Pick an odd positive integer, n. Draw the “complete graph” with n vertices (you may think of the vertices as dots, and between any two of them we draw a line. The lines do not need to be straight lines. The lines are called *edges*. The complete graph with n vertices is denoted K_n). In figure 1 we see two drawings of K_5, the complete graph with 5 vertices. Notice that both drawings (figure 1a and figure 1b) have an odd number of crossings.

In figure 2a there is drawing of K_5 with 4 crossings. However figure 2a has something that’s deemed a bit silly (not the curved edges—they are allowed in this game). The silly aspect is that two of the crossing edges (the red edges) both share a vertex (the bottom right vertex).

In general, there is a technical definition of a “good drawing” that prevents exactly this. There is a reason for this restriction: if two crossing edges are incident with the same vertex, then the drawing can be slightly modified to remove that crossing, without introducing any new crossings in the rest of the drawing, shown abstractly in figure 3a. It is common to ignore a drawing if it is “not good” (a drawing that is not good cannot be part of a drawing of K_n with the *fewest number of crossings*, for example).

Goodness also requires that edges don’t touch each other tangentially (figure 3b), a pair of edges don’t cross each other more than once (figure 3c), and that no edge crosses itself. If three edges happen to cross at the same point we consider it to be 3 crossings (some will simply insist that 3 edges do not cross at the same point, via a slight transition shown in figure 4, that would not affect any other crossings in the drawing).

Figure 2b shows what happens to the drawing in figure 2a after removing the crossing that is not “good,” and guess what? There are an *odd number* of crossings in the *good* drawing shown in figure 2b.

We will soon show that every good drawing of K_5 has an odd number of crossings. First, an easy fact: Every good drawing of K_3 must be without any edge crossings. Furthermore every good drawing of K_3 has an “inside” and an “outside”. If beyond the K_3 there are two other vertices, u and v, and u and v are both on the same side of the K_3 (in figure 5 they are both outside the K_3), then an edge joining u to v must cross the edges of the K_3 an even number of times in total (possibly zero times). Similarly, if u and v are on opposite sides of the K_3 then the edge uv crosses the K_3 an odd number of times.

As a simple test of your understanding of the goodness criteria, convince yourself that every good drawing of K_4 has at most 1 crossing. If you try to draw K_4 with two or more crossings, there will be a goodness violation (probably there will be one vertex incident with 2 crossing edges—there are only 4 vertices, so check them all!)

In any case, the upshot is that each crossing requires 4 vertices and actually, the crossings in a drawing of K_5 can be counted by *counting the K_4’s that have a crossing*.

This may seem much harder than just counting the crossings, *but it works well on an abstract level*. In particular, if all five of the K_4’s have a crossing, then the K_5 has 5 crossings, which is an odd number.

To make this abstract approach more concrete, look at figure 1b–if we remove the top vertex and its incident edges, we end up with a K_4 with a crossing in it (figure 1c). If on the other hand, we start with figure 1b and remove the bottom vertex and its incident edges, we get a K_4 that does not have a crossing (figure 1d).

Now consider figure 1a. If we delete the bottom vertex and all its incident edges, we are left with a K_4 with 1 edge crossing–the crossing highlighted in red.

As there are 5 vertices you could have deleted, there are 5 different drawings you need to make from figure 1a, each with a different crossing, so there are 5 crossings in figure 1a.

Now we know a good drawing of K_5 either has 5 crossings or else we can find a K_4 inside it so that *if we just look at that K_4*, it does not have crossings. This motivates:

**Lemma:** *Suppose a drawing D of K_4 has no edge crossings. Suppose that D is part of a good drawing D’ of K_5. Then D’ has an odd number of crossings.*

We have an example of a drawing of K_4 without crossings (see figure 6). There is an extra vertex v that we wish to use to extend the K_4 to a drawing of K_5. Although this is a *specific picture*, there are features of it that are *always true*. In particular, deleting the edges incident with a specified vertex of the K_4 will result in a K_3 with an inside and an outside (we obtain figure 7a from figure 6 by deleting the edges incident with the vertex ‘a’. Both ‘a’ and ‘v’ are outside the remaining K_3 with vertices b,c and d).

If we wish to count the total number of crossings of a K_5 that involve the edge va, then we should not look at the other edges incident with ‘a’, since the edge va can’t cross them anyway (by goodness–two edges incident with ‘a’ cannot cross each other). Also, va can’t cross the other edges incident with ‘v’ (also by goodness–two edges incident with ‘v’ cannot cross each other), so we won’t worry that we can’t see the edges incident with ‘v’ in figure 7a.

When we count the number of times va crosses anything in the K_5, it must be an even number since ‘v’ and ‘a’ are both on the outside of the K_3 with vertices b,c and d. A similar argument (looking at figure 7b and the K_3 with vertices a,c and d) shows that vb crosses an even number of times. (Note that the va crossings are all different from the vb crossings, by goodness again!).

The number of times vc crosses in D’ is similarly even (see figure 7c); but the number of times vd crosses in D’ is odd, since v and d are on opposite sides of the K_3 made up of a,b and c as shown in figure 7d.

To summarize: All of the crossings of D’ will involve v, since D has no crossings. We add together the number of times each of the edges va, vb, vc and vd cross anything in D’ to get the total number of crossings in D’. Exactly three of these four numbers are even, so the sum is odd. By “goodness,” these numbers represent distinct crossings, (no single crossing can involve two edges incident with ‘v’) and as such these crossings can be added together meaningfully.

All of these facts depended only on the fact that we started with a drawing of K_4 with no crossings, and not the particular drawing in figure 6. (Imagine ‘v’ had been in a different location, for example, very close to ‘d’ inside the K_3 bounded by a,b and c. Does the argument still work?)

This argument essentially proves the lemma. Together with the discussion before the lemma, we now know that every good drawing of K_5 has an odd number of crossings. We’ve also seen that every good drawing of K_3 has zero crossings (and zero is even, so “every good drawing of K_3 has the same parity as every other good drawing of K_3” is a true statement) and we know that K_4 has a good drawing with 0 crossings and another good drawing with 1 crossing (so no parity theorem for good drawings of K_4).

Based on these 3 (measly) examples, n=3, n=4 and n=5, we are now ready to state the theorem. For it’s short proof, miraculously, no more geometry is needed:

**Theorem** (Kleitman): *Let n be an odd integer, and D be a good drawing of K_n. Then the number of crossings in D has the same parity as the binomial coefficient “n choose 5”.*

Here’s an example with n=7 to illustrate what this means. If you have 7 objects, there are 21 ways to select exactly 5 of them (try it). So we say “7 choose 5” is 21. In our context, we count crossings in a K_7 by separately counting the number of crossings in each of the twenty-one K_5’s in the K_7. (This counts each crossing more than once—hmm…but we don’t want the *number* of crossings, we want the *parity of the number* of crossings… More explanation to follow!)

Each of these twenty-one numbers is odd (since each of the twenty-one K_5’s has an *odd number* of crossings). So the sum of the 21 numbers is also odd. When we add these 21 numbers, *every crossing would have been counted 3 times* (a crossing requires 4 vertices; when you look at a single crossing it therefore belongs to 7-4=3 different K_5’s). So our odd sum of 21 numbers, divided by three, is the number of crossings in the K_7. *We don’t know what the number of crossings is (that’s why we didn’t try to count the exact number of crossings)*, but we know the number is *odd*, since an integer that is obtained from odd integer divided by 3 is necessarily odd.

In general, to find the number of crossings in any good drawing of K_n, we add “n choose 5” different (unknown, but definitely odd) numbers together and divide by the odd number (n-4) to get the exact number of crossings…And we don’t know what the exact number is but we also don’t care; the sum of “n choose 5” odd numbers must have the same parity as “n choose 5” (regardless of what integer “n choose 5” turns out to be). If this sum is divided by an odd integer and the quotient happens to be another integer, then this quotient also has the same parity as “n choose 5.” But we divide by n-4, which is odd since n is odd. So this completes the proof.

Remarks:

- This presentation is a first attempt at a response to the criticism: “Why don’t mathematicians or scientists ever write the way they think?” In fact, this “unnecessarily” long exposition does not mean that it is difficult; my hope is quite the opposite. The length of this post is an attempt make it as lucid as possible, for as large an audience as possible. I welcome feedback to improve the clarity, and to reduce the prior expectations on the reader.
- This discussion is much different from Kleitman’s original approach. For a full discussion, please see the paper: A Parity Theorem for Drawings of Complete and Complete Bipartite Graphs, by Dan McQuillan and R. Bruce Richter, in the American Mathematical Monthly, Vol. 117, No. 3 (March 2010), pp. 267-273. Furthermore, there are several other topics in the paper not covered in this post. Please check it out and enjoy!

Except that this strategy lowers your chances of scoring a run. We can prove it, and we did so in the April 2016 edition of Math Horizons, a publication of the Mathematical Association of America (MAA). (Our article, *Run for Third! A Defense of Aggressive Base Running* is free to members of the MAA, and most university libraries will subscribe to the journal *Math Horizons*).

The fact is that at the major league level, your chances of scoring from third base with one out are so much higher than your chances of scoring from second base with one out, that you should run even if you don’t know whether or not the ball will be caught.

To be more specific, if I (an aggressive base runner) see the ball hit to right field, and don’t know if the outfielder will get there on time, I will run and I might make it to third base instead of second base. Sometimes this aggressiveness will result in a double play. If you play the conventional way, you may end up at second base. Sometimes (if the ball is caught) you will end up at first base with TWO outs.

The analysis is therefore simple. Compare the advantage of my way versus the advantage of your way. If the ball falls in for a hit, I have the advantage. I’m at third base, where major leaguers score (with one out) more than 60% of the time. You are at second, where major leaguers score (with one out) less than 40% of the time.

If the ball is caught, you have the advantage. But it turns out this advantage is small. At the major league level, a base runner scores from first base with two outs about 14% of the time (only 13% in high leverage situations, like a tie game in the bottom of the ninth).

Each strategy has an advantage—mine when the ball falls in for a hit, and yours when the ball is caught. We can compare these two advantages by using a little bit of undergraduate mathematics (hence our choice of a famously student-accessible journal). It turns out that my advantage is greater than yours. It’s not close. As long as the game situation suggests that playing for one run is the correct strategy, then this version of aggressive base running is also the correct strategy.

This has been missed by modern statisticians simply because it hasn’t been tried. So please try it! *Please* let me know if you see a game where it was used, or a game where it should have been used and wasn’t.

Thanks to my coauthors, Peter Macdonald and Ian McQuillan. We’ve been talking about this for some time. We’ve tried posts before (like this one), as well as other preliminary preprint versions of the paper that were not as clear as this publication. So please read this article for the details!

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Our example for this post appeared on the 2014 Collegiate Mathematics Competition, sponsored by the Northeastern Section of the Mathematical Association of America:

*A husband and wife invited four married couples to a housewarming party. Just before everyone left, the husband asked each person (including his wife) how many people they had shaken hands with. Their responses were 8, 7, 6, 5, 4, 3, 2, 1, and 0. Given that no one shook hands with his or her own spouse and no pair of people shook hands more than once, determine how many times the husband must have shaken hands with a guest.*

In fact this little puzzle is not too difficult, and you may wish to try it for a few minutes before reading further! (Can you do it in your head?)

— Empty space to encourage playing/thinking —

*Congratulations!* By now you’ve either solved the problem or else realized that it’s not that easy. Either way it’s a win—in both cases the next question is:

Can we make up an analogous question with fewer than four married couples invited to the party? Does the question make sense with just one married couple? With two married couples? With n married couples?

(If you haven’t solved the problem, the one-couple version is a very nice playground, which will hopefully help with the original. If you have solved it, then you may be in store for a better theorem and a different proof).

We don’t know yet whether or not this problem will actually generalize nicely. Nevertheless, for the one-couple version, let’s say the responses to the handshake question are 2, 1, and 0. With only 4 people present at this party, the person shaking 2 hands—not shaking the hand of the 0-handshake person—must shake everyone else’s hand. Hence the 2-handshake and 0-handshake people are married to each other. The 1-handshake person shakes hands with the 2-handshake person and nobody else. Now we have a complete picture of everyone’s handshakes except the husband. Fortunately that’s enough to recover all needed information about the husband, who shakes the hand of one person. His wife also shakes hands with one person.

Now for the two-couple version of the problem: assume the responses to the handshake question are 4, 3, 2, 1 and 0. We could proceed in the same way, forming a complete picture of who must shake everyone else’s hand, starting from the 4-handshake and 0-handshake people, who must be married to each other, etc. But here’s a different tact: once we realize that the 4-handshake and 0-handshake people form a married couple, we remove this couple—induction! Note that this removal necessarily reduces the number of handshakes by 1 for each remaining person. We are in fact reduced to the one-couple version of the problem. This suggests both a general theorem and its proof:

**Theorem**: *Let G be a simple graph with 2n+2 vertices. Assume the degrees of 2n+1 of the vertices are given by the numbers 0, 1, 2, … , 2n. Then the last vertex must have degree n. Moreover, for i=1,2,…,n, the vertex of degree n−i is not adjacent to the vertex of degree n+i, and the two vertices of degree n are not adjacent to each other.*

There are a couple of important objections. The first one is that this doesn’t look like the original problem; unless a vertex represents a person and an edge represents a handshake. We’ll address the other objection after the proof.

**Proof**: Let x be the unknown degree. We wish to prove x=n. We proceed by induction on n, the case n=1 being not too hard (and discussed above). We now assume the result true for n−1. Consider a graph G with 2n+2 vertices as in the hypothesis of the theorem. Delete the vertices of degrees 0 and 2n. This reduces the degrees of each remaining vertex by 1, leaving 2n vertices with degrees:

0, 1, … , 2n−2 and x−1.

By the inductive hypothesis x−1=n−1, the two vertices of degree n−1 are not adjacent to each other, and a vertex of degree n−1+i is not adjacent to the vertex of degree n−1−i (for i=1, 2, … n−1). Replacing the two deleted vertices from G gives the desired result. QED.

Now for the other objection: the hypothesis that no one shook the hand of his or her own spouse seems to have disappeared, suggesting this assumption was not necessary to solving the problem. Yet the conclusion of the theorem includes a description of vertices not adjacent to each other. So our theorem is better—it has a weaker hypothesis and it guarantees that the n-couple version of the problem describes a situation that could actually happen. If we do assume that no one shakes the hand of his or her own spouse, then one can show—by induction on i—that the person who shakes 2n-i hands is married to person who shakes i hands.

So this was much more involved then it had to be. Yet it is satisfying on many levels. We made a small (n=1) toy version of the problem, we provided a generalization and we showed that the hypothesis of the general problem is realizable.

**Remarks**:

1. The n=3 version of this problem appears in Loren C. Larson’s outstanding book, Problem-Solving Through Problems, in a section entitled, “Draw a Figure.” We chose this complementary, algebraic approach. It’s certainly instructive to compare the two approaches, to see exactly how the use of induction on n gets the same conclusion without *explicitly* drawing the entire picture.

2. Another written solution (closer to the one in Larson’s book) is available from the Collegiate Mathematics Competition webpage by looking up the practice problems from 2014, question 4. Many thanks to my colleague Rob Poodiack for bringing my attention to these excellent problems.

3. This problem is perfectly adaptable to many different levels of difficulty and discussion. So it’s a perfect question to consider, even on a lazy summer afternoon. Enjoy!

]]>In the post on the Mean Value Theorem (MVT), we proved as a warm-up example that:

7^(1/3)+9^(1/3) < 4

As a quick review, MVT states that if f(x) is a differentiable function on the open interval (a,b) and continuous on the closed interval [a,b], then there is at least one point z in (a,b) such that f′(z)=[f(b)−f(a)]/(b−a). We rewrite this as:

f(b)−f(a)=(b−a)f′(z) (MVT)

Now set f(x)=x^(1/3) observe that f′(x) is decreasing when x>0 and use MVT twice to get:

f(9)−f(8)<f(8)−f(7). Since f(8)=2, this last inequality gives the desired answer.

These two applications of MVT use different choices for b and a, but in both applications

b−a= 1. As a generalization, we imagine

b−a=d>0, with a+d≤b−d and we consider any function whose derivative is decreasing:

**Which is bigger, f(a)+f(b) or f(a+d)+f(b−d)?**

By comparison, our original question had:

a=7, b=9 and d=1.

From two applications of MVT we see that

f(b)−f(b−d) < f(a+d)−f(a)

Hence we have our generalization:

f(a)+f(b)<f(a+d)+f(b−d)

**But is this generalization useful?**

Here’s where the post, “On the value of believing that you know the answer” comes in to play. The goal of that post was to prove the arithmetic-geometric mean inequality. However, instead of directly proving that the arithmetic average of n numbers was bigger than the n-th root of their product, we instead replaced two of the n numbers, a and b say, with two new numbers a+d and b−d (choosing d so that one of these new numbers was the arithmetic average). This change maintained the same arithmetic average, and we showed that the n-th root of the product increased. Finitely many of these two-at-a-time changes produces n equal numbers, and the result follows from comparisons.

The same process works for a very analogous application; given n numbers, whose (arithmetic) average is A, then Jensen’s inequality states that

if f′(x) is decreasing then f(A)≥Z

where Z is the arithmetic average of the n outputs

i.e. Z = [f(x_1)+f(x_2)+…+f(x_n)]/n and

A=[x_1 + x_2 +…+x_n]/n).

If the n numbers are all equal, there is nothing to prove. Otherwise, let a be the smallest one, a<A and there must be another one b say, which is bigger than A. If A−a<b−A then set d=A−a. Otherwise set d=b−A. This guarantees that either a+d=A or b−d=A. To prove Jensen’s inequality, replace the two numbers a and b with a+d and b−d. The average of the n new numbers will still be A, but the average of the n outputs will be larger—*Our generalization is useful!* Hooray!

Fewer of the n numbers will be different from A. We repeat, similarly altering numbers different from A until there are none remaining.

**Remarks**

- This proof is well-known, but this discussion suggests a natural way to find it.
- It is also well-known (and by now probably not surprising) that setting f(x)=log x in Jensen’s inequality immediately yields the AM-GM inequality.
- In “It’s a Mean Value Theorem” the main example was to show that in any acute triangle with angles A, B and C, that sinA+sinB+sinC>2. Let f(x)=sin x. Notice f′(x) is decreasing for acute x. Hence by Jensen’s inequality, f((A+B+C)/3)≥(sinA+sinB+sinC)/3. Rearranging slightly gives us:

sinA+sinB+sinC≤3sin(π/3).

This is a nice upper bound to go with the lower bound of the previous post.

]]>f′(z)=[f(b)—f(a)]/(b—a). In our applications below, we prefer to write this last equation as:

f(b)—f(a)=(b—a)f′(z) (MVT).

The purpose of this post is to see the Mean Value Theorem at work in a couple of places where we may not expect to see it. The second example is much more interesting than the first.

**Example 1. (Warm-up example):** *Which is bigger*

*7^(1/3) + 9^(1/3) or 4?*

If you have a calculator or a phone handy, it will probably tell you that the first expression is approximately 3.993. But that’s cheating. Instead consider the function

f(x)=x^(1/3). A straightforward application of the power rule shows that f′(x) is decreasing for x>0. Hence (using MVT twice):

f(9)—f(8)<f(8)—f(7).

But since f(8)=2, this last inequality gives the desired answer.

**Example 2:** *Prove that in any acute triangle with angles A, B and C, that*

*sinA+sinB+sinC>2.*

I have a vague recollection of trying this problem as part of a Putnam training session as an undergraduate student. Recently I saw it again as a post on Stephen Cavadino’s blog.

In fact I recognize Stephen’s clever solution as being the same one presented at my Putnam club decades ago. It’s such a slick solution that it’s definitely worth reading.

Nevertheless, I’m not sure that I could have come up with it. So here’s an alternative, using MVT.

This solution hopefully also highlights the value of chosing good examples to help understand the problem.

**How to solve it:**

I experimented with the conditions of the problem, quickly realizing why the “acute” hypothesis is necessary (otherwise sinA+sinB+sinC can be made arbitrarily close to 0). However, the biggest angle could be a right angle and still have the inequality holding.

*In fact, a right triangle with hypotenuse 1 makes for an extremely interesting special case.* Indeed if C=π/2 (radian measure) then sinC=1, and our required inequality reduces to a special case of the triangle inequality.

Inspired by the simplicity of this example, I decided to compare an arbitrary acute triangle to it. Since we will be comparing different triangles T and T′, we

let f(T) be the function that adds the sines of the three angles of a triangle.

First let’s rank the acute angles; without loss of generality A≤B≤C. Also, let

D=(π/2)—C. Observe that D<A. (This is because D+C+B<π, but A+C+B=π).

So consider a new triangle T′ with angles A—D, B and C+D (the side lengths won’t matter). The new triangle has a right angle (C+D=π/2) and so by our previous discussion f(T′)>2. It remains to show that

f(T) ≥ f(T′). This follows from:

**Lemma:** *Let D>0 and assume that D ≤ A ≤ π/2—D. Then:*

*sin(A—D)+1≤sinA+sin(π/2—D) —D[cosA—cos(π/2—D)]*

Assuming the lemma is true, we have:

sin(A—D)+1≤sinA+sin(π/2—D)

i.e., we imagine replacing two of the summands in f(T′) with the corresponding summands from f(T).

Hence f(T′) ≤ f(T).

To prove the lemma, we use the mean value theorem twice. Once to show:

sin(A)—sin(A—D) ≥ DcosA (exercise!)

and the other time to show:

sin(π/2)—sin(π/2—D) ≤ Dcos(π/2—D).

Combining these last two inequalities yields the lemma.

Notice that in our application of the lemma, we disregarded the term:

—D[cosA—cos(π/2—D)]. This extra term could easily be used to sharpen the inequality.

Another way to sharpen it is to compare our right triangle T′ with another

right triangle T′′ where the angle B in T′ is replaced by an angle approaching π/2.

This means that f(T′′) is approaching 2.

At the beginning, searching for an example where f would approach 2 was part of the process. Realizing how the inequality may fail to hold for obtuse triangles also suggested having a right triangle. So I see this solution as a natural consequence of playing with the conditions of the problem and experimenting with the right examples.

]]>*Unwritten Rule: With fewer than 2 outs, a base-runner on first base shall be certain that a ball hit in the air will fall in for a hit before straying too far away from first base.*

This is a natural consequence of a written rule allowing for double plays when the ball is caught and the base-runner doesn’t “tag up” before advancing.

So…the purpose of this post is to identify one situation where the unwritten rule is wrong—so wrong that it costs the game.

**One runner, on first base with 1 out**

Imagine there is a runner on first base with 1 out. It’s a tie game in the bottom of the ninth inning and the outfielders are playing back to prevent a double. The ball is hit in the air, but off the end of the bat—a blooper into right field. The right fielder runs; no one can tell if he’ll get there in time to catch it, but if he does the base-runner better get back to first base to prevent a double play. If the ball falls in for a hit there will only be 1 out but the runner only makes it to second base, since he was standing and waiting to see if the outfielder would catch it.

**Should the base-runner stand and wait to see if the ball is caught?**

The answer is “no”, and it’s not even close. This is the risk/reward analysis that is apparently not understood. But it’s a really simple argument. When the ball is hit in the air, let’s say the runner makes a snap judgment that it’s “better than even” that the blooper will fall in for a hit. In other words, let p be the probability that the blooper will be a hit and the runner quickly decides p>0.5. Our analysis involves two possible base-runners, A and B. Player A (A for “aggressive”) breaks protocol and just runs without waiting to see if the ball is caught. His aggressiveness either gets him to third base or else gets him thrown out. Player B (for “baseball professional”) follows protocol, and waits.

**Who wins more often, A or B?**

Remember it’s tied in the bottom of the ninth, so one run wins the game. If the probability of scoring from third base with 1 out is T (T for “third”) then the product pT is the probability that player A scores. This product captures the situation that the blooper falls in for a hit, and then somehow the base-runner eventually scores. (We don’t care how this happens, we just assume it happens with probability T). He won’t score if the blooper is caught since he’ll be doubled up and the inning will be over.

For player B there are two cases. The first case is that the blooper is a hit, putting him at second base. He could eventually score from second—again we don’t care how he might score from second, but we assume he does so with probability S. The second case is that B could score if the blooper is caught (although it would be much harder since there would be two outs and he would be at first base). His probability of scoring is therefore pS+(1-p)F where F is the probability of eventually scoring from first base with TWO outs. So what’s bigger pT or pS+(1-p)F?

**The punchline**

The quantities T,S and F above can all be calculated! (We know how likely it is that a runner will score from third base with 1 out, etc). The short answer is that

pT>pS+(1-p)F whenever p>0.38.

So if you think there is a 38% chance or better that the blooper will be a hit, RUN!

Many more details are available from our preprint:

http://arxiv.org/pdf/1505.00456.pdf

**Rally killers and Mariano Rivera**

Here is the general justification for running like base-runner B (from http://baseball.isport.com/baseball-guides/cardinal-rules-of-base-running):

*“There’s no worse rally killer than a double play. Moreover, there are few plays more devastating for an offense than a line drive with runners on that turns into a double play. To avoid such a huge momentum shift, you need to make it a priority not to get doubled off on the bases.”*

Well I don’t think it’s much of a rally if there are 2 outs and only one runner, on first base. In other words the rally is on life-support as soon as the blooper is caught. Furthermore, if it’s a tie game in the bottom of the ninth, you just need one run, not a rally. In http://arxiv.org/pdf/1505.00456.pdf we consider “high-leverage innings” defined as the 8^{th} or 9^{th} inning where the difference in score is at most 1. Between the years of 2003 and 2008, Mariano Rivera pitched in 58 different high-leverage innings where there were 2 outs and one runner, on first. None of those runners scored. Unlike most pitchers (where 38% certainty is enough) one only needed to be 14% sure that the blooper would be a hit before running aggressively when Rivera pitched. This is not just because Rivera’s pitching record is better than other pitchers (although it certainly is!) but more importantly it is because he was only slightly better than average at preventing a runner scoring from third with 1 out. This is the point—you don’t need a hit to score from third with one out. In high-leverage situations, one is often facing the pitchers against whom hits are most difficult to obtain, and these are the situations that determine the outcomes of games.

**Why don’t they do it this way?**

The simple answer is that there is no *direct* data to show that player A wins more games, because *no one ever does it that way*. If a team tried it this way, then data could be collected to bear out the conclusions of our indirect arguments. We continue to wait for someone brave…

UPDATE: I’ve written a new post about this topic, in part because we’ve improved it enough that it ended up published in the April 2016 edition of the journal Math Horizons. The new blog post, and even better, the published paper, do a better job of focussing on the main point. Enjoy!

]]>The average of the three numbers 4, 5, 6 is clearly 5. A consequence of AM-GM is that the cube root of the product (4)(5)(6)=120 must be less than 5. Since the cube root of 120 is less than the 5 (i.e. the cube root of 125), we gain some confidence that it works. As a natural-born-devil’s-advocate, we try to change the example so as to give the geometric mean a better chance. Keeping the average at 5, the best choice for maximizing the geometric mean seems to be: 5, 5, 5. Thus, all the numbers are equal. In this extreme case, both the arithmetic and the geometric means must be the same. In fact this suggests a proof.

Keep the arithmetic average the same and make the geometric average as big as possible. We suspect that the only way to achieve this is to make all the numbers equal. Let’s just *believe* this and see what happens. Notice that “keeping the average at 5” really means keeping the sum at 15; also “maximizing the geometric mean” actually suggests maximizing the product. When we replaced “4, 5, 6” with “5, 5, 5” we really needed to compare the product (4)(6) with (5)(5). This is a consequence of the difference of squares formula

**Difference of Squares**: *(x — a)(x + a) = x^2 — a^2*

with *x=5* and *a=1*. In this example we effectively only replace two numbers—the 4 and the 6—with their average—5 with another 5.

So let’s try a more challenging example—three numbers 4, 4 and 7. If we try the same trick, perhaps we replace the 4 and 7 with a couple of 5.5’s. This gets us a bigger product, just as in the above example, but we don’t seem to be getting closer to proving that having all 5’s yields the maximum product.

*Here’s where the belief in the answer helps us:* Instead of moving the 4 and the 7 to

There are many ways to see why the product (5)(6) should be bigger than the original (4)(7), but we defer to the difference of squares formula once again. This time let *x=5.5* and compare what we get with *a=1.5* versus *a=0.5* to see exactly what’s happening. (Don’t skip over this step—it’s the heart of the matter).

Now we have three numbers 5, 4 and 6 with the same sum, but a bigger product. This triple should look familiar (our first example!)—we remember that the next step is to replace the 4 and 6 with a couple of 5’s.

The experienced mathematician now instantaneously fills in all the details of a proof, arguing that *if* there is any number other than the arithmetic average, then there must be one both less than the average and another one more than average. So if any of the numbers are other than average we can make the product bigger, and bigger and bigger…etc…

However the main idea can be understood (perhaps at a different level) by elementary school students, provided they are comfortable with the difference of squares formula.

Toward that end, let’s run through a little game on five numbers 8, 8, 9, 9, 10. The total is 44 and the average is 8.8. We repeatedly replace this list of five numbers with another list, always with total 44, and at each stage the product is bigger. Also at each stage, at least one of the numbers is replaced with the average. Thus:

**8**, 8, 9, 9, **10**—>**8.8**, 8, 9, 9, **9.2**

We replaced the first 8 and the last 10. Now the product is bigger—why? What choice of *x*? What choices for *a*?

Keep going, replacing the two bold numbers by moving the closer one to *the average* 8.8 and moving the other one the same amount in the other direction:

8.8,** 8**, 9, 9,** 9.2**—>8.8, **8.4**, 9, 9, **8.8 **

8.8, **8.4, 9,** 9, 8.8—>8.8, **8.6, 8.8**, 9, 8.8

8.8, **8.6**, 8.8, **9**, 8.8—>8.8, **8.8**, 8.8, **8.8**, 8.8

It’s just crazy! If any of the numbers are other than average, we can compare it to the situation where all the numbers are average, in a small number of steps. It’s slick, it’s fun and it works.

**Remarks:**

1. We just had our last Putnam session at Norwich this semester. We solved a problem using the AM-GM and just for fun tried to prove AM-GM. This post came out of that discussion. Thanks to Wise D. and Jessica F. for their role in this discussion! I asked Prof. Rob P. afterwards if he knew of it and he directed me towards this article:

*An inductive proof of the AM-GM inequality*, by Kong-Ming Chong, *American Mathematical Monthly*, Vol. 83, No. 5 (May, 1976), p. 369.

As is often the case with mathematical expositions, the problem is that you need to be something of an expert in mathematics to even realize that the article is saying the same thing as our exposition in this post! So we hope you enjoyed it and can do something like this with your students.

2. Originally in our Putnam group, we used calculus, instead of the difference of squares formula. We considered two positive numbers with a fixed sum of *S*, trying to maximize the product by using the function *f(x)=x(S-x)*. The breakthrough came when we realized that it’s not just about maximizing *f*, but rather knowing the intervals of increase, since our “belief” in this post led us to increasing *f *rather than maximizing it. In this discussion I replace calculus with the most elementary idea possible, the difference of squares, to make it as clear, accessible and fun as possible.

3. “Fix the arithmetic average and make the geometric average as big as possible” means the same thing as “Keep the arithmetic average the same and make the geometric average as big as possible.” At the last minute I opted for the latter phrasing.

]]>**Mathematical induction-the champion of efficiency.**

*The beautiful idea:*

Induction is often used (and usually taught) to prove formulas that hold for all non-negative integers. For example, the sum of the first n non-negative powers of 2, is equal to one less than the next power of 2. If n=4 it looks like this:

1+2+4+8=16-1

Our next instructions are to add 16 to both sides:

1+2+4+8+16=(16+16)-1

and we see that the pattern continues!

Let’s do it one more time. Start with our newly discovered equation:

1+2+4+8+16=32-1

add 32 to both sides, and voila:

1+2+4+8+16+32=64-1.

It’s pretty neat, and also really clear from this kind of discussion…unless of course…

*Efficiency:*

Unless of course, our discussion only uses one equation, which corresponds to n=1, the so-called *base-case*, and you write:

1=2-1.

The base case is just too small to see what’s going on or to notice any patterns. Your students may even feel stupid writing this down, as if it’s a secret handshake and they really don’t know why it’s necessary.

You guessed it; this is the way it’s almost always taught. But wait!–it gets worse; the instructions for seeing how the “pattern” continues are too cryptic for many beginners to follow. They are hidden in the formal details of the so-called *inductive step*. So it’s a pedagogical lose-lose.

It’s done this way because the base case is the smallest example that will make the proof work both completely and rigorously. After all, a full proof would have to include the case where n=1, and a formal proof works just fine with that as our main example. So let’s be efficient and … actually let’s not.

*Focus on the beautiful idea*

The beautiful idea behind the principle of mathematical induction should be given priority. That’s the first part of this post, and the first thing to be communicated regarding mathematical induction. Formal proof-writing etiquette can come a little later at the university level, or not at all in elementary school.

(Yes—I’m suggesting that elementary teachers should introduce math induction!)

So let’s use n=4 and n=5 for our examples, then look at the inductive step, keeping in mind how they relate to these examples, and then rewrite the proof the next day.

In other words, let’s give up some of the efficiency requirement.

]]>*A little calculation *

“10 is the greatest common divisor, or greatest common factor, of the two numbers 130 and 50.” We may abbreviate this as:

gcd(130,50)=10.

We see this by inspection, by factoring our numbers. There is a harder way to find the answer, called the “Euclidean Algorithm.” It turns out that this method will usually be much easier for large numbers where factoring is difficult. Let us use this example to introduce the Euclidean Algorithm. Consider this little calculation:

130=2(50)+30 … gcd(130,50)=gcd(50,30) (why?)

50=(30)+20 … gcd(50,30)=gcd(30,20) (why?)

30=(20)+10 … gcd(30,20)=gcd(20,10) (why?)

and finally, gcd(20,10)=10, since 10 divides 20.

*What is the main pedagocial point?*

The question “why?” appears three times in the little calculation. If we’re clever, we can answer all three questions all at the same time, with a little lemma—and we should do this (if you don’t know the answer, try it!). The answer constitutes the central idea behind the Euclidean Algorithm. The pedagogical problem is that the required lemma is usually discussed first; so that the chain of equalities: gcd(130,50)=gcd(50,30)=gcd(30,20)=gcd(20,10) can be suppressed!

*Conclusion*

My suggestion to the teacher is to introduce the lemma only after a few examples have been worked by the student in this inefficient way. Thus, the lemma is an *eventual recognition of a pattern*—a celebration of just how useful it appears to be, rather than an efficient shortcut in order to avoid it.

Last semester I ran a mathematical problem solving course. I did a few specific things that are not usually done in a mathematics class. Several small assignments consisted of answering a single question, which the student chose from a list of five. Students therefore would read more than one question before choosing a question to answer. This was my first victory—thinking about several mathematical questions without necessarily writing about them or handing them in is already part of the math-loving process!

This brings us to the next trick—participation grade. When I was a student, any such notion was offensive. It usually meant the extroverts would get more points, or even worse, that teacher preference would play a role in the final grade. I corrected all of this by giving a mathematically precise definition of participation, with several options for obtaining the points. One option was to do five extra questions beyond the minimum at a convenient time during the course. So if students read questions and liked them, they could get a little credit. Another option for participation points came from short presentations of problems to the class. Since these points were for participation only, students lacking in confidence, who were normally too afraid to present, would still do so. Some of these presentations were admittedly not very good—but here’s the key—they got much better. The motivation for doing a good job changed from getting a good grade, to wanting to do a better job, and ultimately to sharing the enjoyment of having done the work. Classmates were quick to point out ways to improve these participation activities, since they knew that it was all about the process-the participation-and not about the presentation’s grade. I certainly don’t claim that all courses should be like this; but I am claiming that not enough courses are like this. Try it!

Extra Remarks: Congratulations to my Ph.D. supervisor, Dr. Jan Minac:

Fortunately there are some classes where love of mathematics can’t help but be taught, just because of the instructor. As a student, the only time I ever remember experiencing extensive choices for homework questions was at Western University, with Dr. Jan Minac as the instructor. So I was in no way surprised to learn that he just won a prestigious national teaching award! From the CMS media release: “Jan Minac shares personal stories of great mathematicians with his students, so that the students become so involved…” How involved? Click here for the full story. Congratulations Jan!

Here is a nice fact:

(1/2)+(1/3)+(1/6)=1

Let’s have some fun with this fact by making it the answer of a question. Think of it as a mystery equation, and it’s our job to point the problem-solver in the right direction to find it. Hmm…

The sum of these 3 denominators is 11. After a bit of thought we see that {2,3,6} is the only set of 3 positive integers with sum 11, such that the sum of reciprocals is equal to 1. So we will tell the solver that we have x numbers whose sum is f(x) and we’ll pick a function f so that f(3)=11. We’ll also mention that the sum of reciprocals is exactly 1.

Now we need a hint so that the solver can be directed towards x=3.

Consider the more general situation where the x numbers are only required to be positive real numbers instead of positive integers. If the sum of reciprocals is 1, then it is well known that the sum f(x) is minimized if the x numbers are equal to each other, and thus equal to x (think of famous inequalities!). Hence x^2<=f(x).

Let’s pick f(x)=5x-4. Then x^2<=5x-4 and this places x between 1 and 4. We are ready to state the problem. Here’s a version of it, close to the way it appeared on the 2005 William Lowell Putnam Mathematical Competition:

Find all positive integers x, k_1, k_2, …, k_x, such that the sum k_1+k_2+…+k_x=5x-4 and such that (1/k_1)+…+(1/k_x)=1.

Our mystery equation corresponds to the case where x=3. There are a couple of other solutions, corresponding to x=1 and x=4.

To be clear, I did not make up this question. I also have no knowledge of what was going through the mind of the person who did make up this question. I was merely having fun with this beautiful problem!

My point is that this question, while incredibly difficult, may not seem so bad from the point of view of the *problem-maker*. It’s only difficult for the problem-solver. More importantly, I believe that the problem-solver has a much better chance of answering this type of question after having experience as a problem-maker. As much as possible, we want to be proactive rather than reactive.

So I propose that we practice making up good questions.

Good luck on the Putnam!

(A version of this post will be included in an up-coming book I’m working on, about mathematical problem-solving).

(Click here for a pdf version of this post, with better equation formatting).

]]>I decided to cover the same material covered in university courses, realizing that I would have to do it in a much different way. Here’s an example:

University version: If a rectangle has perimeter p, what is its maximum possible area?

Here’s what happened in the math club:

I asked them if they knew what a rectangle was, and how to find its perimeter and area. We discussed examples. It turns out they knew quite a bit. Then I asked each of them to produce their own rectangle, with perimeter 12 units. They each had a favorite! We calculated the area of each one. By the end of the meeting, we were convinced that the square provided the maximum area. By the end of the next meeting, we *owned* the difference of squares formula. In the third meeting, we used this formula to understand the formula for the area of a circle, partly by considering an appropriate annulus—this meeting was only a couple of days after a solar eclipse.

So what does this have to do with superheroes?

Special transformations, accompanied by superpowers, can really capture the imagination. It might be Clark Kent removing his glasses; but probably it’s a tadpole transforming into a frog or a caterpillar becoming a butterfly! In this video (link below), we introduce Rover the rectangle, Squeaky the square and Ellise the ellipse. Rover can change his shape as long as he stays rectangular and keeps the same perimeter—he catches circles by flattening himself! The intuitive connection between superpowers and transformations illustrates how a mathematician may think and *feel* about the subject. It is a human activity. It is fun, it is personal—it is mathematics!

So far we’ve made three of these videos. Each one represents portions of a different, hour-long meeting. Perhaps the best way to use them is to watch them with children, pause to discuss certain points, and then go off on a tangent. For example, in our first club meeting we eventually changed the fixed perimeter from 12 units to 14 units, which led to a wonderful discussion about fractions.

I tell the math club members how our discussions fit into the context of a university course. They LOVE this! At the end of our first meeting, we wrote down the formula for the maximum area “in terms of p” and they hurriedly copied it down as if it were a treasure. This was one of many pleasant surprises. Perhaps you are already doing something like this; if not, please try it. Become a superhero!

http://www.youtube.com/watch?v=Ix03mrGrCwE

Note added: I was delighted to see that this video has been used by Bruce Ferrington, who keeps an excellent blog. Here is a recent post, “Same Perimeter, Different Area.”

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