# Proof that doctors could relate to [closed]

I am supposed to present a mathematical proof to a lecture hall full of doctors in order to show them how mathematicians think. I'm having trouble picking a proof that will be easily followed by people who haven't seen anything beyond basic calculus in years. Does anyone have any suggestions? I'm trying to think of something visual to keep their attention.

• See this – taninamdar Sep 9 '15 at 18:01
• Could you find a copy of Tim Gowers's "Mathematics: A Very Short Introduction"? I think this book would give you very good examples for this kind of talk. He for example gives a great proof of why it's "obvious" that the oeis.org/A000127 sequence is not the sequence of powers of 2 (in the 3rd chapter, "Proofs", which might very well interest you ;)) – PseudoNeo Sep 9 '15 at 18:09
• A well-meaning friend tells me that you could anyhow find the book on some Russian websites who claim to provide a kind of "General Library" for the world. I'm reluctant to pass on this advice because this kind of websites are usually illegal in most countries. But I reckon it would be rude to make assumptions; for all I know, you might as well be the dictator of a colony on Mars where copyright laws are more permissive. – PseudoNeo Sep 9 '15 at 18:18
• Not a proof, but Simpson's Paradox is something all (medical) doctors should understand. – BlueRaja - Danny Pflughoeft Sep 9 '15 at 21:33
• See "An amazing card trick" p. 126 in Martin Gardeners charming book, "Aha, Gotcha." Includes proof by induction showing why it works. Several different versions presented. Could use several decks, each specially sorted beforehand. Better practice it thoroughly in small group before presenting it. A TV camera for a colour close-up of the cards projected on a large screen would be helpful. Or maybe make a video if you are inclined to be fumble-fingered. mlook.mobi/files/month_1203/… – Dan Christensen Sep 11 '15 at 3:52

My go-to example of this type is the following problem: It's easy to see how to cover a chessboard with 32 dominoes, each of which covers two adjacent squares. Suppose the chessboard has two diagonally opposite corners deleted. Is it possible to cover the remaining 62 squares with 31 dominoes?

If you try it, you quickly find that it is much more difficult than covering the full chessboard, although you might not be convinced that it was impossible. But it isn't possible. Each domino covers one black and one white square, so the 31 dominoes must cover 31 black and 31 white squares. But the mutilated chessboard has 32 squares of one color and only 30 of the other color.

I think anyone can understand this—and I have presented it to administrative assistants with no mathematical training who understood it right away—but I think it perfectly captures the essential feature of mathematics, which is that we don't consider all the possible ways of placing the dominoes one at a time, but instead we find an underlying regularity of structure and show that the solution, if it existed, would require a different kind of structure.

• I feel that doctors, or non-mathematicians generally, might struggle to understand that this is mathematics and not (non-mathematical) logic. – Kirk Broadhurst Sep 9 '15 at 22:16
• @KirkBroadhurst If so then that's an excellent reason for using this example! To give them a better idea what mathematics is. – David C. Ullrich Sep 9 '15 at 23:47
• @DavidC.Ullrich: Joel Feinstein might have something to say about that. As I understand it (and I was in his G11FPM class this year, though I didn't make such comments as are referenced in his blog), he has repeatedly presented the idea of mathematics being less about numbers and more about logic/reasoning to several audiences, and hasn't quite perfected the argument. That's not to say it's a bad idea (I think it's good, in principle), but it isn't straightforward as it might seem. – Will R Sep 10 '15 at 0:16
• @DavidC.Ullrich: (cont., opinion here independent of Dr Feinstein) I guess another way of approaching it is: we don't want the audience to think that mathematicians sit around all day thinking about mutilated chessboards that have seemingly no value in the real world; it would be much better if we could at least follow up this example with a more applicable example (or even just an example that is more complicated on the surface) for which a similar thought process provides a quick solution. Note that, even then, both examples may be necessary to have the desired effect. – Will R Sep 10 '15 at 0:31

Bayes' theorem!

If a test for a disease is $99$% accurate, and $1$% of the population has a disease, what is the probability that someone who tests positive for the disease actually has the disease?

• That actually is a very relevant problem, aside from being interesting. +1 – triple_sec Sep 10 '15 at 0:17

Why not stick with a classic proof by contradiction of the irrationality of $\sqrt 2$? It is short and illustrates nicely the logical point.

It is also relatively straightforward to draw a parallel to the scientific method and the 'Popper/Quine criterion': a hypothesis tested and then disproved by experiment; or in medical terms, a tentative diagnosis of a patient disproved by a further measurement or test.

What is interesting about the analogy is how it breaks down. Unlike in a science, in mathematics we can make such an argument on strictly logical grounds alone.

• I gave this proof to a group of chemical engineers, who were appreciative. – Cheerful Parsnip Sep 9 '15 at 20:07
• I've seen this mentioned so often in so-called popular mathematics but rarely a complete proof. Are we supposed to hide the details and rely on the audience's intuition about fraction in lowest terms and parity of a product of integers? If so, then there is no point in introducing logic via this example. – user21820 Sep 15 '15 at 2:45

Claim: $1+1/2+1/3+1/4+\ldots=\infty$. [sic!]

Proof:

\begin{align*} &\,1+\frac{1}{2}+\underbrace{\frac{1}{3}+\frac{1}{4}}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}+\underbrace{\frac{1}{9}+\ldots\frac{1}{15}+\frac{1}{16}}+\underbrace{\frac{1}{17}+\ldots+\frac{1}{31}+\frac{1}{32}}+\ldots\\ \geq&\,1+\frac{1}{2}+\underbrace{\frac{1}{4}+\frac{1}{4}}_{\text{2 terms}}+\underbrace{\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}}_{\text{4 terms}}+\underbrace{\frac{1}{16}+\ldots\frac{1}{16}+\frac{1}{16}}_{\text{8 terms}}+\underbrace{\frac{1}{32}+\ldots+\frac{1}{32}+\frac{1}{32}}_{\text{16 terms}}+\ldots\\ =&\,1+\frac{1}{2}+\left(2\times\frac{1}{4}\right)+\left(4\times\frac{1}{8}\right)+\left(8\times\frac{1}{16}\right)+\left(16\times\frac{1}{32}\right)+\ldots\\ =&\,1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\ldots\\ =&\,\infty. \end{align*} Bam! Lecture hall full of doctors shocked! $\quad\blacksquare$

Then show numerical examples to further convince them on the intuitive level that the harmonic series actually does diverge, albeit very slowly. For example: \begin{align*} \sum_{k=1}^{10}\frac{1}{k}\approx&\,2\mathord.93,\\ \sum_{k=1}^{100}\frac{1}{k}\approx&\,5\mathord.19,\\ \sum_{k=1}^{1000}\frac{1}{k}\approx&\,7\mathord.49,\\ \vdots&\,\\ \sum_{k=1}^{10^6}\frac{1}{k}\approx&\,14\mathord.39,\\ \vdots&\,\\ \sum_{k=1}^{10^{100}}\frac{1}{k}\approx&\,230\mathord.84\\ \vdots&\,\\ \sum_{k=1}^{10^{(10^6)}}\frac{1}{k}\approx&\,2\mathord,302\mathord,586\\ \vdots&\, \end{align*}

• I think most untrained people assume that all infinite series diverge. It may be more interesting to show them why $\sum \frac{1}{2^n}$ converges. – Steven Gubkin Sep 9 '15 at 21:54
• @StevenGubkin When I was an untrained high-school student, for what it’s worth, I was having a hard time actually believing that the harmonic series diverged. After all, the extra terms that keep being added are so small! – triple_sec Sep 9 '15 at 22:55
• @StevenGubkin Afterthought: How about phrasing the problem like this?  The hospital is trying to deal with overcrowding in its clinic. Dr. Cuddy orders that from now on, each doctor (even Dr. House) must stay one extra minute on duty on the first week, 30 extra seconds (on top of the one minute) on the second week, 20 extra seconds (on top of all overtime accumulated so far) on the third week, 15 extra seconds on the fourth, and so forth. Assuming that the hospital looks forward to remaining in business practically forever, how does this policy affect future doctors’ workload? – triple_sec Sep 9 '15 at 23:03
• Before you start manipulating the infinite series, you have to check that it goes somewhere. If not $1 - 1 + 1 - ...$ is both one and zero by the same kind of reasoning. Bam! 'Mathematics' is useless! – user21820 Sep 15 '15 at 2:49
• @user21820 In general, you’re right. In a serious formal proof, one needs to be careful about whether convergence (in the wider sense, to a point in $\mathbb R\cup\{-\infty,+\infty\}$) occurs. However, it may not be necessary to emphasize all the important yet abstruse details to a non-professional, non-student audience. Besides, the partial sums of a non-negative series constitute a monotonic non-decreasing sequence, which always either converges to a finite non-negative limit or “converges” to $+\infty$. (This is actually not too difficult to grasp intuitively without the formalities.) – triple_sec Sep 15 '15 at 4:00

I like Euclid's proof that there are an infinitude of primes. It demonstrates a couple useful ideas such as "proof by contradiction" but also allows you to make the point that a proof of existence doesn't necessarily give you an (easy) way to generate all the primes. I think medical doctors can appreciate the fact that just because something is theoretically possible (like curing cancer) does not mean that there are effective or practical methods to do so.

If these are medical doctors, I think something to do with Simpson's paradox, especially as it relates to the (mis)interpretation of medical studies, would be appropriate.

You could pose a problem similar to the seven bridges of Königsberg. Think about some nice description of a problem that translates into finding an Eulerian path in a graph. Then choose a graph that has $4$ vertices of odd degree and prove that such a graph doesn't admit a Eulerian path. Alternatively, you could also show that a certain graph doesn't admit a Eulerian cycle.

There are also a lot of problems that can easily be solved by finding a suitable invariant. For example, consider this problem that was posed in the first round of the Bundeswettbewerb Mathematik 2011:

Ten bowls are placed in a circle. We fill these bowls clockwise with $1, 2, \ldots, 10$ marbles. In each turn you are allowed to either remove or add two marbles into two adjoint bowls. Is it possible to get exactly $2011$ marbles into each bowl after applying finitely many turns?

It is not possible. Note that there are $1 + 2 + \ldots + 10 = 55$ marbles in the beginning, but in each turn you can only add or remove an even number of marbles. This means the total number of marbles after each move is always odd, but if each bowl contains $2011$ marbles, we have a total of $20110$ marbles. Since this is even, this state can't be reached in finitely many turns.

To keep it more visual, you can of course adjust the number of marbles and bowls.

Another interesting example might be simple games. There are a lot of games that can be explained simply, but have a "hidden mathematical twist" which gives one of the players a winning strategy. Nim is an example of such a game, though the math behind it might take a while to explain.

Last but not least, you might impress your doctors with some "mathematical magic".

A result which is visual, simple but non-trivial is the theorem in elementary geometry that says

The angle subtended by a chord of a circle at a point on the circumference is invariant for those points which are located on the same side as the chord.

Stick to the easiest case! Don't bewilder your audience with complexity.

Explain how initially one cannot see how to get a grip on this. Then bring in the relevance of the unique center of symmetry which makes the circle a very special figure. Join the points to the center, and draw attention to the isosceles triangles. Apply the exterior angle theorem, which you have already introduced as an easy lemma during a discussion of the measurement of angles.

The above analysis shows that the angle subtended at the circumferences is a half the angle subtended at the center. We have found the required invariant.

Finally invoke Euclid's axiom "Things which are equal to the same thing are equal to one another". This proof shows that Euclid's axiom is not merely a rigorous definition but also often is the source of a useful heuristic approach to finding a route towards a proof of equality, isomorphism etc.

As an introduction doctors might warm up to the idea that both mathematics and medicine were respected disciplines in ancient Greece.

How bout the good ol' introductory proof by induction: can you get to the top of a ladder if you start at the bottom and take an additional steps until you reach the top? (The base case is going from step 1 to step 2; inductive assumption that you can get to step k; you can get to step k + 1 from step k by taking one step.)

It's basic, tangible, and requires minimal math (adding one), yet shows you the logic behind a certain type of proof.