# Re: The seven step series

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

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.
>
>
> 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/
>
>
>
>
> >
>

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