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	<description>Mathematics related discussions</description>
	<pubDate>Sat, 05 Apr 2008 17:52:32 +0000</pubDate>
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		<title>Open problems concerning card games</title>
		<link>http://gowers.wordpress.com/2008/04/05/open-problems-concerning-card-games/</link>
		<comments>http://gowers.wordpress.com/2008/04/05/open-problems-concerning-card-games/#comments</comments>
		<pubDate>Sat, 05 Apr 2008 17:52:32 +0000</pubDate>
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		<description><![CDATA[I&#8217;m glad to say that editorial work on the Princeton Companion is within a whisker of being completed (about three articles remain to be edited), so although I don&#8217;t quite feel that I have the leisure to give proper attention to this blog, which will be obvious from some of the messages that I haven&#8217;t [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>I&#8217;m glad to say that editorial work on the Princeton Companion is within a whisker of being completed (about three articles remain to be edited), so although I don&#8217;t quite feel that I have the leisure to give proper attention to this blog, which will be obvious from some of the messages that I haven&#8217;t got round to deleting, I can at least write a quick post. It starts with a conversation I had a couple of years ago. I was waiting for a plane to take me from Mykonos to Athens. The plane was severely delayed, but the situation could have been a lot worse as I had Persi Diaconis for company. He told me the not very surprising fact that it was not known what the probability of a win is in the game known as Patience in the UK and Solitaire in the US. (I&#8217;m talking about the one where you start by putting down a row of seven cards with just the first one face up, then on top of all but this first one a row of six cards with just the first one face up, and so on.) To be clear about the probability he is asking for, he simplifies the game by letting you see what all the cards are, so that you can play optimally and don&#8217;t have to worry about probabilities.<span id="more-22"></span></p>
<p>One thing he said that particularly struck me was, &#8220;If anyone can solve this I can guarantee them a front-page spread in the New York Times,&#8221; presumably an allusion to the attention he received for showing that you need seven riffle shuffles to shuffle a pack of cards properly. He also told me of someone he knew who had tried to estimate the probability by actually playing the game several times, but whose empirical evidence was flawed because he had a far from optimal strategy. Incidentally, I think Persi would be pretty impressed by any non-trivial rigorous bounds for the problem. (I&#8217;d count as trivial something like an upper bound that resulted from showing that you couldn&#8217;t win if a given arrangement of two or three cards occurred somewhere, but there may be more sophisticated &#8220;trivial&#8221; upper bounds too.)</p>
<p>That brings me to the main purpose of this post, which is to throw open a problem that I have idly wondered about for a long time. It concerns the game that I know as Beggar My Neighbour.</p>
<p>Well well. I thought I&#8217;d see if I could save myself some time by finding the rules for this game on the web. Sure enough, there they are on Wikipedia, but to my slight disappointment I see that what I thought was a nice problem I&#8217;d thought of myself (which in a sense it is) is in fact &#8220;a longstanding question in combinatorial game theory&#8221;. Let me nevertheless say what the question is and give a few thoughts about it. I&#8217;d also like to discuss a view that the article attributes to Conway. The article, by the way, is <a href="http://en.wikipedia.org/wiki/Beggar-My-Neighbour">here </a> if you don&#8217;t know this particular game (which also has various other names &#8212; another I had heard of is Strip Jack Naked).</p>
<p>Beggar My Neighbour is a game of pure luck: you deal out the cards and the outcome of the game is completely determined by the initial distribution of the cards. The question is simple: experience suggests that the game always finishes, but does it? </p>
<p>One can express the rules of the game as a fairly simple set of rules for a deterministic walk through the space of all possible configurations of cards, where a &#8220;configuration&#8221; could be defined as a permutation of the 52 cards, together with an integer k between 1 and 52 that tells you that the first k cards belong to the player who is about to play. Then the question is whether this walk has any cycles.</p>
<p>According to Wikipedia, Conway describes this as an &#8220;anti-Hilbert problem,&#8221; by which he means a problem that should definitely <em>not</em> set the direction for future mathematical research. While I of course see what he means, I can&#8217;t help finding the problem interesting (and I expect he can&#8217;t either). In particular, it has at least one good feature, which is that one can also give fairly well-motivated variants of it, as I&#8217;ll explain in a moment.</p>
<p>First, let me make one simple (and, I see from Wikipedia, well-known) observation. An obvious line of attack on the problem would be to generalize it and show that <em>every</em> Beggar-My-Neighbour type game had to terminate. But this is false. For instance, consider the reduced game where my hand is (2,2,jack,2) and yours is (jack,2). Suppose also that I start. Then I&#8217;ll put down a 2, you&#8217;ll put down a jack, I&#8217;ll put down a 2, and you&#8217;ll pick up. Now our hands are (jack,2) and (2,2,jack,2) and it&#8217;s your turn. Since this is exactly the position I was in at the start, the game goes on for ever.</p>
<p>Thus, there isn&#8217;t some general principle that tells you that a certain quantity always goes down as you play this sort of game.</p>
<p>For me that&#8217;s quite a compelling reason to agree with Conway&#8217;s judgement that the problem itself is an anti-Hilbert one. However, what about trying to explain the phenomenon that the game does in practice always seem to terminate? To do this it would be sufficient to prove that it terminates with massively high probability, or at least with high enough probability that on the rare occasions when the initial configuration is part of a cycle the players just get bored and stop without actually noticing that their game wouldn&#8217;t terminate if they continued.</p>
<p>Even this statistical question seems to be very hard, but there is a further simplification that makes it neater without reducing its mathematical interest. That is to take a pack of cards that just contains jacks and non-jacks. The rule is that if you play a jack and your opponent follows it with a non-jack then you pick up. I tried this with my daughter: all black cards had the role of jack, and all red cards had the role of non-jacks. I was quite surprised to discover that this was a reasonable game &#8230; and it terminated.</p>
<p>If, as I suspect, even this game is very hard to analyse statistically, it is still quite tempting to try to devise a convincing non-rigorous model for it that would &#8220;explain&#8221; the observed phenomenon that the game appears to terminate with very high probability, and perhaps even predict the distribution of how long it takes. One of the nice features of the game is that the procedure of putting down the cards alternately has a shuffling effect. This is highly relevant to the game, because it seems that a player gets a big advantage from a longish run that is dense with royal cards, but these runs get naturally dispersed unless both players have runs at the same time. Perhaps there is a way of modelling this dispersal effect probabilistically. Similarly, in the real game of Beggar-My-Neighbour, it makes a big difference who has the jacks, since it is quite hard to lose a jack. This is because after you play a jack, your opponent can win it only by playing a royal card and then going on to be the person who next picks up. In practice, this feels like a random event: you play a jack and then either breathe a sigh of relief if you don&#8217;t lose it or curse your luck if you do. And if, like most mortals, you&#8217;ve forgotten the order of the cards, then in a certain sense it <em>is</em> a random event, despite its completely deterministic nature. Perhaps one could devise a realistic &#8220;randomized Beggar-My-Neighbour&#8221; that was easier to study. Even something rather simple might help, such as occasionally inserting or removing random non-royal cards (which I myself know as &#8220;chicken-feed&#8221; for some reason) into hands. Perhaps if this was done at the right rate it would make certain events into ones that could be analysed probabilistically, but it would not destroy too badly the important features of the game such as the gradual dispersal of significant runs of cards. Of course, it might well be possible to prove that after a randomization procedure the game terminated with probability 1 because just occasionally something incredibly unlikely happened that caused it to terminate for trivial reasons. So the question would have to be to show that with high probability it terminates after not too ridiculously long.</p>
<p>And then of course one could try randomizing the simpler version of the game with just jacks and non-jacks.</p>
<p>To summarize, I think it is an interesting challenge (not Hilbert-interesting but certainly interesting given the current level of interest in probabilistic processes with simply described dependencies) to devise some simplified model of Beggar-My-Neighbour and prove rigorously that it terminates reasonably quickly with high probability. Indeed, I find this a more interesting question than the question of whether the actual game always terminates. </p>
<p>All these are very natural reactions to the original question, so it is probable that others have already had them. If you know of their having been raised, then I&#8217;d be interested to hear about it.</p>
<p>Wikipedia has a reference to an American Mathematical Monthly article on the subject. I can&#8217;t get hold of that, but have found a discussion elsewhere (by Richard Guy) that gives a few salient facts that I think are contained in it: there are no cycles if you halve the standard pack (this was discovered by a computer search), but there are if you add or take away two non-royal cards; and an arrangement has been found of the standard pack that takes 5790 goes before it terminates. </p>
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		<title>Probability paradox II</title>
		<link>http://gowers.wordpress.com/2008/02/03/probability-paradox-ii/</link>
		<comments>http://gowers.wordpress.com/2008/02/03/probability-paradox-ii/#comments</comments>
		<pubDate>Sun, 03 Feb 2008 13:20:29 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[Probability]]></category>

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		<description><![CDATA[I had an email from Noga Alon a couple of days ago, who told me about a much better version of the paradox I discussed in an earlier post. Some of the comments relating to that post also allude to this better version. The reason it is better is that one can no longer object [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>I had an email from Noga Alon a couple of days ago, who told me about a much better version of the paradox I discussed in an earlier post. Some of the comments relating to that post also allude to this better version. The reason it is better is that one can no longer object to it on the grounds that it assumes the existence of a probability distribution with impossible properties. <span id="more-21"></span></p>
<p>Here is the version as Noga described it (except that if I say anything accidentally stupid, then that is my own personal contribution). You are presented with two envelopes, but now you are told that the amount of money in the envelopes is <img src='http://l.wordpress.com/latex.php?latex=10%5En&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='10^n' title='10^n' class='latex' /> dollars and <img src='http://l.wordpress.com/latex.php?latex=10%5E%7Bn%2B1%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='10^{n+1}' title='10^{n+1}' class='latex' /> dollars, where the positive integer <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> is chosen with probability <img src='http://l.wordpress.com/latex.php?latex=2%5E%7B-n%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2^{-n}' title='2^{-n}' class='latex' />.</p>
<p>Suppose that you <i>open</i> one of the envelopes. If it contains 10 dollars, then trivially you should switch: the other contains 100 dollars. Now suppose that it contains 100 dollars. This could have happened in one of two ways: with probability <img src='http://l.wordpress.com/latex.php?latex=1%2F4&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/4' title='1/4' class='latex' /> we have <img src='http://l.wordpress.com/latex.php?latex=n%3D1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n=1' title='n=1' class='latex' />, so the other envelope contains 10 dollars and you chose the envelope with more money; and with probability <img src='http://l.wordpress.com/latex.php?latex=1%2F8&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/8' title='1/8' class='latex' /> we have <img src='http://l.wordpress.com/latex.php?latex=n%3D2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n=2' title='n=2' class='latex' />, so the other envelope contains 1000 dollars and you chose the envelope with less money. So the conditional probabilities are <img src='http://l.wordpress.com/latex.php?latex=2%2F3&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2/3' title='2/3' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=1%2F3&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/3' title='1/3' class='latex' />. But the amount you gain by switching is so great in the <img src='http://l.wordpress.com/latex.php?latex=1%2F3&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/3' title='1/3' class='latex' /> case that your expected gain if you switch is certainly positive.</p>
<p>It is not hard to see that this argument works for any amount of money you find in the envelope, as long as it is not 10 dollars. But in that case you still switch&#8212;it&#8217;s just that the reasons are different. So <i>whatever</i> amount you discover in the envelope when you open it, you improve your expected gain if you switch. So it should surely follow that you don&#8217;t need to look in the envelope before deciding to switch. And that is the paradox.</p>
<p>You might object by saying that the expected amount in the envelopes is infinite, so in practice you would always know that the situation was not truly as I have just described it. But that objection will not do: the situation is at least <i>logically</i> possible, and one can easily modify the scenario. For example, suppose that just after your death you find that you are still conscious, but that none of the world&#8217;s major religions have got it quite right about life after death. Instead, you are greeted by the great god <img src='http://l.wordpress.com/latex.php?latex=%5Cmu%5Calpha%5Calpha%5Ctheta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mu\alpha\alpha\theta' title='\mu\alpha\alpha\theta' class='latex' /> (at which point you finally understand why it was that you had been so mysteriously obsessed by mathematics). <img src='http://l.wordpress.com/latex.php?latex=%5Cmu%5Calpha%5Calpha%5Ctheta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mu\alpha\alpha\theta' title='\mu\alpha\alpha\theta' class='latex' /> tells you that you will live a life of eternal bliss if and only if you manage to solve all the Clay millennium problems within a time that is specified in one of two envelopes. Moreover, the amount of time (in centuries) is chosen randomly according to the distribution described above for the dollars. And you are warned that one of the problems is pretty hard, even for someone with mathematical training and ten thousand years to do it, so you are advised to try to maximize your expected time. Then, once again, one argument says you should switch envelopes and another says it makes no difference.</p>
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		<title>Removing the magic from Stirling&#8217;s formula</title>
		<link>http://gowers.wordpress.com/2008/02/01/removing-the-magic-from-stirlings-formula/</link>
		<comments>http://gowers.wordpress.com/2008/02/01/removing-the-magic-from-stirlings-formula/#comments</comments>
		<pubDate>Fri, 01 Feb 2008 13:04:34 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[Probability]]></category>

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		<description><![CDATA[Another topic on the syllabus for the probability course I am giving is Stirling&#8217;s formula. This was lectured to me when I was an undergraduate but I had long since forgotten the proof completely. In fact, I&#8217;d even forgotten the precise statement, so I had some mugging up to do. It turned out to be [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Another topic on the syllabus for the probability course I am giving is Stirling&#8217;s formula. This was lectured to me when I was an undergraduate but I had long since forgotten the proof completely. In fact, I&#8217;d even forgotten the precise statement, so I had some mugging up to do. It turned out to be hard to find a proof that didn&#8217;t look magic: the arguments had a &#8220;consider the following formula, do such-and-such, observe that this is less than that, bingo it drops out&#8221; sort of flavour. I find magic proofs impossible to remember, so I was forced to think a little. As a result, I came up with an argument that was mostly fully motivated, but there is one part that I still find mysterious. On the other hand, it looks so natural that I&#8217;m sure somebody can help me find a good explanation for why it works. When I say &#8220;I came up with an argument&#8221; what I mean is that I came up with a way of presenting an existing argument that doesn&#8217;t involve the pulling of rabbits out of hats, except in the place I&#8217;m about to discuss. <span id="more-20"></span></p>
<p>The first idea in proving Stirling&#8217;s formula is a very natural one: estimate <img src='http://l.wordpress.com/latex.php?latex=%5Clog%28n%21%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log(n!)' title='\log(n!)' class='latex' /> instead. Since that equals <img src='http://l.wordpress.com/latex.php?latex=%5Clog+2%2B%5Cdots%2B%5Clog+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log 2+\dots+\log n' title='\log 2+\dots+\log n' class='latex' />, it is natural to approximate the sum by the integral <img src='http://l.wordpress.com/latex.php?latex=%5Cint_1%5En%5Clog+x+dx&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\int_1^n\log x dx' title='\int_1^n\log x dx' class='latex' />. However, if you do that then the error will be around <img src='http://l.wordpress.com/latex.php?latex=%5Clog+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log n' title='\log n' class='latex' />. This gives you a good rough-and-ready approximation to <img src='http://l.wordpress.com/latex.php?latex=n%21&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n!' title='n!' class='latex' /> but nothing like as good as Stirling&#8217;s formula.</p>
<p>The next idea is to turn things round and try to approximate the integral by a sum rather than the other way round. The obvious next approximation to try is approximating <img src='http://l.wordpress.com/latex.php?latex=%5Clog+x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log x' title='\log x' class='latex' /> by the function <img src='http://l.wordpress.com/latex.php?latex=g%28x%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='g(x)' title='g(x)' class='latex' />, where this is defined to equal <img src='http://l.wordpress.com/latex.php?latex=%5Clog+x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log x' title='\log x' class='latex' /> when <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> is an integer and to be linear in between consecutive integers. Then you can expect a much smaller error, and indeed it is not hard to prove that the difference between the integrals of the two functions is bounded. (The proof I gave for this dropped out rather nicely: if <img src='http://l.wordpress.com/latex.php?latex=r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r' title='r' class='latex' /> is a positive integer then by looking at derivatives it is easy to show that between <img src='http://l.wordpress.com/latex.php?latex=r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r' title='r' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=r%2B1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r+1' title='r+1' class='latex' /> we have the inequalities <img src='http://l.wordpress.com/latex.php?latex=%5Clog+r%2B%28x-r%29%2F%28r%2B1%29%5Cleq+g%28x%29%5Cleq%5Clog+x%5Cleq%5Clog+r%2B%28x-r%29%2Fr&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log r+(x-r)/(r+1)\leq g(x)\leq\log x\leq\log r+(x-r)/r' title='\log r+(x-r)/(r+1)\leq g(x)\leq\log x\leq\log r+(x-r)/r' class='latex' />. Therefore, the difference between the integrals of <img src='http://l.wordpress.com/latex.php?latex=%5Clog+x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log x' title='\log x' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=g%28x%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='g(x)' title='g(x)' class='latex' /> between <img src='http://l.wordpress.com/latex.php?latex=r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r' title='r' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=r%2B1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r+1' title='r+1' class='latex' /> is at most <img src='http://l.wordpress.com/latex.php?latex=%281%2Fr-1%2F%28r%2B1%29%29%2F2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(1/r-1/(r+1))/2' title='(1/r-1/(r+1))/2' class='latex' />. This gives a nice telescoping sum as an upper bound for the sum of all those areas.)</p>
<p>Another important observation is that the integral of <img src='http://l.wordpress.com/latex.php?latex=g%28x%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='g(x)' title='g(x)' class='latex' /> between 1 and <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> works out to be <img src='http://l.wordpress.com/latex.php?latex=%5Clog%28n%21%29-%5Clog+n%2F2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log(n!)-\log n/2' title='\log(n!)-\log n/2' class='latex' /> (because the integral between <img src='http://l.wordpress.com/latex.php?latex=r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r' title='r' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=r%2B1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r+1' title='r+1' class='latex' /> is <img src='http://l.wordpress.com/latex.php?latex=%28%5Clog+r%2B%5Clog%28r%2B1%29%29%2F2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(\log r+\log(r+1))/2' title='(\log r+\log(r+1))/2' class='latex' />). Thus, we have a rather close relationship between <img src='http://l.wordpress.com/latex.php?latex=%5Clog%28n%21%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log(n!)' title='\log(n!)' class='latex' /> and an integral that we can calculate: <img src='http://l.wordpress.com/latex.php?latex=%5Cint_1%5En%5Clog+x+dx%3Dn%5Clog+n-n%2B1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\int_1^n\log x dx=n\log n-n+1' title='\int_1^n\log x dx=n\log n-n+1' class='latex' />.</p>
<p>It is not hard to see that this argument shows that the ratio of <img src='http://l.wordpress.com/latex.php?latex=n%21&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n!' title='n!' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=%28n%2Fe%29%5En%5Csqrt+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(n/e)^n\sqrt n' title='(n/e)^n\sqrt n' class='latex' /> tends to a limit. The one remaining magic part is the calculation of that limit. To do this, we define <img src='http://l.wordpress.com/latex.php?latex=c_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='c_n' title='c_n' class='latex' /> to be the integral <img src='http://l.wordpress.com/latex.php?latex=%5Cint_1%5En%28%5Clog+x-g%28x%29%29dx&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\int_1^n(\log x-g(x))dx' title='\int_1^n(\log x-g(x))dx' class='latex' />. We have established that the sequence <img src='http://l.wordpress.com/latex.php?latex=%28c_n%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(c_n)' title='(c_n)' class='latex' /> converges; our problem is equivalent to working out what the limit is. Here there is a very nice trick, which is to consider instead the sequence <img src='http://l.wordpress.com/latex.php?latex=a_n%3D2c_n-c_%7B2n%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a_n=2c_n-c_{2n}' title='a_n=2c_n-c_{2n}' class='latex' />, which clearly converges to the same limit. But if you calculate <img src='http://l.wordpress.com/latex.php?latex=a_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a_n' title='a_n' class='latex' /> you find that you can get a very good estimate for it if and only if you can get a very good estimate for <img src='http://l.wordpress.com/latex.php?latex=%5Cbinom%7B2n%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\binom{2n}{n}' title='\binom{2n}{n}' class='latex' />, essentially because inside what you write for <img src='http://l.wordpress.com/latex.php?latex=a_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a_n' title='a_n' class='latex' /> you get <img src='http://l.wordpress.com/latex.php?latex=%5Clog%282n%21%29-2%5Clog%28n%21%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log(2n!)-2\log(n!)' title='\log(2n!)-2\log(n!)' class='latex' />, which is <img src='http://l.wordpress.com/latex.php?latex=%5Clog+%5Cbinom%7B2n%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\log \binom{2n}{n}' title='\log \binom{2n}{n}' class='latex' />. This trick is I think reasonably non-magic: if you know in advance that you can calculate <img src='http://l.wordpress.com/latex.php?latex=%5Cbinom%7B2n%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\binom{2n}{n}' title='\binom{2n}{n}' class='latex' /> very accurately, then it is natural to try to use that information, and there is essentially only one way of doing that.</p>
<p>Even the fact that one can get a very good handle on <img src='http://l.wordpress.com/latex.php?latex=%5Cbinom%7B2n%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\binom{2n}{n}' title='\binom{2n}{n}' class='latex' /> is not as magic as all that, since it is equivalent to estimating the probability that a subset of a set of size <img src='http://l.wordpress.com/latex.php?latex=2n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2n' title='2n' class='latex' /> has exactly <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> elements. And that one can think about in all sorts of ways. For example, one can use the central limit theorem to estimate the probability that a random set has size between <img src='http://l.wordpress.com/latex.php?latex=n-c%5Csqrt+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n-c\sqrt n' title='n-c\sqrt n' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=n%2Bc%5Csqrt+n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n+c\sqrt n' title='n+c\sqrt n' class='latex' /> and an elementary argument to show that the probabilities of the sizes within that range are all roughly equal.</p>
<p>However, we have not yet got to the central limit theorem in the course, so at this point I copied another proof, which goes roughly like this. You look at the integrals <img src='http://l.wordpress.com/latex.php?latex=I_n%3D%5Cint_0%5E%7B%5Cpi%2F2%7D%5Ccos%5En%28x%29dx&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_n=\int_0^{\pi/2}\cos^n(x)dx' title='I_n=\int_0^{\pi/2}\cos^n(x)dx' class='latex' />. By integrating by parts in a standard way, you can get the recurrence <img src='http://l.wordpress.com/latex.php?latex=nI_n%3D%28n-1%29I_%7Bn-2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='nI_n=(n-1)I_{n-2}' title='nI_n=(n-1)I_{n-2}' class='latex' />. Since the <img src='http://l.wordpress.com/latex.php?latex=I_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_n' title='I_n' class='latex' /> clearly decrease, we get from this that the ratio of <img src='http://l.wordpress.com/latex.php?latex=I_%7B2n%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_{2n}' title='I_{2n}' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=I_%7B2n-1%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_{2n-1}' title='I_{2n-1}' class='latex' /> tends to 1. But we can also give an exact formula for this ratio by using the recurrence and calculations of <img src='http://l.wordpress.com/latex.php?latex=I_0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_0' title='I_0' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=I_1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_1' title='I_1' class='latex' />. If you do this and rearrange things a bit, then you find that you can show that <img src='http://l.wordpress.com/latex.php?latex=2%5E%7B-2n%7D%5Csqrt+n%5Cbinom%7B2n%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2^{-2n}\sqrt n\binom{2n}{n}' title='2^{-2n}\sqrt n\binom{2n}{n}' class='latex' /> tends to <img src='http://l.wordpress.com/latex.php?latex=%5Cpi%5E%7B-1%2F2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\pi^{-1/2}' title='\pi^{-1/2}' class='latex' />.</p>
<p>Finally, my question: is it just a coincidence that looking at the ratio of <img src='http://l.wordpress.com/latex.php?latex=I_%7B2n%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_{2n}' title='I_{2n}' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=I_%7B2n-1%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='I_{2n-1}' title='I_{2n-1}' class='latex' /> gives you an expression that involves <img src='http://l.wordpress.com/latex.php?latex=%5Cbinom%7B2n%7D%7Bn%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\binom{2n}{n}' title='\binom{2n}{n}' class='latex' />? Surely it isn&#8217;t, but for the moment this part of the argument still looks to me like a convenient piece of magic. It was clean enough to be easily memorable, so this did not present a problem for the lecture, but I&#8217;d still like to know why those integrals are so closely related to the binomial coefficient.</p>
<p>It would be remiss of me not to include a statement of Stirling&#8217;s formula in this account of the proof: it says that the ratio of <img src='http://l.wordpress.com/latex.php?latex=n%21&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n!' title='n!' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%5Cpi+n%7D%28n%2Fe%29%5En&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2\pi n}(n/e)^n' title='\sqrt{2\pi n}(n/e)^n' class='latex' /> tends to 1.</p>
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		<title>A paradox in probability</title>
		<link>http://gowers.wordpress.com/2008/02/01/a-paradox-in-probability/</link>
		<comments>http://gowers.wordpress.com/2008/02/01/a-paradox-in-probability/#comments</comments>
		<pubDate>Fri, 01 Feb 2008 13:04:24 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[Probability]]></category>

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		<description><![CDATA[ This is the first of what I hope will be several posts related to the course I am giving this term on probability.
The following is a well-known paradox. You are presented with two envelopes and told that one contains a sum of money and the other contains twice as much. You are invited to [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p> This is the first of what I hope will be several posts related to the course I am giving this term on probability.</p>
<p>The following is a well-known paradox. You are presented with two envelopes and told that one contains a sum of money and the other contains twice as much. You are invited to choose an envelope but are not told which is which. You choose an envelope, and are then given the chance to change your mind if you want to. Should you?</p>
<p>One argument says that it cannot possibly make any difference to the expected outcome, since either way your expected gain will be the average of the amounts in the two envelopes (so the expected change by switching is zero). But there is another argument that goes as follows. Suppose that the amount of money in the envelope you first choose is <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' />. Then the other envelope has a 50% chance of containing <img src='http://l.wordpress.com/latex.php?latex=2x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2x' title='2x' class='latex' /> and a 50% chance of containing <img src='http://l.wordpress.com/latex.php?latex=x%2F2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x/2' title='x/2' class='latex' />, so your expectation if you switch is <img src='http://l.wordpress.com/latex.php?latex=5x%2F4&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='5x/4' title='5x/4' class='latex' />&#8212;so you should switch.</p>
<p>I tried this out for real in my first lecture, and the student who was given the choice decided to switch. Rather irritatingly, he got more money as a result. Of course, the second argument is incorrect, but the reasons are somewhat subtle. My purpose in putting up a post about it is not so much to invite solutions to the paradox as to see whether it prompts anyone to give me their favourite probabilistic paradoxes. (I&#8217;ve just done Simpson&#8217;s paradox, so that one wouldn&#8217;t be new.)</p>
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		<title>General news&#8212;February 2008</title>
		<link>http://gowers.wordpress.com/2008/02/01/general-news-february-2008/</link>
		<comments>http://gowers.wordpress.com/2008/02/01/general-news-february-2008/#comments</comments>
		<pubDate>Fri, 01 Feb 2008 12:13:25 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[General]]></category>

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		<description><![CDATA[Not entirely surprisingly, my hibernation is going to go on for longer than I had hoped. The reason for this is quite simple: there is a major push to get the Princeton Companion to Mathematics finished within the next couple of months or so, and if I spend time blogging then it won&#8217;t happen. There [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Not entirely surprisingly, my hibernation is going to go on for longer than I had hoped. The reason for this is quite simple: there is a major push to get the Princeton Companion to Mathematics finished within the next couple of months or so, and if I spend time blogging then it won&#8217;t happen. There has also been a delay with the Tricks Wiki, but that may be less severe because I am not the main bottleneck for that (though there are a few things I need to do before it can be up and running). This term I am giving a first course on probability. I had planned a few blog entries on that too, and I hope I&#8217;ll have time for some in due course.</p>
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		<title>General news &#8212; December 2007</title>
		<link>http://gowers.wordpress.com/2007/12/19/general-news-december-2007/</link>
		<comments>http://gowers.wordpress.com/2007/12/19/general-news-december-2007/#comments</comments>
		<pubDate>Tue, 18 Dec 2007 23:12:02 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[General]]></category>

		<guid isPermaLink="false">http://gowers.wordpress.com/2007/12/19/general-news-december-2007/</guid>
		<description><![CDATA[As I predicted in an earlier post, my rate of posting has (temporarily) gone right down. This is partly for the reasons I said&#8212;I am very busy with a final push to finish the Princeton Companion to Mathematics, and busy in general&#8212;but also for another reason. I was going to keep personal matters rigorously excluded [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>As I predicted in an earlier post, my rate of posting has (temporarily) gone right down. This is partly for the reasons I said&#8212;I am very busy with a final push to finish the Princeton Companion to Mathematics, and busy in general&#8212;but also for another reason. I was going to keep personal matters rigorously excluded from this blog but since my secret is out (see the comments on Examples First II on November 17th) I may as well also mention that I have a five-week-old son, Octave, who doesn&#8217;t leave much time for blogging given that my other commitments won&#8217;t go away. So this isn&#8217;t a proper post but just a way of saying that my blog hasn&#8217;t died: it&#8217;s just hibernating. Meanwhile, I can at least briefly mention that a &#8220;Tricki&#8221;&#8212;that is, a Wiki-style website devoted to theorem-proving techniques&#8212;will almost certainly exist in the near future. Remarkably, my <a href="http://gowers.wordpress.com/2007/09/11/what-might-an-expository-mathematical-wiki-be-like/">earlier post on this idea </a> led to an offer of technical help that will be enough to turn it from a fantasy into a reality. And that&#8217;s saying something, since my own technical ability in this area is basically zero. I&#8217;ve seen a prototype and it looks great. Probably we&#8217;ll get a small site up and running and I&#8217;ll then ask for comments about how it could be improved before we throw it open. (We still haven&#8217;t decided what policy to adopt about who can edit what, but we are actively thinking about it.) And that&#8217;s it from me until 2008.</p>
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		<title>Examples first II</title>
		<link>http://gowers.wordpress.com/2007/10/24/examples-first-ii/</link>
		<comments>http://gowers.wordpress.com/2007/10/24/examples-first-ii/#comments</comments>
		<pubDate>Wed, 24 Oct 2007 11:35:31 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[Mathematical pedagogy]]></category>

		<guid isPermaLink="false">http://gowers.wordpress.com/2007/10/24/examples-first-ii/</guid>
		<description><![CDATA[It&#8217;s what blogging is all about I suppose, but I have been surprised in several different ways by the comments on my previous post. To begin with, I was so sure of the principle I was advocating that I thought that all I&#8217;d have to do was explain it briefly and then anybody who read [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>It&#8217;s what blogging is all about I suppose, but I have been surprised in several different ways by the comments on my previous post. To begin with, I was so sure of the principle I was advocating that I thought that all I&#8217;d have to do was explain it briefly and then anybody who read it would instantly agree with it. That was clearly pretty naive of me, and I certainly didn&#8217;t expect that some people would be actively hostile to the idea (though I suspect that their real target was not precisely the same as what I was putting forward). But I was also surprised by the number of interesting further points and qualifications that were made, which I will now try to use to articulate a more nuanced version of the principle.<span id="more-16"></span><br />
Amongst these further points were the following. If one is sufficiently used to a particular style of definition then it may well not be necessary to give examples first: for instance, if you know the definition of a field, then you can easily grasp the definition of a ring without having a chat about polynomials or something first. (Of course, if you want to understand the <em>point</em> of defining rings, then such a chat is essential, but it&#8217;s not so important to have the chat <em>first</em>.) JSE (who, despite his denials, gave a beautiful demonstration of the principle of examples first in his PCM article) makes the point that some mathematicians find examples confusing unless they already know what they are supposed to be illustrating, and the further point that promising one kind of explanation while one gives another can be very reassuring, whichever way round you do it.</p>
<p>One small point in response to JSE: if you don&#8217;t want to confuse the reader/listener when you discuss examples, one approach is not to give away what you are doing. (See your own PCM article for an instance of this.) For example, in the second explanation of fields in the previous post, there is a discussion of number systems. Since it is stating some fairly obvious and familiar facts about those number systems, there can&#8217;t really be much reason for confusion. But if one began by saying, &#8220;By the way, I&#8217;m leading up to a definition of some things called fields here,&#8221; then some people might be distracted by wondering what they were supposed to be getting out of the examples.</p>
<p>Another point that comes out of several comments is that a lot depends on the circumstances of a presentation. I think the principle applies most strongly when the presentation is not fully formal &#8212; e.g. in an expository article, or a conversation with another mathematician, or a colloquium talk, or in a seminar where you can&#8217;t expect too much of your audience. When it comes to a formal lecture course, I think my practice would be to write up fairly traditional notes on the blackboard, but to give a lot of accompanying chat: the preliminary examples would be part of the chat rather than part of the notes. As for textbooks, here there may well be disagreement, but I would argue for something similar to the lecture course approach, except that now the preliminary chat would be written.</p>
<p>On that last point, one person made the interesting comment that they were so used to reading papers and books in a non-linear way that they actually preferred papers and books that did not try to present themselves linearly (which is essentially what one is trying to do with the examples-first approach). My implicit suggestion of clearly distinguishing between the chat and the &#8220;real content&#8221; could perhaps lead to expositions that gave the best of both worlds.</p>
<p>More generally, one might take the attitude that, since it is an essential mathematical skill to be able to read and digest mathematics that is presented in a very formal way, and since part of that skill is to be able to supply one&#8217;s own examples, if you the author provide the examples yourself (whether before or after the generalities) then you are denying the reader the chance to develop that skill. To which I&#8217;d say: if you do not provide that chance, there will always be others who are more than happy to do so.</p>
<p>Now let me look at another mathematical concept and consider how it might be explained. This time I want to discuss a theorem rather than a definition, just to emphasize (as I didn&#8217;t in the previous post) that the examples-first principle is quite general and doesn&#8217;t just refer to places where you first introduce an abstract definition.</p>
<p>The theorem I&#8217;ve gone for is the orbit-stabilizer theorem, and I want to discuss how it might be presented to somebody who was already comfortable with the idea of a group action (though it&#8217;s quite an interesting question in itself how to explain group actions &#8212; in a funny way the examples are all too &#8220;obvious&#8221; for it to be easy to make clear what the real use of the concept is).</p>
<p>The approach that I&#8217;ll label &#8220;traditional&#8221; for the purposes of discussion is something like this. Let <img src='http://l.wordpress.com/latex.php?latex=G&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G' title='G' class='latex' /> be a finite group and let <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' /> be a set on which <img src='http://l.wordpress.com/latex.php?latex=G&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G' title='G' class='latex' /> acts. Let <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> be an element of <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' />. Then we define the orbit <img src='http://l.wordpress.com/latex.php?latex=O_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='O_x' title='O_x' class='latex' /> of <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> to be the set <img src='http://l.wordpress.com/latex.php?latex=%5C%7Bgx%3Ag%5Cin+G%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\{gx:g\in G\}' title='\{gx:g\in G\}' class='latex' /> and the stabilizer <img src='http://l.wordpress.com/latex.php?latex=S_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='S_x' title='S_x' class='latex' /> of <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> to be the set <img src='http://l.wordpress.com/latex.php?latex=%5C%7Bg%5Cin+G%3Agx%3Dx%5C%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\{g\in G:gx=x\}' title='\{g\in G:gx=x\}' class='latex' />. The orbit-stabilizer theorem states that <img src='http://l.wordpress.com/latex.php?latex=%7CO_x%7C%7CS_x%7C%3D%7CG%7C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='|O_x||S_x|=|G|' title='|O_x||S_x|=|G|' class='latex' />. (I stress that I am summarizing the approach here rather than giving it in full: if I did it properly I&#8217;d state the theorem more formally and distinguish it much more clearly from the surrounding discussion, which itself would be a bit longer.)</p>
<p>To prove the theorem, we define a map from <img src='http://l.wordpress.com/latex.php?latex=O_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='O_x' title='O_x' class='latex' /> to the left cosets of <img src='http://l.wordpress.com/latex.php?latex=S_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='S_x' title='S_x' class='latex' /> by sending <img src='http://l.wordpress.com/latex.php?latex=gx&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='gx' title='gx' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=gS_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='gS_x' title='gS_x' class='latex' />. One must check that this map is well-defined and that it is a bijection, which is an easy exercise. The result then follows from the fact that all the left cosets of <img src='http://l.wordpress.com/latex.php?latex=S_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='S_x' title='S_x' class='latex' /> have the same size.</p>
<p>Once one has given this proof in a lecture course, a typical thing to do is to &#8220;test the understanding&#8221; of the theorem by means of some exercises, of which quite a common one is to count symmetries of Platonic solids. For instance, to count the number of rotational symmetries of an icosahedron, one lets <img src='http://l.wordpress.com/latex.php?latex=G&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G' title='G' class='latex' /> be the group of all these symmetries and lets <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> be a vertex of the icosahedron. Then the orbit of <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> (under the obvious action) has size 12, since the icosahedron has twelve vertices that all &#8220;look the same&#8221; and the stabilizer has size 5 (since neighbouring vertices go to neighbouring vertices and you can&#8217;t reflect), so <img src='http://l.wordpress.com/latex.php?latex=G&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G' title='G' class='latex' /> has size 60.</p>
<p>Now here&#8217;s an alternative approach. You begin by asking how many rotational symmetries an icosahedron has (as part of an informal discussion, say, before you &#8220;get down to business&#8221;). Most people will come up for themselves with the argument that a single vertex has 12 choices of where to go, and one of its neighbours then has 5, after which the rotation is determined: hence, there are 60 rotations.</p>
<p>At that point, one can say, &#8220;Now we are going to prove a theorem that shows that this type of argument works very generally.&#8221; And as you go through the proof outlined above, you can say, &#8220;Notice that in the example we looked at earlier, the orbit was the set of all vertices and the stabilizer was the set of all rotations that fixed a particular vertex.&#8221; Then the student will see that what you really need to know is that the set of transformations that send <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> to <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' /> always has the same size (which it does, as it&#8217;s a coset of the stabilizer). In fact, one is led to a better proof, I think: the result is true because what it is saying is that we partition <img src='http://l.wordpress.com/latex.php?latex=G&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='G' title='G' class='latex' /> according to what the elements do to a fixed element <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> of <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' />. The cells of this partition all have the same size since they are cosets of <img src='http://l.wordpress.com/latex.php?latex=S_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='S_x' title='S_x' class='latex' />, and obviously the number of them is the size of <img src='http://l.wordpress.com/latex.php?latex=O_x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='O_x' title='O_x' class='latex' />.</p>
<p>One could have given that last argument as the proof of the orbit-stabilizer theorem, but its true simplicity is much more obvious if you&#8217;ve already experienced it by counting symmetries.</p>
<p>I&#8217;ll probably add to this post in due course &#8212; perhaps giving a list of circumstances where it may be better not to put examples first.</p>
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		<title>My favourite pedagogical principle: examples first!</title>
		<link>http://gowers.wordpress.com/2007/10/19/my-favourite-pedagogical-principle-examples-first/</link>
		<comments>http://gowers.wordpress.com/2007/10/19/my-favourite-pedagogical-principle-examples-first/#comments</comments>
		<pubDate>Fri, 19 Oct 2007 08:03:14 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[Mathematical pedagogy]]></category>

		<guid isPermaLink="false">http://gowers.wordpress.com/2007/10/19/my-favourite-pedagogical-principle-examples-first/</guid>
		<description><![CDATA[This post is about a very simple idea that can dramatically improve the readability of just about anything, though I shall restrict my discussion to the question of how to write clearly about mathematics. The idea is more or less there in the title: present examples before you discuss general concepts. Before I go any [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>This post is about a very simple idea that can dramatically improve the readability of just about anything, though I shall restrict my discussion to the question of how to write clearly about mathematics. The idea is more or less there in the title: present examples before you discuss general concepts. Before I go any further, I want to make very clear what the point is here. It is not the extremely obvious point that it is good to illustrate what you are saying with examples. Rather, it is to do with where those examples should appear in the exposition. So the emphasis is on the word &#8220;first&#8221; rather than on the word &#8220;examples&#8221;.</p>
<p>If this too seems pretty obvious, I invite you to consider how common it is to do the opposite. <span id="more-15"></span>Open a textbook about some general concept in mathematics &#8212; Banach algebras, say &#8212; and the chances are very high that it will start with a formal definition of Banach algebras and only then give you a few examples. I myself became consciously aware of the principle as a result of editing the Princeton Companion to Mathematics: over and over again I found that I could make an article clearer by putting the authors&#8217; well-chosen examples earlier in their discussion.</p>
<p>Why should it be better to do it that way round? Well, if a general definition is at all complex, then you will have quite a lot to hold in your head. This can be difficult, but it is much easier if the various aspects of the definition can be related to an example with which you are familiar. Then the words of the definition cease to be free-floating, so to speak, and instead become labels that you can attach to bits of your mental picture of the example.</p>
<p>By now the alert reader will have noticed that I have not practised what I have preached. So let&#8217;s forget all about the discussion so far and start again, this time doing things properly.</p>
<p><strong>My favourite pedagogical principle: examples first!</strong></p>
<p>Which of the following two explanations do you find clearer and easier to read? They are intended to introduce the concept of a field to a reader who knows what a binary operation is and knows basic definitions such as those of commutativity and identity elements.</p>
<p><em>Explanation 1.</em> A field is a set X together with two binary operations, for which one conventionally uses the notation of addition and multiplication, that has the following properties. Both operations are commutative and associative and have identity elements. Every element of <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' /> has an inverse under the operation +, and every element other than 0 (the name given to the identity of the operation +) has an inverse under <img src='http://l.wordpress.com/latex.php?latex=%5Ctimes&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\times' title='\times' class='latex' /> as well. Finally, we have the rule known as the distributive law: <img src='http://l.wordpress.com/latex.php?latex=x%28y%2Bz%29%3Dxy%2Bxz&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x(y+z)=xy+xz' title='x(y+z)=xy+xz' class='latex' /> for every <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=z&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z' title='z' class='latex' /> in <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' />.</p>
<p>If we interpret + and <img src='http://l.wordpress.com/latex.php?latex=%5Ctimes&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\times' title='\times' class='latex' /> to be the usual operations of addition and multiplication, then we readily see that the familiar number systems <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{R}' title='\mathbb{R}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' /> are fields. (If one goes right back to first principles, then these statements cease to be obvious, but we shall take facts such as the commutativity of multiplication of complex numbers as already established.) By contrast, the number systems <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BN%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{N}' title='\mathbb{N}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Z}' title='\mathbb{Z}' class='latex' /> are not fields, since there is no additive identity in <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BN%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{N}' title='\mathbb{N}' class='latex' /> and not every element of <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Z}' title='\mathbb{Z}' class='latex' /> has a multiplicative inverse. Less obvious examples of fields are number fields (subfields of <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' /> that contain <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />) such as the field of all complex numbers of the form <img src='http://l.wordpress.com/latex.php?latex=a%2Bb%5Csqrt%7B-3%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a+b\sqrt{-3}' title='a+b\sqrt{-3}' class='latex' /> where <img src='http://l.wordpress.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='b' title='b' class='latex' /> are rational, which is denoted <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D%28%5Csqrt%7B-3%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}(\sqrt{-3})' title='\mathbb{Q}(\sqrt{-3})' class='latex' />. (All the field properties are very easily verified, with the exception of the existence of multiplicative inverses: but even that is a simple exercise.) Another important source of examples is the collection of finite fields, of which the simplest cases are obtained by taking a prime p and the set of all integers modulo p. (Here again the only field axiom that is not almost trivial to verify is the existence of multiplicative inverses &#8212; for that one needs Euclid&#8217;s algorithm.)</p>
<p><em>Explanation 2</em>. The five main number systems, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BN%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{N}' title='\mathbb{N}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Z}' title='\mathbb{Z}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{R}' title='\mathbb{R}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' />, though different from each other, have many features in common. For example, if <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' /> is one of these number systems and <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' /> are numbers in <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' />, then one can add or multiply x and y together. One may also be able to subtract <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' /> from <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> or divide <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> by <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' />, but this is not always possible, at least if one wants to stay in the same number system. For example, if we are confined to <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BN%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{N}' title='\mathbb{N}' class='latex' />, then we cannot subtract 5 from 3, and if we are confined to <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Z}' title='\mathbb{Z}' class='latex' /> then we cannot divide 5 by 3.</p>
<p>It is noticeable that there are far fewer problems of this kind in <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{R}' title='\mathbb{R}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' /> than there are in <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BN%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{N}' title='\mathbb{N}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Z}' title='\mathbb{Z}' class='latex' />. In these larger number systems subtraction is always possible (as it is in <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BZ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Z}' title='\mathbb{Z}' class='latex' />), and so is division, provided only that one does not try to divide by 0.</p>
<p>Returning to the properties that these number systems share, we notice that in all five of them addition and multiplication are commutative and associative, and they obey the distributive law: <img src='http://l.wordpress.com/latex.php?latex=x%28y%2Bz%29%3Dxy%2Bxz&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x(y+z)=xy+xz' title='x(y+z)=xy+xz' class='latex' /> for every <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=z&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='z' title='z' class='latex' /> in the number system in question.</p>
<p>A field is a mathematical structure that has the basic properties of the larger number systems <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{R}' title='\mathbb{R}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' />. In other words, it is a set <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' /> on which two binary operations, which we think of as addition and multiplication and therefore denote using conventional notation for these operations, are defined. These operations must both be commutative and associative, and they must obey the distributive law. Also, both operations must have identities, every element must have an inverse under the additive operation, and every element other than 0 (which is defined more formally as the additive identity) must have an inverse under the multiplicative operation as well. Once we have these inverses, we can easily define subtraction and division: <img src='http://l.wordpress.com/latex.php?latex=x-y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x-y' title='x-y' class='latex' /> is the sum of <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> and the additive inverse of <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=x%2Fy&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x/y' title='x/y' class='latex' /> is the product of <img src='http://l.wordpress.com/latex.php?latex=x&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='x' title='x' class='latex' /> and the multiplicative inverse of <img src='http://l.wordpress.com/latex.php?latex=y&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='y' title='y' class='latex' />.</p>
<p>Thus, a field is basically an algebraic structure that &#8220;behaves like <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{R}' title='\mathbb{R}' class='latex' /> or <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' />,&#8221; in the sense that it has two binary operations that obey the algebraic rules that one observes in those number systems.</p>
<p>The concept of a field is important because besides <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />, <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BR%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{R}' title='\mathbb{R}' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' /> there are some less obvious examples that play a central role in number theory. Most notable are number fields and finite fields. The former are fields such as <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D%28%5Csqrt%7B-3%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}(\sqrt{-3})' title='\mathbb{Q}(\sqrt{-3})' class='latex' />, which consists of all complex numbers of the form <img src='http://l.wordpress.com/latex.php?latex=a%2Bb%5Csqrt%7B-3%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a+b\sqrt{-3}' title='a+b\sqrt{-3}' class='latex' /> where <img src='http://l.wordpress.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='b' title='b' class='latex' /> are rational. (In general a number field is a subfield of <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BC%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{C}' title='\mathbb{C}' class='latex' /> that contains <img src='http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BQ%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\mathbb{Q}' title='\mathbb{Q}' class='latex' />.) In a number field it tends to be very easy to verify all the field properties, with the exception of the existence of multiplicative inverses: but even that is usually a simple exercise. The simplest examples of finite fields are obtained by taking the set of all integers modulo a prime p. Here again the only field axiom that is not almost trivial to verify is the existence of multiplicative inverses &#8212; for that one needs Euclid&#8217;s algorithm. <em>End of explanation 2</em>.</p>
<p>I hope very much that you found the second explanation vastly preferable, or that if you didn&#8217;t, then at least you found that it had some quality of speaking directly to the reader that the first explanation lacked. What is the main difference between the two explanations? The content is more or less the same. But there is an important difference in the way that content is organized: in the first explanation the abstract definition of a field is given first and is then followed by some examples, whereas the second starts with the examples (or at least some of them) and uses them as a springboard for a more general discussion. Why should it be an advantage to put the examples first? Well, try to imagine the reaction of a reader who does not know what a field is. At the beginning of the first explanation she [I decided on the sex of the reader by tossing a coin, by the way] is presented with a list that is not related to her previous mathematical experience. Therefore, it is extremely forgettable. Probably she will go on to read about the examples and then look back at the definition to check that it really does apply to them &#8212; a clear sign that the order is pedagogically unnatural. By contrast, if she reads the second explanation then the field axioms are describing something, namely her mental picture of a few fields that she already knows. So she has no need to commit anything to memory or to look back on parts of the text that she has not fully understood.</p>
<p>The comparison may seem unfair because the second explanation was longer, and spent a bit longer discussing the number systems. But that was such a natural thing to do when one started the discussion with the number systems that I think of it as almost a consequence of the policy of putting the examples first.</p>
<p><strong>Back to the top level of this post.</strong></p>
<p>And so, which one of those two explanations of the pedagogical principle did you prefer? I hope very much that you preferred the second. It was certainly much easier for me to explain why I think it is better to put examples first when I had an example to use to illustrate what I was talking about. (I&#8217;m referring here to the theory that the examples give you a mental picture of the concept that allows you to treat the abstract definition as a set of labels attached to concepts you already know rather than as a set of meaningless words with relationships that you just have to learn off by heart.)</p>
<p>When this principle occurred to me, I realized that I had sometimes put it into practice, but I had never been fully conscious of what I was doing. Now that I am, I always either put examples first, or make a conscious decision not to (perhaps because I judge that the reader can cope without). If you are not already conscious of what you are doing in this way, then try thinking about it for a while: if this post does not persuade you, then your own experience surely will &#8212; both of your own writing and of other people&#8217;s.</p>
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		<title>A paper on the ArXiV</title>
		<link>http://gowers.wordpress.com/2007/10/17/a-paper-on-the-arxiv/</link>
		<comments>http://gowers.wordpress.com/2007/10/17/a-paper-on-the-arxiv/#comments</comments>
		<pubDate>Wed, 17 Oct 2007 08:48:46 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
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		<description><![CDATA[Today I did something for the first time ever that I should have done many times before: I put a paper on the ArXiV. Since I&#8217;ve got a blog I thought I&#8217;d use it to give the paper a small plug and, more importantly, to let anybody who might already be familiar with the paper [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>Today I did something for the first time ever that I should have done many times before: I put a paper on the ArXiV. Since I&#8217;ve got a blog I thought I&#8217;d use it to give the paper a small plug and, more importantly, to let anybody who might already be familiar with the paper know that I have revised it quite a bit recently, for the better.</p>
<p>The paper itself is called &#8220;Hypergraph regularity and the multidimensional Szemerédi theorem.&#8221; At the bottom level, the basic idea of the paper is due to Ruzsa, Szemerédi and Rödl. Ruzsa and Szemerédi started the ball rolling with a short and very clever argument that showed that Szemerédi&#8217;s famous theorem on arithmetic progressions, in the case of progressions of length 3, could be deduced from Szemerédi&#8217;s almost as famous <em>regularity lemma</em>, a remarkable result that allows any graph to be partitioned into a bounded number of pieces, almost all of which &#8220;behave randomly.&#8221; <span id="more-13"></span></p>
<p>Rödl then took the further step of noticing that a corresponding proof would work for the general case of Szemerédi&#8217;s theorem if one could generalize the regularity lemma from graphs to hypergraphs. (Whereas a graph consists of pairs of elements from some set, a k-<em>uniform hypergraph</em> consists of k-tuples.) However, turning this observation into a formal conjecture was not completely straightforward because the obvious generalizations of the regularity lemma from graphs to hypergraphs are either too strong to be true or too weak to be useful. Rödl formulated a generalization that avoided these defects (and which, while less obvious, is more natural in a Gromov-type sense).</p>
<p>This led to a clear programme for obtaining a new proof of Szemerédi&#8217;s theorem. Unfortunately, carrying out the details of this programme was much harder than one might have expected given the simplicity of the argument of Ruzsa and Szemerédi. Nevertheless, Rödl, with the help of Frankl, managed to do it for progressions of length 4.</p>
<p>Interest in the problem was heightened when Jozsef Solymosi noticed that Rödl&#8217;s programme would also have the multidimensional Szemerédi theorem as a consequence. This states that, for any finite set <img src='http://l.wordpress.com/latex.php?latex=X%5Csubset%5Cmathbb%7BZ%7D%5Ed&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X\subset\mathbb{Z}^d' title='X\subset\mathbb{Z}^d' class='latex' /> and any <img src='http://l.wordpress.com/latex.php?latex=%5Cdelta%3E0&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\delta&gt;0' title='\delta&gt;0' class='latex' />, there is an <img src='http://l.wordpress.com/latex.php?latex=N&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='N' title='N' class='latex' /> such that every subset of <img src='http://l.wordpress.com/latex.php?latex=%5C%7B1%2C2%2C%5Cdots%2CN%5C%7D%5Ed&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\{1,2,\dots,N\}^d' title='\{1,2,\dots,N\}^d' class='latex' /> of density at least <img src='http://l.wordpress.com/latex.php?latex=%5Cdelta&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\delta' title='\delta' class='latex' /> contains a subset of the form <img src='http://l.wordpress.com/latex.php?latex=aX%2Bb&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='aX+b' title='aX+b' class='latex' />.</p>
<p>A few years ago I managed to complete the Rödl programme, and so, independently, did Rödl, together with Nagle, Schacht and Skokan. Our proofs were interestingly different, however. Their generalization was more directly inspired by Szemerédi&#8217;s regularity lemma as it is usually stated, whereas mine replaced the notion of a &#8220;regular pair&#8221; by that of a quasirandom bipartite graph. For dense graphs, these two notions are equivalent, but when one generalizes to sparse hypergraphs, which necessarily occur in the argument, then they diverge. I came upon the idea of using quasirandomness as a result of thinking about generalizing to hypergraphs some of the analytic notions that had arisen in an earlier proof I had given of Szemerédi&#8217;s theorem.</p>
<p>The story of how the paper reached the form it is in now is an interesting one. While it can be exhilarating to obtain a result that needs a long and complicated proof, it is also pretty depressing to write such a result up. I myself find it almost impossible to do so without introducing errors of what one might call medium seriousness: they don&#8217;t cast doubt on the proof but they do necessitate a boring amount of rewriting. (Why do they not cast doubt on the proof? That is an interesting question and I can do no better than quote Gian-Carlo Rota: &#8220;There are two kinds of mistakes. There are fatal mistakes that destroy a theory, but there are also contingent ones, which are useful in testing the stability of a theory.&#8221; I would substitute the words &#8220;robustness&#8221; and &#8220;proof&#8221; for &#8220;stability&#8221; and &#8220;theory&#8221; but otherwise the second sentence captures very well the idea that small errors have an important positive side to them.)</p>
<p>Something like a year after I put the preprint on my home page, Yoshiyasu Ishigami emailed me to let me know that he had found a fairly serious mistake &#8212; that is, patchable, but not trivially patchable. I was very busy at the time and couldn&#8217;t face thinking about it for a while, but then Terence Tao came up with another proof (more along the lines of that of Nagle, Rödl, Schacht and Skokan but quite a lot simpler) and I realized I had better hurry up with the task.</p>
<p>The result was a big rewriting: when one writes a complicated paper, the last thing one wants ever to have to do is read it over a year later. But if one does, one often finds that one can improve it a lot. In this case, I realized that I could do much better than patching up the old proof: I could give a new and simpler proof. It was based on the same ideas I mentioned earlier, and the same ideas for making those ideas work, but at about the third or fourth level of the hierarchy it was very different. (But I would call it the same proof.)</p>
<p>Once I had done that I thought I had better submit it (something I had not done previously because I had been too lazy to do the long and tedious task of carefully checking it). I can&#8217;t remember exactly when, but quite some time ago it came back from the journal, accepted, but with large numbers of comments from two referees. Another purpose of this post is to thank them very much for the effort they put in. (If you are one of the two referees, then thank <em>you</em>. You have helped to make the paper far better than it would otherwise have been.)</p>
<p>Again I hoped I could get away with minor patches here and there, but this time the biggest problem with the paper was that my &#8220;new and simpler proof&#8221; was clearly not found simple by the referees. And, on rereading it myself after having had enough time to forget most of it once again, I too found it far from easy. Somehow I had not managed to convey the basic simplicity of the argument. So I have ended up substantially rewriting the paper yet again. I have made the notation less neat but easier to interpret. I have also added, at the suggestion of one of the referees, a new section, which turned out to be rather long, in which I illustrated the main arguments on a small example. This was a very good thing to do, not just for the reader&#8217;s sake but also for my own, as it hugely helped me to understand my own argument again and led to a substantial further simplification of the presentation.</p>
<p>In fact, out of the experience comes a tip that I wish I had followed. The main difficulty in writing up the proof was that I had a method that I knew how to apply in any given case, but I needed to describe it in full generality. That meant that I had to formulate a rather complicated inductive hypothesis. That is one of those tasks that &#8220;ought to be routine&#8221; and there is a very strong temptation to proceed as follows: guess the hypothesis, try to prove it, find that the hypothesis isn&#8217;t quite strong enough or general enough, modify the hypothesis, try to prove it, and so on until you finally hack out a proof. Instead, one should proceed as follows: do a small but sufficiently general case and then read off the hypothesis from that; if you&#8217;re not confident that your case is sufficiently general, then do a bigger one. There&#8217;s a kind of laziness that makes one not want to do things the second way, but my experience on this paper has taught me that a bit of effort at this stage can save a huge amount of effort at a later stage. With this paper, it is only very recently that I have gone about things the right way and the result is a small but extremely helpful reorganization of the induction that makes it work far more smoothly and comprehensibly.</p>
<p>The paper, by the way, can be found <a href="http://arxiv.org/abs/0710.3032">here</a>. I&#8217;m not asking anyone to do my dirty work for me, but if you happen to notice an error then I&#8217;d rather know sooner (while the argument is once again fresh in my mind) than later (when the thought of looking at the paper will have gone back to being depressing &#8212; though I hope less depressing now that the paper is better organized).</p>
<p>One final comment: if you look it up on the ArXiV, you will see that the paper is 53 pages long. If that puts you off, bear in mind that a large proportion of that length is not strictly needed for the proof: it is just there to make the proof easier to understand than it would have been if the paper had been half the length.</p>
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		<title>When are two proofs essentially the same?</title>
		<link>http://gowers.wordpress.com/2007/10/04/when-are-two-proofs-essentially-the-same/</link>
		<comments>http://gowers.wordpress.com/2007/10/04/when-are-two-proofs-essentially-the-same/#comments</comments>
		<pubDate>Thu, 04 Oct 2007 09:11:59 +0000</pubDate>
		<dc:creator>gowers</dc:creator>
		
		<category><![CDATA[Somewhat philosophical]]></category>

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		<description><![CDATA[A couple of years ago I spoke at a conference about mathematics that brought together philosophers, psychologists and mathematicians. The proceedings of the conference will appear fairly soon&#8212;I will give details when they do. My own article ended up rather too long, because I found myself considering the question of &#8220;essential equality&#8221; of proofs. Eventually, [...]]]></description>
			<content:encoded><![CDATA[<div class='snap_preview'><br /><p>A couple of years ago I spoke at a conference about mathematics that brought together philosophers, psychologists and mathematicians. The proceedings of the conference will appear fairly soon&#8212;I will give details when they do. My own article ended up rather too long, because I found myself considering the question of &#8220;essential equality&#8221; of proofs. Eventually, I cut that section, which was part of a more general discussion of what we mean when we attribute properties to proofs, using informal (but somehow quite precise) words and phrases like &#8220;neat&#8221;, &#8220;genuinely explanatory&#8221;, &#8220;the correct&#8221; (as opposed to merely &#8220;a correct&#8221;), and so on. It is an interesting challenge to try to be as precise as possible about these words, but I found that even the seemingly more basic question, &#8220;When are two proofs the same?&#8221; was pretty hard to answer satisfactorily. Since it is also a question on which we all have views (since we all have experience of the phenomenon), it seems ideal for a post. You may have general comments to make, but I&#8217;d also be very interested to hear of your favourite examples of different-seeming proofs that turn out, on closer examination, to be based on the same underlying idea (whatever that means).<span id="more-12"></span></p>
<p>A general remark that becomes clear quickly when one thinks about this is that there are fairly standard methods for converting one proof into another, and when we apply such a method then we tend to regard the two proofs as not interestingly different. For example, it is often possible to convert a standard inductive proof into a proof by contradiction that starts with the assumption that <img src='http://l.wordpress.com/latex.php?latex=X&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='X' title='X' class='latex' /> is a minimal counterexample. In fact, to set the ball rolling, let me give an example of this kind: the proof that every number can be factorized into primes.</p>
<p>The usual approach is the minimal-counterexample one: if there is a positive integer that cannot be factorized, let <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> be a minimal one; <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> is not prime, so it can be written as <img src='http://l.wordpress.com/latex.php?latex=ab&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='ab' title='ab' class='latex' /> with both <img src='http://l.wordpress.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='b' title='b' class='latex' /> greater than 1; by minimality <img src='http://l.wordpress.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='b' title='b' class='latex' /> can be factorized; easy contradiction.</p>
<p>Now let&#8217;s give the inductive version. We need the &#8220;strong&#8221; principle of induction. Suppose, then, that we have proved that every integer up to <img src='http://l.wordpress.com/latex.php?latex=n-1&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n-1' title='n-1' class='latex' /> can be factorized and are trying to factorize <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' />. If <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> is already prime, then we are done. Otherwise, we can write <img src='http://l.wordpress.com/latex.php?latex=n%3Dab&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n=ab' title='n=ab' class='latex' /> with both <img src='http://l.wordpress.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='b' title='b' class='latex' /> greater than 1. But then <img src='http://l.wordpress.com/latex.php?latex=a&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='a' title='a' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=b&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='b' title='b' class='latex' /> are less than <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' />, so by induction they can be factorized. Putting together those factorizations gives a factorization for <img src='http://l.wordpress.com/latex.php?latex=n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='n' title='n' class='latex' /> and the inductive step is complete.</p>
<p>(I missed out the proof that the induction starts, just as I did not check in the first proof that there exists a number that can be factorized.)</p>
<p>That&#8217;s a rather boring example of sameness of proofs&#8212;boring because they are so obviously the same, and one can even point to a mechanical procedure that converted one into the other (which can be applied to many other proofs). More interesting are examples where the sameness becomes apparent only after a more complicated process of transformation. At this point, I&#8217;d like to mention the theorem that I discussed in great detail in the bit that I removed from my article: the irrationality of <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}' title='\sqrt{2}' class='latex' />.</p>
<p>Very briefly, here&#8217;s the standard proof. If <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}' title='\sqrt{2}' class='latex' /> is rational, then we can write it as <img src='http://l.wordpress.com/latex.php?latex=p%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p/q' title='p/q' class='latex' />. Let us do so in such a way that <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> are not both even. We know that <img src='http://l.wordpress.com/latex.php?latex=p%5E2%3D2q%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p^2=2q^2' title='p^2=2q^2' class='latex' />, so <img src='http://l.wordpress.com/latex.php?latex=p%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p^2' title='p^2' class='latex' /> is even, so <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> is even, so <img src='http://l.wordpress.com/latex.php?latex=p%3D2r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p=2r' title='p=2r' class='latex' /> for some integer <img src='http://l.wordpress.com/latex.php?latex=r&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='r' title='r' class='latex' />. This gives us <img src='http://l.wordpress.com/latex.php?latex=4r%5E2%3D2q%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='4r^2=2q^2' title='4r^2=2q^2' class='latex' />, so <img src='http://l.wordpress.com/latex.php?latex=q%5E2%3D2r%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q^2=2r^2' title='q^2=2r^2' class='latex' />, so <img src='http://l.wordpress.com/latex.php?latex=q%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q^2' title='q^2' class='latex' /> is even, so <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> is even, which is a contradiction.</p>
<p>Now, equally briefly, here is another proof. Suppose again that <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D%3Dp%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}=p/q' title='\sqrt{2}=p/q' class='latex' /> and let <img src='http://l.wordpress.com/latex.php?latex=p%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p/q' title='p/q' class='latex' /> be written in its lowest terms. Now <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2%7D-1%7D-1%3D%5Cfrac%7B2-%5Csqrt%7B2%7D%7D%7B%5Csqrt%7B2%7D-1%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}=\frac{1}{\sqrt{2}-1}-1=\frac{2-\sqrt{2}}{\sqrt{2}-1}' title='\sqrt{2}=\frac{1}{\sqrt{2}-1}-1=\frac{2-\sqrt{2}}{\sqrt{2}-1}' class='latex' />. Substituting <img src='http://l.wordpress.com/latex.php?latex=p%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p/q' title='p/q' class='latex' /> for <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}' title='\sqrt{2}' class='latex' /> and tidying up, this gives us that <img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7Bp%7D%7Bq%7D%3D%5Cfrac%7B2q-p%7D%7Bp-q%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\frac{p}{q}=\frac{2q-p}{p-q}' title='\frac{p}{q}=\frac{2q-p}{p-q}' class='latex' />. But <img src='http://l.wordpress.com/latex.php?latex=p%3C2q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p&lt;2q' title='p&lt;2q' class='latex' />, so the denominator of the right-hand side is less than <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' />, which contradicts the minimality of <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> (and hence <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> too, since their ratio is determined).</p>
<p>Now there are some similarities between those two arguments: both assume that <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D%3Dp%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}=p/q' title='\sqrt{2}=p/q' class='latex' /> and aim for a contradiction, assuming that <img src='http://l.wordpress.com/latex.php?latex=p%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p/q' title='p/q' class='latex' /> is in its lowest terms. But there are also definite differences. For example, the first proof doesn&#8217;t actually care whether <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> are minimized: it just wants them not both to be even. The second proof doesn&#8217;t care at all about the factorizations of <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> but does care about their sizes. So I&#8217;d maintain that they are different proofs. Having said that, I once put that to Terence Tao in a conversation and he immediately adopted a more general perspective from which one could regard the two arguments as different ways of carrying out the same essential programme. It had something to do with <img src='http://l.wordpress.com/latex.php?latex=SL_2%28%5Cmathbb%7BZ%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='SL_2(\mathbb{Z})' title='SL_2(\mathbb{Z})' class='latex' /> if I remember correctly. Terry, if you felt like reminding me of exactly what you said, that would be a perfect illustration of &#8220;non-obvious equivalence&#8221; between proofs.</p>
<p>Here, though, is a third argument for the irrationality of <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}' title='\sqrt{2}' class='latex' />. You just work out the continued-fraction expansion. It comes out to be <img src='http://l.wordpress.com/latex.php?latex=1%2B%7B%5Cfrac%7B1%7D%7B2%2B%5Cfrac%7B1%7D%7B2%2B%5Cfrac%7B1%7D%7B2%2B%5Cdots%7D%7D%7D%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1+{\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}}' title='1+{\frac{1}{2+\frac{1}{2+\frac{1}{2+\dots}}}}' class='latex' />. Since the continued-fraction expansion of any rational number terminates, <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}' title='\sqrt{2}' class='latex' /> is irrational.</p>
<p>Here&#8217;s a fourth. Let <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' /> be coprime and suppose that <img src='http://l.wordpress.com/latex.php?latex=p%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p^2' title='p^2' class='latex' /> does <em>not</em> equal <img src='http://l.wordpress.com/latex.php?latex=2q%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='2q^2' title='2q^2' class='latex' />. Then <img src='http://l.wordpress.com/latex.php?latex=p%2B2q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p+2q' title='p+2q' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=p%2Bq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p+q' title='p+q' class='latex' /> are also coprime. We also find that <img src='http://l.wordpress.com/latex.php?latex=%28p%2B2q%29%5E2-2%28p%2Bq%29%5E2%3D-%28p%5E2-2q%5E2%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(p+2q)^2-2(p+q)^2=-(p^2-2q^2)' title='(p+2q)^2-2(p+q)^2=-(p^2-2q^2)' class='latex' />. From this observation we can build a succession of fractions <img src='http://l.wordpress.com/latex.php?latex=p_n%2Fq_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p_n/q_n' title='p_n/q_n' class='latex' />, each in their lowest terms, with <img src='http://l.wordpress.com/latex.php?latex=q_n&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q_n' title='q_n' class='latex' /> tending to infinity and with <img src='http://l.wordpress.com/latex.php?latex=%7Cp_n%5E2-2q_n%5E2%7C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='|p_n^2-2q_n^2|' title='|p_n^2-2q_n^2|' class='latex' /> all equal. From that we find that <img src='http://l.wordpress.com/latex.php?latex=%7C2-p_n%5E2%2Fq_n%5E2%7C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='|2-p_n^2/q_n^2|' title='|2-p_n^2/q_n^2|' class='latex' /> has order of magnitude <img src='http://l.wordpress.com/latex.php?latex=1%2Fq_n%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/q_n^2' title='1/q_n^2' class='latex' />, and from that it is easy to verify that <img src='http://l.wordpress.com/latex.php?latex=%7C%5Csqrt%7B2%7D-p_n%2Fq_n%7C&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='|\sqrt{2}-p_n/q_n|' title='|\sqrt{2}-p_n/q_n|' class='latex' /> has order of magnitude <img src='http://l.wordpress.com/latex.php?latex=1%2Fq_n%5E2&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/q_n^2' title='1/q_n^2' class='latex' /> as well. But it is impossible to find a sequence of rationals with denominators tending to infinity that approach a rational this quickly. Indeed, <img src='http://l.wordpress.com/latex.php?latex=%5Cfrac%7Bp%7D%7Bq%7D-%5Cfrac%7Br%7D%7Bs%7D%3D%5Cfrac%7Bps%2Bqr%7D%7Bqs%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\frac{p}{q}-\frac{r}{s}=\frac{ps+qr}{qs}' title='\frac{p}{q}-\frac{r}{s}=\frac{ps+qr}{qs}' class='latex' />, which for fixed <img src='http://l.wordpress.com/latex.php?latex=s&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='s' title='s' class='latex' /> has order of magnitude <img src='http://l.wordpress.com/latex.php?latex=1%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='1/q' title='1/q' class='latex' />. (For large and coprime <img src='http://l.wordpress.com/latex.php?latex=p&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p' title='p' class='latex' /> and <img src='http://l.wordpress.com/latex.php?latex=q&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='q' title='q' class='latex' />, the difference cannot be zero.)</p>
<p>I won&#8217;t demonstrate it here, but it&#8217;s not too hard an exercise to see that the second, third and fourth proofs <em>are</em> all essentially the same. (At some point, perhaps I&#8217;ll put a link to a more detailed account of exactly why, at least for the second and third. The fourth has only just occurred to me.) For instance, the construction of the sequence <img src='http://l.wordpress.com/latex.php?latex=%28p_n%2Cq_n%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(p_n,q_n)' title='(p_n,q_n)' class='latex' /> in the fourth proof is the same as the construction of the continued-fraction expansion of <img src='http://l.wordpress.com/latex.php?latex=%5Csqrt%7B2%7D&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\sqrt{2}' title='\sqrt{2}' class='latex' /> (as lovers of Pell&#8217;s equation will know). Also, the way that we produced <img src='http://l.wordpress.com/latex.php?latex=%28p_n%2Cq_n%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(p_n,q_n)' title='(p_n,q_n)' class='latex' /> from <img src='http://l.wordpress.com/latex.php?latex=%28p_%7Bn-1%7D%2Cq_%7Bn-1%7D%29&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='(p_{n-1},q_{n-1})' title='(p_{n-1},q_{n-1})' class='latex' /> is just the reverse of the way that we produced a smaller fraction from <img src='http://l.wordpress.com/latex.php?latex=p%2Fq&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='p/q' title='p/q' class='latex' /> in the second proof. The fourth proof is perhaps very slightly different, in that it involved inequalities, but that was not a fundamental difference. (It would be interesting to be precise about why not&#8212;this is what I haven&#8217;t got round to thinking about yet.)</p>
<p>To repeat: for the time being I am interested in responses of two kinds. First, I&#8217;d like to see lots of examples of proofs that turn out to be essentially the same when they might not initially seem that way. Secondly, I&#8217;m keen to see examples of &#8220;conversion techniques&#8221;&#8212;that is, methods for transforming a proof into another that is not interestingly different. See <a href="http://gowers.wordpress.com/2007/10/03/the-exchange-lemma-and-gaussian-elimination/#comment-192">this comment</a> for some interesting links, though here I am not so much looking for a formal theory right down at the level of logical formulae. Rather, I would like as good a picture as possible of high-level equivalence of proofs.</p>
<p>Some general questions are quite interesting. For instance, if two proofs are essentially the same, must there always be some more general perspective from which one can see that the only differences between them consist in arbitrary choices that do not affect the argument in an important way? (A simple example of what I mean is something like replacing &#8220;Let <img src='http://l.wordpress.com/latex.php?latex=%5Cepsilon%3D%5Cdelta%2F5&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\epsilon=\delta/5' title='\epsilon=\delta/5' class='latex' />&#8221; by &#8220;Let <img src='http://l.wordpress.com/latex.php?latex=%5Cepsilon%3D%5Cdelta%2F3&amp;bg=ffffff&amp;fg=000000&amp;s=0' alt='\epsilon=\delta/3' title='\epsilon=\delta/3' class='latex' />&#8221; in an argument in real analysis. But there are much more interesting examples of this.) Also, is &#8220;essential equivalence&#8221; really an equivalence relation, or could one morph in several stages from one proof to another that was &#8220;genuinely different&#8221;? (My own feeling, by the way, is that the morph would itself be a demonstration of essential equivalence, but perhaps a really good example might change my mind on that.) Is it ever possible to give a completely compelling argument that two proofs are genuinely different? How would one go about such a task? Could one attach an invariant to a proof, perhaps?</p>
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