Philip Potter wrote:
> Ben Bacarisse wrote:
>
>> Done to death now but yes. Though "irrational" is often used
>> for numbers like sqrt(2) the term in not very ful since
>> transcendental numbers are also irrational. The modern term
>> is "algebraic" giving the hierarchy:
>>
>> naturals + integers + rational + algebraics + transcendentals = reals
>>
>> (with each class including those to the right).
>
> Not the way I see it. Using < for the subset-of relation, we have:
>
> Naturals < integers < rationals < algebraics
>
> But it is certainly not the case that algebraics < trancendentals!
> In fact
>
> algebraics I trancendentals = 0
>
> where I is intersection and 0 is the empty set.

I vaguely remember a proof that between any two rationals we can
find an infinity of transcendentals. Also that rationals are
countable (1 to 1 correspondence with integers), while
transcendentals are not. This leads to the various classes of
infinity. The details are all lost to bit rot.

--
Chuck F (cbfalconer at maineline dot net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home.att.net>



--
Posted via a free Usenet account from http://www.teranews.com

CBFalconer wrote:
> Philip Potter wrote:
>> Naturals < integers < rationals < algebraics
>>
>> But it is certainly not the case that algebraics < trancendentals!
>> In fact
>>
>> algebraics I trancendentals = 0
>>
>> where I is intersection and 0 is the empty set.
>
> I vaguely remember a proof that between any two rationals we can
> find an infinity of transcendentals. Also that rationals are
> countable (1 to 1 correspondence with integers), while
> transcendentals are not. This leads to the various classes of
> infinity. The details are all lost to bit rot.

We're getting highly OT here, but between any two rationals lie an
infinity of rationals. But it's a smaller infinity than the infinity of
trancendentals.

Rationals are countable because they can be expressed as a pair of
integers (numerator, denominator) and the set of pairs of integers forms
a bijection with the set of integers (proof omitted here).

Trancendentals are uncountable due to cantor's diagonal argument.

Phil

In article <4720A429.3B52FF0@yahoo.com>,
CBFalconer <cbfalconer@maineline.net> wrote:

>I vaguely remember a proof that between any two rationals we can
>find an infinity of transcendentals. Also that rationals are
>countable (1 to 1 correspondence with integers), while
>transcendentals are not. This leads to the various classes of
>infinity. The details are all lost to bit rot.

The algebraic numbers are also countable, since there are only a
countable number of polynomials with integer coefficients, and each
has only a finite number of roots. Since the transcendentals are the reals
excluding the algebraics, the transcendentals must be uncountable.

Furthermore, since there are only countably many algebraics, *any*
uncountable set of reals must contain uncountably many
transcendentals.

It's then obvious that there are uncountably many transcendentals
between any two distinct reals, since any non-empty interval contains
uncountably many reals.

-- Richard
--
"Consideration shall be given to the need for as many as 32 characters
in some alphabets" - X3.4, 1963.

Philip Potter <pgp@see.sig.invalid> writes:

> Ben Bacarisse wrote:
>> Done to death now but yes. Though "irrational" is often used for
>> numbers like sqrt(2) the term in not very ful since transcendental
>> numbers are also irrational. The modern term is "algebraic" giving
>> the hierarchy:
>>
>> naturals + integers + rational + algebraics + transcendentals = reals
>>
>> (with each class including those to the right).
>
> Not the way I see it. Using < for the subset-of relation, we have:
>
> Naturals < integers < rationals < algebraics
>
> But it is certainly not the case that algebraics < trancendentals! In fact
>
> algebraics I trancendentals = 0
>
> where I is intersection and 0 is the empty set.

Yes, of course. The trancendentals are defined to not include any of
the "simpler" sets.

--
Ben.

Philip Potter <pgp@see.sig.invalid> writes:

> CBFalconer wrote:
>> Philip Potter wrote:
>>> Naturals < integers < rationals < algebraics
>>>
>>> But it is certainly not the case that algebraics < trancendentals!
>>> In fact
>>>
>>> algebraics I trancendentals = 0
>>>
>>> where I is intersection and 0 is the empty set.
>>
>> I vaguely remember a proof that between any two rationals we can
>> find an infinity of transcendentals. Also that rationals are
>> countable (1 to 1 correspondence with integers), while
>> transcendentals are not. This leads to the various classes of
>> infinity. The details are all lost to bit rot.
>
> We're getting highly OT here,

snap!

> Rationals are countable

as are the algebraic numbers (as I am sure you know). Un-countability
only come with chucking in the "rest"!

--
Ben.

On Oct 25, 10:59 am, Philip Potter <p...@see.sig.invalid> wrote:
> CBFalconer wrote:
> > Philip Potter wrote:
> >> Naturals < integers < rationals < algebraics
>
> >> But it is certainly not the case that algebraics < trancendentals!
> >> In fact
>
> >> algebraics I trancendentals = 0
>
> >> where I is intersection and 0 is the empty set.
>
> > I vaguely remember a proof that between any two rationals we can
> > find an infinity of transcendentals. Also that rationals are
> > countable (1 to 1 correspondence with integers), while
> > transcendentals are not. This leads to the various classes of
> > infinity. The details are all lost to bit rot.
>
> We're getting highly OT here, but between any two rationals lie an
> infinity of rationals. But it's a smaller infinity than the infinity of
> trancendentals.
>
> Rationals are countable because they can be expressed as a pair of
> integers (numerator, denominator) and the set of pairs of integers forms
> a bijection with the set of integers (proof omitted here).
>
> Trancendentals are uncountable due to cantor's diagonal argument.
>
> Phil

OKAY lets make it topical. I came up with this algorithm to solve a
problem in Oracle PLSQL (which only has one dimensional arrays), but
it can be useful in C also.

To make it more relevant to C let me set the stage.

You need an extensible 2D array. You can simulate this with a linear
array and this routine.

/**************
For a linear array X[]
Given 2 indices A,B
return an index value that maps to a location in X[]
With first element at (0,0) and X[] is zero based.
*********************************************/
int twod( a, b )
{
int k,n,ndx;
k=a+b+1;
n=k*(k+1);
n=n/2;
ndx = n-b-1; /* make zero based for C */
return ndx;
}



and the caller uses this within the array, like

X[twod(3,5)]++ ;

/* and for extensible array situations, check the index first */
if ( twod(aa,bb) >numelementsin(X) )
{
/* need to reallocate to have room for the next entry */
<realloc code>
/* wouldn't need this in Oracle, which has sparse arrays */
/* and allocates elements for you */
}
X[twod(aa,bb)]=nextvalue;

A final comment:
this routine basically fills the array along diagonals. So it really
is best if your application needs a triangular array rather than a
square one. For example if you want to be able to use X[twod(10,10)],
then X must have 221 elements.

Enjoy.
Ed

user923005 wrote:
>
> On Oct 24, 5:03 pm, pete <pfil...@mindspring.com> wrote:
> > CBFalconer wrote:
> > > Are you claiming sqrt(2) is not transcendental?
> >
> > The square root of two is irrational.
> > Transcendental numbers are also not the roots of any equation.
> > pi is transcendental.
>
> A transcendental number is not the root of any integer polynomial.
> There are equations which have pi as root(s).

Thank you.

--
pete

CBFalconer wrote:
> Philip Potter wrote:
>> Ben Bacarisse wrote:
>>
>>> Done to death now but yes. Though "irrational" is often used
>>> for numbers like sqrt(2) the term in not very ful since
>>> transcendental numbers are also irrational. The modern term
>>> is "algebraic" giving the hierarchy:
>>>
>>> naturals + integers + rational + algebraics + transcendentals = reals
>>>
>>> (with each class including those to the right).
>> Not the way I see it. Using < for the subset-of relation, we have:
>>
>> Naturals < integers < rationals < algebraics
>>
>> But it is certainly not the case that algebraics < trancendentals!
>> In fact
>>
>> algebraics I trancendentals = 0
>>
>> where I is intersection and 0 is the empty set.
>
> I vaguely remember a proof that between any two rationals we can
> find an infinity of transcendentals. Also that rationals are
> countable (1 to 1 correspondence with integers), while
> transcendentals are not. This leads to the various classes of
> infinity. The details are all lost to bit rot.
>

Countable and uncountable infinity are ideas put forth
by Georg Cantor. Cantor spent a lot of time in mental
institutions. I do *not* think these two facts are
unrelated...

--
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