"In essence, the arch is an architectural method for spanning space. The nature of the arch allows the stress to flow more evenly throughout, thereby avoiding concentration on the center."
We're on the second page of the section on the arch. Here Pappas tells us exactly why the arch is such a stable structure.
According to the author, the first group to use the arch extensively was the Romans. They used the semicircular arch -- what we normally think of when we hear the word "arch." Pappas writes:
"Add to the arch their discovery and use of concrete and brick, and an architectural revolution took place."
This goes back to what she writes earlier -- new geometrical shapes (the arch) and new building materials (bricks and concrete) go hand-in-hand.
Here is the caption of the only photo found on this page:
"The Roman aqueduct at Segovia, Spain, constructed with 148 arches 90 feet high."
Of course you need a link for the photo:
http://www.history.com/topics/ancient-history/ancient-rome/pictures/roman-architecture-and-engineering/roman-aqueduct-segovia-spain
Chapter 2 of Stanley Ogilvy's Excursions in Number Theory is called "Number Patterns." Ogilvy begins with an interesting question:
"Given the number 123456789, in how many ways can the digits (numerals) of this number be rearranged to form new numbers; and how many are divisible by three -- meaning evenly divisible, with remainder zero?"
First, Ogilvy calculates how many different numbers there are -- it's 9! (that is, nine factorial):
9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 362,880
Then he tells that, much to our surprise, all 362,880 of these are multiples of three. The reason is the divisibility rule for three -- a number is divisible by three if the sum of its digits is. And no matter the order of the digits, the sum is the same:
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
As an aside, Ogilvy shows us the Gauss trick for finding this sum:
1 + 9 = 10
2 + 8 = 10
3 + 7 = 10
4 + 6 = 10
5 = 5 (no mate, because the collection contained an odd number of objects)
Total 45
After telling us the story of how the young Gauss showed us his teacher by adding up all the whole numbers from 1 to 100 in his head, Ogilvy returns to the divisibility rules, and indeed, he shows us why they work. He tells us that a four-digit number abcd really means:
1000a + 100b + 10c + d
He writes this as the sum of two numbers:
999a + 99b + 9c
+ a + b + c + d
The first row is clearly divisible by 3 and 9, and so the original number is divisible by 3 or 9 exactly when the second row, the sum of the digits, is. Ogilvy also proves the divisibility rule for 11:
1001a + 99b + 11c
- a + b - c + d
As we may recall from our discussion of number bases and the website Dozens Online, the rule for 3 and 9 is called "omega," while the rule for 11 is called "alpha."
The author now moves on to induction, or inductive reasoning. He warns us that inductive reasoning isn't always valid -- a statement may be true for a few cases but not in general. His example is the sum of the first few even numbers:
2 + 4 + 6 + ... + 2n
Is this sum equal to the following function:
Y(n) = n^4 - 10n^3 + 36n^2 - 49n + 24
Let's try it:
Y(1) = 1 - 10 + 36 - 49 + 24 = 2
and likewise Y(2) = 6, Y(3) = 12, and Y(4) = 20. And this does match:
2 =
2 + 4 = 6
2 + 4 + 6 = 12
2 + 4 + 6 + 8 = 20
But the next sum is 30, while Y(5) = 54. So the pattern doesn't work. Ogilvy mentions a formula that gives primes for the first 79 cases, while the 80th is composite. There's also a famous example -- the number of regions formed by joining n points around a circle with lines. The sequence starts 1, 2, 4, 8, 16, but the next number is 31, not 32. Here's a reference to that sequence at the home of all sequences, the OEIS:
https://oeis.org/A000127
On the other hand, what is the sum of the first n odd numbers?
1 = 1
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
The sum seems to be n^2 -- but "seems to be" isn't enough -- we need to prove it. We actually talked about this pattern earlier in Pappas, but now Ogilvy will actually prove it. He uses mathematical induction -- which has a initial and inductive cases. The initial case is trivial since 1^2 = 1, so all our work is for the inductive case. We start with case n = k and use this to prove case n = k + 1 -- notice that the kth odd number is 2k - 1:
1 + 3 + 5 + ... + (2k - 1) = k^2
1 + 3 + 5 + ... + (2k - 1) + (2k + 1) = k^2 + (2k + 1)
= k^2 + 2k + 1
= (k + 1)^2
which is exactly what we want to prove. Therefore:
1 + 3 + 5 + ... + (2n - 1) = n^2. QED
Ogilvy tells us that it's possible to write other powers as sums of consecutive odds as well. He tells us that even powers are themselves squares, so the same pattern works for even powers:
81 = 3^4 = 1 + 3 + 5 + ... + 17
For odd powers, the odd numbers to be added don't start with 1. For example, for cubes like m^3, the author tells us to start with m^2 - m + 1 and end with m^2 + m + 1, for a total of m odd numbers:
5^3 = 21 + 23 + 25 + 27 + 29
According to Ogilvy, mathematical induction is a powerful technique, but it only allows us to prove theorems -- it doesn't construct theorems from scratch. In other words, we can't use mathematical induction to come up with our own conjectures.
The author gives us a few more formulas to prove via induction. The sum of the first n positive integers -- also known as the nth triangular number -- is:
1 + 2 + 3 + ... + n = (n^2 + n)/2
The sum of the first n squares is:
1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6
And the sum of the first n cubes is:
1^3 + 2^3 + 3^3 + ... + n^3 = ((n^2 + n)/2)^2
Ogilvy points out that the sum of the first n cubes is the square of the first n integers. As for induction itself, recall that there are some theorems in Geometry that can be proved using induction. The most notable example is the sum of the angles of a polygon.
The author wraps up the chapter with a discussion of perfect numbers -- numbers equal to the sum of their proper divisors. He tells us that the ancient Greeks knew of four perfect numbers:
P_1 = 6 = 2(2^2 - 1)
P_2 = 28 = 2^2(2^3 - 1)
P_3 = 496 = 2^4(2^5 - 1)
P_4 = 8128 = 2^6(2^7 - 1)
He writes that medieval mathematicians made two conjectures about perfect numbers -- that the nth perfect number P_n had n digits, and that the last digit of P_n is 6 for odd n and 8 for even n. The first conjecture is completely false:
P_5 = (2^12)(2^13-1) = 33550336
Also, while all known perfect numbers end in either 6 or 8, they don't simply alternate. But all known perfect numbers are of the form:
2^(p - 1)(2^p - 1)
(I admit that this doesn't look good in ASCII.) According to Ogilvy, it was Euclid who first proved that if 2^p - 1 is prime, then N = 2^(p - 1)(2^p - 1) is perfect. Here is the proof: S_1 is the sum of all the powers of two dividing the number N, and S_2 is the sum of all the powers of two multiplied by the one odd factor 2^p - 1. Since this odd factor is prime, these are all the factors:
S_1 = 1 + 2 + 2^2 + ... + 2^(p - 1)
S_2 = (1 + 2 + 2^2 + ... + 2^(p - 1))(2^p - 1) = S_1(2^p - 1)
Therefore S_1 + S_2 = S_1 + S_1(2^p - 1) = S_1[1 + (2^p - 1)] = S_1(2^p)
Ogilvy now uses the formula for the sum of a geometric sequence, often learned in Algebra II (a 1st term, l last term, r ratio):
S = (lr - a)/(r - 1)
S_1 = (2^(p - 1)*2 - 1)/(2 - 1) = 2^p - 1
Thus our total is:
S_1 + S_2 = (2^p - 1)2^p = 2^p(2^p - 1)
= 2[2^(p - 1)(2^p - 1)] = 2N
And since we must add only the proper divisors (that is, the ones less than N), we must subtract N, leaving us with N as the sum of the proper divisors. Therefore N is perfect. QED
Euler, of Konigsberg fame, proved a partial converse of Euclid's theorem. If N is an even perfect number, then N must be of the form:
N = 2^(p - 1)(2^p - 1), where 2^p - 1 is prime
The converse doesn't work for potential odd perfect numbers. As of now, no odd perfect number is known -- Ogilvy gives 10^25 as a lower bound on the smallest odd perfect. This was in 1966 -- the lower bound on the smallest odd perfect in 2017 is 10^1500.
Also, at the time this book was published, only 19 perfect numbers were known. In the half-century since the book was written, 30 more have been discovered for a total of 49. Notice that we know as many perfect numbers as we know primes of the form 2^p - 1 -- Mersenne primes. I mentioned the largest known Mersenne prime in my January 10th post -- it has over 22 million digits. The largest known perfect is in fact the triangular number of this Mersenne -- it has over 44 million digits.
Ogilvy ends this chapter as follows:
"More data has thus been obtained concerning certain problems of very large numbers during the past decade [before 1966 -- dw] than in all the previous years of the history of these problems."
He was writing about odd perfect numbers, but he could have been talking about Goldbach's conjecture -- which I just happened to mention two years ago in today's Geometry lesson:
Lesson 2-2 of the U of Chicago text is called "If-then Statements." This is what I wrote two years ago about today's lesson:
Lesson 2-2 of the U of Chicago text continues the study of logic by focusing on "if-then" statements. I certainly agree with the text when it writes:
"The small word 'if' is among the most important words in the language of logic and reasoning."
There are a few changes that I will make to the text. First of all, the text refers to the two parts of a conditional statement as the antecedent and the consequent -- although it does mention hypothesis and conclusion as acceptable alternatives. I'm going to follow what the majority of texts do and just use the words hypothesis and conclusion. Actually, Dr. Franklin Mason doesn't even use the word hypothesis -- he simply uses the word given -- since after all, the hypothesis of a theorem corresponds to the "given" statement in a two-column proof.
When I teach or tutor students in geometry, one of my favorite examples is "if a pencil is in my right hand, then it is yellow." So I pick up three yellow pencils, and we observe that the conditional is true. But let's suppose that I pick up a blue pencil in addition to the three yellow pencils. Now the conditional is false, since we can find a counterexample -- the blue pencil, since that's a pencil in my right hand yet isn't yellow.
Notice that I decided to replace the word instance with the word example -- so that the connection between examples and counterexamples becomes evident.
The text has to go back to an example from that dreaded algebra again. Of course, it's an important example, since students often forget that 9 has two square roots, 3 and -3. But I decided to include it anyway since it's simple -- it's not as if I'm making students use the quadratic formula or anything like that.
Then the book moves on to a famous mathematical statement: Goldbach's conjecture, named after the German mathematician Christian Goldbach who lived 300 years ago:
If n is an even number greater than 2, then there are always two primes whose sum is n.
At the time the book was written, the conjecture had been verified up to 100 million, but the conjecture had yet to be proved. But what about now -- has anyone proved Goldbach's conjecture yet? As it turns out, the answer is still no -- but now the conjecture has been verified up to four quintillion -- that is, the number 4 followed by 18 zeros.
But there has been work on a similar statement, called Goldbach's weak conjecture:
If n is an odd number greater than 5, then there are always three primes whose sum is n.
This is called weak because if the better-known (or strong) conjecture is true, the weak is automatically true because we can always let the third prime just be 3. Ironically, when Goldbach himself actually stated his conjecture, he stated the weak version of the conjecture. It was a letter from Euler -- you know, the same Euler who solved the bridge problem that we discussed as an Opening Activity -- that convinced Goldbach to state the strong conjecture instead.
Now as it turns out, someone has claimed a proof of Goldbach's weak conjecture -- namely the Peruvian mathematician Harald Helfgott. Last year, Helfgott's proof was still being peer-reviewed -- that is, checked by other mathematicians to find out whether the proof is correct. By now, Helfgott's proof has finally been verified. Yes, mathematicians are still proving new theorems everyday.
Dr. M also mentions Goldbach's conjecture, on a worksheet for his Lesson 2-1. Often students are fascinated when they hear about conjectures that take centuries to prove, such as Goldbach's conjecture or Fermat's Last Theorem. I often use these examples to motivate students to be persistent when trying to come up with proofs in geometry -- if mathematicians Helfgott and Wiles didn't give up even after centuries of trying to prove these conjectures, then why should they give up after minutes?
The final example in this section has students rewrite statements into if-then form. With the newly released Common Core scores still fresh on my mind, I still can't help but think of the conditionals:
If a school is an elementary school, then its math scores are higher than its ELA scores.
If a school is a high school, then its math scores are lower than its ELA scores.
These two statements undoubtedly have counterexamples, but these hold more often than not, judging by the released scores.
I've found that oftentimes, English learners struggle with this part of the lesson. The teacher must point out why, for example, the "something" in the example "all triangles have three sides" must be a figure: "if a figure is a triangle, then it has three sides." So not only must we appease algebra haters when we include algebra in the geometry lesson, but we must also consider English learners when including English in the geometry lesson (again, no wonder that so many students performed poorly on the Common Core tests).
In addition to today's worksheet, I'm restoring my old pattern of posting a weekly activity. In a way, nearly all of Chapters 0 and 1 are activities, so I resume this tradition here in Chapter 2. Our first activity from two years ago is a list of logic puzzles, to go along with the logic that we learn here in this chapter. Yes, I know that for Illinois State, I do projects on Tuesdays and assessment on Fridays, but for now I'm posting tests on Wednesdays and activities on Fridays.
Here's a little of what I wrote two years ago about the logic puzzles:
As it turns out, I've seen a version of this puzzle before last spring. It is a similar brain teaser known as the "Sum and Product Puzzle." The next link contains a statement and solution of the puzzle:
http://www.qbyte.org/puzzles/p003s.html
Notice that in describing the solution, the author actually uses Goldbach's conjecture -- the unproved conjecture that I mentioned earlier in this post. Of course, the numbers involved in this problem are much too small to be counterexamples to Goldbach.
I'll repeat the same activity worksheet from last year, although it might be interesting to replace the old Puzzle #10 with Cheryl's birthday problem. The sum and product version of this puzzle might be suitable in an algebra class, especially near the lesson on factoring quadratic polynomials.
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