can we define all of transcendent numbers via rational numbers?

[ahmet]

e.g.: [math] \pi = \frac{22}{7} [/math]

(only 4 operation is allowed)

Edited October 18, 2020 by ahmet

[Markus Hanke]

1 hour ago, ahmet said:

e.g.: π=227

(only 4 operation is allowed)

\(\pi\) is not equal to 22/7, this is just a convenient approximation. So no, you can’t express any arbitrary transcendental number via rationals.

[HallsofIvy]

The whole point of "irrational numbers", and the set of transcendal numbers is a subset of the irrational numbers, is that they are NOT  rational numbers and cannot be written as a fraction with integer numerator and denominator.  As Markus Hanke said, 22/7 is an approximation to [math]\pi[/math], it is NOT equal to [tex]\pi[/tex].

(We can "use" rational numbers to define the irrational numbers.  We say that two sequences of rational numbers, [tex]\{a_1, a_2, a_3, \cdot\cdot\cdot \}[/tex] and [tex]\{b_1, b_2, ...\}[/tex] are "equivalent" if and only if the sequence [tex]\{a_1- b_1, a_2- b_2, a_3- b_2, \cdot\cdot\cdot\}[/tex] and then define the real numbers to be the set of equivalence classes of such sequences.  [So the sequence 3, 3.1, 3.14, 3.141, 3.1415, 3.14156, ... is in the eqivalence class we identify as "[tex]\pi[/tex]]).

Edited October 19, 2020 by HallsofIvy

[studiot]

On 10/18/2020 at 5:07 AM, ahmet said:

e.g.: π=227

(only 4 operation is allowed)

Numbers are more fun than just  N  → Z  →  Q  →  R 

Consider the equation   x² - 2 = 0

This has no solution in  N, Z, or Q

In other words there is no rational solution

But let us propose a solution and represent it by the character  £

Now taking inspiration from the form of complex numbers  (a + ib)

let us consider  expressions of the form  (a + £b)

We will find that we have a complete number system that follows all the four rules you mention and the equation has a solution in this number system!

This number system is the usual rational numbers plus one irrational number, £.

I have disguised the fact that we usually call £ the square root of 2.

So number systems exist partway between Q and R.

Our simple system does not contain a solution to the equation x² - 3 = 0 as it cannot write the irrational number square root of 3 in terms of root 2.

 

 

Edited October 19, 2020 by studiot

[ahmet]

On 10/18/2020 at 8:13 AM, Markus Hanke said:

π is not equal to 22/7, this is just a convenient approximation. So no, you can’t express any arbitrary transcendental number via rationals.

yes, this is right. Unfortunately sometimes it is probable for me or maybe someone like me as a specialised one in the area to make very simple failures.

in fact, yes here [math] \pi [/math] is an irrational number.

Edited October 22, 2020 by ahmet

[joigus]

Pi is even worse than irrational. Garden-variety irrationals like sqr(2) are solutions of polynomial equations with rational coefficients.

As @studiot said, you can get some kind of "restricted number system" with them, similar to complex numbers. I think those are called "ideals" --correct me if I'm wrong.

Transcendental numbers cannot even be obtained in that way. Generally you need limits to define them.

The funny thing is pi can be very well approx'd by fractions, yet such humble irrational number as the golden ratio,

\[\varphi=\frac{1+\sqrt{5}}{2}\]

which is the solution to,

\[x^{2}=x+1\]

stubbornly resists approximations by rationals. Its approximations by continued fractions are really, really bad.

https://en.wikipedia.org/wiki/Continued_fraction

It's sometimes referred-to as the most irrational number. Generally irrationals are approached by an expression like,

\[a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\ddots+\frac{1}{a_{n}}}}}\]

For rationals, the expansion ends at finite order. But for the golden ratio it is this bad:

\[\varphi=1+\frac{1}{1+\frac{1}{1+\frac{1}{\ddots}}}\]

All the a_n's are 1.

If you mean "approached", all can be approached; some better than others. But if you really mean "defined," then, as @Markus Hanke says, it's impossible.

Edited October 22, 2020 by joigus

[ahmet]

28 minutes ago, joigus said:

you can get some kind of "restricted number system" with them, similar to complex numbers. I think those are called "ideals" --correct me if I'm wrong.

to begin with,I would thank to you for your post and while I am not sufficiently sure for other branches of science (maybe in physcis) , the keyword given in bold is wrong in mathematics.

And I will try to provide the exact description of that keyword.

what does "ideal" mean in mathematics?

Let E be a ring, and F be another ring with the condition of F< E , 

if;

i)for every x ∈  E and  for every y  ∈ F  x.y ∈ F and 

ii) for every k ,t ∈ F k-t ∈ F. 

then F is called as "ideal" of E. 

the most simple example is presumably this one:  Odd numbers (T) and Even numbers (C) (sets) are each subsets of Z

But C subset is specifically satsfies the given conditions above. Thus C is an ideal of Z (Z shows integer numbers (set))

 

 

 

 

Edited October 22, 2020 by ahmet
spelling error detected, telling errors

[joigus]

9 minutes ago, ahmet said:

to begin with,I would thank to you for your post and while I am not sufficiently sure for other branches of science (maybe in physcis) , the keyword given in bold is wrong in mathematics.

And I will try to provide the exact description of that keyword.

what does "ideal" mean?

Let E be a ring, and F be another ring with the condition of F< E , 

if;

i)for every x ∈  E and and for every y  ∈ F  x.y ∈ F and 

ii) for every k ,t ∈ F k-t ∈ F. 

then F is called as "ideal" of E. 

 

 

 

Oh, I see. Thank you.

It must be some other term.

Do you happen to remember the name of the closed system that Studiot mentioned?

[ahmet]

44 minutes ago, joigus said:

Oh, I see. Thank you.

It must be some other term.

Do you happen to remember the name of the closed system that Studiot mentioned?

hi again,

I commonly believe that I was not good in algebra.But I shall try to represent basic contexts as well as I can. 

first, studiot's comment seems like a solution for some problems.

because to my knowledge , we are allowed to say I < R and Q < R (also,even R<C (here C shows complex numbers))

but in spite of all of the details given above, in existing methods I have not seen at anywhere or have not come across any expression ,claiming Q < I , I was thinking this would be correct from the notes I taken in the university. but generally as; there is no doubt or contradictions between mathematical thorems in general or it is not a common action,  I taught or said to 12nd grade students that any rational number will never be an irrational number. But how to make a clear comparison between I and Q sets. Or how are the mathematical characters of such sets ?

thus, studiot's comment seems like a good solution and may provide a new set (under assumption; all (of (my) ) the claims appearing above was right) between I and R ,which is bigger than I and Q and also includes both of these sets.

 

(some contexts that might potentially be relevant to this subject. 1) yes again under assumption that we would be able to say Q was a subset of I, studiot's solution shows an ideal of R, but he presumably created a new set.  (if so, congratulations) 2) gauss integer numbers ring 3) some other algebraic contexts (E.g. H/K where K is a prime or maximal  ideal of H,not sure)

 

 

Edited October 22, 2020 by ahmet
prefix to a word, adjective / adverb correction on a word. And/or such other changes.Thanks for reading the note.

[joigus]

11 minutes ago, ahmet said:

(some contexts that might potential be relevant to this subject. 1) yes again under assumption that we would be able to say Q was a subset of I, studiot's solution shows an ideal of R, but he presumably created a new set.  (if so, congratulations) 2) gauss integers number region 3) some other algebraic contexts (E.g. H/K where K is a prime or maximal  ideal of H,not sure)

I've known the set for years. There is closure under products:

\[\left(a+b\sqrt{2}\right)\left(c+d\sqrt{2}\right)=ac+2bd+\left(ad+bc\right)\sqrt{2}\]

\[\frac{1}{a+b\sqrt{2}}=\frac{1}{a^{2}+b^{2}}\left(a-b\sqrt{2}\right)\]

for a, b, c, d rational.

My question was: What's its name?

[wtf]

2 hours ago, joigus said:

My question was: What's its name?

https://en.wikipedia.org/wiki/Field_extension#Extension_field

[joigus]

8 hours ago, wtf said:

https://en.wikipedia.org/wiki/Field_extension#Extension_field

Thanks a lot.

[ahmet]

14 hours ago, joigus said:

I've known the set for years. There is closure under products:

 

(a+b2–√)(c+d2–√)=ac+2bd+(ad+bc)2–√

 

 

1a+b2–√=1a2+b2(a−b2–√)

 

for a, b, c, d rational.

My question was: What's its name?

hi,

algebra is not as easy as it appears/stands. Some notions even could be stricter than you might imagine. I have not come across with any well described such expresion. But of course some new groups might be defined with some expressions (like you provided). Also ,gauss ring may be assessed in this regard.but if you provide more contexts, maybe I and some other mathematicians might help you. 

mm,now,among the contexts and when I would overview the big picture under this thread, studiot's this sentence seems like so much pretty good or cute (quoted below). 

 

On 10/20/2020 at 12:11 AM, studiot said:

Numbers are more fun than just  N  → Z  →  Q  →  R 

 

Edited October 23, 2020 by ahmet

[studiot]

28 minutes ago, ahmet said:

hi,

algebra is not as easy as it appears/stands. Some notions even could be stricter than you might imagine. I have not come across with any well described such expresion. But of course some new groups might be defined with some expressions (like you provided). Also ,gauss ring may be assessed in this regard.but if you provide more contexts, maybe I and some other mathematicians might help you. 

mm,now,among the contexts and when I would overview the big picture under this thread, studiot's this sentence seems like so much pretty good or cute (quoted below). 

 

 

Yes there is a great deal more to algebra than is taught in school, indeed there is more than one 'algebra'.

As regards number systems,

N and Z do not form algebraic fields

N fails for additive inverses and Z fails for multiplicative inverses.

Q constitutes a field, that does not contain any irrational or transcental numbers.
Since Q is closed under both the binary operations of multiplication and addition, it is impossible to find a sequence of operations that will lead to an irrational or trancendental number.

R contains all real numbers, including all the irrational and trancendental ones. It is, of course, a field.

As such it also contains (all numbers of the form a + b√2), which therefore correspond to a single number in R

The point that Joigus was moving towards is that these 'compound numbers' is that they form a subfield of R

wtf is quite correct that the name for this is an extension field.

There are many extension fields.

 

Hope this helps clear things up.

[ahmet]

1 hour ago, studiot said:

Yes there is a great deal more to algebra than is taught in school, indeed there is more than one 'algebra'.

As regards number systems,

N and Z do not form algebraic fields

N fails for additive inverses and Z fails for multiplicative inverses.

ahahaha @studiot ,you seem like you criticize the sets pahaha   rather than criticizing people ,well ,very well.   I really appreciate the tone of your tongue. haha.

(like saying; Z is not a successfull set for multiplicative inverses and N already seems like just a stupid set, he even cannot contain additive inverses    )

Meanwhile, 

I saw @wtf  ' s explanations (provided in the link) but I also thought that there would be many extension fields.

mmm may I ask a question , with what are you expresing a ring satisfying all the conditions given below in english? (in fact bourbaki has provided that definition but I could not immediately /quickly find)

---->> with no zero divisor.

---->> commutative

--->> contains unitary element of multiplication

one more question: what do you mathematically call a specific ring with, that has not had any else ideals except its trivial ideals?

Edited October 23, 2020 by ahmet

[studiot]

45 minutes ago, ahmet said:

one more question: what do you mathematically call a specific ring with, that has not had any else ideals except its trivial ideals?

All of this categorisation goes a lot further into pure maths than I normally like to.

But I know you work more on the pure side.

Here are two pages from one of my most worn books, I cannot recommend it highly enough.
I have made the scans largr than usual becasue they have packed a lot of text into each page.
Not only does it provide definitions, but it provides cross references to associated or contrasting terminology to help the reader avoid mixing them up.

So see what it has to say about both ideals and rings.

Collins Reference Dictionary of Mathematics

E J Borowski and J M Borwein

 

[ahmet]

12 minutes ago, studiot said:

All of this categorisation goes a lot further into pure maths than I normally like to.

But I know you work more on the pure side.

hi, I do not know what exactly caused you to think so.

but no problem for now.

meanwhile I do not think that that description was available here. 

a reminding: with trivial ideals I meant for instance in <H,+,.> ring, H and {0H} are trivial ideals. (There should be no more ideals for this ring)

Edited October 23, 2020 by ahmet

[wtf]

2 hours ago, ahmet said:

a reminding: with trivial ideals I meant for instance in <H,+,.> ring, H and {0H} are trivial ideals. (There should be no more ideals for this ring)

If [math]x[/math] is nonzero then [math]x H[/math] is an ideal. If it's not a proper ideal then it's all of the ring, including 1. Hence every nonzero element is invertible. So a commutative ring with no proper ideals is a field. If the ring is not commutative a similar proof would show that it must be a division ring, but you'd have to check both left and right ideals for the proof.

Note that our ring is required to have 1. If not, I'm not sure if this result is still true.

Edited October 23, 2020 by wtf

[ahmet]

1 hour ago, wtf said:

If x is nonzero then xH is an ideal. If it's not a proper ideal then it's all of the ring, including 1. Hence every nonzero element is invertible. So a commutative ring with no proper ideals is a field. If the ring is not commutative a similar proof would show that it must be a division ring, but you'd have to check both left and right ideals for the proof.

Note that our ring is required to have 1. If not, I'm not sure if this result is still true.

welcome @wtf,

.I think I have done a mistake. (But some terms changing between languages, so I am not sure whether I would express all terms correctly (or equiavlently)

Description: if <H,+,.> has 1H  and  has no zero divisors, and is commutative, then we call this ring with " completeness field" (I translated from turkish into english that term)

Description: if H is commutative and has 1_H and for every x element H-{0_H}, at least y element H-{0_H} such that x.y=1H  conditions are being satisfied then we call this completeness field specifically with "object" (but not sure about the correctness of this term)( or equivalently this ring should satisfy group criteria for the second operation (multiplication))

 

All Q,R,C are objects. 

also, all finite completeness fields are also objects. 

Q,R and C should have no more than their trivial ideals

 

 

Edited October 23, 2020 by ahmet

[studiot]

All this seems now to be going well beyond the original posted question, which I believe I have answered within the specified number systems (which do not go as far a C).

I will leave you in wtf's capable pen hand to discuss rarified algebraic theory and vocabulary.

[wtf]

1 hour ago, ahmet said:

if <H,+,.> has 1H  and  has no zero divisors, and is commutative, then we call this ring with " completeness field" (I translated from turkish into english that term)

I don't know what you mean by "completeness" field. Can't even guess, I'm afraid. A commutative ring with unity in which every nonzero element is invertible is a field. Sometimes a "corpus" in French, I believe. 

A noncommutative ring with unity and multiplicative inverses is a division ring, called in older books a skew field.

 

 

1 hour ago, ahmet said:

Description: if H is commutative and has 1_H and for every x element H-{0_H}, at least y element H-{0_H} such that x.y=1H  conditions are being satisfied then we call this completeness field specifically with "object" (but not sure about the correctness of this term)( or equivalently this ring should satisfy group criteria for the second operation (multiplication))

Same remark. Don't know what you mean by object or completeness field.

An ordered field with the least upper bound property is complete. So the reals are complete but not the rationals, if that is what you mean. 

1 hour ago, ahmet said:

All Q,R,C are objects. 

also, all finite completeness fields are also objects. 

Don't know what you mean by objects, unless perhaps you mean sets. Or fields. Completeness field is a bad translation and I can't even make a guess as to what it is intended to mean.

 

51 minutes ago, studiot said:

I will leave you in wtf's capable pen hand to discuss rarified algebraic theory and vocabulary.

FWIW this is first-semester undergrad abstract algebra. Even the computer science majors need to know some of it, because finite fields are a big thing in CS. 

I agree that we're way past the original question.

Edited October 23, 2020 by wtf

[studiot]

3 minutes ago, wtf said:

FWIW this is first-semester undergrad abstract algebra. Even the computer science majors need to know some of it, because finite fields are a big thing in CS. 

I agree that we're way past the original question.

Thanks for your support for my colourful language, by supplying formally correct terminology.  +1

Edited October 23, 2020 by studiot

[ahmet]

12 hours ago, wtf said:

Don't know what you mean by objects, unless perhaps you mean sets. Or fields. Completeness field is a bad translation and I can't even make a guess as to what it is intended to mean.

as I remember Z and Q are not in the same category. One of them should be different.

Menwhile, to you; which type of fields do you claim? there are many types of fields (E.g. P.F.F. , Euclid field or completeness field etc.)

Anyway, It seems disccussing such contexts under another thread will be better. 

 

[studiot]

1 hour ago, ahmet said:

as I remember Z and Q are not in the same category. One of them should be different.

Depends what you mean.

I have already noted that Z is not a field, but Q is.

 

One problem I find is that so many English words have a specialized meaning in some part of Maths it is difficult know whcih is meant and how to find a substitute when you wnat the general English meaning.

Category is just such a word.

https://en.wikipedia.org/wiki/Category_theory

[wtf]

9 hours ago, studiot said:

Category is just such a word.

https://en.wikipedia.org/wiki/Category_theory

Who's rarified now LOL.