A set related to the algebraic numbers: What is it?
Ted Shoemaker
I used to think, mistakenly, that an algebraic number was:
a number that could be obtained by a finite application of
arithmetic operations (adding, subtracting, multiplying,
dividing, and taking roots) on the integers.
Then someone on this newsgroup corrected me. Actually, an
algebraic number is: a number that is a solution of a
polynomial equation.
But I'm still left with that first definition. I don't know
what kind of numbers are defined by it. For lack of the
correct name, I will call them the "arithmetic numbers"
until I learn better.
Three questions:
1. What is the correct name for the "arithmetic numbers"?
2. Can someone give an example of an arithmetic number that
is not an algebraic number?
3. Can someone give me an example of an algebraic number that
is not an arithmetic number?
Thank you very much!
Ted Shoemaker
shoema...@yahoo.com
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Mike Oliver
>
> I used to think, mistakenly, that an algebraic number was:
> a number that could be obtained by a finite application of
> arithmetic operations (adding, subtracting, multiplying,
> dividing, and taking roots) on the integers.
> correct name, I will call them the "arithmetic numbers"
> until I learn better.
>
> Three questions:
> 1. What is the correct name for the "arithmetic numbers"?
They are called "surds".
> 2. Can someone give an example of an arithmetic number that
> is not an algebraic number?
No. All surds are algebraic.
> 3. Can someone give me an example of an algebraic number that
> is not an arithmetic number?
Take some fifth-degree polynomial which has no solution in surds
(Galois proved there are such; I don't remember a specific example
but no doubt someone here can provide one.) Any root of that
polynomial is an algebraic number, but not a surd.
John Savard
<shoema...@yahoo.com> wrote, in part:
>2. Can someone give an example of an arithmetic number that
>is not an algebraic number?
No.
>3. Can someone give me an example of an algebraic number that
>is not an arithmetic number?
The roots of x^5 + x + 1 would be examples.
Gerry Myerson
<shoema...@yahoo.com> wrote:
=> ... an algebraic number is: a number that is a solution of a
=> polynomial equation.
Careful - it's a number that is a solution of a polynomial equation,
the polynomial having integer coefficients.
Gerry Myerson (ge...@mpce.mq.edu.au)
Herman Jurjus
"Mike Oliver" <oli...@math.ucla.edu> wrote in message
news:3BEF2BC2...@math.ucla.edu...
> Ted Shoemaker wrote:
> >
> > I used to think, mistakenly, that an algebraic number was:
> > a number that could be obtained by a finite application of
> > arithmetic operations (adding, subtracting, multiplying,
> > dividing, and taking roots) on the integers.
> > [...] For lack of the
> > correct name, I will call them the "arithmetic numbers"
> > until I learn better.
> >
> > Three questions:
> > 1. What is the correct name for the "arithmetic numbers"?
>
> They are called "surds".
Are *all* roots are allowed, or only square roots?
For the latter variant I heard the term "constructible numbers"; and a
(rather famous) counter example would simply be 2^(1/3). The term
constructible is kind of shorthand for 'constructible with ruler and
compass'.
Regards,
Herman
Jon Haugsand
> (Galois proved there are such; I don't remember a specific example
> but no doubt someone here can provide one.) Any root of that
> polynomial is an algebraic number, but not a surd.
My national pride demands me to insert a minor correction. In fact,
it was the
Norwegian mathematician Niels Henrik Abel who proved the insolubility
of polynomials of fifth degree. Galois knew very well of Abel's
results, he even wrote letters from prison where he criticised the
French establishment for the treatment they gave Abel.
Of course, Galois's theory is much wider and is what we use today.
--
Jon Haugsand
Jon-H.H...@norges-bank.no
Gérard Nin
If C is the set of these numbers, C is the smallest sub-field of R (the real
numbers) which is invariant by the square-root function.
Geometricaly , C is the set of coordinates of points which are obtained by
lines and circles intersections.
The set of the algebraic numbers A contains C, so it's not possible to find
a constructible number which is not algebraic.
A is the set of all numbers which are solutions of polynomial equations,
where the polynomials have rationnel (or integer its equivalent)
coefficients.
A famous theorem established by a german mathematicien Wantzel, says that a
necessary condition for a number b to be a constructible number is that is
minimal polynomial (ie the polynomial of Q[X] with smallest degree which
vanished at b) have a degree of the form 2^n. So, cubic-root of 2 which
minimal polynomial is X^3 - 2, is not a constructible number, but, of course
an algebraic number. But the condition is not suffisant.
The fact that 2^(1/3) is not a constructible number is the deep reason why
it's not possible with a ruler and a compass to draw a segment which lenght
is 2^(1/3). For the same reason it's not possible to construct with ruler
and compass the side of a cube which volum is twice the volum of a given
cube. It was a famous Greek problem the problem of the "cube duplication"
(Greeks wanted to construct a temple for their god Apollon which would be
twice biggest than another temple).
Excuse my bad english, Nico.
"Ted Shoemaker" <shoema...@yahoo.com> a écrit dans le message news:
3BEF250F...@yahoo.com...
Dave Seaman
>The name of your "arithmetic numbers" is "constructible numbers".
No. The cube root of 2 is an example of an "arithmetic number" that is not
constructible.
--
Dave Seaman dse...@purdue.edu
Amnesty International calls for new trial for Mumia Abu-Jamal
<http://www.amnestyusa.org/abolish/reports/mumia/>
Michael Barr
> >
> > I used to think, mistakenly, that an algebraic number was:
> > a number that could be obtained by a finite application of
> > arithmetic operations (adding, subtracting, multiplying,
> > dividing, and taking roots) on the integers.
> > [...] For lack of the
> > correct name, I will call them the "arithmetic numbers"
> > until I learn better.
> >
> > Three questions:
> > 1. What is the correct name for the "arithmetic numbers"?
>
> They are called "surds".
I didn't know that name. I had always called them radicals, since we
always talked about solutions by radicals. I guess that means that
the numbers with non-solvable Galois group should be called absurd.
Someone below mentions the roots of x^5 + x + 1 as absurd. As a
matter of fact, in a certain precise sense, almost every polynomial of
degree n > 4 has either S_n or A_n as its Galois group and therefore
is not solvable by radicals. So the binomial theorem (known to the
Babylonians), Cardano's formula for the cubic and Tartaglia's for the
quartic, are the end of the line as far as solving polynomial
equations by radicals.
A N Neil
> => ... an algebraic number is: a number that is a solution of a
> => polynomial equation.
>
> Careful - it's a number that is a solution of a polynomial equation,
> the polynomial having integer coefficients.
If you insist on being careful, then say it is a nonzero polynomial.
Jon Miller
> Someone below mentions the roots of x^5 + x + 1 as absurd. As a matter of fact, in a certain
> precise sense, almost every polynomial of degree n > 4 has either S_n or A_n as its Galois
> group and therefore is not solvable by radicals. So the binomial theorem (known to the
> Babylonians)
I assume you mean the quadratic formula
> , Cardano's formula for the cubic and Tartaglia's for the quartic, are the end of the line as
> far as solving polynomial equations by radicals.
Well, not entirely. Some higher degree polynomials are solvable by radicals, and for those there
are formulas. I am sure some work has been done on this (meaning I remember having seen some
stuff on it, but I'm not going to look it up). It is, however, not considered important.
Jon Miller
Chan-Ho Suh
"Jon Miller" <jonatha...@home.com> wrote in message
news:3BF0085D...@home.com...
I don't think Michael was asserting that no higher degree polynomials are
solvable by radicals, but that there is no _general_ formula for higher
degree (5 or greater) polynomials. The formulas you speak of are not
considered important for the precise reason Michael mentioned above: almost
every polynomial of degree greater than 4 is not solvable by radicals.
Chan-Ho Suh
"A N Neil" <ann...@nym.alias.net> wrote in message
news:121120011150165128%ann...@nym.alias.net...
To be even more careful, one should say rational integer coefficients. :)
Gerry Myerson
ba...@barrs.org (Michael Barr) wrote:
=> I guess that means that
=> the numbers with non-solvable Galois group should be called absurd.
=>
=> Someone below mentions the roots of x^5 + x + 1 as absurd.
Someone chose a bad example.
x^5 + x + 1 = (x^2 + x + 1) (x^3 - x^2 + 1).
Gerry Myerson (ge...@mpce.mq.edu.au)
Bill Taylor
|> Are *all* roots are allowed, or only square roots?
|> For the latter variant I heard the term "constructible numbers";
*I* think they oughta be called "geometric numbers"!
Accurate, and goes nicely with "alegbraic numbers".
------------------------------------------------------------------------------
Bill Taylor W.Ta...@math.canterbury.ac.nz
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Exam: The Projective Plane
Candidates must write on one side of the paper only
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.
Chan-Ho Suh
"Gerry Myerson" <ge...@mpce.mq.edu.au> wrote in message
news:gerry-FEF31B.13001113112001@[137.111.1.11]...
> In article <19daab49.01111...@posting.google.com>,
> ba...@barrs.org (Michael Barr) wrote:
>
> => I guess that means that
> => the numbers with non-solvable Galois group should be called absurd.
> =>
> => Someone below mentions the roots of x^5 + x + 1 as absurd.
>
> Someone chose a bad example.
>
> x^5 + x + 1 = (x^2 + x + 1) (x^3 - x^2 + 1).
>
Try x^5 -x -1. Good job Gerry for spotting this!
Ted Shoemaker
Thank you to all who replied. Because I did not make myself clear
enough
the first time, I will re-state part of the question.
Here is the definition of a set:
All numbers which can be obtained as the result of a finite
number of applications of adding, subtracting, multiplying,
dividing, exponents, and root-taking on the integers.
(Note: There are many integers other than "2". Therefore,
taking the nth root is permissible under this definition, as
long as
(1) n is an integer, or derivable herein from the integers,
(2) we don't violate any conventional rules, e.g. 1/0, 0^0 , etc.,
and
(3) we stop after a finite number of steps.)
If I am correctly reading the posts, this describes the set of surds,
but not the set of constructibles.
Consider a number of the form
( ... (( a^b + c^d )^z + f^g)^y + ... + m^n) ^ p
with all given variables being rational.
In other words, a messy mix of roots and sums. In general, is such a
thing going to be an algebraic number? My guess is that it isn't.
But in case it is, let's make matters worse, and now use this horrendous
number as a divisor or exponent. (This is still permitted in the
definition above.) Are all such things going to be algebraic?
Now: Is the given set (defined above), which I think is the surds,
a subset of the algebraics, a superset of the algebraics, or neither?
Thank you again,
Denis Feldmann
>
> Thank you to all who replied. Because I did not make myself clear
> enough
> the first time, I will re-state part of the question.
>
> Here is the definition of a set:
> All numbers which can be obtained as the result of a finite
> number of applications of adding, subtracting, multiplying,
> dividing, exponents, and root-taking on the integers.
> (Note: There are many integers other than "2". Therefore,
> taking the nth root is permissible under this definition, as
> long as
> (1) n is an integer, or derivable herein from the integers,
Try to express yourself *even* more clearly.
> (2) we don't violate any conventional rules, e.g. 1/0, 0^0 , etc.,
0^0 is a conventional rule (it is defined as 1 : read the FAQ)
> and
> (3) we stop after a finite number of steps.)
>
> If I am correctly reading the posts, this describes the set of surds,
> but not the set of constructibles.
But, actually , "the set of surds" is not a standard denomination
>
> Consider a number of the form
> ( ... (( a^b + c^d )^z + f^g)^y + ... + m^n) ^ p
> with all given variables being rational.
> In other words, a messy mix of roots and sums. In general, is such a
> thing going to be an algebraic number? My guess is that it isn't.
Read the replies again. This is necessarily algebraic. And stop guessing.
> But in case it is, let's make matters worse, and now use this horrendous
> number as a divisor
Doesn't change anything
or exponent. (This is still permitted in the
> definition above.)
Not clearly. Anyway, this changes a lot. It is well known that 2^sqrt2 is
transcendant (Gelfond-Schneider)
>Are all such things going to be algebraic?
> Now: Is the given set (defined above), which I think is the surds,
No . 2^(sqrt 2) was never a surd. Anyway, how do *you* define it?
> a subset of the algebraics, a superset of the algebraics, or neither?
Probably neither, but this is a guess (albeit a *very* good bet). I am not
sure it is possible to prove that no number in this construction is a root
of x^5-x-1. OTOH, a lot of those numbers are provably transcendant.
But those numbers (the smallest real field closed under exponentiation) are
probably not really interesting : exponentiation by irrational exponents is
not a *nice* operation (and not an algebraic one, ayway).
derek Kirsch
how about 2^(1/2) is it Constructible ?
Dave Seaman
derek Kirsch <dki...@datelnet.net> wrote:
>so, 2^(1/3) isn't consructible.
>how about 2^(1/2) is it Constructible ?
It's the diagonal of a unit square.
Gerry Myerson
mat...@math.canterbury.ac.nz (Bill Taylor) wrote:
=> "Herman Jurjus" <hhju...@hetnet.nl> writes:
=>
=> |> Are *all* roots are allowed, or only square roots?
=> |> For the latter variant I heard the term "constructible numbers";
=>
=>
=> *I* think they oughta be called "geometric numbers"!
=>
=> Accurate, and goes nicely with "alegbraic numbers".
Many years ago, I had the thought that if roots of algebraic functions
with rational coefficients are algebraic numbers then roots of analytic
functions with integer coefficients (that it, power series with
rational coefficients) should be called analytic numbers, and that's
what analytic number theory should be about. E.g., pi is an analytic
number since sin pi = 0.
Somehow, the idea never got anywhere.
Gerry Myerson (ge...@mpce.mq.edu.au)
Bill Taylor
|> Many years ago, I had the thought that if roots of algebraic functions
|> with rational coefficients are algebraic numbers then roots of analytic
|> functions with integer coefficients (that it, power series with
|> rational coefficients) should be called analytic numbers, and that's
|> what analytic number theory should be about. E.g., pi is an analytic
|> number since sin pi = 0.
|>
|> Somehow, the idea never got anywhere.
No smiley, so I guess this is a serious post? If not, I'm the sucker.
But surely - aren't *all* reals analytic that way? Certainly all reals
in [-1,1] are.
Given t with -1 < t < 1 just construct SUM(r_i x^i) with the r_i
chosen so that the partial sums of SUM(r_i t^i) go above and below 0
as often as possible. It can even be done with all the coefficients
being 1 or -1. And for any real at all, just use tiny coefficients.
Oh this is all too obvious... of course I'm the sucker. Silly me!
I wonder though - can any real t at all be achieved this way using only
*integer* coefficients? Hmmm... no; no convergence outside [-1,1] then.
Yeah - nothing comes of it. Silly me.
====
Anon
====
default
Denis Feldmann
"default" <no...@none.org> a écrit dans le message news:
3BF2011B...@none.org...
> Ruffini was earlier.
Yes. But nobody seems to know this.
The whole story (Ruffini, Abel, Galois, say) reads like a mystery novel (try
http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Abel.html ,
http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Ruffini.html
, etc.)
Somehow , it feels like Something (E.T. ?) didn't want humankind to learn
about the unsolvabiliy of the quintic. Those three outstanding
mathematicians had incredibly romantic sad lives, and their work was lost,
not understood, rejected, etc. in almost unbelievable circumstances.
John R Ramsden
>
> "A N Neil" <ann...@nym.alias.net> wrote:
> >
> > > => ... an algebraic number is: a number that is a solution of a
> > > => polynomial equation.
> > >
> > > Careful - it's a number that is a solution of a polynomial equation,
> > > the polynomial having integer coefficients.
> >
> > If you insist on being careful, then say it is a nonzero polynomial.
>
> To be even more careful, one should say rational integer coefficients. :)
There's no need to be that careful, because a root of a polynomial with
algebraic integer coefficients is also a root of some polynomial with
rational integer coefficients.
(To be extra careful though, I guess one might need to add that only a
finite number of the coefficients, but at least one, must be non-zero.)
Cheers
---------------------------------------------------------------------------
John R Ramsden (j...@redmink.demon.co.uk)
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The new is in the old concealed, the old is in the new revealed.
St Augustine.
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Neo 1061
<ge...@mpce.mq.edu.au> jacked into the Matrix and remarked thus:
I'd conjecture that these "analytic numbers" are probably the whole
real line. If not, probably the non-Liouville numbers or something
"large" like that. There are countably many polynomials with integer
coefficients, but uncountably many power series with integer or with
rational coefficients, as there are uncountably many infinite
sequences of these. (There are uncountably many infinite sequences of
0s and 1s!) Also, not all rational power series can be turned into
integer ones. The limit of the lcm of the denominators of the first n
terms as n goes to infinity might not exist -- the denominators in
lowest terms might become arbitrarily large.
Proving the above conjecture seems difficult, but the best approach is
probably to take an arbitrary real number x in [0, 1] and write it as
a0 + a1*2 + a2*4 + ..., ai in {0, 1}, and try to construct a power
series in terms of the ai that has x as a root. If this works for any
x, the conjecture is proved, since there is surely some manipulation
of power series to shift the roots by any given integer that preserves
having rational coefficients. (If not, the so-called analytic numbers
don't form a subfield and are of considerably lesser interest anyway.)
--
In my short time on earth, I've come to realize that male/female
attraction ultimately comes down to anatomy. I've found that most
men are attracted to breasts and behinds while most women are
attracted to assholes. -- Anthony Fehring