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Pattern in Irrational Numbers's digits
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Is there any way by which we can see the pattern of digits of irrational numbers(only those natural numbers whose square root is not a natural number)? Square root of 2 is an irrational number so, is there any way or theorem by which we can see a pattern in its digits after decimal place?
number-theory irrational-numbers pattern-recognition
edited Jul 22 at 0:39
tinlyx
90811118
asked Jul 22 at 0:34
Adarsh Kumar
317
 |Â
show 7 more comments
up vote
3
down vote
favorite
Is there any way by which we can see the pattern of digits of irrational numbers(only those natural numbers whose square root is not a natural number)? Square root of 2 is an irrational number so, is there any way or theorem by which we can see a pattern in its digits after decimal place?
number-theory irrational-numbers pattern-recognition
edited Jul 22 at 0:39
tinlyx
90811118
asked Jul 22 at 0:34
Adarsh Kumar
317
3
What denotes a “pattern†in a series of digits?
– Chandler Watson
Jul 22 at 2:08
Pattern means a general formula which can tell you any digit after decimal .
– Adarsh Kumar
Jul 22 at 2:13
1
Not to be pedantic, but couldn’t you just define a formula that gives you the nth decimal place of the number by definition? It seems like you want the definition to be some way of having the digits be computable from scratch. Maybe something like computable numbers?
– Chandler Watson
Jul 22 at 2:16
Yaa just like computable numbers.
– Adarsh Kumar
Jul 22 at 2:20
It looks like the computable numbers are countable, but not “completely enumerable,†meaning there is no such Turing machine that can generate all of them one by one. I’m not sure if there exists a Turing machine that could tell you whether a number is computable (has a pattern), but I’m sure someone else here might have an idea.
– Chandler Watson
Jul 22 at 2:36
 |Â
show 7 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Is there any way by which we can see the pattern of digits of irrational numbers(only those natural numbers whose square root is not a natural number)? Square root of 2 is an irrational number so, is there any way or theorem by which we can see a pattern in its digits after decimal place?
number-theory irrational-numbers pattern-recognition
edited Jul 22 at 0:39
tinlyx
90811118
asked Jul 22 at 0:34
Adarsh Kumar
317
Is there any way by which we can see the pattern of digits of irrational numbers(only those natural numbers whose square root is not a natural number)? Square root of 2 is an irrational number so, is there any way or theorem by which we can see a pattern in its digits after decimal place?
number-theory irrational-numbers pattern-recognition
edited Jul 22 at 0:39
tinlyx
90811118
asked Jul 22 at 0:34
Adarsh Kumar
317
edited Jul 22 at 0:39
tinlyx
90811118
edited Jul 22 at 0:39
tinlyx
90811118
edited Jul 22 at 0:39
tinlyx
90811118
90811118
asked Jul 22 at 0:34
Adarsh Kumar
317
asked Jul 22 at 0:34
Adarsh Kumar
317
asked Jul 22 at 0:34
Adarsh Kumar
317
317
3
What denotes a “pattern†in a series of digits?
– Chandler Watson
Jul 22 at 2:08
Pattern means a general formula which can tell you any digit after decimal .
– Adarsh Kumar
Jul 22 at 2:13
1
Not to be pedantic, but couldn’t you just define a formula that gives you the nth decimal place of the number by definition? It seems like you want the definition to be some way of having the digits be computable from scratch. Maybe something like computable numbers?
– Chandler Watson
Jul 22 at 2:16
Yaa just like computable numbers.
– Adarsh Kumar
Jul 22 at 2:20
It looks like the computable numbers are countable, but not “completely enumerable,†meaning there is no such Turing machine that can generate all of them one by one. I’m not sure if there exists a Turing machine that could tell you whether a number is computable (has a pattern), but I’m sure someone else here might have an idea.
– Chandler Watson
Jul 22 at 2:36
 |Â
show 7 more comments
3
What denotes a “pattern†in a series of digits?
– Chandler Watson
Jul 22 at 2:08
Pattern means a general formula which can tell you any digit after decimal .
– Adarsh Kumar
Jul 22 at 2:13
1
Not to be pedantic, but couldn’t you just define a formula that gives you the nth decimal place of the number by definition? It seems like you want the definition to be some way of having the digits be computable from scratch. Maybe something like computable numbers?
– Chandler Watson
Jul 22 at 2:16
Yaa just like computable numbers.
– Adarsh Kumar
Jul 22 at 2:20
It looks like the computable numbers are countable, but not “completely enumerable,†meaning there is no such Turing machine that can generate all of them one by one. I’m not sure if there exists a Turing machine that could tell you whether a number is computable (has a pattern), but I’m sure someone else here might have an idea.
– Chandler Watson
Jul 22 at 2:36
3
3
What denotes a “pattern†in a series of digits?
– Chandler Watson
Jul 22 at 2:08
What denotes a “pattern†in a series of digits?
– Chandler Watson
Jul 22 at 2:08
Pattern means a general formula which can tell you any digit after decimal .
– Adarsh Kumar
Jul 22 at 2:13
Pattern means a general formula which can tell you any digit after decimal .
– Adarsh Kumar
Jul 22 at 2:13
1
1
Not to be pedantic, but couldn’t you just define a formula that gives you the nth decimal place of the number by definition? It seems like you want the definition to be some way of having the digits be computable from scratch. Maybe something like computable numbers?
– Chandler Watson
Jul 22 at 2:16
Not to be pedantic, but couldn’t you just define a formula that gives you the nth decimal place of the number by definition? It seems like you want the definition to be some way of having the digits be computable from scratch. Maybe something like computable numbers?
– Chandler Watson
Jul 22 at 2:16
Yaa just like computable numbers.
– Adarsh Kumar
Jul 22 at 2:20
Yaa just like computable numbers.
– Adarsh Kumar
Jul 22 at 2:20
It looks like the computable numbers are countable, but not “completely enumerable,†meaning there is no such Turing machine that can generate all of them one by one. I’m not sure if there exists a Turing machine that could tell you whether a number is computable (has a pattern), but I’m sure someone else here might have an idea.
– Chandler Watson
Jul 22 at 2:36
It looks like the computable numbers are countable, but not “completely enumerable,†meaning there is no such Turing machine that can generate all of them one by one. I’m not sure if there exists a Turing machine that could tell you whether a number is computable (has a pattern), but I’m sure someone else here might have an idea.
– Chandler Watson
Jul 22 at 2:36
 |Â
show 7 more comments
1 Answer
1
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oldest
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up vote
5
down vote
No (or at least not that I know of. Somebody might find some pattern tomorrow). One can construct irrational numbers that have patterns, for example the Liouville numbers like $0.1010010000001ldots$ where the number of zeros is the next factorial number. The pattern just can't be repeating one sequence of digits because that would make the number rational. The continued fractions of square roots also have simple patterns.
answered Jul 22 at 0:43
Ross Millikan
276k21186352
1
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
2
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
No (or at least not that I know of. Somebody might find some pattern tomorrow). One can construct irrational numbers that have patterns, for example the Liouville numbers like $0.1010010000001ldots$ where the number of zeros is the next factorial number. The pattern just can't be repeating one sequence of digits because that would make the number rational. The continued fractions of square roots also have simple patterns.
answered Jul 22 at 0:43
Ross Millikan
276k21186352
1
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
2
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
add a comment |Â
up vote
5
down vote
No (or at least not that I know of. Somebody might find some pattern tomorrow). One can construct irrational numbers that have patterns, for example the Liouville numbers like $0.1010010000001ldots$ where the number of zeros is the next factorial number. The pattern just can't be repeating one sequence of digits because that would make the number rational. The continued fractions of square roots also have simple patterns.
answered Jul 22 at 0:43
Ross Millikan
276k21186352
1
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
2
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
add a comment |Â
up vote
5
down vote
up vote
5
down vote
No (or at least not that I know of. Somebody might find some pattern tomorrow). One can construct irrational numbers that have patterns, for example the Liouville numbers like $0.1010010000001ldots$ where the number of zeros is the next factorial number. The pattern just can't be repeating one sequence of digits because that would make the number rational. The continued fractions of square roots also have simple patterns.
answered Jul 22 at 0:43
Ross Millikan
276k21186352
No (or at least not that I know of. Somebody might find some pattern tomorrow). One can construct irrational numbers that have patterns, for example the Liouville numbers like $0.1010010000001ldots$ where the number of zeros is the next factorial number. The pattern just can't be repeating one sequence of digits because that would make the number rational. The continued fractions of square roots also have simple patterns.
answered Jul 22 at 0:43
Ross Millikan
276k21186352
answered Jul 22 at 0:43
Ross Millikan
276k21186352
answered Jul 22 at 0:43
Ross Millikan
276k21186352
answered Jul 22 at 0:43
Ross Millikan
276k21186352
276k21186352
1
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
2
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
add a comment |Â
1
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
2
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
1
1
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
I am puzzled by the downvote to this question---it answers the question as completely as reasonably possible given that OP did not specify much what sort of pattern they had in mind.
– Travis
Jul 22 at 0:55
2
2
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@Travis Did you mean "to this answer"? ...
– user202729
Jul 22 at 6:27
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@user202729 Yes, of course; thanks for the correction (though now I am puzzled by the downvotes [plural]).
– Travis
Jul 22 at 9:14
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
@Travis I downvoted (and then had to drive my wife to work). The reason? "No" is trivially the opposite of the right answer. Pattern, with such level of generality, trivially exist. Like $a_n=[10^nsqrta]-10[10^n-1sqrta]$, which is just an in-homogeneous, linear recurrence with constant coefficients, or order zero. Take into account that the person asking the question has also clarified.
– user574889
Jul 22 at 15:39
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3
What denotes a “pattern†in a series of digits?
– Chandler Watson
Jul 22 at 2:08
Pattern means a general formula which can tell you any digit after decimal .
– Adarsh Kumar
Jul 22 at 2:13
1
Not to be pedantic, but couldn’t you just define a formula that gives you the nth decimal place of the number by definition? It seems like you want the definition to be some way of having the digits be computable from scratch. Maybe something like computable numbers?
– Chandler Watson
Jul 22 at 2:16
Yaa just like computable numbers.
– Adarsh Kumar
Jul 22 at 2:20
It looks like the computable numbers are countable, but not “completely enumerable,†meaning there is no such Turing machine that can generate all of them one by one. I’m not sure if there exists a Turing machine that could tell you whether a number is computable (has a pattern), but I’m sure someone else here might have an idea.
– Chandler Watson
Jul 22 at 2:36