Representing positive integers as floor of integer powers of real number.
Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?
exponentiation
asked Nov 27 '18 at 10:06
user4201961
629411
add a comment |
Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?
exponentiation
asked Nov 27 '18 at 10:06
user4201961
629411
1
You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09
I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11
add a comment |
Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?
exponentiation
asked Nov 27 '18 at 10:06
user4201961
629411
Does there exist real $a$ such that for every positive integer $c$ there exists integer $b$ such that $lfloor{a^b}rfloor = c$?
exponentiation
exponentiation
asked Nov 27 '18 at 10:06
user4201961
629411
asked Nov 27 '18 at 10:06
user4201961
629411
edited Nov 27 '18 at 10:11
asked Nov 27 '18 at 10:06
user4201961
629411
asked Nov 27 '18 at 10:06
user4201961
629411
asked Nov 27 '18 at 10:06
user4201961
629411
629411
1
You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09
I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11
add a comment |
1
You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09
I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11
1
1
You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09
You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09
I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11
I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11
add a comment |
1 Answer
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votes
No.
The simplest way to see why the answer is no is the fact that, for $a>1$, you have
$$lim_{btoinfty} (a^b-a^{b-1}) = infty$$
(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)
The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$
answered Nov 27 '18 at 10:21
5xum
89.5k393161
add a comment |
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
No.
The simplest way to see why the answer is no is the fact that, for $a>1$, you have
$$lim_{btoinfty} (a^b-a^{b-1}) = infty$$
(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)
The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$
answered Nov 27 '18 at 10:21
5xum
89.5k393161
add a comment |
No.
The simplest way to see why the answer is no is the fact that, for $a>1$, you have
$$lim_{btoinfty} (a^b-a^{b-1}) = infty$$
(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)
The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$
answered Nov 27 '18 at 10:21
5xum
89.5k393161
add a comment |
No.
The simplest way to see why the answer is no is the fact that, for $a>1$, you have
$$lim_{btoinfty} (a^b-a^{b-1}) = infty$$
(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)
The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$
answered Nov 27 '18 at 10:21
5xum
89.5k393161
No.
The simplest way to see why the answer is no is the fact that, for $a>1$, you have
$$lim_{btoinfty} (a^b-a^{b-1}) = infty$$
(which you can see is true since $a^b-a^{b-1} = a^{b-1}(a-1)$)
The limit tells you that, for large enough values of $b$, you will have $a^b-a^{b-1}>2$ which means that there must be at least one integer $n$ between $lfloor a^{b-1}rfloor$ and $lfloor a^brfloor$ (that is, $lfloor a^{b-1}rfloor<n<lfloor a^brfloor$. Couple that with the fact that $lfloor a^brfloor$ is an increasing sequence, and you are done, because $$forall b: lfloor a^brfloorneq n$$
answered Nov 27 '18 at 10:21
5xum
89.5k393161
answered Nov 27 '18 at 10:21
5xum
89.5k393161
answered Nov 27 '18 at 10:21
5xum
89.5k393161
answered Nov 27 '18 at 10:21
5xum
89.5k393161
89.5k393161
add a comment |
add a comment |
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1
You mean a single $a$ for all $b,c$?
– gammatester
Nov 27 '18 at 10:09
I've edited the question, sorry for my mistake. Yes, one $a$.
– user4201961
Nov 27 '18 at 10:11