Real irrational algebraic numbers “never repeat”











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An oft-used phrase describing irrational numbers is that their (decimal) expansions "never repeat".



The sense of "never repeating" intended is, of course, that their expansions don't repeat forever. And it's straight-forward to show that rational numbers do repeat forever.



It is also easy to show that irrational numbers never repeat forever because that would make them rational.



Now, the question: Suppose I define "never repeating" as simply meaning that the first N digits of the expansion (in whatever base you like) are not repeated. I.e. positions 1..N are not the same as positions (N+1)..2N. Clearly I can construct transcendental numbers with that property. But if we restrict ourselves to irrational algebraic numbers, can it be shown that there are any numbers in that set that "never repeat" in this sense?



It seems to me that even for a randomly chosen irrational algebraic number, the probably that it fulfils this property very very quickly becomes infinitessimally small. I.e. the first billions digits will not match the next billion digits. However, having looked at the first N digits, the next N digits could still be anything, and even though the probability shrinks as p^-N, there are sill infinitely many opportunities as N grows.



So, my question is: can it be shown that there exists a (real) irrational algebraic number for which it is never true that digits 1..N are the same as digits (N+1)..2N for any N?




Related:How to know that irrational numbers never repeat?






decimal-expansion transcendental-numbers algebraic-numbers







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  • Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well.
    – vidyarthi
    yesterday






  • 2




    It's conjectured that every irrational algebraic number is Normal, which would imply that no example of what you want can occur. Very little has been proven, though.
    – lulu
    yesterday












  • @vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want.
    – ThePopMachine
    yesterday








  • 1




    Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified).
    – lulu
    yesterday






  • 1




    @lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all.
    – ThePopMachine
    yesterday

















up vote
6
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3












An oft-used phrase describing irrational numbers is that their (decimal) expansions "never repeat".



The sense of "never repeating" intended is, of course, that their expansions don't repeat forever. And it's straight-forward to show that rational numbers do repeat forever.



It is also easy to show that irrational numbers never repeat forever because that would make them rational.



Now, the question: Suppose I define "never repeating" as simply meaning that the first N digits of the expansion (in whatever base you like) are not repeated. I.e. positions 1..N are not the same as positions (N+1)..2N. Clearly I can construct transcendental numbers with that property. But if we restrict ourselves to irrational algebraic numbers, can it be shown that there are any numbers in that set that "never repeat" in this sense?



It seems to me that even for a randomly chosen irrational algebraic number, the probably that it fulfils this property very very quickly becomes infinitessimally small. I.e. the first billions digits will not match the next billion digits. However, having looked at the first N digits, the next N digits could still be anything, and even though the probability shrinks as p^-N, there are sill infinitely many opportunities as N grows.



So, my question is: can it be shown that there exists a (real) irrational algebraic number for which it is never true that digits 1..N are the same as digits (N+1)..2N for any N?




Related:How to know that irrational numbers never repeat?






decimal-expansion transcendental-numbers algebraic-numbers







share|cite|improve this question









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  • Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well.
    – vidyarthi
    yesterday






  • 2




    It's conjectured that every irrational algebraic number is Normal, which would imply that no example of what you want can occur. Very little has been proven, though.
    – lulu
    yesterday












  • @vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want.
    – ThePopMachine
    yesterday








  • 1




    Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified).
    – lulu
    yesterday






  • 1




    @lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all.
    – ThePopMachine
    yesterday















up vote
6
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up vote
6
down vote

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An oft-used phrase describing irrational numbers is that their (decimal) expansions "never repeat".



The sense of "never repeating" intended is, of course, that their expansions don't repeat forever. And it's straight-forward to show that rational numbers do repeat forever.



It is also easy to show that irrational numbers never repeat forever because that would make them rational.



Now, the question: Suppose I define "never repeating" as simply meaning that the first N digits of the expansion (in whatever base you like) are not repeated. I.e. positions 1..N are not the same as positions (N+1)..2N. Clearly I can construct transcendental numbers with that property. But if we restrict ourselves to irrational algebraic numbers, can it be shown that there are any numbers in that set that "never repeat" in this sense?



It seems to me that even for a randomly chosen irrational algebraic number, the probably that it fulfils this property very very quickly becomes infinitessimally small. I.e. the first billions digits will not match the next billion digits. However, having looked at the first N digits, the next N digits could still be anything, and even though the probability shrinks as p^-N, there are sill infinitely many opportunities as N grows.



So, my question is: can it be shown that there exists a (real) irrational algebraic number for which it is never true that digits 1..N are the same as digits (N+1)..2N for any N?




Related:How to know that irrational numbers never repeat?






decimal-expansion transcendental-numbers algebraic-numbers







share|cite|improve this question









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ThePopMachine is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











An oft-used phrase describing irrational numbers is that their (decimal) expansions "never repeat".



The sense of "never repeating" intended is, of course, that their expansions don't repeat forever. And it's straight-forward to show that rational numbers do repeat forever.



It is also easy to show that irrational numbers never repeat forever because that would make them rational.



Now, the question: Suppose I define "never repeating" as simply meaning that the first N digits of the expansion (in whatever base you like) are not repeated. I.e. positions 1..N are not the same as positions (N+1)..2N. Clearly I can construct transcendental numbers with that property. But if we restrict ourselves to irrational algebraic numbers, can it be shown that there are any numbers in that set that "never repeat" in this sense?



It seems to me that even for a randomly chosen irrational algebraic number, the probably that it fulfils this property very very quickly becomes infinitessimally small. I.e. the first billions digits will not match the next billion digits. However, having looked at the first N digits, the next N digits could still be anything, and even though the probability shrinks as p^-N, there are sill infinitely many opportunities as N grows.



So, my question is: can it be shown that there exists a (real) irrational algebraic number for which it is never true that digits 1..N are the same as digits (N+1)..2N for any N?




Related:How to know that irrational numbers never repeat?






decimal-expansion transcendental-numbers algebraic-numbers




decimal-expansion transcendental-numbers algebraic-numbers






share|cite|improve this question









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  • Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well.
    – vidyarthi
    yesterday






  • 2




    It's conjectured that every irrational algebraic number is Normal, which would imply that no example of what you want can occur. Very little has been proven, though.
    – lulu
    yesterday












  • @vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want.
    – ThePopMachine
    yesterday








  • 1




    Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified).
    – lulu
    yesterday






  • 1




    @lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all.
    – ThePopMachine
    yesterday




















  • Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well.
    – vidyarthi
    yesterday






  • 2




    It's conjectured that every irrational algebraic number is Normal, which would imply that no example of what you want can occur. Very little has been proven, though.
    – lulu
    yesterday












  • @vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want.
    – ThePopMachine
    yesterday








  • 1




    Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified).
    – lulu
    yesterday






  • 1




    @lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all.
    – ThePopMachine
    yesterday


















Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well.
– vidyarthi
yesterday




Since any irrational algebraic number is an irrational number, therefore, whatever applies to irrational numbers applies to them as well.
– vidyarthi
yesterday




2




2




It's conjectured that every irrational algebraic number is Normal, which would imply that no example of what you want can occur. Very little has been proven, though.
– lulu
yesterday






It's conjectured that every irrational algebraic number is Normal, which would imply that no example of what you want can occur. Very little has been proven, though.
– lulu
yesterday














@vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want.
– ThePopMachine
yesterday






@vidyarthi: Irrational numbers include transcendentals, and it's clear that I can construct a transcendental with this property anytime I want.
– ThePopMachine
yesterday






1




1




Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified).
– lulu
yesterday




Study the definition. It says that any specified $N$ digits occurs infinitely often (and the density is specified).
– lulu
yesterday




1




1




@lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all.
– ThePopMachine
yesterday






@lulu: I didn't ask if a particular string of N digits occurs again. I asked if the first N digits match the next N digits, for any N at all.
– ThePopMachine
yesterday

















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