Conference rusure::math

    There does not seem to be a general way to do this.  The proof that
    the square root of a particular prime, say 2, is irrational is not
    very hard - just assume the root is p/q and you soon get a
    contradiction.
    
    Pi and e are much more difficult although they were known to be
    irrational before they were shown to be non-algebraic.
    
    The transcendence of pi is given a whole chapter in "A History of
    Pi," by Petr Beckmann, 1971, St. Martin's Press.
    
    Chuck

    Re .0:
    
    Here is the proof that the square root of two is irrational.
    
    Assume sqrt(2) is rational.  Then there exist relatively prime integers
    p and q such that sqrt(2) = p/q.
    
    Take that equation:		sqrt(2) = p/q.
    Square both sides:		      2 = p^2 / q^2.
    Multiply by q^2:		  2 q^2 = p^2.
    
    Now since p^2 = 2 q^2, p^2 is a multiple of 2.  If p were odd, p^2
    would be odd, so p must be even.  Therefore, there is a k such that p =
    2k.  So p^2 = (2k)^2 = 4 k^2.
    
    Substitute for p^2:		  2 q^2 = 4 k^2.
    Divide by two:		    q^2 = 2 k^2.
    
    Now we see q^2 = 2 k^2, so q^2 is a multiple of two.  As before, this
    means q is even.  So both p and q are even.  But this is impossible,
    because they are relatively prime.
    
    Because there is a contradiction, our assumption is false.  There do
    not exist relatively prime p and q such that p/q = sqrt(2).  By
    definition, sqrt(2) is irrational.
    
    In general, that's the proposition you need to prove to show a number
    is irrational.  How you prove it depends on the number. 
    
    
    				-- edp 

     re .2

>>    In general, that's the proposition you need to prove to show a number
>>    is irrational.  How you prove it depends on the number. 

     For the special case of an integral root of an integer: if n > 1 is
     a positive integer and M is an integer, then the n-th root of M is
     rational if and only if it is an integer.  Stated another way, the
     n-th root of M is irrational if and only if it is not an integer.

     The proof is a suitably generalized version of the one in .2.

     Dan

Edp, I followed your proof well throughout, but the following summary
threw me:

    Because there is a contradiction, our assumption is false.  There do
    not exist relatively prime p and q such that p/q = sqrt(2).  By
    definition, sqrt(2) is irrational.


To me, "by the above proof, sqrt(2) is irrational".  Why do you say
"By definition, sqrt(2) is irrational" ?  If we didn't have a proof,
we might want to say "by definition", but if we've just proved something,
we no longer need to merely agree to a definition, right ?

/Eric
   

    Re .4:
    
    Eric, I think it make sense if you replace:
    
>	By definition, sqrt(2) is irrational.
    
    with
    
>	By the definition of 'irrational' then, sqrt(2) is irrational.

    --  Barry

    Re .4:
    
    .5 is correct; a full wording would be:
    
    We've just proven there do not exist relatively prime p and q such that
    p/q = sqrt(2).  By the definition of "irrational", if there do not
    exist relatively prime p and q such that p/q = x, x is irrational.
    Therefore, sqrt(2) is irrational.
    
    
    				-- edp 

    To further clarify - a *transcendental* number is NOT the root of
    *any* polynomial equation with rational numbers for coefficients. The
    class of such roots is called the *algebraic* numbers, and it includes
    all the *rationals*, which in turn includes all the *integers*.

    Proving that a number is not algebraic is generally fairly difficult,
    but although there are infinitely many integers and rationals and
    algebraics, nearly all numbers are transcendental. Pi and e are
    transcendental; all nth roots are algebraic.
    
    I have pointed out in other contexts that it is impossible to represent
    a random transcendental number in a computer except as a *function* of
    some more basic numbers, e.g. integers, since they all have
    nonterminating and nonperiodic n-ary expansions, i.e. are infinite in
    length. 

Does anybody want to mention hyper-radicals?

     re .7    I liked your title.

     re .8    Hmm. Okay.

     But what about hyper-radicals?

     Dan

    Hyper-radicals?  I've run onto hyper-reals but what are hyper-radicals
    or at least the context so I can look 'um up?
    
    Chuck

     I don't know.  I don't think I've ever heard of them.
     
     Maybe he's just being irrational.
     
     Dan

     Just trying for a time stamp of 00:00. :-)
     
     Dan

    Re .7:
    
    > I have pointed out in other contexts that it is impossible to represent
    > a random transcendental number in a computer except as a *function* of
    > some more basic numbers, e.g. integers, since they all have
    > nonterminating and nonperiodic n-ary expansions, i.e. are infinite in
    > length. 
    
    That's a bit harsh; representation can accomplish quite a bit.  The
    HP-28 represents pi with a symbol named "pi", and it can make some
    limited use of that symbol without convering it to a decimal
    representation.  For example, it knows "sin(pi/2)" is 1. 
    
    On the other hand, I would say the floating point representation of a
    number as a mantissa and exponent, m and e such that the number is
    m*2^e, is a representation of the number as functions.  Even the
    representation of integers as bits corresponds to functions:  The
    addition of powers of two.
    
    When we want to add, multiply, or otherwise work with numbers
    represented in binary, we have to write detailed routines that depend
    upon the representation we have chosen.  We could do the same for other
    representations of numbers.
    
    
    				-- edp 

    I believe the notion of hyper-radicals, or maybe ultra-radicals, comes
    from the problem of solving Nth-degree equations in radicals.
    
    We all know (or _should_ know :-) that you can solve in radicals all
    univariate polynomial equations with rational coefficients of the first
    through fourth degree, for example the quadratic formula is a solution
    in radicals for second degree equations.  But there is no general
    solution for polynomial equations of degree 5 or higher; this is proved
    in Galois theory.
    
    Hyper-radicals are a way of generalizing the notion of radicals in some
    manner that totally escapes me, and is why I asked the question in the
    first place.  The idea, I think, is that ALL polynomial equations with
    rational coefficients can be solved in hyper-radicals.  For example,
    the quintic:
    
         5
    	x  +  x  +  a  =  0
    
    has as one solution the value 
    
        5 ___
    x =  U a    loosely translated as "x equals 5th hyperroot of a"
    
    								        ___
    Note the curved "hyper-radical sign" instead of the pointed symbol V  
    This is how you distinguish hyper-radicals from radicals.  Exactly what
    that difference MEANS is what I'm asking about.  What is the 5th hyper-
    root of 29, anyway?
    
    Since we were talking about irrationals and transcendentals, I wondered
    if somebody know about hyper-radicals.  They are algebraic numbers, but
    are evidently more difficult to deal with than numbers you can express
    in radical form.
    
      John

Please note that although transcendental numbers may lack recognizable
patterns when expressed in decimal, for example:

	e = 2.718281828459045...

they can have very recognizable patterns when expressed in other, just
as valid ways.  For example,

	e = [2 1 2 1 1 4 1 1 6 1 1 8 1 1 10 1 1 12 1 1 14 1 ... ]

That is to say e is 2 + 1/(a+1/(b+1/(c+1/(d+...
where a,b,c,d... are 1,2,1,1,4,1,1,6,1,1,8,1...

However, pi is NOT recognizable even in this form (which, by the way
is called "regular continued fraction" form)

Interesting question:  How can we characterize which transcendental numbers
will have simple patterns in theire regular continued fraction
representations, and which will not ?

Can we classify transcendental numbers into two classes, just as we
classify decimal numbers into rationals and irrationals according to
whether we see patterns in their representation ?

Of course, you can also pose such puzzles in reverse.  For instance,
can anyone figure out any "nice" ways of saying what this number is:

	n = [1 2 3 4 5 6 7...]

/Eric

> Interesting question:  How can we characterize which transcendental numbers
> will have simple patterns in theire regular continued fraction
> representations, and which will not ?
    
    Just what is a "simple pattern"?  Until you _precisely_ define what you
    mean, how can you hope to work with it?
    
> Of course, you can also pose such puzzles in reverse.  For instance,
> can anyone figure out any "nice" ways of saying what this number is:
>
> 	n = [1 2 3 4 5 6 7...]
    
    For lack of a more historical term, let's call it "Osman's constant".
    That way we can use the mnemonic O and confuse everybody.  (Isn't that
    a "nice" way of defining the number?  :-)
    
    On the same topic, do hyper-radicals have repeating continued fraction
    representations?
    
      John

>    On the same topic, do hyper-radicals have repeating continued fraction
>    representations?
    
    If a number has a repeating continued fraction representation, it is
    the root of a quadratic with integer coefficients.

    
    John, I think you missed the essence of what I was saying.
    
    Someone could specify
    
    	n = [2 1 2 1 1 4 1 1 6 1 1 8 1 1 10 1 1 12 1 1 14 1 ...]
    
    and it would not mean much, but if someone else demonstrates
    that
    
    	n = e
    
    then that's worth noting.
    
    I specify
    
    	n = [1 2 3 4 5 6 7 ...]
    
    Perhaps someone else could show that this is something "neat", like
    cos(13/7) or pi^e/4 or something else as surprising.
    
    As for your question about how do we define "patterns", that's an
    interesting one.
    
    For rational numbers in decimal, we talk strictly about repeating
    exact patterns.  In the example of e, we have
    
    	e = [s0 s1 s2 s3...]
    
    where
    	
    	s(0) = 2
    	s(n) = 2*(n+1)/3	n=3,6,9,12...
    	s(n) = 1		all other n
    
    I'm not sure how to define "patterns".  Perhaps it's any sequence
    that we can easily define the nth term.
    
    I suspect that although you might argue that
    
    	2,1,2,1,1,4,1,1,6,1,1,8,1,1...
    
    is too different from
    
    	1,2,3,4,5,6,7...
    
    to warrant treating them together, if you look at the sequence for
    pi's continued fraction (sorry, I don't have it HAKMEM here, anyone
    have it?), it looks as unpatterned as the decimal representation of
    pi itself.
    
    What's the SAME in both of the above sequences is that we can easily
    answer the question "what's the 1000th term ?", which I suspect
    can't be done with the one for pi.
    
    /Eric

     Eric,
     
     Have you computed the first few convergents to "your"
     number [1, 2, 3, 4, 5, ...]?  Maybe it will look familiar.
     
     Dan
     
     P.S.  A few things about numbers expressed by infinite
     regular continued fractions:
     
     If you represent real numbers (irrational) by the c.f.
     [a0, a1, a2, ...] then:
     
          The set of reals for which the sequence a0,a1,a2,... is
          bounded is null (i.e., has Lebesgue measure zero).
     
     or even this:
     
          If f(n) is an increasing function of n for which
          sum(n = 1,...,oo) 1/f(n) diverges, then the set of
          reals for which an <= f(n) for all sufficiently
          large n, is null.  On the other hand, if the sum
          sum(n = 1,...,oo) 1/f(n) converges, then an <= f(n)
          for almost all reals [i.e., except on a null set]
          and sufficiently large n.

RE: .18
    
    Seems to me that you are playing around the edges of Kolmogorov
    complexity theory.  Some numbers, including all rationals and
    algebraic irrationals and some transendentals are the output of
    finite programs executing a countable number of steps.  There
    are only a countable number of these out of the "uncountable"
    number of real numbers, and thus they form a set of measure zero
    within the reals (e.g., if you select a real number "randomly"
    it is not "impossible" that you will choose one, but the probability
    that you will is zero).
    
    Some of those programs will produce sequences of integers which are
    therefore "patterned" and are used in the continued fraction
    representation you mention to produce (in aleph-null steps) the
    numbers you are interested in.
    
    For more details see the articles cited in the reposting from USENET
    I'm about to make (synchronicity strikes again! :-) or check out
    "Randomness in Arithmetic" by Gregory J. Chaitin, Sci. Am. July 1988.
    
    					Topher

    re:  Note 1028.18 by HANNAH::OSMAN
    
    No, Eric, I understood the essence of what you were saying, but you
    were speaking with an artist's voice.  When you say:
    
>    What's the SAME in both of the above sequences is that we can easily
>    answer the question "what's the 1000th term ?", which I suspect
>    can't be done with the one for pi.
    
    you have taken a step in the right direction of identifying those
    "interesting" numbers.  Unfortunately, the word "easily" is not clearly
    defined above.  There are ways to rigorously define things to allow the
    "e" representation but I doubt that you can create a definition that
    will admit all the "neat" patterns.  Do you understand the essence of
    what I'm saying?
    
    My favorite cf representation is given by the constant [1,1,1,1,1,...]
    which turns out to be a very "interesting" number.  No Fibbing!
    (Even though Dan might consider it to be a dull, boring number.
    Well, he has no taste.  :-)
    
    MACSYMA has some good capabilities to deal with cfs, does anybody have
    a copy I can play with?
    
      John

     John,

>>    My favorite cf representation is given by the constant [1,1,1,1,1,...]

     (Au + Au + ... + Au)/n

>>    which turns out to be a very "interesting" number.  No Fibbing!

     Yes.  I have read about it, and have even watched the movie.

     Dan

re: .18

>    	n = [1 2 3 4 5 6 7 ...]
>    
>    Perhaps someone else could show that this is something "neat", like
>    cos(13/7) or pi^e/4 or something else as surprising.

	You might be interested in:

	             z    -z
	            e  - e
	tanh(z)  =  --------  =  [ 1/z, 3/z, 5/z, 7/z, ... ].
	             z    -z
	            e  + e

	See Knuth Vol.2, problem 4.5.3.16 for hints on proving this.
	It may offer ideas for finding the value of [ 1, 2, 3, ... ].

By the way, perhaps other people besides me didn't realize this, but
ANY nonending continued fraction, for instance:

	a = [1 1 1 1 1 1...]

or

	b = [2 3 5 7 11 13 17...]

is IRRATIONAL.

This is alot stronger a statement than the parallel one for the decimal
numbers, which says that only those that don't repeat are irrational.

/Eric

Expanding on what Mr. Gilbert said, which was:

                     z    -z
                    e  - e
        tanh(z)  =  --------  =  [ 1/z, 3/z, 5/z, 7/z, ... ].
                     z    -z
                    e  + e
   

If we let z be 1, we have


	e^2  -  1
       -----------   =  [1 3 5 7 9 . . .]
	e^2  +  1


So, if we want to find out what [1 2 3 4 5 6 7 . . .] is, maybe we
can first find out what [2 4 6 8 10 . . .] is.

But then, how does one "weave" two continued fractions ?  For instance,
if

	f1 = [a1 b1 c1 d1 . . .]

and

	f2 = [a2 b2 c2 d2 . . .]

can we easily express f1+f2 ?

/Eric

> For instance, if
> 	f1 = [a1 b1 c1 d1 . . .]
> and
> 	f2 = [a2 b2 c2 d2 . . .]
> can we easily express f1+f2 ?

	No.

	Here's a suggestion on how to proceed.  First, a comment on notation:

	                          1
	// a1, a2, a3, ... // = ----------
	                               1
	                        a1 + ----------
	                                    1
	                             a2 + ----------
	                                  a3 + ...

	Let f (z) = // 1/z, 2/z, 3/z, ... //, and f  (z) = 1/f (z) - (n+1)/z.
	     0                                     n+1        n

	Now we want to find a differential equation:

		d f (z) / dz = D( n, f (z), z )
		   n                  n

	such that the above recurrence for f  (z) gives us:
	                                    n+1

		d f  (z) / dz = D( n+1, f  (z), z )
		   n+1                   n+1

	(we need to deterine the function D).  Given this, we solve the
	differential equation

		d f (z) / dz = D( 0, f (z), z ),
		   0                  0

	and this presumably gives us our solution.

	                                                     2
	In the case of tanh(z), we had D(n, f (z), z) = 1 - f (z) - 2 n f (z)/z.
	                                     n               n           n
	I'd expect the above problem to have a similar function D.

	Gee, this topic has covered a lot of areas.   Let me contribute
my 2�.


Back to the base note.   Proving pi irrational.   I believe that this
was originally done by continued fractions!   But let me sketch out the
proof I know.

Let:
	I_n = integral(from: -1, to +1) cos(�x)*(1-x�)^n dx

Now by integrating by parts, and then induction, can show that:
	I_n * �^(2*n+1) / n! = cos(�)*P_n + sin(�)*Q_n
where P_n and Q_n are polynomials in � with integral coefficients.

Now imagine that pi = a/b, with a & b integral.   Set � = pi/2.

Then define:
	J_n = I_n * a^(2*n+1) / n! 

Make three observations:

	(i)  J_n is an integer, for all n >= 0
	(ii) J_n > 0, for all n >= 0
	(iii) J_n -> 0 as n -> oo (infinity)

These are mutually inconsistent, therefore pi is irrational.


Andrew.

re .26
	Let f  (z) = 1/f (z) - (n+1)/z.
	     n+1        n

	The differential equation is:
		                      2
		d f (z) / dz = 2 - 2 f (z) - (2 n + 1) f (z) / z
		   n                  n                 n

	We want to solve the this equation when n = 0.  Note that

			   2z      -2z        2z      -2z
		f(z) = (a e   + b e   ) / (c e   + d e   )

	(with appropriate values of a, b, c, and d) gives a solution to the
	similar equation:
		                      2
		d f (z) / dz = 2 - 2 f (z)
		   n                  n