Finding all rationals in a given range using Stern-Brocot tree
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I know the Stern-Brocot tree lists out all the possible fractions. But how do I enumerate the fractions that are present in $[a, b]$ where $a$ and $b$ are two fractions.
fractions
asked Mar 29 at 18:39
Free Radical
18311
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show 3 more comments
up vote
1
down vote
favorite
I know the Stern-Brocot tree lists out all the possible fractions. But how do I enumerate the fractions that are present in $[a, b]$ where $a$ and $b$ are two fractions.
fractions
asked Mar 29 at 18:39
Free Radical
18311
1
Between any two distinct real numbers, it would be impossible to list out all of the possible fractions due to their density in the real numbers (vaguely, there are infinitely many fractions between any two distinct numbers). Perhaps you can provide more context within your post?
– Clayton
Mar 29 at 18:41
@Clayton They can be enumerated with an infinite list.
– Mike Earnest
Mar 29 at 18:44
Say I want to find all rationals between 1 and 1/2. Is there a particular pattern or design that we have to follow in the tree structure to list them out? I understand that the pattern would be infinitely long.
– Free Radical
Mar 29 at 18:44
Do you mean that you would like a formula that would iterate through all the rationals number in this interval? It is possible because the set of rationals is countable. You could look at proofs of: $mathbb Q$ is countable
– Max Ft
Mar 29 at 18:45
1
Yeah, something like that if it is possible. I was wondering if all rationals in a particular range follow some pattern in the Stern-Brocot tree/series
– Free Radical
Mar 29 at 18:47
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know the Stern-Brocot tree lists out all the possible fractions. But how do I enumerate the fractions that are present in $[a, b]$ where $a$ and $b$ are two fractions.
fractions
asked Mar 29 at 18:39
Free Radical
18311
I know the Stern-Brocot tree lists out all the possible fractions. But how do I enumerate the fractions that are present in $[a, b]$ where $a$ and $b$ are two fractions.
fractions
fractions
asked Mar 29 at 18:39
Free Radical
18311
asked Mar 29 at 18:39
Free Radical
18311
edited Mar 29 at 18:53
asked Mar 29 at 18:39
Free Radical
18311
asked Mar 29 at 18:39
Free Radical
18311
asked Mar 29 at 18:39
Free Radical
18311
18311
1
Between any two distinct real numbers, it would be impossible to list out all of the possible fractions due to their density in the real numbers (vaguely, there are infinitely many fractions between any two distinct numbers). Perhaps you can provide more context within your post?
– Clayton
Mar 29 at 18:41
@Clayton They can be enumerated with an infinite list.
– Mike Earnest
Mar 29 at 18:44
Say I want to find all rationals between 1 and 1/2. Is there a particular pattern or design that we have to follow in the tree structure to list them out? I understand that the pattern would be infinitely long.
– Free Radical
Mar 29 at 18:44
Do you mean that you would like a formula that would iterate through all the rationals number in this interval? It is possible because the set of rationals is countable. You could look at proofs of: $mathbb Q$ is countable
– Max Ft
Mar 29 at 18:45
1
Yeah, something like that if it is possible. I was wondering if all rationals in a particular range follow some pattern in the Stern-Brocot tree/series
– Free Radical
Mar 29 at 18:47
|
show 3 more comments
1
Between any two distinct real numbers, it would be impossible to list out all of the possible fractions due to their density in the real numbers (vaguely, there are infinitely many fractions between any two distinct numbers). Perhaps you can provide more context within your post?
– Clayton
Mar 29 at 18:41
@Clayton They can be enumerated with an infinite list.
– Mike Earnest
Mar 29 at 18:44
Say I want to find all rationals between 1 and 1/2. Is there a particular pattern or design that we have to follow in the tree structure to list them out? I understand that the pattern would be infinitely long.
– Free Radical
Mar 29 at 18:44
Do you mean that you would like a formula that would iterate through all the rationals number in this interval? It is possible because the set of rationals is countable. You could look at proofs of: $mathbb Q$ is countable
– Max Ft
Mar 29 at 18:45
1
Yeah, something like that if it is possible. I was wondering if all rationals in a particular range follow some pattern in the Stern-Brocot tree/series
– Free Radical
Mar 29 at 18:47
1
1
Between any two distinct real numbers, it would be impossible to list out all of the possible fractions due to their density in the real numbers (vaguely, there are infinitely many fractions between any two distinct numbers). Perhaps you can provide more context within your post?
– Clayton
Mar 29 at 18:41
Between any two distinct real numbers, it would be impossible to list out all of the possible fractions due to their density in the real numbers (vaguely, there are infinitely many fractions between any two distinct numbers). Perhaps you can provide more context within your post?
– Clayton
Mar 29 at 18:41
@Clayton They can be enumerated with an infinite list.
– Mike Earnest
Mar 29 at 18:44
@Clayton They can be enumerated with an infinite list.
– Mike Earnest
Mar 29 at 18:44
Say I want to find all rationals between 1 and 1/2. Is there a particular pattern or design that we have to follow in the tree structure to list them out? I understand that the pattern would be infinitely long.
– Free Radical
Mar 29 at 18:44
Say I want to find all rationals between 1 and 1/2. Is there a particular pattern or design that we have to follow in the tree structure to list them out? I understand that the pattern would be infinitely long.
– Free Radical
Mar 29 at 18:44
Do you mean that you would like a formula that would iterate through all the rationals number in this interval? It is possible because the set of rationals is countable. You could look at proofs of: $mathbb Q$ is countable
– Max Ft
Mar 29 at 18:45
Do you mean that you would like a formula that would iterate through all the rationals number in this interval? It is possible because the set of rationals is countable. You could look at proofs of: $mathbb Q$ is countable
– Max Ft
Mar 29 at 18:45
1
1
Yeah, something like that if it is possible. I was wondering if all rationals in a particular range follow some pattern in the Stern-Brocot tree/series
– Free Radical
Mar 29 at 18:47
Yeah, something like that if it is possible. I was wondering if all rationals in a particular range follow some pattern in the Stern-Brocot tree/series
– Free Radical
Mar 29 at 18:47
|
show 3 more comments
1 Answer
1
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up vote
2
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Repeatedly interleaving mediants (as in the original Stern Brocot tree) should work. See https://arxiv.org/pdf/1301.6807.pdf
answered Nov 23 at 23:37
ttw
26124
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Repeatedly interleaving mediants (as in the original Stern Brocot tree) should work. See https://arxiv.org/pdf/1301.6807.pdf
answered Nov 23 at 23:37
ttw
26124
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
add a comment |
up vote
2
down vote
Repeatedly interleaving mediants (as in the original Stern Brocot tree) should work. See https://arxiv.org/pdf/1301.6807.pdf
answered Nov 23 at 23:37
ttw
26124
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
add a comment |
up vote
2
down vote
up vote
2
down vote
Repeatedly interleaving mediants (as in the original Stern Brocot tree) should work. See https://arxiv.org/pdf/1301.6807.pdf
answered Nov 23 at 23:37
ttw
26124
Repeatedly interleaving mediants (as in the original Stern Brocot tree) should work. See https://arxiv.org/pdf/1301.6807.pdf
answered Nov 23 at 23:37
ttw
26124
answered Nov 23 at 23:37
ttw
26124
answered Nov 23 at 23:37
ttw
26124
answered Nov 23 at 23:37
ttw
26124
26124
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
add a comment |
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
right, we have just to use $a$ and $b$ as the starting "parents" and then the mediant chain will provide the Stern-Brocot tree in between, i.e. all the rationals in the interval $[a,b]$.
– G Cab
Nov 23 at 23:50
add a comment |
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Between any two distinct real numbers, it would be impossible to list out all of the possible fractions due to their density in the real numbers (vaguely, there are infinitely many fractions between any two distinct numbers). Perhaps you can provide more context within your post?
– Clayton
Mar 29 at 18:41
@Clayton They can be enumerated with an infinite list.
– Mike Earnest
Mar 29 at 18:44
Say I want to find all rationals between 1 and 1/2. Is there a particular pattern or design that we have to follow in the tree structure to list them out? I understand that the pattern would be infinitely long.
– Free Radical
Mar 29 at 18:44
Do you mean that you would like a formula that would iterate through all the rationals number in this interval? It is possible because the set of rationals is countable. You could look at proofs of: $mathbb Q$ is countable
– Max Ft
Mar 29 at 18:45
1
Yeah, something like that if it is possible. I was wondering if all rationals in a particular range follow some pattern in the Stern-Brocot tree/series
– Free Radical
Mar 29 at 18:47