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Well, welcome everybody to the first lecture on Alan Turing on computer ability and intelligence.

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I'm Peter Milliken. As you see at Harvard College and as most of you probably know, I'm quite keen on computer science and philosophy,

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which is the degree programme towards which these lectures around.

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Anyone here who's not doing computer science philosophy? You're very welcome to.

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It's a very interesting topic, and I hope you enjoy the lectures.

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As you'll be aware, they're being recorded this year. That's so that they're available in the future.

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So for that reason, I'm not going to be taking any questions during the lectures.

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But if any problems arise, I will be organising classes during the term for anyone who needs them.

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Okay, those doing computer science philosophy later in the term will be having tutorials on the stuff to do with philosophy of mind.

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And so when you'll be writing essays and I'll give you details about that later.

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The role of the course is part of the introduction to philosophy for the prelims course in computer science and philosophy.

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And just to get this out of the way, you have a three hour examination and it contains six either all questions on general

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philosophy and six questions on curing from which you are expected to choose for.

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And you have to do at least one from each section, so you could end up doing one question on curing.

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You could end up doing three. In practise, I have to say computer science philosophy students tend to do more Turing papers than gentle questions,

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so you'll probably find that the course is quite fruitful from that point of view.

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Here's the overall plan of the lectures. Today we'll be talking about types of number about cantor infinities diagonal argument.

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Next lecture We're going to be talking about Hilbert programme and girdles theorem.

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We won't be going through Girdles Theorem in all detail, but we will be seeing how it works.

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The ingenious method by which girdle proved his result, and by then, you'll understand why it's important.

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Lecture three to five Focus on the the really main core of the lectures,

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which is Alan Turing's 1936 paper on computable numbers with an application to the incidence problem.

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Now that was the paper that introduced the Turing machine and hence virtually invented the computer.

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We have a set text for that, which makes it feasible to do this.

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It's a fiercely difficult paper, but with the help of this book by Charles Petzold, you should, I hope, find it entirely manageable.

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Those doing computer science and philosophy, it's probably a good idea to buy this book.

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The reason is that you might very well find as you go through it, that you want to make notes and highlights and so forth on it.

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And of course, you shouldn't be doing that with library books. Okay, so that's lecture three to five.

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And then in the last three lectures, we'll be looking at Alan Turing's other very famous paper,

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the 1950 paper computing machinery and intelligence, and that's the paper that introduced the Turing test.

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So that's where we get into stuff from philosophy of mind and so on.

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Okay, so I've mentioned the textbook. Here are the details about it.

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It really does contain everything you need to know on the background of the 1936 paper that it contains the

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actual text of the paper if you see highlighted here in grey so that you can see exactly what Turing wrote.

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But it's surrounded by a lot of commentary, which makes it all comprehensible, explains difficulties, explains corrections and so forth.

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What I'll be doing in these lectures is giving you, I hope, enough material that will make that paper much more comprehensible.

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Even with petzold help, you might find it difficult sometimes to see the wood for the trees in the lectures.

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I'm going to be focussing on bits that are particularly difficult and setting the background for that.

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OK? Highly recommended reading a couple of books by Andrew Hodges, which are biographical books about Turing,

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but obviously also contain quite a lot of stuff that's relevant to what we're going to be talking about.

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His mammoth biography from 1983 and very well known.

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If you want to read something during the long vacation, maybe that's a good idea.

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A much shorter book, just 58 pages is very well worth reading to get a feel for what Turing was about,

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and I'll be mentioning this book, which is a huge collection on Turing remarkable value.

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It's actually surprisingly cheap for a book of over 900 pages.

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So if you're a real enthusiast and you want an absolute doorstop of a book, that's worth considering that you should.

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Most of you have it in college libraries. And if you don't get them to order it and this small book, it's a classic book on girdles proof.

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I'll be mentioning next lecture. Okay, so who was Alan Turing?

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Well, he was born in 1912, educated at Hazelhurst and Sherburne schools.

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His parents were in India, so he was at boarding school.

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He formed a close relationship with another lad at the school called Christopher Malcolm.

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And both of them applied to read maths at Trinity Cambridge.

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Malcolm was accepted. Turing wasn't. But the next year, Malcolm died of tuberculosis.

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And it seems they had a very intimate affair. Alan Turing was gay,

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losing his intimate friend had a profound effect on him and got him thinking a lot seems about the nature of mind where the mind can survive death,

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the relation between mind and body and so forth.

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I mean, so some of this is speculative, but it seems very plausible later in his life, as is well known.

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He played a huge role in the war in the Second World War.

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He was largely responsible for the techniques by which the Nazi Enigma codes were decrypted,

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thus enabling allied shipping to escape the U-boats in the Atlantic without cheering.

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It's entirely possible that the war would have gone on a lot longer and been a lot more

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difficult to win because without being able to divert the ships around the U-boat packs,

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Britain's position would really have been quite perilous in the early 1940s.

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This was completely unknown until relatively recently because it was all official secrets,

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and Turing was prosecuted for homosexuality in the early 1950s.

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And again, nobody knew what a hero he was.

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He was forced to take hormone treatment, which had very bad effects on him, and he ended up probably committing suicide in 1954.

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He seems to have set it up in such a way that there was a little bit of ambiguity.

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He he apparently took a bite out of an apple on which he'd committed suicide cyanide.

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But it is possible that he killed himself by mistake.

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So a plausible suggestion is that he deliberately set it up that way so that his

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mother would be able to believe that he hadn't actually killed himself deliberately.

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Anyway, back to his early life, he reapplied to Cambridge.

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In 1930, he was accepted for King's College, went up in 1931.

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He took his part to try pass exams in 1934, passed with distinction and got a research student ship.

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Then the next year, he was elected to a three year fellowship, so he obviously did extremely well in maths at Cambridge.

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By 1935, he was a fellow and he attended a Part three course given by Max Neumann on the foundations of mathematics.

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And this, again, is something that seems to have had quite a profound effect on him to understand why we

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need to look a bit at one of the greatest mathematicians of the early 20th century.

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David Hilbert Hilbert was born in Konigsberg, the city with which Immanuel Kant is greatly associated.

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He was perhaps the most influential mathematician of the early 20th century, partly because he set the agenda.

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He made a point of saying, Here are big problems on which we should focus.

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He also developed the formalised approach to philosophy of maths. So those of you who study philosophy of maths you will come across, you know,

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the the politeness, the approach, the intuition is the approach, the formalist approach.

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And Hilbert is very much a founder of the last of these, the formalised approach CS mathematics as involving manipulations of symbols.

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So we discuss proofs and things in terms of their syntactic properties, and we'll see that playing a role later,

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especially next time at the 19:00 Paris International Conference of Congress of Mathematicians.

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He famously listed prominent unsolved problems of mathematics, including this one,

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the tenth problem determination of the solve ability of a Die-Off and time equation.

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Now I put this in, partly because Petzold, in his book,

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discusses these at some length and very interestingly, so I recommend that you go and look at his book.

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The question is, suppose you have an equation which has a number of unknowns and with integer coefficients, can there be a solution to that equation?

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And trying to devise a process by which you can tell whether there is a solution or not is a major problem.

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The the equations were named after an Alexandrian mathematician, Dave Fanta.

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The most famous example of a die Fantine equation is associated with them as last Theorem X to the end, plus Y to the end.

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And does this have any integer solutions when N is greater than two?

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I mean, it clearly does when any equals one.

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It clearly does. When any calls to, for example, three squared plus four squared equals five squared.

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Does it have any when an equal three or four or more than that?

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And for many years, this was a huge puzzle.

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The great French mathematician Thelma famously wrote in the margin of one of his notebooks that he had discovered a wonderful proof of this theorem,

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but I don't have room to write it down in the margin. And this intrigued future generations of mathematicians.

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They all tried to discover or rediscover this proof that Thelma had supposedly discovered.

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It actually took a huge amount of work by lots of mathematicians and eventually was proved by Andrew Wiles in 1995.

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I think most people think that Sam probably found an ingenious proof that applied to certain cases,

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but not to all because proving it as a general theorem is actually extraordinarily difficult requires very sophisticated mathematics.

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OK, let's have a quick look at some of Gilbert's other problems, apart from the one concerning Di Fantine equations.

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We've got the continuum hypothesis that's a very important problem in philosophy of mathematics.

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Again, here on, I'm not going to discuss it, but I'm just going to refer you to Petzold.

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That's something he discusses and is well worth knowing about proving the

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consistency of the axioms of arithmetic that something will be coming back to,

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particularly next time. The Riemann hypothesis and Goldmark conjecture gold, but conjecture is a favourite of philosophers.

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It's the claim that any even number greater than two can be expressed as the sum of two prime numbers.

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Philosophers love it because it hasn't yet been proved.

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So when I started out studying maths at Oxford quite a long time ago, there were three of these very famous unsolved problems,

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and one of them was the full-color theorem, which got proved the subsequent year.

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One of them was Fermat's Last Theorem, which, as I say, was proved in 1995.

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I'm hoping that Girlboss conjecture will last a bit longer so that we philosophers can continue to give.

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It is a nice example. OK.

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Let's come to here. But the challenge is that Hilbert set, he set out a programme in philosophy of mathematics.

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He wanted to establish foundations for mathematics, which were provably consistent.

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So starting from axioms which were ideally self-evident, clearly true and manifestly consistent.

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So even if you can't necessarily see immediately that they're all true, you want to be able to prove that they all hang together.

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There's no inconsistency there and a set of axioms from which in principle,

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all mathematical truths can be deduced by the standard rules of First Order predicate logic.

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So I assume you've all learnt, first of all, the predicate logic, standard logic that we inherited from Fraga.

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Is it possible to get a set of basic axioms from which by those rules, you can deduce all of mathematics or at least all of arithmetic?

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A related problem, the incidents problem,

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the problem that Alan Turing was going to be concerned with the decision problem, can an effective procedure be devised,

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which would demonstrate in a finite time with any given mathematical proposition is or is not provable from a given set of axioms?

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OK, so we want these axioms. We want them to be provably consistent and complete, such that all of mathematics can be deduced from them.

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We also want to have a procedure which in a finite time is going to tell us as it were,

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mechanically simply by applying this procedure in a mechanical way to tell us

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whether any given proposition is or is not provable from the set of axioms.

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Okay. Some of the relations between those will become much clearer next time when we talk about girdle.

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The 1936 paper, which is said, the sort of central topic of these lectures.

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Second, the decision problem, the entitlements problem.

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It showed, in fact, that there could be no such procedure.

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But the paper starts, as the title suggests, from the concept of a computable number.

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So then the paper the ultimate aim of Turing's paper is to settle the incidence problem.

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It gets there by talking about a certain kind of number, a computable number.

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Now, in order to understand what those are and how they fit in with everything else, we're going to need to talk first about types of number.

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So that's actually going to be the main focus of the rest of this lecture. In the next lecture, as I've said, we'll move on to girdle.

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We'll talk about completeness and consistency.

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And then by the third lecture, we'll have a good framework for understanding where Turing's paper comes in and and why it takes the form it does.

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Okay, so what we we're going to look at various types of number and find interesting results about some of these sets of numbers.

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So here are some set standards used in mathematics the set of natural numbers one two three four You're studying computer science.

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You've probably been told the first natural number zero. I'm going to take the first natural number to be one because zero.

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Isn't that natural? Is it real? OK? It's actually pretty harmless for most contexts.

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Whether you take the natural numbers are starting from zero or one.

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We're going to go with one the set of integers from Xylem and German of the symbol Zaidi's commonly used for that.

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The set of integers obviously is both positive and negative, and zero the set of rational numbers.

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A rational number is simply a fraction of integers, so you've got an integer on the top, you've got an integer on the bottom.

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Obviously not zero on the bottom. And that set is commonly denominated Q, probably for quotient.

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And then there are real numbers, say R, and real numbers essentially correspond to to all the non complex numbers,

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all the numbers that we normally handle when we're doing things like calculus or trigonometry and so forth.

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Those are all real numbers now. Real numbers include irrational numbers, numbers that cannot be written as a fraction of integers,

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algebraic numbers which are roots of algebraic equations and transmittal numbers, which on either of those?

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OK, I'll be coming to those a little bit later in the lecture. We're not going to be talking here about the set of complex numbers.

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Some of what we do could apply equally to complex numbers, but they're not really relevant to the business here.

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OK. Now, these are obviously all infinite sets.

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There's an infinite number of natural numbers. There's an infinite number of rational numbers and so forth.

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But what Georg Cantor did was show that not all of these infinities are the same.

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We're going to end up with more than one type of infinity.

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So that's the theory of Tehran's finite cardinal numbers.

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And the key to understanding how this works is a principle that's to my mind, somewhat ironically known as Hume's principle.

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As you probably know, I'm rather a fan of David Hume, but it has to be said that philosophy of mathematics wasn't exactly Hume's strongest point.

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So there's an irony in this absolute key principle in the philosophy of mathematics being named after him.

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But the reason is that Fraga, in foundations of arithmetic, pointed out that Hume had enunciated this principle when two numbers are so combined,

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as the one has always a unit answering to every use of the other, we pronounce them equal.

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OK, so let's see what's being said here.

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The idea is that if you have two sets of objects and you can put those two sets in one to one

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correspondence so that each unit in one of the sets corresponds to one and only one unit in the other set,

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then you can say that the two sets are equally numerous.

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They have the same number of elements.

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Mathematically, we well, first of all, we use a technical term called finality for the number of elements in a set.

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And this is meant to be a generalisation so that we can extend it to infinite numbers, not just finite.

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So two sets of the same card inanity if and only if Abhijit two function were a one to one, correspondence can be defined between them.

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So let's just get clear what objective function is a function that is both injected and surge active.

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So and this is an example of an injected function here.

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We've got the domain here, we've got the code domain and you can see that this object is the function of this.

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This is the function of this or, as we say, the image of it.

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And you can see that there is no object in the code domain, which is the image of two or more objects in the domain.

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You never get two arrows going to the same point. OK.

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That's an injected function. Here's an example of a subjective function.

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Here you can see this one is not injected. We sometimes do have two arrows going at the same point.

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But there is no object within the code domain, which is not hit by at least one arrow.

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Got that? Now, do you agree that if we have an injected function, if not, we never have two arrows going to the same point here,

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then this set must have at least as many elements as this one.

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Yeah. Because there has to be a separate element for every act.

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Whereas if we have a subjective function, that means every element here gets hit by at least one arrow,

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then this set must have at least as many elements as that one agree.

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So if we have objective function, a function that is both injected and surgically, if a one to one correspondence,

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it follows that the two sets must have the same commonality, the same number of elements.

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Now, essentially, what we do when we talk about infinite sets is we take that principle, which is so natural with finite sets,

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and we extend it and apply it to infinite sets and cantors were basically comes from doing that.

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OK, now when we're dealing with infinite sets rather than attempting to define objection,

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we normally make do with a search action, a search active function if we can define a function from set to set B, which is A.

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In other words, every element of B gets hit by an arrow.

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Then we can say that the commonality of pay must be at least as great as the code Nancy at the OK.

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And that's that's normally enough for our purposes.

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We don't go to the trouble of trying to define objection a one to one correspondence that in many cases,

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would be a lot more complicated and unnecessarily so we'll see that quite shortly.

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Now, clearly, if you can define a subjection both ways, then on Hume's principle, the two commonalities should be the same.

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OK, now one very important special case, if a suggestion can be defined from the set of natural numbers to any set,

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then we say that that stuff is countable or innumerable, in other words.

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So we've got a function that goes from the natural numbers. Each natural number is mapped onto an element of this set.

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Every single element of that set is hit by at least one natural number, at least one of these arrows from a natural number.

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Right. So you can imagine in principle,

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writing down the elements of this set in order start according to the natural number which mapped onto them the first, the second or the 4.0.

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Now, some of those elements may occur multiple times in the list.

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That doesn't matter. We've said we're going to make do with a surge action.

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But every single element of that set should be there somewhere in the list if we have correctly defined as objective function.

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So we can enumerate the list in the sense of listed in principle because it's infinite.

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So we can't actually in practise, but in principle we can. OK.

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I've already thought about that. So one of the results that Cantor proved, rather surprisingly,

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is that the set of rational numbers and the set of algebraic numbers are both innumerable.

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So we're going to see how that works. OK, a rational number is a number that can be expressed as a fraction of integers expressed as a decimal.

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If you take a decimal number, a number is rational if and only if that decimal eventually recurs.

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So, for example, three point not not not not not not more recurring.

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That's the rational 7.5 no more recurring nought point six 666, etc. nought point three six seven three six seven three six seven.

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All of these are rational numbers. Now, it's actually pretty easy to show this.

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Take that last example nought point three six seven three six seven three six seven,

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etc. You can see that that is actually the sum of the geometric series.

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OK, you've got 367 over a thousand. That's that Mate Plus three, six seven over a million.

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That's that plus three six seven over a billion, which is that the and so on.

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So you've got a series in which each each item in the series is 1000 less than the item before.

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And you can easily that some geometric geometric progression using this formula, the first element divided by one minus the common ratio.

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And that means that some of that will be 367 over 9-9-9.

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OK. So any recurring decimal basically is going to be a rational number.

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When we're talking about irrational numbers, we're talking about decimals that go on and on forever, but do not actually recur.

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OK, now you might then think that the number of rational numbers will be hugely greater than the number of integers taken at the set of integers.

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Let's just talk about positive integers for the moment.

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One two three four five six seven And so between one and two, there are zillions of rational numbers, right?

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One followed by any proper fraction you get a mention is going to be a rational number.

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You know, one and seven twelfths or whatever is going to be a rational when you've got all these huge number of rational

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numbers between one and two and then you've got the same number between two and three and three and four.

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How can it be as Cantor claims that the set of rational numbers, all rational numbers has no more elements than the set of integers?

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Well,

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let's apply whose principle and very famous argument we set out all the rational numbers in an infinite array of clearly infinite in both directions.

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OK. So you can see that the first row has a denominator of one.

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The second row has a denominator of two. Third row has to nominate the three, and so the numerator is are in the column.

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So the first column has a numerator of one.

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The second column has a numerator to the third column as a new range of three.

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And so every fraction of positive integers is going to occur somewhere in that array.

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Agreed. And moreover, whatever fraction you care to name, it's actually pretty easy to say where in that array it look.

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So there's no mystery here. OK, now what we do, we put these fractions in a sequence starting from here.

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Then that one, then that one in that one, then that one in that one following the red line.

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And you can see that if we carry on in that pattern sooner or later, every single fraction in that array will be taken.

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Agree. So in and again, it's not that difficult actually to work out a formula which will tell you at what point in the sequence it will arise.

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So we've defined a suggestion, a suggestion from the natural numbers to the fractions.

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I've not spelt it out, but it's implicit here.

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You can see f of one is one over one effort to two over one after three is one half so that the F function is

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simply taking any natural number and mapping it to whichever fraction occurs at that point along the the red line.

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So it turns out you can put all the rational numbers in a list.

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An infinite list, of course, but that means there are no more natural numbers, so no more rational numbers than there are natural numbers you can pet.

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You can define this objection, so we list all of the fractions.

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And that's the start of it. Our suggestion maps one to the first of these two to the second three to the third.

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As we've seen. A curious fact about this is that every single fraction in that list occurs not just once, but an infinity of times.

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I mean, consider the fraction one over three. That's the same as two over six three overnight for over 12.

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So you've got this amazing result that that a list in which every single possible fraction

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occurs an infinity of times still contains no more elements than the number of natural numbers.

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You might be wondering what happens with negatives? Well, if you want to include negative fractions, you just alternate.

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Okay? It just means the list will be spaced out twice as much.

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OK, so a natural thought at this point would be, well, maybe it's not so surprising.

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OK. I mean, infinity is infinity. You've got an infinite number of natural numbers, an infinite number of rational numbers.

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Why should it be a surprise that these come out to be the same?

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Well, to see the impact of countless result, we need to look at irrational numbers, and I'm just going to introduce the most famous of these,

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perhaps it's a beautiful proof, I think really quite elegant, and I think it dates back to Euclid.

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Square two cannot be expressed as a fraction of integers. The square root of two is not a rational number.

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Amazingly, it's very easy to prove. Suppose that the square root of two is equal to m over n where N and N integers with no common factors.

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Now, do you agree? But if f if Root two can be written as a fraction of integers,

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then it is possible to write it as a fraction of integers in which the numerator and the denominator have no common factors, right?

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You just cancel them. OK. Square both sides of this and you get two equals n squared over in square.

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Multiply both sides by n squared. You get to square and then square.

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So it follows that and square. There's an even number. Yeah.

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Now, if the square is even the square root must be even two because any old number has an odd square.

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If you take an old number, we're talking about integers here, say to K +1 and you square it,

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you get for K squared plus Fourcade plus one, which is creating an odd number.

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So if you've got an even square, you must have an even number that's being squared.

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So it follows that n we had n squared is even now we know that Emmys even but his name is even n can be written as two K for some K integer.

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And that means that m squared is four k squared and we know that m squared is to n squared.

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It follows then that n squared is going to be even as well, from which it follows by the previous reasoning.

284
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The end is even so, we end up with the conclusion Emmys even and then is even which is a contradiction

285
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because we started with the assumption that men have no factor in common.

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Turns out they have to have two income.

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So if we start with the assumption that the square root of two can be written as a fraction of integers, we end up contradicting that assumption.

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Therefore, Root two is not a rational number.

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OK, so that shows that the fractions of integers do not exhaust all the numbers, there are numbers like root to which cannot be written in that form.

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But the square root of two is at least an algebraic number.

291
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The square root of two is a number that can be written as a solution of an equation in one variable with integer coefficients.

292
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Right. So x squared equals to the solution.

293
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Root two is called an algebraic number.

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It's not rational, but it's algebraic. An irrational number is straightforwardly algebraic and over.

295
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B is the solution of the equation. B X equals a any square root or n through to the rational number is algebraic.

296
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As we've seen in the case of square root of two. So we've been forced to acknowledge that there are numbers beyond rational numbers.

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Not every number can be written as a fraction of integers.

298
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OK, let's accept then, that there are these other numbers which are solutions two equations with integer coefficients.

299
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And will that suffice? Well, maybe it will.

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Maybe it won't. We'll come to that shortly. But let's show for good measure that the algebraic numbers are innumerable numeral.

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So just like the rational numbers, we can actually put all the algebraic numbers, which include the rational numbers into a list to prove this.

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Let's suppose we got an algebraic equation. Here's a general form of an algebraic equation.

303
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And what are we going to do? Is define the rank of the equation as this.

304
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OK, now don't worry about this. All right, just just take this as a useful way of listing all these equations.

305
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So what we want is a number which indicates, as it were, how big the numbers are in this equation.

306
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So what we're doing, we taking them the constant turn and taking it to absolute value so we don't care whether it's positive or negative.

307
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The positive value of the coefficient of X, the positive value of the coefficient of X squared, etc., etc.,

308
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etc. We're adding the absolute value of all the coefficients together and then adding to that the highest degree of X.

309
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OK. That gives us a number. It's clearly going to be a positive number.

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Yeah. Very sensible equation can't be negative because we're taking absolute advantage of time.

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Now here's the crucial thing. Suppose you take all the equations that have rank three.

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Say there will be a finite number of those and you can list what they are.

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So and we can order them first by end, then by the various coefficient.

314
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So this is easiest understood if we take examples. So these equations have rank two because this has coefficients minus one and zero.

315
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So the absolute value of minus one is one at zero.

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We get one and the highest degree of X is one, so we get one plus one to rank two.

317
00:38:31,820 --> 00:38:36,980
This has rank two as well. Those are the only two equations of rank two, right?

318
00:38:36,980 --> 00:38:42,050
One of them has a coefficient of minus one before x one has a coefficient of one.

319
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For equations of rank three, there are more of those, including a couple of quadratic.

320
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In order to get rank three here, since the highest degree of X is two, we're only allowed one as the coefficient, so it better be minus one or one.

321
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So those are the only two quadratic that get rank three.

322
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But you can see we're getting more of the linear equations.

323
00:39:10,730 --> 00:39:22,230
You get the idea. So, so the rank simply measures as it were the size of the coefficients of the equation linked with the degree of the equation.

324
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And I haven't. I've just started off with rank four. OK.

325
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So we've got a way of listing all the algebraic equations.

326
00:39:35,360 --> 00:39:43,880
First of all, we list those with rank two, then those with rank three, then those with rank four, then between five and within each group.

327
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We list them in the order just specified, awarded first by and then by and then by eight minus one and so on, and finally by a zero.

328
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OK, so imagine we've got this infinite list of all the all the possible algebraic equations.

329
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Now, every algebraic equation has it most in solutions.

330
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So if any is the highest power of X in the equation, you know, quadratic equation can have at most two solutions.

331
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Imagine solving all these equations, those that are solvable at any rate and listing the roots of the equation in numerical order.

332
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I've mentioned here you could put complex numbers into this if you want, but we're not thinking about complex numbers here.

333
00:40:33,860 --> 00:40:42,770
So now we can make a list that will include all the algebraic numbers because we've got this list of all the algebraic equations.

334
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And then for each of those equations, in order, we list all of the roots of that equation again in numerical order and every single algebra number,

335
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every number, which is a root as of any one of those equations will appear somewhere in that infinite list.

336
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So as before, we can define a suggestion and QED quod erat demonstrated to me proves what we set out to prove.

337
00:41:09,650 --> 00:41:18,950
You can actually produce an infinite list of algebraic numbers, which contains every single algebraic number.

338
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So there are no more algebraic numbers than there are integers.

339
00:41:25,070 --> 00:41:31,280
OK, so again, you might think maybe this isn't a surprise.

340
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Infinity is infinity.

341
00:41:33,230 --> 00:41:44,120
Maybe all infinite sets are of the same size, but now we come to transcendental numbers, and these are irrational numbers that are not algebraic.

342
00:41:44,120 --> 00:41:53,360
Okay, piety of the most famous examples. But if trigonometric functions often yield them as well.

343
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Notice, by the way, that when we're dealing with trigonometric functions,

344
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we standardly think of the angle in radians, not in degrees, so that that fundamentally changes things.

345
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OK. So there are transcendental numbers.

346
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There are numbers that are not roots of a algebraic equations.

347
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Surprisingly, although Pi and E may be the only familiar ones or the most familiar ones.

348
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It turns out that to a rough approximation, all numbers are transcendental.

349
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So this is due to a proof by Cantor, which we now going to see famous dive in the truth.

350
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So Cantor showed that real numbers in general cannot be put into an infinite list.

351
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Rational numbers can, algebraic numbers can, real numbers can't.

352
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And again, we've got a proof by reductio out absurdum.

353
00:43:00,700 --> 00:43:07,810
We start with the assumption that they can be put into a list and then we show that that's impossible.

354
00:43:07,810 --> 00:43:14,860
So let's suppose that the set of real numbers were in fact innumerable could be put into an infinite list.

355
00:43:14,860 --> 00:43:22,270
In that case, it would A44 why it would be true that all the real numbers between zero and one could be put into an infinite list.

356
00:43:22,270 --> 00:43:28,810
Yes. Let's just restrict our attention to that small subset of the real numbers between zero and one.

357
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Ignore all the infinite others. Imagine that they have been put into an infinite list.

358
00:43:36,730 --> 00:43:40,120
So here's the list. That's the first of them.

359
00:43:40,120 --> 00:43:49,630
Second, third, fourth, fifth and so on. And one is the first digit after the decimal point of the first number in the list.

360
00:43:49,630 --> 00:43:54,700
8.4 is the fourth digit off the decimal point of the second number in the list.

361
00:43:54,700 --> 00:44:02,320
And so everyone happy with that. So now look down the long diagonal there.

362
00:44:02,320 --> 00:44:06,400
What I've done, I've chosen the first digit of the first number in the list.

363
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The second digit of the second number of the list. The third digit at the third number in the list and so on.

364
00:44:11,860 --> 00:44:22,090
What we're now going to do is construct a number, a decimal number, which differs from each of those.

365
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So in the fourth place, we're going to choose a digit, which is different from eight four four in the fifth place.

366
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We're going to choose a digit different from a five, five and so on.

367
00:44:36,370 --> 00:44:43,780
Do you agree that if we do that, we will end up with a number which differs from every single number in the list

368
00:44:43,780 --> 00:44:51,820
because it will differ from the number in the eighth place after the decimal point?

369
00:44:51,820 --> 00:44:57,790
OK, so we can define this number, we can call it number C in honour of Cantor.

370
00:44:57,790 --> 00:45:06,670
Let's just say that if the end digit of the ninth number in that list is five,

371
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then the digit of our specially defined number is going to be made zero.

372
00:45:12,030 --> 00:45:19,060
But otherwise, we're going to make it five. So we're going to we're defining a number here.

373
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Every single digit is either going to be zero or five.

374
00:45:22,840 --> 00:45:32,860
But we fixed it so that if the f number in the list happens to have a five at the crucial point, we have a zero.

375
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And if that number has a zero, we have a five. In fact, if it has anything else, we have a thought.

376
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So the number that we defined is going to differ from the entry number in the list at the ninth decimal point.

377
00:45:46,170 --> 00:45:54,000
So effectively what we're doing, we're writing out a number where if you go to say the fourth digit of that number,

378
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you can be quite sure that it will be different from the fourth digit of the fourth number on the list.

379
00:46:00,450 --> 00:46:07,560
It will therefore in the same way it will differ at some point from every single number in that list.

380
00:46:07,560 --> 00:46:12,930
But we started off with the assumption that every single real number is in that list.

381
00:46:12,930 --> 00:46:18,930
We've just defined a number which a real number, which is not in that list.

382
00:46:18,930 --> 00:46:28,220
So we've got a contradiction. This is a for obvious reasons, called a diagonal argument.

383
00:46:28,220 --> 00:46:35,480
It's an argument that works by taking two dimensions of variation and providing some construction that runs down the diagonal.

384
00:46:35,480 --> 00:46:42,080
We're going to be looking at some other diagonal arguments. A very famous one is the proof of cantors theorem.

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For any set, the powerset of a that is the set of subsets of a always has a greater carnality than a itself.

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So suppose, for example, you've got three people, how many different sets of those three people can you make?

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Well, the first person can either be in or out of the set.

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Yeah. So that's two possibilities for each of those possibilities.

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The second person can be in or out of the set. So that gives you two times two.

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And for each one of those possibilities, the third person can either be in or out of the set, which gives you eight.

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Yeah. So you get eight subsets. So in general, with a finite set, if you've got an elements,

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the number of subsets is two to the end because basically every new element doubles up the possible subsets that you can have.

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Bear in mind, the null set is a subset of the set itself is a subset. OK.

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So with finite sets, it's very obvious that the set of subsets in this case, the set of those eight subsets has more elements than the initial set.

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The set of three is that true for infinite sets?

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Will Cantor showed that it was the very nice proof, so let's suppose, first of all, as in innumerable sets.

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In other words, it's a set where we can list all the elements of the set in a list, an infinite list fine.

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But those are the elements of the set nested along the top.

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And let's suppose that we can define a subjective function from each element of that list,

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which which hits as it were every single subset of that set.

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So if a value one will be some subset of that set and it will be the one to which A1 is mapped by the suggestion.

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So we've got a subjective function. It maps each element of the set onto some subset of the set, and it's a subjective function,

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which means that every single subset gets hit at least once.

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Is that OK? Now imagine then that we set up a table like this where here we've got the subset that to which A1 matches.

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Here's the subset to which a two maps. Here's the subset to which a three maps.

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And so and we've put a tick or a cross to indicate which elements of the set are actually in that subset.

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OK. Don't imagine there's any authority in the ticks and crosses I've put here.

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It's entirely arbitrary. OK? I've just said, let's suppose that the first four to have this pattern, they might have a completely different pattern.

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It doesn't matter what we're we going to do is go down the diagonal,

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look at that diagonal and suppose we consider the set of elements that have crosses down the long diagonal.

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OK, that's in this case. That's an A3 and A4.

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They're the ones that have crosses down the long diagonal.

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In other words, those are the ones that do not belong to the set to which they are mapped by our surge action.

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Imagine that we take all of those elements which have a cross in the long diagonal of the former subset of them,

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so we form the set of all the elements which are not contained in the subset to which they are mapped.

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By similar reasoning to cantors, diagonal of the the diagonal arguments about real numbers,

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and we have thus constructed a subset which is not present in the list.

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So again, we get a contradiction. It is not possible to have a subjective function from an infinite set when the any set of innumerable,

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any innumerable set of objects onto the subsets of that set.

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That's quite tricky. This is something that to it's worth thinking about.

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You know, the the text of the slides actually explains it.

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If you find it difficult as some eyes suggest you do, which isn't surprising to try with some examples with small sets, see how they work out.

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See how you fill in the grid. And that will probably help.

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Now, in fact, the proof of counsel's theorem doesn't rely on innumerable.

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Here we did.

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I mean, the only I was only able to set up this grid on the assumption that our objects here, the set of a was innumerable, but quite generally,

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where a is any set at all innumerable or not,

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we can think of the set of elements of a which are not elements of the subset to which they have been mapped.

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And in fact, we get exactly the same kind of contradiction.

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It's this this subset, the set of elements which do not belong to the subset to which their map cannot be in the range of function f,

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because if X is fixed for any X, then if X is it a subset is an element of X.

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It turns out that X also cannot be a sub in an element to this.

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So this implies that the power sets of an infinite set must have a card analogy strictly greater than that set.

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And I'm going to refer you here to Petzold, who draws some lessons from this about the commonality of the continuum and other infinite numbers.

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I mean, effectively, this means that if you start with some set and then you take the set of subsets,

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you have to move to a larger number, whether that set was finite or infinite.

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So if you start with a set of natural numbers and then you take the set of some subsets of natural numbers, you must get to a greater infinity.

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And if you then take the set of subsets of that set, you must get to a greater infinity again and so on and so on and so on.

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So not only do we have at least two infinities, namely the number of natural numbers and the number of rational numbers.

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Actually, there's an infinite number of infinity.

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You can see why some people were rather dubious about cantors theory, but turns out to be to all make sense.

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You know, you can't generate contradictions from it. So it's very widely accepted.

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Finally, I'm going to end with an important consequences consequence of this for philosophy of logic and mathematics.

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Really, very, very important indeed. Bertrand Russell contemplated cantors arguments and you can see how close cantors argument gets to this,

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and hence Russell thought about the set of all sets that are not members of themselves.

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So, for example, take the set of cats. The set of cats is not itself a cat, so it's not a member of itself.

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Great. Whereas the set of all sets is itself a set and is therefore an element of itself.

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Take all the sets that are like the set of cats,

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all the sets that are not members of themselves think the set of everything that is not an element of itself doesn't mean it can't be a set.

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But if it is a set, it's not an element of itself, is it?

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Or is it not an element of itself? Well, if it is an element of itself, then it's not an element of itself.

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If it's not an element of itself, then it is an element of it. So you get a contradiction.

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It's rather similar to the paradox of the village barber who shapes all and only those men who do not shave themselves.

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Does he shave himself? OK, well, you can get round that one because you can say either there just is no such barber.

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It's an impossibility. Or you can more coming in to say, Well, maybe the barber was a woman.

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But the problem with the set of all sets that are not members of themselves is that it seems to be defined in such a straightforward way that to say,

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Oh, there's no such thing makes us wonder, Well, why not?

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We know what it is for a set to be a member of itself.

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Why can't we simply take all the assets that are not members themselves and put them in a big set?

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What stops us doing that? Well, it leads to a contradiction. Yeah, OK.

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It leads to a contradiction. But if that leads to a contradiction based on such straightforward ideas, where else might there be contradictions?

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And you can see the effect that this had on Friday, Russell wrote to Fraser in 1982.

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Explaining the paradox, Schrager wrote back to Russell six days later.

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So he probably only just received it. Your discovery of the contradiction has surprised me beyond words, and I should almost like to say,

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let me thunderstruck because it is rocked the ground on which I'm meant to build arithmetic.

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So there we're going to end today. Next lecture, we will see another phenomenal rocking of the ground in the form of girdles theorem.

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See you then. Thank you.