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Prove $a_n = lfloor ni rfloor$ for some irrational $i$ has no pattern
$begingroup$
I have a sequence $a_n = lfloor ni rfloor$ for an irrational $i$ and I want to prove that there is no 'pattern' to these terms.
In particular, I have $1 < i <2$ and I want to prove that $forall a,b in mathbb{N}, exists k in mathbb{N}$ such that $a +bk neq a_n$ for any $n$.
This makes sense to me intuitively because $a_n - a_{n-1}$ will be 1 or 2 'randomly', so the sequence must eventually 'skip over' one of the terms of $a+bk$.
My approach has been to assume that it holds, then show that $i$ must be rational for a contradiction. However, the floor is messing up the simple divisibility techniques I would use.
sequences-and-series irrational-numbers
asked Jan 20 at 23:43
MannerPotsMannerPots
234
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add a comment |
$begingroup$
I have a sequence $a_n = lfloor ni rfloor$ for an irrational $i$ and I want to prove that there is no 'pattern' to these terms.
In particular, I have $1 < i <2$ and I want to prove that $forall a,b in mathbb{N}, exists k in mathbb{N}$ such that $a +bk neq a_n$ for any $n$.
This makes sense to me intuitively because $a_n - a_{n-1}$ will be 1 or 2 'randomly', so the sequence must eventually 'skip over' one of the terms of $a+bk$.
My approach has been to assume that it holds, then show that $i$ must be rational for a contradiction. However, the floor is messing up the simple divisibility techniques I would use.
sequences-and-series irrational-numbers
asked Jan 20 at 23:43
MannerPotsMannerPots
234
$endgroup$
add a comment |
$begingroup$
I have a sequence $a_n = lfloor ni rfloor$ for an irrational $i$ and I want to prove that there is no 'pattern' to these terms.
In particular, I have $1 < i <2$ and I want to prove that $forall a,b in mathbb{N}, exists k in mathbb{N}$ such that $a +bk neq a_n$ for any $n$.
This makes sense to me intuitively because $a_n - a_{n-1}$ will be 1 or 2 'randomly', so the sequence must eventually 'skip over' one of the terms of $a+bk$.
My approach has been to assume that it holds, then show that $i$ must be rational for a contradiction. However, the floor is messing up the simple divisibility techniques I would use.
sequences-and-series irrational-numbers
asked Jan 20 at 23:43
MannerPotsMannerPots
234
$endgroup$
I have a sequence $a_n = lfloor ni rfloor$ for an irrational $i$ and I want to prove that there is no 'pattern' to these terms.
In particular, I have $1 < i <2$ and I want to prove that $forall a,b in mathbb{N}, exists k in mathbb{N}$ such that $a +bk neq a_n$ for any $n$.
This makes sense to me intuitively because $a_n - a_{n-1}$ will be 1 or 2 'randomly', so the sequence must eventually 'skip over' one of the terms of $a+bk$.
My approach has been to assume that it holds, then show that $i$ must be rational for a contradiction. However, the floor is messing up the simple divisibility techniques I would use.
sequences-and-series irrational-numbers
sequences-and-series irrational-numbers
asked Jan 20 at 23:43
MannerPotsMannerPots
234
asked Jan 20 at 23:43
MannerPotsMannerPots
234
edited Jan 21 at 3:00
MannerPots
asked Jan 20 at 23:43
MannerPotsMannerPots
234
asked Jan 20 at 23:43
MannerPotsMannerPots
234
asked Jan 20 at 23:43
MannerPotsMannerPots
234
234
add a comment |
add a comment |
1 Answer
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This is a case of the equidistribution theorem. If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
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1 Answer
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1 Answer
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$begingroup$
This is a case of the equidistribution theorem. If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
$endgroup$
add a comment |
$begingroup$
This is a case of the equidistribution theorem. If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
$endgroup$
add a comment |
$begingroup$
This is a case of the equidistribution theorem. If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
$endgroup$
This is a case of the equidistribution theorem. If the sequence repeats, subdivide $[0,1]$ finely enough and some pieces will get nothing.
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
answered Jan 21 at 3:14
Ross MillikanRoss Millikan
298k24200374
298k24200374
add a comment |
add a comment |
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