Conference rusure::math

A rational number is defined as an integer or the quotient of two integers.
You haven't shown that either the diameter or the circumference of the circle
is an integer.

-Mike

    re .0,
    
    I think you are confusing "mutability" with "expressibility" as
    explained in .1.  Of course PI is a fixed number, but it is not
    expressible with ratia of integers.
    
    Eugene

    Reminds me of a fellow who once said that probability is
    meaningless, for an event E either happens or it doesn't.
    Therefore everything has a chance of 50% to be...
     

Hmmm,

This also reminds me of the proof that all triangles are isoceles.
From which you can conclude that all triangles must also be equilateral, 
that 1=0, .... etc. etc.

In case someone isn't familiar with the distinction, here are classes of 
(positive) real numbers in increasing order of complexity:

Integers :	1,2,3,... None have fractional parts.

Rationals :	Includes integers and (integer / integer), e.g. 1/3. All
		Have repeating/cyclic fractional parts, e.g. .3333...

Algebraics :	Includes all the zeroes of polynomials of finite degree
		with rational coefficients, e.g. sqrt(3). Except for the
		rationals, all have nonterminating, acyclic fractional
		parts. 

Transcendentals: Everything else. Pi and e are transcendentals, as are all 
		their rational powers, e.g. Pi**(6/7). 

One can also define intermediate classes, e.g. that class of all reals 
with terminating fractional parts, which is that subset of the rationals
whose cyclic fraction is ...0000.... 

re .5,
    
    But then there are also zillions of extension fields in the middle.
    
    Eugene

There are the Cantorian infinitesimals and transfinite numbers to consider.

Wook

    i wrote this befor i saw .5 defintions, and .6, .7 
    but here we go:
    
i received this in mail today, and it is about a book called irrational
numbers by ivan Niven, i have a problem with a statement in its outline,
may be one have shed a light too. it says

"Irrational number
A master expositor gives us the properties of irrational number, among other
topics, Niven treats algebraic irrationals, transcendental, and normal
numbers. he concludes with the solution of Hilbert's 7th problem.
this result , proved by Gelfond and Schneider (independently) states, that
a^b is transcendental, where a and b are algebraic over the rationales, with
0 != a != 1, and b is either irrational or not a real number"

the problem i have with last 2 lines, i think it should have said
"a^b is transcendental, where a is an algebraic rational integer, and b
is an algebraic irrational integer, with 0 != a != 1"

because he said first that "a and b are algebraic over the rationales" and then
said "..and b is either irrational or not a real number" .

any ways, i picked a book on number theory and got some definitions out to
know what is all this about, some relevant definitions are:

1) A complex number theta is called ALGEBRAIC NUMBER if it satisfies
some polynomial equation f(x)=0 where f(x) is a polynomial
over Q (rational numbers).

2) an ALGEBRAIC NUMBER theta is an ALGEBRAIC INTEGER if it satisfies
some monic (i.e. one variable) polynomial equation
 f(x)= x^n+ b1 x^n-1 + ... + bn = 0 with integral coeffients.

3) among the rational numbers, the only ones that are ALGEBRAIC INTEGERS
are the integers 0, +- 1, +- 2, ...

4) There are 2 types of ALGEBRAIC INTEGERS, the RATIONAL ALGEBRAIC INTEGERS
and the IRRATIONAL ALGEBRAIC INTEGERS (like SQRT(2)).


so i think a simplified picture is like this:

       +----------------------------------------------------------+
       | complex numbers                                          |
       |     +-------------------------+   +---------------------+|
       |     | Algebraic numbers       |   |non Algebraic numbers||
       |     |  +--------------------+ |   | +----------------+  ||
       |     |  |Algebraic integers  | |   | |Transcendental  |  ||
       |     |  |  +------------+    | |   | |PI, EXP         |  ||
       |     |  |  | Rational   |    | |   | +----------------+  ||
       |     |  |  +------------+    | |   +---------------------+|
       |     |  |  +------------+    | |                          |
       |     |  |  |irrationals |    | |                          |
       |     |  |  +------------+    | |                          |
       |     |  +--------------------+ |                          |
       |     +-------------------------+                          |
       +----------------------------------------------------------+

any ways, having said all that, i think to say that a^b is transcendental
when a is algebraic rational integer (not 0 or 1) and b algebraic irrational
integer is really a neat theory, so this means that 2^SQRT(2) is a 
transcendental number.

is there an infinite number of transcendental numbers? from this it is so:

2^SQRT(2), 3^SQRT(2), 4^SQRT(2), ..... are all transcendental..

/nasser
    

    Got any examples of non-algebraic, non-transcendental numbers?

re .8,

>1) A complex number theta is called ALGEBRAIC NUMBER if it satisfies
>some polynomial equation f(x)=0 where f(x) is a polynomial
>over Q (rational numbers).

Correct; this is also equivalent to saying it satisfies some
polynomial equation g(x)=0 where g(x) is a polynomial over Z
(the (signed) integers, ..., -2, -1, 0, 1, 2, ...).  Just let
g(x) = N f(x) where N is a common denominator.

>2) an ALGEBRAIC NUMBER theta is an ALGEBRAIC INTEGER if it satisfies
>some monic (i.e. one variable) polynomial equation
> f(x)= x^n+ b1 x^n-1 + ... + bn = 0 with integral coeffients.

Monic means the coefficient of the leading (highest degree) term
is 1.  Monic does not mean one variable.  (The form you show for
f is correct.)

So algebraic corresponds to a polynomial with integral coefficients,
and algebraic integer to a monic polynomial with integral coefficients.
Rational corresponds to a polynomial with integral coefficients having
degree one.

Later, your diagram includes the rationals and irrationals in the
algebraic integers.  That is incorrect.  1/2 is rational but is not
an algebraic integer (2x-1 is not monic, nor is any other polynomial
with integral coefficients with 1/2 as a root).  Likewise sqrt(1/2)
is irrational and algebraic but not an algebraic integer.  And pi
and e are irrational, so not all irrationals are algebraic.

The diagram also makes some identities look like proper subsets.
For example, non-algebraic and transcendental are the same (see .9)

The statement of the theorem was correct.

Dan

>Monic means the coefficient of the leading (highest degree) term
>is 1.  Monic does not mean one variable.  (The form you show for
    correct offcourse, too many definitions, i got spaced out for
    a second.
    
>Later, your diagram includes the rationals and irrationals in the
>algebraic integers.  That is incorrect.  1/2 is rational but is not
>an algebraic integer (2x-1 is not monic, nor is any other polynomial
>with integral coefficients with 1/2 as a root).  
    
    yes, i have to correct diagram, there are rationals that are not
    algebraic. one day i'll make big diagram that has everything in it.
    
>The diagram also makes some identities look like proper subsets.
>For example, non-algebraic and transcendental are the same (see .9)
    
     right, it was misleading, i did it like that, thinking there might be
    some that are non-algebraic and but transcendental.
    
     /nasser
    
    

nice theory about how to find if a number is transcendental or not:

if  log(a)
    ------   is not rational then this number is transcendental
    log(b)


where a,b are ALGEBRAIC NUMBERS different from 1 and 0

to use this to show that 2^sqrt(2) is transcendental:
by contradiction, assume 2^SQRT(2) is algebraic, then log(2^SQRT(2))
                                                      --------------
                                                      log 2

must be transcendental, but the above equals to  SQRT(2) ,after simplifications,
and SQRT(2) this is algebraic (irrational), so assumption is wrong, 
and 2^SQRT(2) is not algebraic, hence it is transcendental.

a similar thing can be done to show that e^PI is transcendental .
    

>    yes, i have to correct diagram, there are rationals that are not
>    algebraic. one day i'll make big diagram that has everything in it.

All rationals are algebraic [numbers].  There are rationals that are not
algebraic *integers*.  "Algebraic" by itself usually means "algebraic
number", not "algebraic integer".

re .12,

>if  log(a)
>    ------   is not rational then this number is transcendental
>    log(b)			   ^^^^
>				   Which number?
>
>where a,b are ALGEBRAIC NUMBERS different from 1 and 0

Dan

    the number that comes out when you divide log(a) by log(b) ?
    iam missing something?

I wasn't sure whether or not you were referring back to the a^b
of the theorem in a previous reply, i.e., if "this number" meant
a^b or log(a)/log(b).  The example had used the form a^b as well.

Dan

Article 27488 of sci.math:
From: dbailey@wk49.nas.nasa.gov (David H. Bailey)
Newsgroups: sci.math
Subject: Simple proof that pi is irrational
Organization: NAS, NASA Ames Research Center, Moffett Field, CA
Lines: 33

I just found the following elementary proof that pi is irrational in
my files.  Unfortunately, I have no reference.  Does someone recognize
this proof from a standard reference?

It is a pity that there is not a similarly succint proof that pi is
transcendental.

David H. Bailey
dbailey@nas.nasa.gov

-----------------------------------------------

Theorem:  pi is irrational

Proof:  Suppose pi = p / q, where p and q are integers.  Consider the
functions f_n(x) defined on [0, pi] by

f_n(x)  =  q^n x^n (pi - x)^n / n!  =  x^n (p - q x)^n / n!

Clearly f_n(0) = f_n(pi) = 0 for all n.  Let f_n[m](x) denote the m-th
derivative of f_n(x).  Note that

f_n[m](0) = - f_n[m](pi) =  0 for m <= n or for m > 2n; otherwise some integer

max f_n(x) = f_n(pi/2) = q^n (pi/2)^(2n) / n!

By repeatedly applying integration by parts, the definite integrals of
the functions f_n(x) sin x can be seen to have integer values.  But
f_n(x) sin x are strictly positive, except for the two points 0 and
pi, and these functions are bounded above by 1 / pi for all
sufficiently large n.  Thus for a large value of n, the definite
integral of f_n sin x is some value strictly between 0 and 1, a
contradiction.