*Mind-Bending Math*is called "Elementary Math Isn't Elementary." In this lecture, Dave Kung explains that even something as simple as numbers -- elementary arithmetic -- is filled with paradoxes. Here are some of the ones that he mentions:

-- Berry's Paradox: "The smallest natural number that cannot be described in English using fewer than one hundred ten characters" describes a certain natural number using only 107 characters. Kung chooses 110 characters since it is just greater than then 107 needed in the definition, but perhaps it makes more sense in this day and age to say 140 characters, because of Twitter and all that.

-- The Banach-Tarski Paradox: Kung says that this paradox implies that 1 + 1 = 1. But there's already a lecture -- the next-to-last lecture -- devoted to this paradox, so I should just wait until we reach that lecture to describe it.

-- Averages: Kung gives several paradoxes involving averages. For example, he states that if you have a hybrid car with a fuel efficiency of 50 mpg and an SUV with an efficiency of 10 mpg, you'd be better off upgrading the SUV from 10 to 12 mpg than upgrading the hybrid from 50 to 100 mpg. He writes that it's easier to see why if we use something called the

*harmonic mean*, 2 / (1/

*a*+ 1/

*b*) -- which can be simplified as 2

*ab*/ (

*a*+

*b*) -- or just by flipping the fractions so that the reference unit, miles, are in the denominator. Kung uses the units of gallons per 100 miles -- I've heard that in metric countries, fuel efficiency is given as liters per 100 kilometers, which avoids this paradox.

I've seen this problem with averaging before. For example, if a friend lives 60 miles away and I drive at 60 mph, then a round trip will obviously take two hours. But if I drive there at only 30 mph, I can't make up for it by driving the return trip at 90 mph. This is because by driving at 30 mph, it already takes two hours just to get there -- so I've have to go back at

*instantaneous*speed in order to make it back in time.

Here's another example to dramatize why naively taking averages of the two speeds doesn't work -- let's say there's a relay race. The goal is to run one lap around a standard 400-meter track. You will run half a lap, and a

*snail*will run the other half. Let's say that you decide just to

*walk*your 200 meters, and it takes two minutes to finish your half of the race. As for the snail, notice that 200 meters is approximately equal to one

*furlong*, and an old joke is that a snail travels at approximately one furlong per fortnight. (Do you remember fortnights from my last blog post?) So the total race time will be 14 days and two minutes.

Now let's ask, what will improve our race time the most -- doubling the snail's speed, or

*tripling*your own speed? If you triple your speed, the total race time is 14 days and 40 seconds. If you quadruple your speed, the total race time is 14 days and 30 seconds. If you have the Jamaican sprinter Usain Bolt run your half of the race, the total race time would be 14 days and 19.19 seconds. Clearly, it's much better to double the snail's speed, so that it will only take about a week to finish the race. In fact, it's much better to increase the snail's speed by about 7% (so that it takes only 13 days) rather than improve your speed by

*any*amount. The total time for the race is dominated by the snail's speed, since for most of the race time, the snail is the one with the baton.

The idea that we need to put the reference units in the denominator appear in various word problems that often appear in Algebra I and II. In these speed problems mentioned above, miles need to be in the denominator. But for some problems, hours should be in the denominator. If you, starting from your house, are driving towards your friend's house at 60 mph, and your friend, starting from your destination, is traveling towards your house at 30 mph, then it is correct to add 30 and 60 mph. Hours also need to be in the denominator in work problems -- if it takes you six hours to paint your house and it takes your friend only three hours, then it's wrong to add six and three (hours per house). We must add 1/6 and 1/3 (houses per hour) to solve the problem.

The Quick Conundrum for today's lecture involves breaking uncooked spaghetti. As it turns out, it's impossible to break spaghetti sticks in two -- a third, albeit smaller, piece always appears. According to Kung, this is because breaking it sends a wave down the larger piece, and as it bounces back, it causes that piece to break again.

Kung wraps up the lesson by discussing class size -- which is relevant to us as teachers. There are three different ways to count the average class size at a school or university. The simplest is just to find the student-to-faculty ratio. But if you ask students how many others are in their classes, as well as ask teachers how many students are in their classes, both will obtain a higher ratio. This is because these ratios are biased towards larger classes -- they are counted more often. This leads to yet another paradox -- the Friendship Paradox: On average, your friends have more friends than you, as defined by counting your friends on a website such as Facebook. (First I mention Twitter in this post, and now I mention Facebook -- what's this blog coming to?)

Oh, and speaking of school -- yes, winter break is over, and it's back to Geometry. From now until the end of the semester, I'm filling in a few topics that aren't covered well in the U of Chicago text. As it turns out, all of these topics fit in a single chapter of the Glencoe Geometry text -- Chapter 5. But in the U of Chicago text, these topics are spread out among different chapters.

Today's Lesson 13-7, on exterior angles, corresponds roughly to Lesson 5-2 in Glencoe. Now the Glencoe text uses the theorems of 5-2 to prove the Triangle Inequality two lessons later, in 5-4. On the other hand, Triangle Inequality is a postulate in the U of Chicago's Lesson 1-9. Since I want to follow the order of Glencoe's Chapter 5 this year, we will cover 13-7 first and then 1-9.

Last year, I combined Lessons 13-7 and 1-9 into a single worksheet. This was because I was actually tutoring a Geometry student who was getting ready to study Glencoe's Lesson 5-4, and I wanted to have a worksheet finished in time to help him. Since I'm not tutoring anyone this year, I can go back and separate 13-7 and 1-9 into different worksheets. So today we are covering 13-7, and then we can cover 1-9 tomorrow.

This is what I wrote last year about Lesson 13-7:

But the first four theorems are proved in U of Chicago's Lesson 13-7. Since the Exterior Angle Theorem (TEAE) is easy enough to decipher, and the TEAI follows almost trivially from TEAE, let's skip directly to the Unequal Sides Theorem. Its proof is given in the two-column format. Here I reproduce that proof, starting with a Given step:

Unequal Sides Theorem (Triangle Side-Angle Inequality, TSAI):

If two sides of a triangle are not congruent, then the angles opposite them are not congruent, and the larger angle is opposite the longer side.

Given: Triangle

*ABC*with

*BA*>

*BC*

Prove: angle

*C*> angle

*A*

Proof:

Statements Reasons

1. Triangle

*ABC*with

*BA*>

*BC*1. Given

2. Identify point

*C'*on ray

*BA*2. On a ray, there is exactly one point at a given distance from

with

*BC'*=

*BC*an endpoint.

3. angle 1 = angle 2 3. Isosceles Triangle Theorem

4. angle 2 > angle

*A*4. Exterior Angle Inequality (with triangle

*CC'A*)

5. angle 1 > angle

*A*5. Substitution (step 3 into step 4)

6. angle 1 + angle 3 = angle

*BCA*6. Angle Addition Postulate

7. angle

*BCA*> angle 1 7. Equation to Inequality Property

8. angle

*BCA*> angle

*A*8. Transitive Property of Inequality (steps 5 and 7)

The next theorem is proved only informally in the U of Chicago. The informal discussion leads to an indirect proof.

Unequal Angles Theorem (Triangle Angle-Side Inequality, TASI):

If two angles of a triangle are not congruent, then the sides opposite them are not congruent, and the longer side is opposite the larger angle.

Indirect Proof:

The contrapositive of the Isosceles Triangle Theorem is: If two angles in a triangle are not congruent, then sides opposite them are not congruent. But which side is opposite the larger angle? Because of the Unequal Sides Theorem, the larger side cannot be opposite the smaller angle. All possibilities but one have been ruled out. The larger side must be opposite the larger angle. QED

My student told me that he wanted to see one more indirect proof before showing him the Triangle Inequality, so why not show him this one? The initial assumption is, assume that the longer side is

*not*opposite the larger angle. Since the angle opposite the longer side is not

*greater*than the angle opposite the shorter side, the former must be

*less than or equal to*the latter. And these are the two cases that lead to contradictions of Isosceles Triangle Contrapositive and Unequal Sides as listed in the above paragraph proof.

The new worksheet that I'm posting today focuses on these four theorems.

Let's conclude this post just as Kung ends today's lecture -- with an anonymous quote. "There are two rules for success: 1) Never tell everything you know."

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