-- "Wrong! Gary," Mr. Mason replied.
That is the teacher's response to the question asked at the bottom of the last page in Pappas. We are in the middle of a murder mystery. Gary is a student who asks Mr. Mason whether Madre's failure to take his medicine was the cause of his heart attack, and he replies that Gary is wrong.
The other students now ask each other whether they've considered all the initial facts of the case. And Pappas continues:
-- "No," Bob, who rarely talked in class, replied. "We haven't considered why the sawdust information was mentioned, what his job was at the circus, and what about the cane, and Madre's wife and the trapeze artist."
The other students agree with Bob. One of them, Tom, thinks that the sawdust was left when someone shortened the cane with a saw. And so Pappas ends the page with his question:
-- "Perhaps Madre got upset because someone wrecked his cane."
-- "NO"
The lone image on this page is a picture of the circus. I assume that there are no clues hidden in this picture -- but who knows? Maybe there are...
TO BE CONTINUED
You may notice that today's post has been labeled "traditionalists." Technically, the traditionalists post shouldn't be until tomorrow, the day of the Chapter 2 Test. But I'm doing traditionalists a day early -- and no, it's not an excuse to bring up politics, race, and sports that have dominate the news lately. No, it's because there are things on today's Pappas and Ogilvy pages that address the traditionalist cause.
Let's start with Pappas. We back up a few pages to the beginning of this mystery, when she writes:
"These logic stories always seem to spark even the shyest student."
And of course, that "shyest student" is Bob. According to Pappas, Bob "rarely talked in class," yet he is definitely engaged by the logic story.
But traditionalists, of course, would consider telling logic stories in class a waste of time. So I ask, what's the likelihood that Bob would be equally engaged by any lesson that the traditionalists would prefer to teach? Bob should consider himself lucky that his teacher, Mr. Mason, isn't a traditionalist.
Last night was the premiere of the CBS sitcom Young Sheldon. It is a spin-off of the long-running series Big Bang Theory. I'm not a BBT fan, and so I won't be watching Young Sheldon either. And so all I know about the show is what I gleaned from watching the commercials and promos all summer.
Apparently, the young Sheldon is a nine-year-old prodigy. He lives with his parents, his older brother, and his twin sister. As the series begins, it's the first day of school, and Sheldon has been allowed to skip several grades and start the year as a high school freshman. One promoted scene asserts that he is now in the same grade as his older brother, and thus no longer in the same grade as his twin sister.
The young Sheldon definitely enjoys learning. One scene shows him watching some program, most likely on PBS. His sister wants to change the channel to Duck Tales, but Sheldon doesn't let her, since her favorite show isn't educational. Her response is, "It's TV. You're not supposed to learn!"
But the one scene I want to focus on is just before school starts. The young Sheldon looks forward to learning at his new school. He imagines how the other high school students will treat him once they find out how smart he is. Indeed, he fantasizes, "Maybe they'll make me their leader."
Now I never watched the actual episode, but it's safe to say that nothing of the sort happens. Here's what probably happens when Sheldon arrives at the school -- the other students are not too impressed by Sheldon's smarts and act hostile towards him. At the least, they make fun of him as a "nerd," and at the worst, they might even beat him up.
The problem, of course, is Sheldon's naivete. He wrongly assumes that the other high school students value book smarts in their friends. The only way for a nine-year-old to be "made the leader" of a group of high school guys is for the youngster to be large for his age and good enough to be made the "leader" -- or captain -- of a sport, especially if it's football or basketball. (Oops, so I did mention sports in this traditionalist post anyway.)
But Sheldon is only nine, so he doesn't know any better. My problem is that many traditionalists seem to think that every student is like Sheldon. They promote the idea that every student is interested in learning, and that it's the "progressive pedagogues" who separate students from what they "want."
Also, we know that traditionalists are enamored with the idea of tracking. Students should be divided not by age but by knowledge, and so students like Sheldon ought to be promoted to high school if they are intelligent enough. But "progressive pedagogues" wish to hold Sheldon back and force him to be in the same class as his age-mates, like his twin sister, in order to spare the latter's "self-esteem."
But there are other reasons to oppose tracking besides "self-esteem." If we promote Sheldon to high school, then we expose him, as I mentioned above, to being beaten up by the others. So it's not just about the broken feelings of the students on the low track but the broken bones of the students on the high track. (Let's hope that Sheldon wasn't beaten up too badly last night!)
Speaking of nine-year-olds, I point out that I actually taught a nine-year-old last year -- of course, he was only in sixth grade, not ninth like Sheldon. His tenth birthday was in late September (so in fact he turns 11 very close to today's date). Of course, this was back when California had a later cut-off date (as the first cohort of transitional kindergartners hasn't quite made it to middle school yet). He told me that when he was four, his parents tried to enroll him in kindergarten due to his fifth birthday being before the old cut-off date -- except he was so bright that they placed him in first grade instead.
I'm sure that a student like Sheldon would be engaged by a traditionalist lesson. But many students would be more engaged by lessons other than the direct instruction of math -- like Bob. And so traditionalist lessons work well for the Sheldons of the world, but we also need nontraditional lessons like logic problems for the Bobs of the world. Some students look forward to working on the problem sets, while others just want to watch Duck Tales when they get home.
Logic problems teach students logic, which is important to math and especially Geometry. Sure, maybe Bob learns very little by solving the logic problem, but he learns nothing at all from a traditional lecture that he doesn't pay attention to. And Sheldon's sister may learn a little from a project in her math class, but not from a problem set -- if she ignores and just watches Duck Tales.
Does skipping all of middle school help the young Sheldon in the long run? I don't know, since his "future" is of course his character on the main series, which I don't watch. I do know that Sheldon is supposed to be the ultimate "nerd" on a program that's all about "nerdity" -- after all, the show is named after a theory in physics. Then again, Sheldon isn't too "nerdy" to have a girlfriend -- a promo for the main series actually shows Sheldon trying to propose marriage to her.
Of course, I must also look at myself when criticizing the young Sheldon's naivete. Last year, I told students that they must do the work in order to maintain a high grade. This assumes that the students are motivated by high grades -- Sheldon probably is, but many others aren't. My mistake, of course, was when I didn't answer "Because I said so!" when answering why they had to do the work.
Chapter 9 of Stanley Ogilvy's Excursions in Number Theory is called "Calculating Prodigies and Prodigious Calculations." It begins:
"From time to time there have come to public notice youngsters with extraordinary powers of performing mental calculations."
Such students are, of course, the exact opposite of "drens." And now you can see why I'm tying today's Ogilvy chapter to traditionalism as well. Traditionalists are worried about the youngsters who lack even ordinary powers of performing mental calculations -- the "drens."
Ogilvy begins by writing about Zerah Colburn, a nineteenth-century whiz-kid from Vermont. He tells us that at the age of nine -- yes, the same age as the young Sheldon -- he was taken by his father to England, where his math skills impressed an audience. (Speaking of Sheldon, I don't watch either BBT or YS enough to know whether he could do hard math problems in his head or not.)
According to Ogilvy, the young Zerah could do all of the following in his head:
-- raised one-digit numbers to the tenth power
-- raised two-digit numbers to the sixth, seventh, and eighth power
-- extracted a square root and a cube root, with the answer in each case a three-digit integer
-- factored the fifth Fermat number, 2^32 + 1, with the factors having three and seven digits
Each time the author tells us an impressive feat of one of these human calculators, let's compare it to something that a "dren" can't do. For example, Zerah could raise one-digit numbers to the tenth power, while some drens can't even raise one-digit numbers to the second power (though generally, even drens often find squares easier than other multiplications). In my own class, I taught my eighth graders the first ten squares and first six cubes, and most of them did well. In fact, in the future I may be bolder and teach them the first sixteen squares and first ten cubes. (The reason I keep stopping with numbers that end in 6 is that all of their powers end in 6, so these should be slightly easier than the powers of numbers ending in 7.)
Ogilvy now moves on to the German Zacharias Dase. While in his twenties, Dase mentally:
-- multiplied two eight-digit numbers in under a minute
-- multiplied two 20-digit numbers in six minutes
-- multiplied two 40-digit numbers in 40 minutes
-- multiplied two 100-digit numbers in under nine hours
-- extracted the square root of a 100-digit number in under one hour
On the other hand, a dren can't even multiply two two-digit numbers. Well, we don't expect students to multiply two-digit numbers mentally anyway, but let's say that our two digit numbers both end in zero, which is indeed something students should be able to do mentally, but some drens can't.
"The most promising young mathematician known to the present writers has some of this ability."
And this is what this mathematician wrote on the blackboard:
e^(pi sqrt(163)) = 262,537,412,640,768,743.999999999999250...
I'm tempted to say that this student must be Srinivasa Ramanujan, whom I mentioned back in my May 1st post. But Ramanujan died in 1920, so he couldn't be the young student known to the authors Ogilvy and John Anderson, who wrote this book in the 1960's. Nonetheless, the number mentioned here is known as Ramanujan's constant, since the famous Indian mathematician first noticed that it is within one-trillionth of an integer.
The next thing mentioned in this chapter is the calculation of the digits of pi. In 1962, pi was calculated to 100,000 digits. Strangely enough, there's no mention of the digit memorization record, which would seemingly belong in a chapter on math prodigies. I don't know what the record was at the time the book was written, but now the record belongs to Akira Haraguchi. He first recited 100,000 digits of pi in 2006 -- that's right, all the digits known in 1962 were recited by one man merely 44 years later!
Ogilvy now moves on to large numbers. He begins with the number described by Edward Kasner, who asked his nine-year-old nephew (young Sheldon's age again!) to name it:
One googol = 10^100.
So the numbers that Dase multiplied and found the square root of were on the order of a googol. Now Kasner went a step further and proposed yet another name:
Googolplex = 10^googol
Ogilvy now compares googol and googolplex to the Fermat numbers. If F_n is the n'th Fermat number, that is, F_n = 2^2^n + 1, then the author obtains the following results, which he includes in his table. Here I include only part of his table, but add on prime-related discoveries made after the book was published:
-- F_8 has 78 digits.
-- Googol has 100 digits. (Well, I guess it's the first 101-digit number -- 1 one, 100 zeros.)
-- 1000! has 2568 digits.
-- 2^11213-1 has 3376 digits. (This is the largest known Mersenne prime at publishing.)
-- F_17 has approx. 40,000 digits. (Smallest Fermat of unknown primality at publishing.)
-- F_36 has approx. 20 billion digits. (Smallest Fermat of unknown primality now.)
-- Googolplex has 10^100 digits.
-- F_1945 has approx. 10^600 digits. (Largest Fermat with known factor at publishing.)
-- F_9448 has approx. 10^3000 digits. (Largest Fermat with known factor now.)
-- Skewes' number has approx. 10^10,000,000,000,000,000,000,000,000,000,000,000 digits.
Again, Ogilvy reminds us that Skewes' number is:
S = e^e^e^79 = approx. 10^10^10^34.
To answer this, Ogilvy provides us with a formula that tells us how many factors of a prime p appear in the factorization of n!:
[n/p] + [n/p^2] + [n/p^3] + ...
Here [x] is the greatest integer function. It is the same function as INT, which we encountered last week in our BASIC programming. Technically, this sum is infinite, but all except finitely many terms are zero, so in practice we stop after the last nonzero term.
Ogilvy finds the number of factors of two in 9! as an example:
[9/2] + [9/2^2] + [9/2^3] + [9/2^4] STOP
= 4 + 2 + 1 + 0 = 7
He checks this by multiplying it out:
9! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9
= 1 * 2 * 3 * (2 * 2) * 5 * (2 * 3) * 7 * (2 * 2 * 2) * (3 * 3)
= 1 * 2^7 * 3^4 * 5 * 7.
So there are indeed seven factors of two. This means that 9! ends in seven zeros in binary. He explains why this works informally as follows -- every even factor provides a factor of two. Every multiple of four provides an additional factor of two (beyond the factor already counted by the number being even), every 8th multiple provides an extra factor of two, and so on. If we replace 2 by p we have the general proof. QED
To find the number of zeros in 1000! (in decimal), we can't use the formula since it only works for prime factors, and 10 isn't prime. Ogilvy explains that the factors of two dominate the factors of five (as there are always more of the former), and so we use 5 in the formula above:
[1000/5] + [1000/5^2] + [1000/5^3] + [1000/5^4] + [1000/5^5]
= 200 + 40 + 8 + 1 + 0 = 249.
Hence 1000! has 249 terminal zeros.
By the way, this question becomes trickier in dozenal. Suppose we wanted to know how many zeros the factorial of 1000 (great gross) ends in dozenal. The problem is that the base is 2^2 * 3, and while there are always more factors of two than three, it's not obvious whether there are twice as many factors of two as three. So we must use the formula for both 2 and 3.
[c] (default dozenal -- that is, all calculations below are in dozenal):
[1000/2] + [1000/2^2] + [1000/2^3] + ... + [1000/2^9] + [1000/2^a] + [1000/2^b]
= 600 + 300 + 160 + 90 + 46 + 23 + 11 + 6 + 3 + 1 + 0 = bb8
Notice that half of this is [bb8/2] = 5ba, which is how many factors of four (that is, pairs of factors of 2, since we're in base 2^2 * 3) we have.
[1000/3] + [1000/3^2] + [1000/3^3]+ [1000/3^4] + [1000/3^5] + [1000/3^6] + [1000/3^7]
= 400 + 140 + 54 + 19 + 7 + 2 + 0 = 5ba
We see that there are exactly twice as many factors of two as three! This isn't a coincidence -- this phenomenon occurs only in certain bases. Dozenal is the first -- the next is 39 (written in dozenal -- the base is 45 in decimal). So the answer is that there are 5ba zeros. (This is 862 in decimal.)
Ogilvy ends the chapter as follows:
"Having stretched our minds and our credulity in the chapter by strolling down a few of the lesser lanes and byways, some of which do not properly belong to number theory, we end our digression and return now to the main highway."
Well for today, the main highway is our second day of the Chapter 2 review. I've decided that when we have these two-day reviews, I'll use the second day to post interesting activities that I find from other sources.
My usual go-to sites, Fawn Nguyen and Sarah Carter, are out for Geometry, since the former is a middle school teacher and the latter is an Algebra I teacher. But hey -- the most well-known Geometry site is probably now Shaun Carter, ever since he married Sarah. (That's right -- just as physics nerd Sheldon is about to marry a fellow nerd, Shaun Carter married a fellow math teacher.)
The following worksheet is all about the three undefined terms (point, line, plane). Recall that the first question on the test asks students to identify these undefined terms. The page didn't print well for me, so you might want to find this worksheet at the original source:
https://blog.primefactorisation.com/2017/09/09/undefined-terms-inb-pages/
Oh, and while I'm posting links, let's get back to the traditionalist Barry Garelick. Here's a link to his most recent post:
https://traditionalmath.wordpress.com/2017/09/26/the-convoluted-logic-of-education-dept/
In this post, Garelick writes about progressive classes that either lack texts altogether, or use texts that are anti-traditionalist. He mentions CPM and the U of Chicago elementary texts, and also EngageNY, an online curriculum that I mentioned years ago on the blog. The EngageNY texts are based on Common Core, and I've seen some suggestions from Dr. Hung-Hsi Wu of Berkeley incorporated into the EngageNY Geometry curriculum.
The "convoluted logic" mentioned in Garelick's title is this -- progressives want to abolish textbooks because they aren't helpful in learning. But from Garelick's perspective, of course their progressive texts aren't helpful. To him, the only helpful texts are the traditionalist texts -- those which contain short explanations and lots of problem sets. Instead of abandoning textbooks for online curricula, teachers should return to traditionalist texts. Then maybe there would be a lot more Zerah Colburns in our generation today -- or at least more young Sheldons.
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