course, this problem will be for tomorrow, the fourteenth.

Rapoport Geometry Problem #3

In Triangle ABC, Angle A = 3x, Angle B = 4.5x, Angle C is right.

OK, so we should be able to solve this one quickly as the acute angles of a right triangle add up to 90:

A + B = 90
3x + 4.5x = 90
7.5x = 90
x = 12

Therefore the desired problem is 12 -- and of course, this problem was for yesterday, the twelfth.

Rapoport Geometry Problem #4

Line AP is tangent to Circle O at Point A. Point Q is between A and P. And these lengths are shown: PQ = 4, BQ = 6, BP = 8, and radius = r. What is r^2?

I'll just say it right now -- this problem has me completely stumped! I just can't figure out what exactly we're supposed to do to solve this one!

Part of the trouble is the diagram -- both lines BP and BQ appear to be tangent to the circle at B. But of course, they can't both be -- the true tangent must be perpendicular to the radius OB due to the Radius-Tangent Theorem, and BP and BQ can't both be perpendicular to OB.

OK then, so let's try it both ways. Before we begin, we notice that all three sides of Triangle BPQ are known -- and so we can use the Law of Cosines to find an angle.

Our first choice is that BQ is the line that's tangent to the circle. Let's find Angle BQP:

cos BQP = (BQ^2 + PQ^2 - BP^2)/(2 * BQ * PQ)
cos BQP = (6^2 + 4^2 - 8^2)/(2 * 6 * 4)
cos BQP = -1/4
BQP = 104.48 degrees

Then Angle BQA, which forms a linear pair with BQP, must be 75.52 degrees. To continue with this problem, we consider Triangle BQO. This triangle must be congruent to Triangle AQO by HL (since legs OA = OB = r, hypotenuses OQ = OQ, and they are both right triangles by Radius-Tangent). This means that Angles BQO = AQO, and since BQO + AQO = BQA = 75.52, this means that BQO must be half of this, or 37.76 degrees.

Now we set up Triangle BQO using the tangent ratio:

tan BQO = OB/BQ
tan 37.76 = r/6
r = 6 tan 37.76 = 4.64758...
r^2 = 21.6

And this looks wrong, since 21.6 obviously isn't a date. Notice that I still haven't revealed to you, the blog readers, what date this problem comes from -- but no, it wasn't April 21.6.

Thus we must make the other choice -- BP is the line that's tangent to the circle. Let's find BPQ:

cos BPQ = (BP^2 + PQ^2 - BQ^2)/(2 * BP * BQ)
cos BPQ = (8^2 + 4^2 - 6^2)/(2 * 8 * 4)
cos BPQ = 11/16
BPQ = 46.57 degrees

Then we need Angle BPO which, by the same reasoning as in the above case, is half of BPQ. And so Angle BPO is 23.28 degrees.

Now we set up Triangle BPO using the tangent ratio:

tan BPO = OB/BP
tan 23.28 = r/8
r/8 = sqrt(0.185185185...)

This time, once I found the tangent of the angle, I decided to square it on my calculator at once. I knew that if the square turned out to be something irrational, then r^2 couldn't possibly be a natural number, much less the date. In this case, 0.185185185... definitely looks rational, and so this route appears promising.

As it turns out, 0.185185185... = 5/27. Let's fill it in and try rationalizing the denominator:

r/8 = sqrt(5/27)
r/8 = sqrt(15/81)
r/8 = sqrt(15)/9

And guess what -- the date of this problem is the fifteenth, so it's great that sqrt(15) appears here.

But hold on a minute -- we don't obtain r = sqrt(15) here, because there's an 8 on the left-hand side and a 9 on the right-hand side. Those don't cancel. Instead, we get:

r = sqrt(15)(8/9)
r^2 = 15(8/9)^2 = 320/27

After this, I went back and tried letting BQ be tangent to the circle again, but this time, I stop after taking the tangent of the angle and rationalizing it again. When I got r^2 as the simple fraction 21.6, I saw no need to do this, but now I want to write the tangent as a radical and then rationalize it:

tan 23.28 = r/6
r/6 = sqrt(3/5) = sqrt(15/25) = sqrt(15)/5

So sqrt(15) does appear on this branch, but once again, it has the wrong denominator. There's a 6 on the left-hand side and a 5 on the right-hand side, so those don't cancel to r = sqrt(15).

You might think that there's a typo in one of the lengths, but it's not that simple. On the BQ tangent branch, we used BQ = 6 in the Law of Cosines, so we can't just change it to BQ = 5. Likewise on the BP tangent branch, we used BP = 8 in the Law of Cosines, so we can't just change it to BP = 9.

As I look at the diagram again, I'm now wondering whether the point I'm calling B is even on Circle O in the first place! This is how Lines BQ and BP can both be tangent to the circle -- the points of tangency are different for each line.

And we know that point B is equidistant from those points of tangency. (This is the Two Tangents Theorem, which appears often on our math calendars but not in the U of Chicago text.) Perhaps that common distance is 1. (It must be small, since I was fooled into thinking that B is on the circle.)

And that little distance of 1 might solve the "5 vs. 6" and "9 vs. 8" problems. Thus on the BQ branch, the distance from Q to B is 6, but from Q to the actual point of tangency is 5. And on the BP branch, the distance from P to B is 8, but from P to the actual point of tangency is 9.

Thus if we let s be the common distance from B to the point of tangency, instead of:

r/6 =  sqrt(15)/5
r/8 = sqrt(15)/9

we have:

r/(6 - s) =  sqrt(15)/5
r/(8 + s) = sqrt(15)/9

And this is a system of equations whose solution is r = sqrt(15), s = 1. Therefore r^2 = 15, so the desired value is 15 -- and the date of this problem is the indeed, the fifteenth.

Even though we've solved the problem, there are a few things inelegant about my solution. First of all, it never occurred to me to write the tangent of those angles as fractions of sqrt(15) -- especially on the BQ branch -- until after we checked that the date is the fifteenth. Using the date to solve a problem on the Rapoport calendar is cheating a bit.

Second, of course, is that I didn't realize that B isn't on the circle until after I solved the problem. In fact, if this had been a regular school day post, I would probably have just presented the problem to you differently, perhaps by letting C, D be the points where tangent lines BP, BQ touch the circle without ever claiming that B is on the circle (or better yet, relabel the whole diagram so that A, B, C are on the circle and P, Q, R are points off the circle). But since I had the luxury of time and not much else to write about due to the coronavirus, I decided to retain my thought process in this post.

And finally, whenever there's a problem where two steps of trig are needed -- first the Law of Cosines and then the tangent ratio -- there's a greater risk of rounding error. I used the Ans key on my calculator so that a rounded value from the first step isn't used in the second. If the rounding error is great enough, then sqrt(15) won't be recognized -- for example:

r/(6 - s)  = 0.7634... (that is, tan 37.36)
r/(8 + s) = 0.4302... (that is, tan 23.28)

Solving this gives r = 3.852 and r^2 = 14.84. Although it still rounds to 15 as the nearest integer, the error here is quite noticeable.

The fact that the solution is the square root of an integer suggests that trig shouldn't be needed to solve this problem -- simple Geometry should suffice. Perhaps there might be a way to set up the Pythagorean Theorem at some point -- after all, we do have two right triangles.

In fact, now that we know that B isn't on the circle, we can use Two Tangents directly from P and Q, to find s = 1. But that still doesn't give us the radius r without trig, although now we only have to use the Law of Cosines and the tangent ratio once each (since either the P or Q branches now give the same answer once we take s = 1 into account).

In general, we should never use trig when Geometry is good enough to solve it. But I wasn't, and still can't, see how to set up the problem using pure Geometry. (Of course, it didn't help me at all that I mistakenly thought point B was on the circle.)

Opposite Pi Day

Today is Easter Monday, not Easter Sunday, but the holiday does have some significance. The annual White House Easter Egg Roll is traditionally held on Easter Monday (though not this year because of the coronavirus).

Both Canada and the UK treat Easter Monday as an official holiday. Our northern neighbors treat Easter as a three-day weekend from Saturday-Monday, so all schools are closed on Monday no matter when spring break is. And across the pond, all schools are closed for at least the four-day weekend from Friday-Monday, just like my old district. (Of course, nearly all schools in all three countries are now closed due to the virus.)

Even though I've never posted on Easter Monday before, I have posted on April 13th or 4/13 -- that is, Opposite Pi Day. Four years ago, I wrote about Opposite Pi Day, including why I made such a big deal about it on the blog:

But it can be argued that today is actually a "Pi Day" of sorts. You see, instead of 3/14, today is April 13th, which is 4/13. As the digits of pi appear in reverse, we can think of this as "Opposite Pi Day."

And now you're thinking -- here we go grasping at straws to come up with another math holiday. We already have Pi Day on March 14th, Pi Approximation Day on July 22nd, and Pumpkin Pi Day on the 314th day of the year in November. We had Square Root Day of the Decade on 4/4/16, Square Root Day of the Century on 4/5/2025, and several Square Root Days of the Month -- including yesterday, April 12th, which can serve as sqrt(17) day. And now I insist on adding yet another Pi Day on April 13th just because 3/14 reversed is 4/13! Do I really think that anyone is actually going to celebrate any of these extra so-called "Pi Days"?

Well, actually I didn't invent "Opposite Pi Day" -- the creator of the following video did:

It's yet another parody of Rebecca Black's "Friday" -- as I mentioned back on the original Pi Day, the fact that Friday and Pi Day rhyme is too irresistible for many math parodists to avoid. The poster of this song, who goes by the username "AsianGlow," probably just missed uploading the song on the original Pi Day, so rather than wait a whole year to post it, he just reversed the digits. He uploaded the video exactly nine years ago today -- April 13th, 2011.

AsianGlow writes:

Of course by now you know of Rebecca Black making her infamous debut with Friday, but what about Pi Day?

How come Fridays get so much more love? Friday usually gets four times a month to party, but Pi Day only gets one... March 14th...

Does this mean I can only eat pies once a year?! OMG NO! Any date containing 1, 3, or 4 should be a piece of Pi Day! hahaha! See what I did there? You don't think it's funny? Well don't be so bitter and get some sweets in yo life! :D

This is the first video in a while that I did allll by myself! Well the filming anyways...
FilmRebelRoby - http://youtube.com/filmrebelroby - helped me with mastering my vocals.

April 13 is like opposite Pi Day! OMG it's still April fools! That's right, that's Pi!

Notice that AsianGlow's proposal that any date containing 1, 3, and 4 should be Pi Day only applies to March 14th and April 13th. We see that January 34th, April 31st, and 13/4 all just barely avoid being valid Gregorian dates, but they could exist in certain versions of Calendar Reform. In particular, all three date exist in a Leap Week Calendar where every third month (including January and April) have 35 days, 28 days in all the rest, and Leap Week labeled as Month 13.

I wouldn't have mentioned Opposite Pi Day here on the blog at all had I not subbed in a class learning about pi today. In some ways, I'm celebrating Opposite Pi Day today for the exact same reason as AsianGlow -- we both already missed the real Pi Day, yet we want to celebrate pi today. Both Pi Day and Square Root Day can be easily manipulated so that they fall during the unit on pi or square roots.

Returning to the present, there's one more thing I want to say about Pi Day and subbing. Recall that my last visit to a middle school was on March 4th-6th. Even though I was in an eighth grade class, I also had a co-teacher who wanted to celebrate Pi Day in her class with a digit reciting contest.

I didn't mention this on the blog at the time because I had so much else to discuss about the actual classes (the ones that weren't co-teaching). I was considering mentioning it on Pi Day itself (or maybe Pi Day Eve), not expecting our worlds to be rocked that day. The only thing I could think about was the impending closure of the schools. (So I wonder whether that co-teacher really did her contest on the last day before the schools were shut down.)

Remembering John Horton Conway

I know that there have been several virus outbreaks that have affected humans lately (including, for example, SARS and ZIKA). But one way I can tell that this coronavirus is much more serious than the others is when I look at a list of recent celebrity passings and see "of coronavirus complications" listed alarmingly often as the cause of death.

This person wasn't a "celebrity," but he was a famous mathematician -- John Horton Conway. Two days ago, on Holy Saturday, mathematician John Horton Conway died of coronavirus complications at age 82.

Today, Numberphile posts its own John Conway tribute with a podcast:

In the background, we see the Game of Life -- one of Conway's most famous creations. The "Game" of Life is what he called a zero-player game -- all we do is set the initial conditions and the game basically plays itself.

Some math teachers have written about Conway and how to incorporate some of his ideas into our classrooms, for example, Math Tango:

https://mathtango.blogspot.com/2015/07/author-at-play-siobhan-roberts-reports.html

In fact, I mentioned Conway's Game of Life on the blog myself, in my November 13th post:

John von Neumann was the originator of cellular automata theory. A few weeks ago on the blog I mentioned John Conway -- his "Game of Life" is based on cellular automata." Of the work of von Neumann and Conway on cellular automata:

"This turns out to be an excellent way to work out which factors are important, and to uncover general insights into why complex systems do what they do. A complex system that defies analysis by traditional modelling techniques is the formation of river basins and deltas."

And the ultimate connection between cellular automata and biological life is profound:

"Complex systems support the view that on a lifeless planet with sufficiently complex chemistry, life is likely to arise spontaneously and to organize itself into ever more complex and sophisticated forms."

John Conway and the Trapezoid

I've mentioned Conway on the blog in several other posts. Most of these are in discussion of the inclusive definition of trapezoid. According to both Conway and the U of Chicago text, all parallelograms are trapezoids.

For example, in my October 30th post, I wrote the following:

Section 5-4 of the U of Chicago text covers kites. The kite is a relatively new quadrilateral classification. Not only did Euclid never define kite, but many texts made no mention of kites -- including my class geometry textbook from 20 years ago. Nowadays most texts define kite, but some include kites only in bonus questions, not in the main text.

Here's what John Conway wrote about the kite -- over 20 years ago, right around the time that I was taking my geometry class:

In fact it's not quite true, either, because "kite" is not
a very traditional name - it was obviously inserted because
this was a type of quadrilateral that SHOULD have received a
traditional name, but didn't, until recently.

Why do we include the kite - plainly because it represents
the one type of symmetry not otherwise mentioned. But this
reason suggests we should also EXCLUDE the non-isosceles
trapezoid.

Unfortunately, this was part of a discussion with Conway, and I can't find the original source. I keep reblogging this excerpt every year when I cover Lesson 5-4 on kites, even though the original discussion was about trapezoids. And since I never posted a hot link, I no longer have any access to Conway's original discussion.

m as the Slope 

Five years ago, during the first year of this blog, I wrote the following:

A question that is often asked during this lesson is, why so we use m to denote slope? One urban legend is that it refers to the French word monter, meaning to climb. An English cognate of this word is "mountain," and of course mountains have slopes. The problem with this theory is that there is no evidence that the mathematician Rene Descartes ever used the letter m. One would think that if m had a French origin, Descartes -- you know, the French creator of the Cartesian plane on which slope is usually measured -- would have been the first to use it. A discussion appears at this thread:

http://mathforum.org/library/drmath/view/52477.html

Notice that John Conway -- the mathematician I previously mentioned as an advocate of the inclusive definition of trapezoid -- is a participant in this 20-year old thread. (That's right -- when Conway wrote in this thread, I was myself a young geometry student!) Conway suggests that m may stand for "modulus of slope." One teacher tells his students that m stands for "move" and b stands for "begin," since this is how students learn to graph lines in slope-intercept form.

Since this time I did include a hot link, let me quote the late mathematician directly:

From: "John Conway"
Date: Sun, 13 Nov 94 19:38:58 EST
Subject: Re: why "m" for slope?
Cc: geometry-pre-college@mathforum.org

I believe (but am not sure) that what we now call just the
"slope" was once called the "modulus of slope", the word
"modulus" being used in its sense of "number used to
measure" (as in "Young's modulus").

    Descartes' "La Geometrie" doesn't use the "m" in this
connection, but I seem to remember that Euler often does.

Descartes makes no use of Greek letters for parameters in
his Geometrie, and most later authors have followed him in
this, if we except the use of theta, phi  for variable
angles.

    This is an interesting question - I'll try to track it back.

               John Conway

Date: Mon, 19 Feb 1996 08:52:40 -0500 (EST)
From: John Conway
Subject: Re: GEOMETRIC TERMS

    When this question came up last year, it emerged that many
people have been taught that it comes from the French, "monter",
to climb.  I think this is an "urban legend".

    I don't have much faith in the theory I put forward then -
that this  m  stands for "modulus of slope".  It is true that
the term "modulus" has often been used for "the essential parameter
determining".

    John Conway

Date: Mon, 19 Feb 1996 12:38:34 -0500 (EST)
From: John Conway
Subject: Re: GEOMETRIC TERMS

     But M. Risi  plainly wasn't the first person to use "m" in this
connection!

     It interests me that on this continent the typical form is
y = mx + b,  whereas in England and in "the" Continent  it is
y = mx + c.   The latter form still seems to me to be more natural,
since this "c" is like the arbitrary constants in indefinite
integrals,  and so it will probably be very hard to date.  But
the "b" usage probably originated with the author of a particular
influential North American textbook, and maybe we can find out
just who it was.

    John Conway

John Conway and theta as the Angle

Another discussion that year was all about the use of theta to denote angles:

Here's an interesting question: why do we use the letter theta to denote angles? I decided to do a Google search to find out why. I noticed that an old Mathforum post from 1996 appeared in my results -- and that's the same site I found the information about Dr. John Conway and the inclusive definition of trapezoid. I clicked on the search result -- and sure enough, it was Conway:

This raises another question : just WHY ARE

theta, phi, psi

the traditional angle-letters? Let's see where they are in the
Greek alphabet, which I usually write in tetrads, because it
"rhymes" best that way :

alpha beta gamma delta
epsilon zeta eta THETA
iota kappa lambda mu
nu xi omicron pi
rho sigma tau upsilon
PHI chi PSI omega

There doesn't seem to be much logic in that particular triple,
but it occurs to me that chi and omega are also reasonably
common angle-letters, so it might well be that it's just
"theta and the last few letters". On the other hand, it might
be that they (or at least some of them) are the initial letters
of some obvious Greek words - I'm not competent enough to know.
I remark that since phi and psi are both written and pronounced
similarly, once you've used one for some purpose, you're quite
likely to use the other for a similar purpose.

Can anyone fluent in Greek (maybe Andreas?) tell us why theta
is THE standard angle name?

John Conway

So Conway doesn't know the answer -- at least he didn't 24 years ago. Just to show how old this post is, I myself was in a Precalc class and just learning about theta when this was posted. Earlier in this post, a (Geometry!) student who knew that his/her teacher spoke to Conway on the Internet said:

"Why don't you write to John Conway. If he doesn't
know, then nobody in this world knows." I replied, "Either that, or if
Prof. Conway doesn't know, then it's probably not worth knowing."

Sadly, we can no longer follow this student's advice, because Conway is no longer with us.

I included a link in this old post, but the link is dead. In this case, I believe that I preserved Conway's entire post, so I don't need a link here.

The Conway Doomsday Algorithm

My most recent reference to Conway was about a month and a half ago, on Leap Day:

The following link explains the origin of Leap Days and the Julian Calendar. It also explains the Doomsday Algorithm -- a method of determining the day of the week given its date, first devised by one of my favorite mathematicians, John Conway:

http://people.cst.cmich.edu/graha1sw/Pub/Doomsday/Doomsday.html

Julius Caesar introduced the concept of the Leap Year; i.e., the idea of adding an extra day to February in every year divisible by 4. His calendar is called the Julian Calendar, and it was used throughout the Western world until 1582. In 1582, Pope Gregory XIII decreed a modification of the Julian Calendar. He declared that Century years (that is years divisible by 100) are leap years if and only if they are divisible by 400. Thus 1700, 1800, and 1900 are not leap years, but 2000 will be a leap year. The resulting calendar is called the Gregorian Calendar.

Let's honor Conway by using his Doomsday Algorithm calculate the day of the week for his last day on this earth, April 11th, and prove that it was really a Saturday.

The link was created in 1998, and so it focuses on 20th century dates. We must use the following chart to find dates in the 21st century:

1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 2500 2600 Wed Tue Sun Fri

So our base Doomsday for this century is Tuesday. To this, we add:

+1 day for the DOZEN years that have passed since 2000

= Wednesday, Doomsday 2012

+10 days for the eight YEARS, plus two LEAP days, since 2012

= Saturday, Doomsday 2020

Now in a given year, the dates 4/4, 6/6, 8/8, 10/10, 12/12 all fall on Doomsday. Thus April 4th this year fell on Saturday. Since April 11th is exactly one week after 4/4, it also falls on Doomsday. And so April 11th, the day of Conway's passing, was indeed a Saturday.

For another example, let's try Conway's birthday -- Boxing Day 1937. According to the chart, the base Doomsday for the 20th century is Wednesday. To this, we add:

+3 days for the three DOZEN years that have passed since 1900

= Saturday, Doomsday 1936

+1 day for the YEAR since 1936

= Sunday, Doomsday 1937

Now 12/12, December 12th, is one of the Doomsday anchor dates. Thus December 12th, 1937 fell on a Sunday. Since December 26th is exactly two weeks after 12/12, it also falls on Doomsday. And so December 26th, the date of Conway's birth, was indeed a Sunday. Leave it to Conway to both enter and leave this world on a Doomsday!

Let me try my own birth, just for an example where the target date isn't Doomsday. As you already know, I was born on December 7th, 1980. Again, we begin with Wednesday. To this, we add:

+6 days for the six DOZEN years that have passed since 1900

= Tuesday, Doomsday 1972

+10 days for the eight YEARS, plus two LEAP days, since 1972

= Friday, Doomsday 1980

Since December 7th is five days before 12/12, we subtract those five days to get to Sunday. And so like Conway, I was born on a Sunday in December.

Conway also gave a formula for an Easter calculation. Let's do it for this year, 2020, in order to prove that yesterday really was Easter:

Easter is the first Sunday strictly later than the Paschal full moon, which is an arithmetical approximation to the astronomical one. The Paschal full moon is given by the formula

(April 19 = March 50)-(11 G + C) mod 30,

except that when the formula gives

April 19 you should take April 18,

and when the formula gives

April 18 and G >= 12, you should take April 17.

In the formula, G is the Golden number. For the year Y, the Golden number is

G = Y mod 19 + 1.

The term C is the Century term. For all Julian years, C = 3. In the Gregorian calendar,

C = -4 for 15xx,16xx -5 for 17xx,18xx, -6 for 19xx,20xx.

For 2020, we divide 2020 by 19 and get 6 as the remainder, so the Golden number G = 7. And for our century, C = -6, so we get:

(11G + C) = 11 * 7 - 6 = 77 - 6 = 71 == 11 mod 30
April 19 - 11 = April 8th

And since Doomsday is Saturday 4/4, April 8th is four days later, a Wednesday. Therefore Easter Sunday must be four more days after this, on April 12th -- which it indeed was.

Notice that for perpetual Calendar Reforms such as the Cotsworth Calendar or Eleven Calendar, algorithms such as Conway Doomsday are no longer necessary. But on the Usher Calendar, while 4/4, 6/6, 8/8, 10/10, and 12/12 are still on the same day in a given year, the more complicated rule for Leap Days means that the rules for determining that Doomsday no longer work.

Instead of counting centuries, dozens, and quadrennia, we're now forced to count cycles of 62 years and 17 years, and the 2nd, 6th, 10th, and 15th years within each cycle has a Leap Day.

Notice that Doomsday 2046 will fall on a Wednesday. The 62- and 17-year cycles were both chosen to be multiples of 17, and so Doomsday in the first year of each cycle will also be Wednesday. And so finding the first year of a cycle is half the battle -- once this is done, Doomsday for the target year will be found easily.

There's also a formula for the Jewish lunisolar holidays of Rosh Hashanah and Passover, but I won't demonstrate them here. Notice that Passover is still going on now -- tonight is the sixth night.

Conway & the Surreal Numbers


OK, that's enough with all these old posts and reblogs -- let me post something new. John Conway also invented a new type of number -- the surreal numbers.

https://www.whitman.edu/Documents/Academics/Mathematics/Grimm.pdf

The paper I linked to above, by Gretchen Grimm, explains Conway's surreal numbers in detail. The process is quite complex. A surreal number is considered to be a pair of sets (the left set and the right set), where the elements of each set are themselves surreal numbers.

This is recursive, so we need a base case -- if both left and right sets are empty, the surreal is 0. The surreal {0|} (0 on the left, empty on the right) is 1, {0, 1|} is 2, {0|1} is 1/2, and so on.

All reals are surreals, but not all surreals are reals. As Grimm explains, there exist infinite reals such as omega:

As well as the reals, we also get some interesting infinite numbers on the ω day. One number is ω itself, which is ω = {1, 2, 3, 4, . . . |}. We can show that it is larger than all the other numbers created through day ω. Suppose there exists some x = {XL|XR} such that x ≥ ω. Then by Axiom 2, for all n ∈ {1, 2, 3, 4, . . .}, n < x. But this is clearly a contradiction since the set {1, 2, 3, 4, . . .} has no upper bound. Thus x < ω for all x. Note that ω has many other forms such as ω = {1, 2, 4, 8, 16, . . . |}, or ω = {all dyadic numbers|} [2]. Similarly, −ω = {| − 1, −2, −3, . . .} is the most negative number created by day ω.

There also exist infinitesimal surreals like epsilon. (Recall that to Rapoport and others, epsilon is a small but real number, but to Conway, epsilon is infinitely small.)

Surreals are similar to standard reals as defined using Dedekind cuts. They can be added, subtracted, multiplied, and divided by manipulating the left and right sets.

Conclusion & Cosmos Episode 11: "Shadows of Forgotten Ancestors"

Somehow, I was able to fill most of this post with math, as Rapoport's Geometry problem and our John Conway tribute took over the page.

But there's still some room left for scientist, specifically an astrophysicist. Since tonight is Monday -- Cosmos night -- let's end my post with a quick summary of Neil DeGrasse Tyson's latest episode. I'm posting this just as Episode 11 ends -- Episode 12 will be saved for my next post.

Here is a summary of Cosmos Episode 11, "Shadows of Forgotten Ancestors":

• How would an extraterrestrial species tell humans apart from all of earth's other species?
• Fire was first tamed in South Africa a million years ago by our ancestors, Homo erectus.
• In Ancient Persia in around 600 BC, Zoroastrian worship was devoted to fire.
• They worshipped Ahura Mazda and believed that good thoughts and deeds keep a devil away.
• Microorganisms are silent predators that can jump, say, from bats to other organisms.
• The rabies virus takes free will away from a dog, making it a ferocious attacker, not a guardian.
• Do honeybees have free will? Oleic acid makes the other bees to remove the body even if alive.
• A mother goose will retrieve a white object and put it in her nest -- an egg or a golf ball.
• The ability of a fly to evade a rolled-up newspaper is clearly hardwired into its brain.
• What is distinctly human about us? Is there anything that is ours alone?
• According to Plato, man is a featherless biped -- but so is a plucked chicken.
• Adam Smith thought that bartering is unique to humans -- except that chimps barter too.
• Are humans the only species that enslaves others? No, a wasp can enslave a cockroach.
• Are we no more than automata, slaves of our DNA. Does our DNA tell our entire story?
• Our DNA tells a story of heroism -- the way a gazelle heroically escapes a predator cheetah.
• Only 2000 years ago (five seconds on the Cosmic Calendar), tyrants took over Persia and India.
• In the 2nd century BC, Indian emperor Ashoka cut off the heads of those who challenged him.
• Ashoka's Hell, a torture chamber, demonstrated how much cruelty humans are capable of -- until a Buddhist monk stood up to him.