CBFalconer said:

<snip>

> Are you claiming sqrt(2) is not transcendental?

Yes.

A transcendental number is one that is not a root of any integer polynomial
equation.

sqrt(2) is a root of the equation: x^2 - 2 = 0

Therefore, sqrt(2) is not transcendental, QED.

> It is a long time
> since I took math classes, and I have been operating under the
> (possibly delusional) assumption that a number that cannot be
> expressed exactly is transcendental.

The word for which you are reaching is "irrational". It is indeed the case
that sqrt(2) is irrational.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

pete said:

> 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.

Close, but no banana.

> pi is transcendental.

It is the root of the equation x - pi = 0. Therefore, according to your
definition of "transcendental", pi is *not* transcendental. Nevertheless,
we know that it /is/. Therefore, your definition must be wrong.

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

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).

Richard Heathfield wrote:
> CBFalconer said:
>
> <snip>
>
>> Are you claiming sqrt(2) is not transcendental?
>
> Yes.
>
> A transcendental number is one that is not a root of any integer
> polynomial equation.
>
> sqrt(2) is a root of the equation: x^2 - 2 = 0
>
> Therefore, sqrt(2) is not transcendental, QED.
>
>> It is a long time since I took math classes, and I have been
>> operating under the (possibly delusional) assumption that a
>> number that cannot be expressed exactly is transcendental.
>
> The word for which you are reaching is "irrational". It is indeed
> the case that sqrt(2) is irrational.

No, I wasn't reaching for 'irrational'. I was misusing
'transcendental'. I had entirely forgotton about roots of integer
polynomial equations.

--
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

In article <471f6cc0.1253142937@news.sbtc.net> cri@tiac.net (Richard Harter) writes:
> On Wed, 24 Oct 2007 10:53:07 GMT, "Dik T. Winter"
> <Dik.Winter@cwi.nl> wrote:
....
> >So 'sqrt(2)' expresses the real number, and there are lots of computers
> >where I can express it that way. 'sqrt(2)' is not different from '123'
> >in that aspect, both give rules on how to calculate the actual number
> >if there is any need.
....
> In practice a real number is any point on the real line; a
> canonical representation is in the form of an infinite sequence
> of digits in any convenient base.

Some practice. Much of what I have done would have given wrong results
if (for instance) sqrt(2) had been given as a finite approximation.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

CBFalconer <cbfalconer@yahoo.com> writes:

> Ben Bacarisse wrote:
>> CBFalconer <cbfalconer@yahoo.com> writes:
<snip>
>>> I agree with Dik. The 'representations' are attempts to express
>>> those values with some combination of digits and numerical base,
>>> which can never be exact, since the values are transcendental.
>>
>> That's a leap! Neither of the number you quote are transcendental
>> although the number that is at the heart of this thread may well be.
>
> Are you claiming sqrt(2) is not transcendental?

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).

--
Ben.

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.

Phil

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...

--
+----------------------------------------------------------------+
| Charles and Francis Richmond richmond at plano dot net |
+----------------------------------------------------------------+