Bruno, I loved your post on the square root of "2"! (I also laughed at it, to stay at the puns).

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You went out of your way and did not save efforts to prove how inadequate and wrong (y)our number system is. (ha ha). Statement: *if square-rooting is right* (allegedly, and admittedly) *then THERE IS such a 'quantity'* (call it number and by this definition it must be natural) *we consider as the square root of '2'*. You gave the plastic elemenary rule, how to get to it. Thankyou. Accepted. I believe the '1' and the sophistication of Pythagoras. *(provided that * *< 1^2 = 1 > which is also 'funny') a n d :* If it is not part of your series *of* - what you call: - *natural numbers*, then *YOUR* series is wrong. We need another system (if we really need it). Your math pupil John, the 'commonsenser'. On Wed, Sep 9, 2009 at 3:21 AM, Bruno Marchal <marc...@ulb.ac.be> wrote: > This is the last post before we proof Cantor theorem. It is an "antic > interlude". We are about 2000 years back in time. > > *The square root of 2.* > > It is a number x such that x^2 = 2. It is obviously smaller than 2 and > bigger than 1. OK? It cannot be a natural number. But could it be a > fraction? > > The square root of two is the length of the diagonal of a square with side > one unity. > > Do you see that? I can't draw, so you have to imagine a square. You may > draw it. And then draw or imagine the square which sides the diagonal (of > you square unity). Draw it with its diagonals and you will see, in your mind > or on the paper that the second square is made of four times the half of > your square unity: meaning that the square which sides the diagonal has an > area twice the area of the square unity. But this means that if you call d > the length of the diagonal of the square unity, you have, by the law of the > area of square, d^2 = 2. OK? > > So the length of the diagonal d = square root of 2. It is a 'natural' > length occurring in geometry (and quantum mechanics!). > > The function or operation "taking the square root of two" is the inverse > operation of taking the square. > (In term of the couples defining the function as set, it means that if (x, > y) belongs-to "taking the square root of two" then (y, x) belongs-to "taking > the square of"). > > Now, please continue to imagine the diagonal, and try to evaluate its > length. Is is not clearly less than two unities, and clearly more than one? > Is it 3/2 i.e. 1.5? Well, 1.5 should give two when squared, but (1.5)^2 = > 2,25. That's too much! Is 1,4? > Oh, 1.4 gives 1.96, not too bad, it is slightly more, may be 1.41? 1.41^2 > gives 1,9881, we get closer and closer, but will such a search ends up with > the best approximation? This would mean we can divide the side of the square > unity in a finite number of smaller unities, and find a sufficiently little > unity which could measure the diagonal exactly. It would mean d is equal to > p * 1/q, where q is the number of divisions of the side of the square unity. > If we find such a fraction the diagonal and the sided would be said > commensurable. > > Is there such a fraction? > > It is not 3/2, we know already. Is it 17/12 (= 1,416666666666...)? (1712)^2 > = 2,0069444444....). Close but wrong. > Is it 99/70? (= 1,414285714...) (99/70)^2 = 2,000204082...). Very close, > but still wrong! > > ... > > is it 12477253282759/8822750406821 ? My pocket computer says that the > square of that fraction is 2. Ah, but this is due to its incompetence in > handling too big numbers and too little numbers! It is still wrong! > > How could we know if that search will end, or not end? > > Sometimes, we can know thing in advance. Why, because things obeys laws, > apparently. > > Which laws of numbers makes the problem decidable? > > Here is one: > > - a number is even if and only if the square of that number is even, > similarly > - a number is odd if and only if the square of that number is odd. > > Taking the square of an integer leaves invariant the parity (even/odd) of > the number. > > Why? suppose n is odd. There will exist a k (belonging to {0, 1, 2, 3, ...} > such that n = 2k+1. OK? So n^2 = (2k+1)^2 = 4k^2 + 4k + 1, OK? and this is > odd. OK. And this is enough to show that if n^2 is even, then n is even. OK? > > And why does this answer the question. > > Let us reason 'by absurdo". > > Suppose that there are number p and q such (p/q)^2 = 2. And let us suppose > we have already use the Euclid algorithm to reduce that fraction; so that p > and q have no common factor. > p^2 / q^2 = 2. OK? Then p^2 = 2q^2 (our 'Diophantine equation). OK? > > Then p^2 is even. OK? (because it is equal to 2 * an integer). > Then p is even. OK? (by the law above). > This means p is equal to some 2*k (definition of even number). OK? > But p^2 = 2q^2 (see above), and substituting p by 2k, we get 4k^2 = 2q^2, > and thus, dividing both sides by 2, we get that 2k^2 = q^2. > So q^2 is even. OK? > So q is even. (again by the law above). OK? > So both p and q are even, but this means they have a common factor (indeed, > 2), and this is absurd, given that the fraction has been reduced before. > > So p and q does not exist, and now we know that our search for a finite or > periodic decimal, or for a fraction, will never end. > > The diophantine equation x^2 = 2y^2 has no solution in the integers, and > the number sqrt(2) = square root of two is not the ratio of two integers/ > Such a number will be said irrational. If we want associate a number to each > possible length of line segment, we have to expand the rational number (the > reduced fraction, the periodic decimal) with the irrational number. > > The numerical value of the sqrt(2) can only be given through some > approximation, like > > sqrt(2) = > 1,414 213 562 373 095 048 801 688 724 209 698 078 569 671 875 376 948 073 176 > 679 737 990 732 478... > > OK? > > I have to go now. > > Please ask any question. > > Does the "beginners" see that (2k+1)^2 = 4k^2 + 4k + 1? This comes from the > 'remarkable product (a+b)^2 = a^2 + 2ab + b^2. Could you prove this > geometrically (in your head or with a drawing)? Hint: search the area of a > rectangle which sides (a+b). > > In proof by 'reduction ad absurdo', sometimes Brent says this proofs only > the result OR the fact that an error occur in the proof. Please comment. > > I have not the time to give easy exercise, so I give one which is not so > easy. try to use the fundamental theorem of arithmetic (saying that all > natural numbers have a unique decomposition into prime factor to generalize > this result: if n is not a square number (like 1, 4, 9, 16, 25, ...) then > sqrt(n) is irrational. Guess who proves this theorem originally? Theaetetus! > (well I read that somewhere). > > This ends the antic interlude. > > Next post: Cantor theorem(s). There is *NO* bijection between N and N^N. I > will perhaps show that there is no bijection between N and {0, 1}^N. The > proof can easily be adapted to show that there is no bijection between N and > many sets. > > After Cantor theorem, we will be able to tackle Kleene theorem and the > 'mathematical discovery of the mathematical universal machine', needed to > grasp the mathematical notion of computation, implementation, etc. > > Bruno > > > On 08 Sep 2009, at 10:43, Bruno Marchal wrote: > > > > On 31 Aug 2009, at 19:31, Bruno Marchal wrote: > > > Next: I will do some antic mathematic, and prove the irrationality > > of the square root of two, for many reasons, including some thought > > about what is a proof. And then I will prove Cantor theorem. Then I > > will define what is a computable function. I will explain why Cantor > > "reasoning" seems to prove, in that context, the impossibility of > > the existence of universal machine, and why actually Cantor > > "reasoning" will just make them paying the big price for their > > existence. > > > Any question, any comment? I guess that I am too quick for some, > > too slow for others. > > > Don't forget the exercise: show that there is always a bijection > > between A+ and N. > > (A+ = set of finite strings on the alphabet A). This is important > > and will be used later. > > > I illustrate the solution on a simple alphabet. > > An alphabet is just any finite set. Let us take A = {a, b, c}. > A+ is the set of words made with the letters taken in A. A "word" is > any finite sequence of letters. > > To build the bijection from N to A+, the idea consists in enumerating > the words having 0 letters (the empty word), then 1 letter, then 2 > letters, then 3 letters, and so on. For each n there is a finite > numbers of words of length n, and those can be ordered alphabetically, > by using some order on the alphabet. In our case we will decide that a > > b, and b> c (a > b should be read "a is before b"). > > > So we can send 0 on the empty word. Let us note the empty word as *. > > 0 ------> * > > then we treat the words having length 1: > > 1 ------> a > 2 ------> b > 3 ------> c > > then the words having length 2: > > 4 -------> aa > 5 ------> ab > 6 ------> ac > > 7 ------> ba > 8 ------> bb > 9 ------> bc > > 10 ------> ca > 11 ------> cb > 12 ------> cc > > Then the words having lenght 3. There will be a finite number of such > words, which can be ranged alphabetically, > > etc. > > Do you see that this gives a bijection from N = {0, 1, 2, 3 ...} to A+ > = {*, a, b, c, aa, ab, ac, ba, bb, bc, ...} > > It is one-one: no two identical words will appears in the enumeration. > It is onto: all words will appear soon or later in the enumeration. > > I will call such an enumeration, or order, on A+, the lexicographic > order. > > Its mathematical representation is of course the set of couples: > > {(0, *), (1, a), (2, b), (3, c), (4, aa), ...} > > OK? No question? > > Next, I suggest we do some antic mathematics. It will, I think, be > helpful to study a simple "impossibility" theorem, before studying > Cantor theorems, and then the many impossibilities generated by the > existence of universal machines. It is also good to solidify our > notion of 'real numbers', which does play some role in the > computability general issue. > > > Here are some preparation. I let you think about relationship between > the following propositions. I recall that: the square root of X is Y > means that Y^2 = X. (It is the 'inverse function of the power 2 > function). The square root of 9 is 3, for example, because 3^2 = 3*3 = > 9. OK? > > - There exists incommensurable length (finite length segment of line > with no common integral unity) > - the Diophantine equation x^2 = 2(y^2) has no solution (Diophantine > means that x and y are supposed to be integers). > - the square root of 2 is irrational (= is not the ratio of integers) > - The square root of 2 has an infinite and never periodic decimal. > - If we want to measure by numbers any arbitrary segment of line, we > need irrational numbers > > Take it easy, explanation will follow. This antic math interlude will > not presuppose the 'modern math' we have seen so far. > > Bruno > > > > > > > > http://iridia.ulb.ac.be/~marchal/ > > > > > > > > http://iridia.ulb.ac.be/~marchal/ > > > > > > > --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Everything List" group. 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