Characterizing a set of numbers that are not dyadic rationals.
We define the set
$S={x: exists c>0 text{ such that} |x-j/2^k| geq c/2^k, forall jinmathbb{Z},forall kin mathbb{N}_0 }$,
where $N_0$ is the set of natural numbers with $0$.
Clearly, this set does not contain any dyadic rational, namely
a rational of the form $j/2^k$.
I want to prove that
this set contains for example the number $1/3$. After doing some experiments
I found out that for all $j in mathbb{Z}, k in mathbb{N}_0$ it should be
true that
begin{equation}
|1/3 - j/2^k| geq 1/(3cdot 2^k)
end{equation}
I tested the above inequality for pretty large $k,j$ using code.
How can I prove this inequality for all $j, k$ ?
Moreover, I think all rationals of the form $a/b$ so that $b$ contains a prime factor
other than $2$ should belong to the set $S$. How can I show this claim?
real-analysis algebra-precalculus number-theory analysis elementary-number-theory
asked Dec 3 '18 at 9:06
elrond
1336
add a comment |
We define the set
$S={x: exists c>0 text{ such that} |x-j/2^k| geq c/2^k, forall jinmathbb{Z},forall kin mathbb{N}_0 }$,
where $N_0$ is the set of natural numbers with $0$.
Clearly, this set does not contain any dyadic rational, namely
a rational of the form $j/2^k$.
I want to prove that
this set contains for example the number $1/3$. After doing some experiments
I found out that for all $j in mathbb{Z}, k in mathbb{N}_0$ it should be
true that
begin{equation}
|1/3 - j/2^k| geq 1/(3cdot 2^k)
end{equation}
I tested the above inequality for pretty large $k,j$ using code.
How can I prove this inequality for all $j, k$ ?
Moreover, I think all rationals of the form $a/b$ so that $b$ contains a prime factor
other than $2$ should belong to the set $S$. How can I show this claim?
real-analysis algebra-precalculus number-theory analysis elementary-number-theory
asked Dec 3 '18 at 9:06
elrond
1336
add a comment |
We define the set
$S={x: exists c>0 text{ such that} |x-j/2^k| geq c/2^k, forall jinmathbb{Z},forall kin mathbb{N}_0 }$,
where $N_0$ is the set of natural numbers with $0$.
Clearly, this set does not contain any dyadic rational, namely
a rational of the form $j/2^k$.
I want to prove that
this set contains for example the number $1/3$. After doing some experiments
I found out that for all $j in mathbb{Z}, k in mathbb{N}_0$ it should be
true that
begin{equation}
|1/3 - j/2^k| geq 1/(3cdot 2^k)
end{equation}
I tested the above inequality for pretty large $k,j$ using code.
How can I prove this inequality for all $j, k$ ?
Moreover, I think all rationals of the form $a/b$ so that $b$ contains a prime factor
other than $2$ should belong to the set $S$. How can I show this claim?
real-analysis algebra-precalculus number-theory analysis elementary-number-theory
asked Dec 3 '18 at 9:06
elrond
1336
We define the set
$S={x: exists c>0 text{ such that} |x-j/2^k| geq c/2^k, forall jinmathbb{Z},forall kin mathbb{N}_0 }$,
where $N_0$ is the set of natural numbers with $0$.
Clearly, this set does not contain any dyadic rational, namely
a rational of the form $j/2^k$.
I want to prove that
this set contains for example the number $1/3$. After doing some experiments
I found out that for all $j in mathbb{Z}, k in mathbb{N}_0$ it should be
true that
begin{equation}
|1/3 - j/2^k| geq 1/(3cdot 2^k)
end{equation}
I tested the above inequality for pretty large $k,j$ using code.
How can I prove this inequality for all $j, k$ ?
Moreover, I think all rationals of the form $a/b$ so that $b$ contains a prime factor
other than $2$ should belong to the set $S$. How can I show this claim?
real-analysis algebra-precalculus number-theory analysis elementary-number-theory
real-analysis algebra-precalculus number-theory analysis elementary-number-theory
asked Dec 3 '18 at 9:06
elrond
1336
asked Dec 3 '18 at 9:06
elrond
1336
asked Dec 3 '18 at 9:06
elrond
1336
asked Dec 3 '18 at 9:06
elrond
1336
asked Dec 3 '18 at 9:06
elrond
1336
1336
add a comment |
add a comment |
1 Answer
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If $2^{k}-1 <3j < 2^{k}+1$ then $3j=2^{k}$ which is a contradiction. Hence either $2^{k}-1 geq 3j$ or $3j geq 2^{k}+1$ . In the first case $frac j {2^{k}} leq frac 1 3 -frac 1 {32^{k}}$ and in the second case $frac j {2^{k}}geq frac 1 3 +frac 1 {32^{k}}$. Your inequality follows from these.
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
If $2^{k}-1 <3j < 2^{k}+1$ then $3j=2^{k}$ which is a contradiction. Hence either $2^{k}-1 geq 3j$ or $3j geq 2^{k}+1$ . In the first case $frac j {2^{k}} leq frac 1 3 -frac 1 {32^{k}}$ and in the second case $frac j {2^{k}}geq frac 1 3 +frac 1 {32^{k}}$. Your inequality follows from these.
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
add a comment |
If $2^{k}-1 <3j < 2^{k}+1$ then $3j=2^{k}$ which is a contradiction. Hence either $2^{k}-1 geq 3j$ or $3j geq 2^{k}+1$ . In the first case $frac j {2^{k}} leq frac 1 3 -frac 1 {32^{k}}$ and in the second case $frac j {2^{k}}geq frac 1 3 +frac 1 {32^{k}}$. Your inequality follows from these.
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
add a comment |
If $2^{k}-1 <3j < 2^{k}+1$ then $3j=2^{k}$ which is a contradiction. Hence either $2^{k}-1 geq 3j$ or $3j geq 2^{k}+1$ . In the first case $frac j {2^{k}} leq frac 1 3 -frac 1 {32^{k}}$ and in the second case $frac j {2^{k}}geq frac 1 3 +frac 1 {32^{k}}$. Your inequality follows from these.
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
If $2^{k}-1 <3j < 2^{k}+1$ then $3j=2^{k}$ which is a contradiction. Hence either $2^{k}-1 geq 3j$ or $3j geq 2^{k}+1$ . In the first case $frac j {2^{k}} leq frac 1 3 -frac 1 {32^{k}}$ and in the second case $frac j {2^{k}}geq frac 1 3 +frac 1 {32^{k}}$. Your inequality follows from these.
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
answered Dec 3 '18 at 9:18
Kavi Rama Murthy
51.2k31855
51.2k31855
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
add a comment |
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
Can you generalize this for any $a/b$ such that $b$ is not a power of 2?
– elrond
Dec 3 '18 at 9:33
add a comment |
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