"Dwellings, be they private or commercial, were viewed by Frank Lloyd Wright as a conglomeration of boxes or cubes. Space in Euclidean geometry is defined as the set of all points."
This is the second page of the section on Frank Lloyd Wright. The definition of space given here should sound familiar, since it appeared two weeks ago, in Lesson 1-6 of the U of Chicago text:
Definitions:
Space is the set of all possible points.
The point Pappas makes here is that just as a square or rectangle is a representation of a plane rather than a plane itself, a cube or box is a representation of space rather than space itself, since a box doesn't contain all points. She writes:
"Wright wanted his works to give the feeling of space -- the flowing of points from the inside to the outside."
And space is not a box. This is why the title of this section is "The Destruction of the Box." Pappas tells us how the architect achieved his illusion of space:
"Wright realized that the potential of certain building materials had not been utilized."
So once again, with a new geometrical structure comes a new material structure. Here, the materials of choice are steel and glass. She explains:
"Wright's designs did away with corners of the box by removing the supports from the corners and relocating them along the walls by using cantilevers."
By doing so, as Pappas concludes:
"Space was allowed to flow."
There are two drawings on this page. One of them, as described by its caption, is traditional "box architecture," while the other is Wright's design. Unfortunately, I couldn't find a good picture similar to this one on the Internet. (I tried a Google image search: frank lloyd wright blueprints.) So you'll just have to visualize his designs on your own.
Chapter 4 of Stanley Ogilvy's Excursions in Number Theory is called "Congruence Arithmetic." This is how this chapter begins:
"Congruence arithmetic is in common use in the building industry. Designers, manufacturers, even carpenters and masons, without having studied number theory, use it every day."
So what exactly is this "congruence arithmetic," anyway? No, it has nothing to do with the concept of congruence in Geometry. Ogilvy's first example of congruence involves a carpenter (maybe he's trying to build to a Frank Lloyd Wright design). He needs two walls of lengths 11 ft. and 19 ft., yet he has only 8-foot sections available. Each wall has a remainder of 3 ft., therefore:
19 == 11 (mod 8).
The symbol "==" should actually be three lines and read "is congruent to." But a double-equal sign is the best way to render this symbol in ASCII.
His next example has a scientist perform a 41-hour experiment beginning at midnight. The end of the experiment is at 5:00, since:
41 == 5 (mod 12).
This is clock arithmetic -- and in fact, I wrote about clock arithmetic in several posts, most recently last year in my October 5th post. Since it's been almost a year since I posted it, let's link again the song from Square One TV associated with clock arithmetic, "Timekeeper":
By the way, this is what I wrote last year about it:
I actually purchased one more math text, but I must admit that my reason for purchasing this book is pure nostalgia. I bought Mathematics Around the Clock, published in 1970. This text is considered part of the "California State Series," even though this was well before California State Standards.
The subject of this text is "clock arithmetic," more formally called "modular arithmetic." Here is a link discussing modular arithmetic:
https://nrich.maths.org/4350
There is no grade level associated with this text. The above link labels this as a "Stage 4" topic, which corresponds roughly to junior and senior years in high school, but I suspect that this text can be used with much younger students (though probably not younger than fourth grade, as these kids are still learning ordinary arithmetic). In fact, I was given a copy of this text when I was younger -- I no longer own that copy, which is why I said that I bought the book for nostalgic reasons.
So I owned this text when I was young, and found it again last year. About halfway between my ownership of the two copies, I bought another book by the same publisher, Probability: The Science of Chance. (This was well before I started this blog.) It was only a couple of months ago when I realized that I owned both texts from this publisher. Incidentally, the Probability book (which is great for seventh grade Common Core despite its being printed in 1968) would have helped us yesterday, when Ogilvy mentions two probability concepts. Both of these (complementary events and independent events) appear in Part III of Probability.
But let's look at Mathematics Around the Clock alongside Ogilvy's book. He tells us that:
If 41 == 5 (mod 12), then 41k == 5k (mod 12)
and in general:
If a == b (mod m), then ak == bk (mod m)
The first proof in this chapter is of the following:
Given a == b (mod m) and c == d (mod m), is it true that ac = bd (mod m)?
In order to prove this, notice that a == b (mod m) is defined to mean that a - b is divisible by m -- that is, there exists an integer j such that a - b = jm.
Proof:
Now c == d (mod m) says that c - d = km, or c = km + d. Therefore:
ac - bd = a(km + d) - bd
= akm + ad - bd
= akm + d(a - b).
The right-hand side is divisible by m since it's the sum of two multiples of m -- akm and d(a - b), the latter a multiple of m since a == b (mod m) implies a - b a multiple of m. Therefore the left hand side, ac - bd, has m as a factor, or ac == bd (mod m). QED
Ogilvy now proves the Addition Property of Congruence. The proof is so simple that I can produce his entire proof right here:
Proof:
Given
a - b = jm for some j
and
c - d = km for some k
(a + c) - (b + d) = (j + k)m = lm for some l, QED
Ogilvy gives us one more theorem, a Power Property of Congruence:
If a == b (mod m), then a^k == b^k (mod m).
Proof:
a^k - b^k is always divisible by a - b. But a - b is divisible by m (given). Therefore so is a^k - b^k. QED
The author tells us that a^k - b^k is divisible by a - b because long division gives us the following, which he calls a "cyclotomic" expression:
a^(k - 1) + a^(k - 2)b + a^(k - 3)b^2 + ... + a^2 b^(k - 3) + ab^(k - 2) + b^(k - 1)
According to Ogilvy, congruence mod m is an equivalence relation with three properties:
(1) Reflexivity: a == a (mod m)
(2) Symmetry: if a == b (mod m), then b == a (mod m)
(3) Transitivity: if a == b (mod m) and b == c (mod m), then a == c (mod m).
Notice that we've just proved that congruence mod m satisfies many of the "Postulates from Algebra" mentioned in Lesson 1-7 of the U of Chicago text. In fact, congruence mod m satisfies all of the Properties of Equality, but none of the Properties of Inequality listed there.
At this point, Ogilvy tells us that we can create addition and multiplication tables mod 12. He doesn't print them in his book, but they appear in Part I of Mathematics Around the Clock. He does provide us with the tables mod 5, which appear in Part III of Clock.
Now Ogilvy gives us some examples of how the Properties of Congruence may be used:
Example 1: Is 999999 divisible by 7? We write:
999999 = 10^6 - 1
10 == 3 (mod 7)
10^6 == 3^6 (mod 7)
But 3^6 = (3^2)^3 = 9^3, and 9 === 2 (mod 7), therefore 9^3 == 2^3 == 1 (mod 7). Finally, substituting from above,
10^6 == 1 (mod 7)
That is to say, 999999 is evenly divisible by 7.
Example 2. Every even power of any odd number is congruent to 1 modulo 8. He proves this by noticing that all odd numbers are congruent to 1, 3, 5, or 7 mod 8, and the squares of these are 1, 9, 25, and 49, which are all congruent to 1 mod 8.
Example 3. Now he returns to an example from an earlier chapter, the proof that the Fermat number:
F_5 = 2^2^5 + 1 = 2^32 + 1
isn't prime, but is divisible by 641. We begin by writing:
640 = 10 * 64 = 5 * 128 = 5 * 2^7 == -1 (mod 641).
Taking the fourth power of both sides,
5^4 * 2^28 == 1 (mod 641)
The author now points out that 5^4 = 625 == -16 (mod 641), which we substitute in as -(2^4):
-(2^4) * 2^28 == 1 (mod 641)
-(2^32) == 1 (mod 641)
2^32 == -1 (mod 641)
2^32 + 1 == 0 (mod 641).
Ogilvy now moves on to a possible Division Property of Congruence. That is:
Given ab == ac (mod m), when is b == c (mod m)?
Again, he reverts to the definition of congruence:
Given (ab - ac) is divisible by m, or a(b - c) is divisible by m, when is (b - c) divisible by m?
A sufficient condition is when m and a are relatively prime. Notice that if the modulus m is itself prime, then a can be any number other than a multiple of m.
Examples: (a) 99 == 9 (mod 10)
and 11 == 1 (mod 10)
We divided by 9, a number relatively prime to the modulus.
(b) 48 == 12 (mod 6)
and (1) 8 == 2 (mod 6)
but (2) 4 =!= 1 (mod 6)
Both times we divided by a number not relatively prime to the modulus. The first time it worked and the second time it didn't. But in the big theorem that we're about to prove, the modulus will be prime.
The theorem is known as Fermat's Little Theorem. Believe it or not, Fermat's Little Theorem is stated (but not proved) in Mathematics Around the Clock. I remember being a young child and reading about this amazing result, which appears near the end of the book "Part VI: For the Adventurous." I point out that the theorem isn't referred to as "Fermat's Little Theorem" in Clock (and in fact, Ogilvy doesn't call it a "Little" Theorem either).
Fermat's Little Theorem:
If a is not divisible by a prime p, then a^(p - 1) == 1 (mod p).
Proof (Ogilvy):
We consider the set of numbers a, 2a, 3a, ... (p - 1)a, and ask how they fall into the residue classes modulo p. But no one of our set is congruent to p itself (because we stopped at (p - 1)a. and a is not divisible by p). Furthermore, no two of the set are congruent to each other. For if xa == ya (mod p) then x == y (mod p) by our division theorem. Therefore each is congruent to some one of 1, 2, 3, ... (p - 1) (not necessarily in order). Therefore the product of all the set is congruent to the (p - 1)!. That is, factoring our all the a's:
a^(p - 1)(p - 1)! == (p - 1)! (mod p)
Since (p - 1)! isn't divisible by p, we can divide by it:
a^(p - 1) == 1 (mod p). QED
The author tells us that Example 1 from earlier becomes much easier now. He writes:
10^6 == 1 (mod 7).
"Fermat's [Little] Theorem says that it is true."
Ogilvy moves on to binomial expansions. He writes:
(x + y)^0 = 1
(x + y)^1 = 1x + 1y
(x + y)^2 = 1x^2 + 2xy + 1y^2
(x + y)^3 = 1x^3 + 3x^2 y + 3xy^2 + 1y^3
(x + y)^4 = 1x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + 1y^4
He then finds (x + y)^5 using a form of "skeletal multiplication":
(x + y)^4 1 4 6 4 1
(x + y) 1 1
1 4 6 4 1
1 4 6 4 1
(x + y)^5 1 5 10 10 5 1
This, of course, is Pascal's Triangle. I don't reproduce it here, since it's so easy to find. It is a staple of Algebra II classes.
Ogilvy gives a well-known formula for the (k + 1)th term in the n'th row:
n! / (k!(n - k)!)
Notice that here the author is referring to the top 1 of Pascal's triangle as the 0th row, but the two 1's in the n'th row are the 1st and (n + 1)th terms in that row. I've seen other ways of indexing the rows and terms in each row (zero vs. one-based indexing).
Ogilvy asks, do there exist two consecutive terms in Pascal's triangle such that one of them is exactly 2/3 of the next term? If a is the ratio between the (k + 1)th and k'th terms, then we write:
n! / (k!(n - k)!) = an! / ((k - 1)!(n - k + 1)!)
which he solves as:
n = ka + k - 1.
Much to our surprise, this equation has infinitely many solutions for any positive rational a. For the example of 2/3 (that is, a = 3/2), he lists the following solutions:
k = 2, n = 4 (values 4, 6)
k = 4, n = 9 (values 84, 126)
k = 6, n = 14 (values 2002, 3003)
and so on. The author also briefly points out that in the p'th row for p a prime, all the values except for the first and last 1 are prime. This is because if we look at the expression:
C = p! / (k!(p - k)!)
no term in the denominator can cancel the p in the numerator (as p!) unless either k! or (p - k)! is p! itself, which correspond to the first and last 1.
Ogilvy uses this to give a second proof of Fermat's Little Theorem using Pascal's Triangle. He says:
If a^(p - 1) == 1 (mod p), then a^p == a (mod p).
That is a^p - a is a multiple of p. This he proves by mathematical induction on a -- the initial case is 1^p == 1 (mod p), which is trivial. The inductive case is, if the formula works for a = b, does the formula work for a = b + 1? He expands, using the binomial theorem:
(b + 1)^p - (b + 1) = {b^p + [all the terms except the first and the last] + 1} - (b + 1)
= b^p - b + [all the terms except the first and the last]
He tells us that all the terms except the last are divisible by p from above, and b^p - b is also divisible by p from the induction assumption (that is, that the formula works for a = b). So all the terms on the right are divisible by p, and so the left side, (b + 1)^p - (b + 1), is also divisible by p. QED
Ogilvy now rewrites Fermat's Little Theorem as a conditional (shades of U of Chicago Lesson 2-2!):
"If p is a prime it divides a^p - a."
And now he asks whether the converse is true:
"If a^p - a is divisible by p, then p is a prime."
If this were true, he adds, it would be a criterion for primality. When I was a young child reading about the theorem in Mathematics Around the Clock, I actually wondered whether we could tell whether large numbers are prime by using the theorem. My guess was that we couldn't, because for large candidates p, a^p would be too hard to calculate.
But I was wrong. Indeed, we don't need to calculate a^p itself, but a^p modulo p. We saw earlier in this chapter that there are tricks to finding powers mod p (just look at how we reduced 2^32 mod 641 earlier today). Ogilvy writes:
"Strangely enough, this converse holds most of the time but not always."
In fact, it holds often enough that the Fermat method is in fact the dominant method for checking primality nowadays. It is modified in order to avoid the exceptions. The author shows us one of the simplest counterexamples, a = 2, p = 341. We see that 341 = 31 * 11 is composite yet satisfies the Fermat primality test, and so it's known as a pseudoprime:
2^5 == 1 (mod 31)
(2^5)^68 == 1^68 (mod 31)
2^340 == 1 (mod 31)
2^5 == -1 (mod 11)
2^340 == 1 (mod 11)
Thus both 11 and 31 are prime factors of 2^340 - 1. And now:
2^341 - 2 = 2 * (2^340 - 1)
= 2 * 11 * 31 * other things
= 2 * 341 * other things.
Fermat's Little Theorem is closely related to number bases. The length of the period of 1/p (p prime) written as a repeating "decimal" (in a base) is related to the theorem. The availability of certain divisibility rules (such as 5 in dozenal) is also related to the theorem.
The final thing in this chapter is the Monkey and Coconuts Problem. I don't want to type it out, so let's just provide a link:
http://mathworld.wolfram.com/MonkeyandCoconutProblem.html
Ogilvy solves the case (for 5 men and 1 leftover coconut) as follows: he lets N be the original number of coconuts and a be the number each sailor gets in the division (just as Wolfram does at the link):
1/5 (4/5 (4/5 (4/5 (4/5 (4/5 (N - 1) - 1) - 1) - 1) - 1) - 1) = a.
Removing the parentheses and regrouping yields
(4/5)^5 N - [1 + 4/5 + (4/5)^2 + (4/5)^3 + (4/5)^4 + (4/5)^5] = 5a.
He then uses the same formula for a finite geometric series from an earlier chapter:
4^5 (N + 4) = 5^6 (a + 1)
Since none of the factors of 5 from the right side are in 4^5, they must be in N + 4. So he writes:
(N + 4) is divisible by 5^6 = 15625
N == -4 (mod 15625).
He explains that -4 coconuts is impossible, so the smallest possible solution is 15621. He remarks:
"That's a lot of coconuts!"
I wonder whether number bases can help us solve this problem as well. After all, I notice that the smallest solution in quinary (base 5) is 444441 -- and in fact, the solution set is the set of all numbers ending in 444441 in quinary. All, we see the fractions 1/5 and 4/5 (five times) in the first equation written above, and all of those numerators in reverse gives 444441.
In fact, I sort of see the first few steps in quinary. The last 1 is for the monkey, and so the 444440 remaining coconuts are divisible by 5. The sailor removes 44444 coconuts leaving 344441 -- another number ending in several 4's followed by a 1. So again the 1 is for the monkey, and so on.
Fermat's Theorem is an example of a theorem whose converse is false. But what exactly do we mean by a "converse"? Well, that's the topic of today's lesson.
Lesson 2-4 of the U of Chicago text is called "Converses." (It appears as Lesson 2-3 in the modern edition of the text.)
This is what I wrote two years ago about today's lesson. There is a little bit of politics near the end of this post, because I'd perceived one of the examples of fallacious reasoning (assuming that a statement and its converse are equivalent) as one often committed by Republicans. So I added a similar fallacy made by Democrats in order for this post to remain politically balanced -- the point being made that both parties are prone to making logical fallacies. But only the example from the text actually appears on the worksheet:
Lesson 2-4 of the U of Chicago text deals with an important concept in mathematical logic -- converses. We know that every conditional statement has a converse, found by switching the hypothesis and conclusion of that conditional.
The conditional "if a pencil is in my right hand, then it is yellow" has a converse, namely -- "if a pencil is yellow, then it is in my right hand." The original conditional may be true -- suppose all the pencils in my right hand happen to be yellow -- but the converse is false, unless every single yellow pencil in the world happens to be in my right hand. A counterexample to the converse would be a yellow pencil that's on the teacher's desk, or in a student's backpack, or even in my left hand -- anywhere other than my right hand.
If converting statements into if-then form can be confusing for English learners, then writing their converses is even more so. Here's an example from the text:
-- Every one of my [Mrs. Wilson's] children shall receive ten percent of my estate.
Converting this into if-then form, it becomes:
-- If someone is Mrs. Wilson's child, then he or she shall receive ten percent of the estate.
Now if a student -- especially an English learner -- blindly switches the hypothesis and the conclusion, then the following sentence will result:
-- If he or she shall receive ten percent of the estate, then someone is Mrs. Wilson's child.
But this is how the book actually writes the converse:
-- If someone receives ten percent of the estate, then that person is Mrs. Wilson's child.
In particular, we must consider the grammatical use of nouns and pronouns. In English, we ordinarily give a noun first, and only then can we have a pronoun referring to that noun. Therefore the hypothesis usually contains a noun, and the conclusion usually contains a pronoun. (Notice that grammarians sometimes refer to the noun to which a pronoun refers as its antecedent -- and of course, the text refers to the hypothesis of a conditional as its antecedent. So the rule is, the antecedent must contain the antecedent.)
And so when we write the converse of a statement, the hypothesis must still contain the noun -- even though the new hypothesis may be the old conclusion that contained a pronoun. So the converse of another conditional from the book:
-- If a man has blue eyes, then he weighs over 150 lb.
is:
-- If a man weighs over 150 lb., then he has blue eyes.
Saying the converse so that it's grammatical may be natural to a native English speaker, but may be confusing to an English learner.
Now let's look at the questions to see which ones are viable exercises for my image upload. An interesting one is Question 9:
A, B, and C are collinear points.
p: AB + BC = AC
q: B is between A and C
and the students are directed to determine whether p=>q and q=>p are true or false. Now notice that the conditional q=>p is what this text calls the Betweenness Theorem (and what other books call the Segment Addition Postulate). But p=>q is the converse of the Betweenness Theorem -- and the whole point of this chapter is that just because a conditional is true, the converse need not be true. (Notice that many texts that call this the Segment Addition Postulate simply add another postulate stating that the converse is true.)
Now the U of Chicago text does present the converse of the Betweenness Theorem as a theorem, but the problem is that it appears in Lesson 1-9, which we skipped because we wanted to delay the Triangle Inequality until it can be proved.
Here, I'll discuss the proof of this converse. I think that it's important to show the proof on this blog because, as it turns out, many proofs of converses will following the same pattern. As it turns out, one way to prove the converse of a theorem is to combine the forward theorem with a uniqueness statement. After all, the truth of both a statement and its converse often imply uniqueness. Consider the following true statement:
-- Donald Trump is currently the President of the United States.
We can write this as a true conditional:
-- If a person is Donald Trump, then he is currently the President of the United States.
The converse of this conditional:
-- If a person is currently the President of the United States, then he is Donald Trump.
This converse is clearly true as well. By claiming the truth of both the conditional and its converse, we're making a uniqueness statement -- namely that Trump is the only person who is currently the President of the United States.
So let's prove the converse of the Betweenness Theorem. The converse is written as:
-- If A, B, and C are distinct points and AB + BC = AC, then B is on
(I explained why segment AC has a strikethrough back in Lesson 1-8.)
Proof:
Let's let AB = x and BC = y, so that AC = x + y. To do this, we begin with a line and mark off three points on it so that B is between A and C, with AB = x and BC = y. This is possible by the Point-Line-Plane Postulate (the Ruler Postulate). By the forward Betweenness Theorem, AC = x + y.
Now we want to show uniqueness -- that is, that B is the only point that is exactly x units from A and y units from C. We let D be another point that is x units from A, other than B. By the Ruler Postulate again, D can't lie on
In the former case, the forward Betweenness Theorem gives us AC + AD = DC. Then the Substitution and Property of Equality give us DC = (x + y) + x or 2x + y, which isn't equal to y (unless 2x = 0 or x = 0, making A and D the same point when they're supposed to be distinct).
In the latter case, with ACD a triangle, we use the Triangle Inequality to derive AD + DC >AC. Then the Substitution Property gives us x + CD > x + y, and then the Subtraction Property of Inequality gives us the statement CD > y, so DC still is not equal to y.
So B is the only point that makes BC equal to y -- and it lies on
(The explanation in the book is similar, but this is more formal.) Later on, we're able to use this trick to prove converses of other theorems. So the converse of the Pythagorean Theorem will be proved using the forward Pythagorean Theorem plus a uniqueness statement -- and that statement is SSS, which tells us that there is at most one unique triangle with side lengths a, b, and c up to isometry. And, more importantly, the converse to the what the text calls the Corresponding Angles Postulate (the forward postulate is a Parallel Test, while the converse is a Parallel Consequence) can be proved using forward postulate plus a uniqueness statement. The uniqueness statement turns out to be the uniqueness of a line through a point parallel to a line -- in other words, Playfair's Parallel Postulate. This explains why the forward statement can be proved without Playfair, yet the converse requires it.
In the end, I decided to throw out Question 9 -- I don't want to confuse students by having a statement whose converse requires the Triangle Inequality that we skipped (and the students would likely just assume that the converse is true without proving it) -- and used Question 10 instead.
But now we get to Question 13 -- and here's where the controversy begins. I like to include examples in this lesson that aren't mathematical -- doing so might engage students who are turned off by doing nothing but lifeless math the entire period. This chapter on mathematical logic naturally lends itself to using example outside of mathematics. But the problem is that Question 13 is highly political. Written in conditional form, Question 13 is:
-- If a country has communist, then it has socialized medicine.
and its converse is:
-- If a country has socialized medicine, then it is communist.
Just like the article at the beginning of the chapter (with discussion of "terrorists" before 9/11), the change in political climate since the publishing of this text has rendered the question controversial. For during the past few years, this country has moved in the direction of socialized medicine with the passage of the Affordable Care Act -- better known by its nickname, "Obamacare."
Now some opponents of Obamacare would argue that moving towards socialized medicine and away from a free-market solution is indeed a step towards socialism or communism. But the intended answer of this question is that this reasoning is incorrect, since it assumes that the converse of a statement is true just because the original conditional is true. But some opponents of Obamacare might argue that the converse is true, anyway. Since defenders of Obamacare tend to be Democrats -- that is, of the same party as the administration that passed the law -- and opponents of the law tend to be Republican, the question may be viewed as having a pro-Democratic slant.
Now I wish this blog to be politically neutral. But unfortunately, the problem is that this is a Common Core blog, and Common Core is itself politically charged. Once again, it was a Democratic administration that passed Common Core, and so once again, many Republicans oppose the standards (indeed, one nickname for the standards is "Obamacore"). One argument against Common Core is that it promotes slanted viewpoints that favor the Democrats -- and by including Question 13 on a blog that purports to be a Common Core blog, I'm lending credence to that argument!
Notice that I have a mixed opinion of the Common Core Standards. I write about CCSS not because I wholeheartedly endorse them, but because I want teachers to know know to teach to them. But I do defend Common Core against invalid arguments. For example, the argument that Common Core endorses a slanted view of history is invalid because the standards cover only English and math (and with the upcoming Next Generation Science Standards, science as well), not history.
Now I don't want Question 13 to convince readers that the Common Core encourages a political view slanted towards the Democrats. (Recall that the question comes from a book that was written years before there even was a Common Core, and that the question has a Chinese immigrant -- not an American Republican -- making the fallacious argument.) It is argued by critics that math classes should be teaching nothing but math, not non-mathematical politics. But I want to include non-mathematical topics because these are more likely to engage the students than teaching nothing but math.
In the end, I will include Question 13 on the worksheet. But I left a space so that if a teacher feels that the question is politically slanted, then he or she could add another question to balance it. For example, that teacher can add a fallacious argument often made by Democrats, for example:
-- If a white person is racist, then he or she opposes Obama.
-- If a white person opposes Obama, then he or she is racist.
Of course, this question adds a new layer of controversy (race) to the mix. Teachers who want to add a balancing question should just write in their own question, or just throw out the question about socialized medicine altogether.
In the review section, I'd have loved to include Question 15, a review of the last lesson on programming (and of course changed it to TI-BASIC). But in deference to those classes that skipped the lesson because not every student has a graphing calculator, I've thrown it out and included only questions from the fully covered Lessons 2-2, 1-8, and 1-6.
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