Defining number set sequences based on the fundamental theorem of arithmetic
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The Fundamental theorem of arithmetic gives a natural way to write natural numbers $mathbb{N}$ uniquely in terms of products in powers of primes:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{N}$ with $n_i in mathbb{N}$
This can be generalized to rational numbers:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{Q}$ with $n_i in mathbb{Z}$
Can this scheme be generalized on more ways?
For example, is not clear that algebraic irrationals are contained in the set of
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ with $n_i in mathbb{Q}$
but clearly this set is contained in the algebraic irrationals union rationals.
In a parallel fashion, if we take imaginary unit $i$ as a new element of the prime set, we can rotate in any dense direction in the complex plane by taking all rational exponents
One in principle could define a family of sets in a recursive way:
$$mathbb{T}_{i+1} = big{ N = 2^{n_2} 3^{n_3} 5^{n_5} ldots n_j in mathbb{T}_i big } $$
with $mathbb{T}_0 = mathbb{Z}$
And ask, do transcendental numbers like $pi$ or $e$ belong to $mathbb{T}_i$ for some $i$?
number-theory complex-numbers irrational-numbers transcendental-numbers
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
$endgroup$
add a comment |
$begingroup$
The Fundamental theorem of arithmetic gives a natural way to write natural numbers $mathbb{N}$ uniquely in terms of products in powers of primes:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{N}$ with $n_i in mathbb{N}$
This can be generalized to rational numbers:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{Q}$ with $n_i in mathbb{Z}$
Can this scheme be generalized on more ways?
For example, is not clear that algebraic irrationals are contained in the set of
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ with $n_i in mathbb{Q}$
but clearly this set is contained in the algebraic irrationals union rationals.
In a parallel fashion, if we take imaginary unit $i$ as a new element of the prime set, we can rotate in any dense direction in the complex plane by taking all rational exponents
One in principle could define a family of sets in a recursive way:
$$mathbb{T}_{i+1} = big{ N = 2^{n_2} 3^{n_3} 5^{n_5} ldots n_j in mathbb{T}_i big } $$
with $mathbb{T}_0 = mathbb{Z}$
And ask, do transcendental numbers like $pi$ or $e$ belong to $mathbb{T}_i$ for some $i$?
number-theory complex-numbers irrational-numbers transcendental-numbers
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
$endgroup$
$begingroup$
I am not giving you a solution but it can be done using programing. Wolfram alpha and other sites are best to check your predictions.
$endgroup$
– user627824
Dec 19 '18 at 10:44
1
$begingroup$
Also the smallest field $F$ containing $mathbb{Q}$ and closed under $x mapsto x^{1/m}$ for every $min mathbb{Z}$ is well-known, this is the largest radical extension of $mathbb{Q}$, whose Galois group is the largest solvable quotient group of $Gal(overline{mathbb{Q}}/mathbb{Q})$. That is $F$ is the field containing all the towers of cyclic extensions of the full cyclotomic field $bigcup_n mathbb{Q}(zeta_{n})$.
$endgroup$
– reuns
Dec 19 '18 at 11:29
add a comment |
$begingroup$
The Fundamental theorem of arithmetic gives a natural way to write natural numbers $mathbb{N}$ uniquely in terms of products in powers of primes:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{N}$ with $n_i in mathbb{N}$
This can be generalized to rational numbers:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{Q}$ with $n_i in mathbb{Z}$
Can this scheme be generalized on more ways?
For example, is not clear that algebraic irrationals are contained in the set of
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ with $n_i in mathbb{Q}$
but clearly this set is contained in the algebraic irrationals union rationals.
In a parallel fashion, if we take imaginary unit $i$ as a new element of the prime set, we can rotate in any dense direction in the complex plane by taking all rational exponents
One in principle could define a family of sets in a recursive way:
$$mathbb{T}_{i+1} = big{ N = 2^{n_2} 3^{n_3} 5^{n_5} ldots n_j in mathbb{T}_i big } $$
with $mathbb{T}_0 = mathbb{Z}$
And ask, do transcendental numbers like $pi$ or $e$ belong to $mathbb{T}_i$ for some $i$?
number-theory complex-numbers irrational-numbers transcendental-numbers
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
$endgroup$
The Fundamental theorem of arithmetic gives a natural way to write natural numbers $mathbb{N}$ uniquely in terms of products in powers of primes:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{N}$ with $n_i in mathbb{N}$
This can be generalized to rational numbers:
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ for all $N in mathbb{Q}$ with $n_i in mathbb{Z}$
Can this scheme be generalized on more ways?
For example, is not clear that algebraic irrationals are contained in the set of
$$N = 2^{n_2} 3^{n_3} 5^{n_5} ldots $$ with $n_i in mathbb{Q}$
but clearly this set is contained in the algebraic irrationals union rationals.
In a parallel fashion, if we take imaginary unit $i$ as a new element of the prime set, we can rotate in any dense direction in the complex plane by taking all rational exponents
One in principle could define a family of sets in a recursive way:
$$mathbb{T}_{i+1} = big{ N = 2^{n_2} 3^{n_3} 5^{n_5} ldots n_j in mathbb{T}_i big } $$
with $mathbb{T}_0 = mathbb{Z}$
And ask, do transcendental numbers like $pi$ or $e$ belong to $mathbb{T}_i$ for some $i$?
number-theory complex-numbers irrational-numbers transcendental-numbers
number-theory complex-numbers irrational-numbers transcendental-numbers
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
edited Sep 11 '18 at 18:53
lurscher
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
asked Sep 11 '18 at 18:48
lurscherlurscher
327215
327215
$begingroup$
I am not giving you a solution but it can be done using programing. Wolfram alpha and other sites are best to check your predictions.
$endgroup$
– user627824
Dec 19 '18 at 10:44
1
$begingroup$
Also the smallest field $F$ containing $mathbb{Q}$ and closed under $x mapsto x^{1/m}$ for every $min mathbb{Z}$ is well-known, this is the largest radical extension of $mathbb{Q}$, whose Galois group is the largest solvable quotient group of $Gal(overline{mathbb{Q}}/mathbb{Q})$. That is $F$ is the field containing all the towers of cyclic extensions of the full cyclotomic field $bigcup_n mathbb{Q}(zeta_{n})$.
$endgroup$
– reuns
Dec 19 '18 at 11:29
add a comment |
$begingroup$
I am not giving you a solution but it can be done using programing. Wolfram alpha and other sites are best to check your predictions.
$endgroup$
– user627824
Dec 19 '18 at 10:44
1
$begingroup$
Also the smallest field $F$ containing $mathbb{Q}$ and closed under $x mapsto x^{1/m}$ for every $min mathbb{Z}$ is well-known, this is the largest radical extension of $mathbb{Q}$, whose Galois group is the largest solvable quotient group of $Gal(overline{mathbb{Q}}/mathbb{Q})$. That is $F$ is the field containing all the towers of cyclic extensions of the full cyclotomic field $bigcup_n mathbb{Q}(zeta_{n})$.
$endgroup$
– reuns
Dec 19 '18 at 11:29
$begingroup$
I am not giving you a solution but it can be done using programing. Wolfram alpha and other sites are best to check your predictions.
$endgroup$
– user627824
Dec 19 '18 at 10:44
$begingroup$
I am not giving you a solution but it can be done using programing. Wolfram alpha and other sites are best to check your predictions.
$endgroup$
– user627824
Dec 19 '18 at 10:44
1
1
$begingroup$
Also the smallest field $F$ containing $mathbb{Q}$ and closed under $x mapsto x^{1/m}$ for every $min mathbb{Z}$ is well-known, this is the largest radical extension of $mathbb{Q}$, whose Galois group is the largest solvable quotient group of $Gal(overline{mathbb{Q}}/mathbb{Q})$. That is $F$ is the field containing all the towers of cyclic extensions of the full cyclotomic field $bigcup_n mathbb{Q}(zeta_{n})$.
$endgroup$
– reuns
Dec 19 '18 at 11:29
$begingroup$
Also the smallest field $F$ containing $mathbb{Q}$ and closed under $x mapsto x^{1/m}$ for every $min mathbb{Z}$ is well-known, this is the largest radical extension of $mathbb{Q}$, whose Galois group is the largest solvable quotient group of $Gal(overline{mathbb{Q}}/mathbb{Q})$. That is $F$ is the field containing all the towers of cyclic extensions of the full cyclotomic field $bigcup_n mathbb{Q}(zeta_{n})$.
$endgroup$
– reuns
Dec 19 '18 at 11:29
add a comment |
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$begingroup$
I am not giving you a solution but it can be done using programing. Wolfram alpha and other sites are best to check your predictions.
$endgroup$
– user627824
Dec 19 '18 at 10:44
1
$begingroup$
Also the smallest field $F$ containing $mathbb{Q}$ and closed under $x mapsto x^{1/m}$ for every $min mathbb{Z}$ is well-known, this is the largest radical extension of $mathbb{Q}$, whose Galois group is the largest solvable quotient group of $Gal(overline{mathbb{Q}}/mathbb{Q})$. That is $F$ is the field containing all the towers of cyclic extensions of the full cyclotomic field $bigcup_n mathbb{Q}(zeta_{n})$.
$endgroup$
– reuns
Dec 19 '18 at 11:29