Mixing
Density of the set ${n^{frac{1}{q}}:qin N}$ in the natural numbers where $n>1$
$begingroup$
I know that any natural number greater than $1$, $n>1$, has an uncountably infinite number of roots such that ${n^{frac{1}{q}}:qin N}$
in the natural numbers and hence no counting function since
$lim_{qtoinfty}{n^{frac{1}{q}}}=1$.
My first question is, what type of density does the set of roots of any $n>1$ have in the natural numbers?
Does density only apply to sets of integers? My second question is, can this set's density be determined or estimated if it exists?
I am stumped because there is no way to count how many roots of a number lie between $1$ and another arbitrarily large natural number. I understand that the asymptotic density of the set of perfect squares is $0$ since the limit of the number of perfect squares less than or equal to a given value over that value tends to $0$ as the value gets arbitrarily large, but I know that such a limit cannot be applied here.
number-theory asymptotics
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
$endgroup$
add a comment |
$begingroup$
I know that any natural number greater than $1$, $n>1$, has an uncountably infinite number of roots such that ${n^{frac{1}{q}}:qin N}$
in the natural numbers and hence no counting function since
$lim_{qtoinfty}{n^{frac{1}{q}}}=1$.
My first question is, what type of density does the set of roots of any $n>1$ have in the natural numbers?
Does density only apply to sets of integers? My second question is, can this set's density be determined or estimated if it exists?
I am stumped because there is no way to count how many roots of a number lie between $1$ and another arbitrarily large natural number. I understand that the asymptotic density of the set of perfect squares is $0$ since the limit of the number of perfect squares less than or equal to a given value over that value tends to $0$ as the value gets arbitrarily large, but I know that such a limit cannot be applied here.
number-theory asymptotics
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
$endgroup$
$begingroup$
What do you mean by root? Is ${n^{frac{1}{q}}:qinmathbb{N}}$ the set of roots, or is it ${n^{frac{1}{q}}:qin(0,infty)}$ or something else altogether? If ${n^{frac{1}{q}}:qinmathbb{N}}$ the answer would be no, because $n^{frac{1}{q}}leq n$ so $n+1$ cannot be a limit point. If ${n^{frac{1}{q}}:qin(0,infty)}$, then the answer is yes, since after all $q=log_m n$ gives $n^{frac{1}{q}}=m$ precisely for each natural $m>1$.
$endgroup$
– nathan.j.mcdougall
Jan 11 at 0:26
$begingroup$
@nathan.j.mcdougall I am referring to the former.
$endgroup$
– Gnumbertester
Jan 11 at 0:41
1
$begingroup$
Your first statement is incorrect. As we know the pairs of naturals are countable, so are the numbers of the form $n^{1/q}$ In the reals, each number $n$ has at most $2 q^{th}$ roots. It is only in the complex numbers that $n^{1/q}$ has $q$ values.
$endgroup$
– Ross Millikan
Jan 11 at 3:28
add a comment |
$begingroup$
I know that any natural number greater than $1$, $n>1$, has an uncountably infinite number of roots such that ${n^{frac{1}{q}}:qin N}$
in the natural numbers and hence no counting function since
$lim_{qtoinfty}{n^{frac{1}{q}}}=1$.
My first question is, what type of density does the set of roots of any $n>1$ have in the natural numbers?
Does density only apply to sets of integers? My second question is, can this set's density be determined or estimated if it exists?
I am stumped because there is no way to count how many roots of a number lie between $1$ and another arbitrarily large natural number. I understand that the asymptotic density of the set of perfect squares is $0$ since the limit of the number of perfect squares less than or equal to a given value over that value tends to $0$ as the value gets arbitrarily large, but I know that such a limit cannot be applied here.
number-theory asymptotics
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
$endgroup$
I know that any natural number greater than $1$, $n>1$, has an uncountably infinite number of roots such that ${n^{frac{1}{q}}:qin N}$
in the natural numbers and hence no counting function since
$lim_{qtoinfty}{n^{frac{1}{q}}}=1$.
My first question is, what type of density does the set of roots of any $n>1$ have in the natural numbers?
Does density only apply to sets of integers? My second question is, can this set's density be determined or estimated if it exists?
I am stumped because there is no way to count how many roots of a number lie between $1$ and another arbitrarily large natural number. I understand that the asymptotic density of the set of perfect squares is $0$ since the limit of the number of perfect squares less than or equal to a given value over that value tends to $0$ as the value gets arbitrarily large, but I know that such a limit cannot be applied here.
number-theory asymptotics
number-theory asymptotics
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
edited Jan 11 at 0:41
Gnumbertester
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
asked Jan 11 at 0:03
GnumbertesterGnumbertester
545112
545112
$begingroup$
What do you mean by root? Is ${n^{frac{1}{q}}:qinmathbb{N}}$ the set of roots, or is it ${n^{frac{1}{q}}:qin(0,infty)}$ or something else altogether? If ${n^{frac{1}{q}}:qinmathbb{N}}$ the answer would be no, because $n^{frac{1}{q}}leq n$ so $n+1$ cannot be a limit point. If ${n^{frac{1}{q}}:qin(0,infty)}$, then the answer is yes, since after all $q=log_m n$ gives $n^{frac{1}{q}}=m$ precisely for each natural $m>1$.
$endgroup$
– nathan.j.mcdougall
Jan 11 at 0:26
$begingroup$
@nathan.j.mcdougall I am referring to the former.
$endgroup$
– Gnumbertester
Jan 11 at 0:41
1
$begingroup$
Your first statement is incorrect. As we know the pairs of naturals are countable, so are the numbers of the form $n^{1/q}$ In the reals, each number $n$ has at most $2 q^{th}$ roots. It is only in the complex numbers that $n^{1/q}$ has $q$ values.
$endgroup$
– Ross Millikan
Jan 11 at 3:28
add a comment |
$begingroup$
What do you mean by root? Is ${n^{frac{1}{q}}:qinmathbb{N}}$ the set of roots, or is it ${n^{frac{1}{q}}:qin(0,infty)}$ or something else altogether? If ${n^{frac{1}{q}}:qinmathbb{N}}$ the answer would be no, because $n^{frac{1}{q}}leq n$ so $n+1$ cannot be a limit point. If ${n^{frac{1}{q}}:qin(0,infty)}$, then the answer is yes, since after all $q=log_m n$ gives $n^{frac{1}{q}}=m$ precisely for each natural $m>1$.
$endgroup$
– nathan.j.mcdougall
Jan 11 at 0:26
$begingroup$
@nathan.j.mcdougall I am referring to the former.
$endgroup$
– Gnumbertester
Jan 11 at 0:41
1
$begingroup$
Your first statement is incorrect. As we know the pairs of naturals are countable, so are the numbers of the form $n^{1/q}$ In the reals, each number $n$ has at most $2 q^{th}$ roots. It is only in the complex numbers that $n^{1/q}$ has $q$ values.
$endgroup$
– Ross Millikan
Jan 11 at 3:28
$begingroup$
What do you mean by root? Is ${n^{frac{1}{q}}:qinmathbb{N}}$ the set of roots, or is it ${n^{frac{1}{q}}:qin(0,infty)}$ or something else altogether? If ${n^{frac{1}{q}}:qinmathbb{N}}$ the answer would be no, because $n^{frac{1}{q}}leq n$ so $n+1$ cannot be a limit point. If ${n^{frac{1}{q}}:qin(0,infty)}$, then the answer is yes, since after all $q=log_m n$ gives $n^{frac{1}{q}}=m$ precisely for each natural $m>1$.
$endgroup$
– nathan.j.mcdougall
Jan 11 at 0:26
$begingroup$
What do you mean by root? Is ${n^{frac{1}{q}}:qinmathbb{N}}$ the set of roots, or is it ${n^{frac{1}{q}}:qin(0,infty)}$ or something else altogether? If ${n^{frac{1}{q}}:qinmathbb{N}}$ the answer would be no, because $n^{frac{1}{q}}leq n$ so $n+1$ cannot be a limit point. If ${n^{frac{1}{q}}:qin(0,infty)}$, then the answer is yes, since after all $q=log_m n$ gives $n^{frac{1}{q}}=m$ precisely for each natural $m>1$.
$endgroup$
– nathan.j.mcdougall
Jan 11 at 0:26
$begingroup$
@nathan.j.mcdougall I am referring to the former.
$endgroup$
– Gnumbertester
Jan 11 at 0:41
$begingroup$
@nathan.j.mcdougall I am referring to the former.
$endgroup$
– Gnumbertester
Jan 11 at 0:41
1
1
$begingroup$
Your first statement is incorrect. As we know the pairs of naturals are countable, so are the numbers of the form $n^{1/q}$ In the reals, each number $n$ has at most $2 q^{th}$ roots. It is only in the complex numbers that $n^{1/q}$ has $q$ values.
$endgroup$
– Ross Millikan
Jan 11 at 3:28
$begingroup$
Your first statement is incorrect. As we know the pairs of naturals are countable, so are the numbers of the form $n^{1/q}$ In the reals, each number $n$ has at most $2 q^{th}$ roots. It is only in the complex numbers that $n^{1/q}$ has $q$ values.
$endgroup$
– Ross Millikan
Jan 11 at 3:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As a set of real numbers,
that set has density zero
since it is countable.
It probably is
everywhere dense,
in the sense that
for every $c > 0$
and integer $m$
there are integers
$n$ and $q$ such that
$|m-n^{1/q}| < c
$.
I'll see if I can prove it.
I'll be back if I succeed,
otherwise I'll leave it at this.
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
Proof of the fact that ${n^{1/q}: n,q in mathbb N}$ is dense in $(1,infty)$: let us first prove that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Given $xin (0,infty)$ and $0< epsilon <x$ choose $n$ so large that the length of the interval $(frac {log, n} {x+epsilon},frac {log, n} {x-epsilon})$ is greater than $1$ and such that $frac {log, n} {x+epsilon} >1$ . This interval then contains a positive integer $k$. From the inequalities $frac {log, n} {x+epsilon} <k <frac {log, n} {x-epsilon}$ we get $x-epsilon <frac {log, n} k <x+epsilon$ or $|frac {log, n} {k}-x|<epsilon$. We have proved that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Now given $y>1$ we can find a sequence of the type ${frac {log, n_j} {q_j}}$ converging to $log, y$ and this implies $n_j^{1/q_j}=e^{frac {log, n_j} {q_j}}$ tends to $e^{log, y}=y$.
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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$begingroup$
As a set of real numbers,
that set has density zero
since it is countable.
It probably is
everywhere dense,
in the sense that
for every $c > 0$
and integer $m$
there are integers
$n$ and $q$ such that
$|m-n^{1/q}| < c
$.
I'll see if I can prove it.
I'll be back if I succeed,
otherwise I'll leave it at this.
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
As a set of real numbers,
that set has density zero
since it is countable.
It probably is
everywhere dense,
in the sense that
for every $c > 0$
and integer $m$
there are integers
$n$ and $q$ such that
$|m-n^{1/q}| < c
$.
I'll see if I can prove it.
I'll be back if I succeed,
otherwise I'll leave it at this.
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
$endgroup$
add a comment |
$begingroup$
As a set of real numbers,
that set has density zero
since it is countable.
It probably is
everywhere dense,
in the sense that
for every $c > 0$
and integer $m$
there are integers
$n$ and $q$ such that
$|m-n^{1/q}| < c
$.
I'll see if I can prove it.
I'll be back if I succeed,
otherwise I'll leave it at this.
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
$endgroup$
As a set of real numbers,
that set has density zero
since it is countable.
It probably is
everywhere dense,
in the sense that
for every $c > 0$
and integer $m$
there are integers
$n$ and $q$ such that
$|m-n^{1/q}| < c
$.
I'll see if I can prove it.
I'll be back if I succeed,
otherwise I'll leave it at this.
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
answered Jan 11 at 3:04
marty cohenmarty cohen
73.5k549128
73.5k549128
add a comment |
add a comment |
$begingroup$
Proof of the fact that ${n^{1/q}: n,q in mathbb N}$ is dense in $(1,infty)$: let us first prove that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Given $xin (0,infty)$ and $0< epsilon <x$ choose $n$ so large that the length of the interval $(frac {log, n} {x+epsilon},frac {log, n} {x-epsilon})$ is greater than $1$ and such that $frac {log, n} {x+epsilon} >1$ . This interval then contains a positive integer $k$. From the inequalities $frac {log, n} {x+epsilon} <k <frac {log, n} {x-epsilon}$ we get $x-epsilon <frac {log, n} k <x+epsilon$ or $|frac {log, n} {k}-x|<epsilon$. We have proved that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Now given $y>1$ we can find a sequence of the type ${frac {log, n_j} {q_j}}$ converging to $log, y$ and this implies $n_j^{1/q_j}=e^{frac {log, n_j} {q_j}}$ tends to $e^{log, y}=y$.
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
add a comment |
$begingroup$
Proof of the fact that ${n^{1/q}: n,q in mathbb N}$ is dense in $(1,infty)$: let us first prove that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Given $xin (0,infty)$ and $0< epsilon <x$ choose $n$ so large that the length of the interval $(frac {log, n} {x+epsilon},frac {log, n} {x-epsilon})$ is greater than $1$ and such that $frac {log, n} {x+epsilon} >1$ . This interval then contains a positive integer $k$. From the inequalities $frac {log, n} {x+epsilon} <k <frac {log, n} {x-epsilon}$ we get $x-epsilon <frac {log, n} k <x+epsilon$ or $|frac {log, n} {k}-x|<epsilon$. We have proved that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Now given $y>1$ we can find a sequence of the type ${frac {log, n_j} {q_j}}$ converging to $log, y$ and this implies $n_j^{1/q_j}=e^{frac {log, n_j} {q_j}}$ tends to $e^{log, y}=y$.
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
add a comment |
$begingroup$
Proof of the fact that ${n^{1/q}: n,q in mathbb N}$ is dense in $(1,infty)$: let us first prove that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Given $xin (0,infty)$ and $0< epsilon <x$ choose $n$ so large that the length of the interval $(frac {log, n} {x+epsilon},frac {log, n} {x-epsilon})$ is greater than $1$ and such that $frac {log, n} {x+epsilon} >1$ . This interval then contains a positive integer $k$. From the inequalities $frac {log, n} {x+epsilon} <k <frac {log, n} {x-epsilon}$ we get $x-epsilon <frac {log, n} k <x+epsilon$ or $|frac {log, n} {k}-x|<epsilon$. We have proved that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Now given $y>1$ we can find a sequence of the type ${frac {log, n_j} {q_j}}$ converging to $log, y$ and this implies $n_j^{1/q_j}=e^{frac {log, n_j} {q_j}}$ tends to $e^{log, y}=y$.
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
$endgroup$
Proof of the fact that ${n^{1/q}: n,q in mathbb N}$ is dense in $(1,infty)$: let us first prove that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Given $xin (0,infty)$ and $0< epsilon <x$ choose $n$ so large that the length of the interval $(frac {log, n} {x+epsilon},frac {log, n} {x-epsilon})$ is greater than $1$ and such that $frac {log, n} {x+epsilon} >1$ . This interval then contains a positive integer $k$. From the inequalities $frac {log, n} {x+epsilon} <k <frac {log, n} {x-epsilon}$ we get $x-epsilon <frac {log, n} k <x+epsilon$ or $|frac {log, n} {k}-x|<epsilon$. We have proved that ${frac {log, n} q:n,q in mathbb N}$ is dense in $(0,infty)$. Now given $y>1$ we can find a sequence of the type ${frac {log, n_j} {q_j}}$ converging to $log, y$ and this implies $n_j^{1/q_j}=e^{frac {log, n_j} {q_j}}$ tends to $e^{log, y}=y$.
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
answered Jan 11 at 7:28
Kavi Rama MurthyKavi Rama Murthy
58.5k42161
58.5k42161
add a comment |
add a comment |
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$begingroup$
What do you mean by root? Is ${n^{frac{1}{q}}:qinmathbb{N}}$ the set of roots, or is it ${n^{frac{1}{q}}:qin(0,infty)}$ or something else altogether? If ${n^{frac{1}{q}}:qinmathbb{N}}$ the answer would be no, because $n^{frac{1}{q}}leq n$ so $n+1$ cannot be a limit point. If ${n^{frac{1}{q}}:qin(0,infty)}$, then the answer is yes, since after all $q=log_m n$ gives $n^{frac{1}{q}}=m$ precisely for each natural $m>1$.
$endgroup$
– nathan.j.mcdougall
Jan 11 at 0:26
$begingroup$
@nathan.j.mcdougall I am referring to the former.
$endgroup$
– Gnumbertester
Jan 11 at 0:41
1
$begingroup$
Your first statement is incorrect. As we know the pairs of naturals are countable, so are the numbers of the form $n^{1/q}$ In the reals, each number $n$ has at most $2 q^{th}$ roots. It is only in the complex numbers that $n^{1/q}$ has $q$ values.
$endgroup$
– Ross Millikan
Jan 11 at 3:28