Understanding Normal Numbers












1












$begingroup$


I am trying to understand what normal numbers are. Just for simplicity I want to talk about base 10. I understand that a number is normal in base 10 if there a probability of $frac{1}{10
} $ such that the numbers 0-9 pop up and a probability of $frac{1}{100}$ that the numbers $0-99$ pop up in the decimal expansion and so on...



However I am wondering, in this case when we talk about normal numbers we are only talking about irrational numbers since numbers like $2$ or $3$ don't have a decimal expansion so there is no sense in talking about them as normal numbers?



Or for example a number like $frac{1}{3} = 0.3333..$ is not normal base 10 since it's decimal expansion only contains the number 3.



I am just wondering if a normal number has to do anything with the normal form of a number.






decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000overline{0}. $$
    $endgroup$
    – Xander Henderson
    Dec 1 '18 at 20:42










  • $begingroup$
    A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:43








  • 1




    $begingroup$
    Normal doesn't just refer to single digits. Yes, each digit should occur with probability $frac 1{10}$ but each pair of digits should occur with probability $frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:46






  • 1




    $begingroup$
    @Sasha Yes. But that does not imply irrational. The rational number $.overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:57








  • 1




    $begingroup$
    Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such.
    $endgroup$
    – lulu
    Dec 1 '18 at 21:15
















1












$begingroup$


I am trying to understand what normal numbers are. Just for simplicity I want to talk about base 10. I understand that a number is normal in base 10 if there a probability of $frac{1}{10
} $ such that the numbers 0-9 pop up and a probability of $frac{1}{100}$ that the numbers $0-99$ pop up in the decimal expansion and so on...



However I am wondering, in this case when we talk about normal numbers we are only talking about irrational numbers since numbers like $2$ or $3$ don't have a decimal expansion so there is no sense in talking about them as normal numbers?



Or for example a number like $frac{1}{3} = 0.3333..$ is not normal base 10 since it's decimal expansion only contains the number 3.



I am just wondering if a normal number has to do anything with the normal form of a number.






decimal-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000overline{0}. $$
    $endgroup$
    – Xander Henderson
    Dec 1 '18 at 20:42










  • $begingroup$
    A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:43








  • 1




    $begingroup$
    Normal doesn't just refer to single digits. Yes, each digit should occur with probability $frac 1{10}$ but each pair of digits should occur with probability $frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:46






  • 1




    $begingroup$
    @Sasha Yes. But that does not imply irrational. The rational number $.overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:57








  • 1




    $begingroup$
    Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such.
    $endgroup$
    – lulu
    Dec 1 '18 at 21:15














1












1








1





$begingroup$


I am trying to understand what normal numbers are. Just for simplicity I want to talk about base 10. I understand that a number is normal in base 10 if there a probability of $frac{1}{10
} $ such that the numbers 0-9 pop up and a probability of $frac{1}{100}$ that the numbers $0-99$ pop up in the decimal expansion and so on...



However I am wondering, in this case when we talk about normal numbers we are only talking about irrational numbers since numbers like $2$ or $3$ don't have a decimal expansion so there is no sense in talking about them as normal numbers?



Or for example a number like $frac{1}{3} = 0.3333..$ is not normal base 10 since it's decimal expansion only contains the number 3.



I am just wondering if a normal number has to do anything with the normal form of a number.






decimal-expansion







share|cite|improve this question











$endgroup$




I am trying to understand what normal numbers are. Just for simplicity I want to talk about base 10. I understand that a number is normal in base 10 if there a probability of $frac{1}{10
} $ such that the numbers 0-9 pop up and a probability of $frac{1}{100}$ that the numbers $0-99$ pop up in the decimal expansion and so on...



However I am wondering, in this case when we talk about normal numbers we are only talking about irrational numbers since numbers like $2$ or $3$ don't have a decimal expansion so there is no sense in talking about them as normal numbers?



Or for example a number like $frac{1}{3} = 0.3333..$ is not normal base 10 since it's decimal expansion only contains the number 3.



I am just wondering if a normal number has to do anything with the normal form of a number.






decimal-expansion




decimal-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 22:25









Andrés E. Caicedo

65.4k8158249




65.4k8158249










asked Dec 1 '18 at 20:39









SashaSasha

657




657








  • 1




    $begingroup$
    Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000overline{0}. $$
    $endgroup$
    – Xander Henderson
    Dec 1 '18 at 20:42










  • $begingroup$
    A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:43








  • 1




    $begingroup$
    Normal doesn't just refer to single digits. Yes, each digit should occur with probability $frac 1{10}$ but each pair of digits should occur with probability $frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:46






  • 1




    $begingroup$
    @Sasha Yes. But that does not imply irrational. The rational number $.overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:57








  • 1




    $begingroup$
    Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such.
    $endgroup$
    – lulu
    Dec 1 '18 at 21:15














  • 1




    $begingroup$
    Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000overline{0}. $$
    $endgroup$
    – Xander Henderson
    Dec 1 '18 at 20:42










  • $begingroup$
    A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal.
    $endgroup$
    – Arthur
    Dec 1 '18 at 20:43








  • 1




    $begingroup$
    Normal doesn't just refer to single digits. Yes, each digit should occur with probability $frac 1{10}$ but each pair of digits should occur with probability $frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:46






  • 1




    $begingroup$
    @Sasha Yes. But that does not imply irrational. The rational number $.overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property.
    $endgroup$
    – lulu
    Dec 1 '18 at 20:57








  • 1




    $begingroup$
    Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such.
    $endgroup$
    – lulu
    Dec 1 '18 at 21:15








1




1




$begingroup$
Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000overline{0}. $$
$endgroup$
– Xander Henderson
Dec 1 '18 at 20:42




$begingroup$
Numbers like 2 and 3 certainly do have decimal expansions. For example, $$2 = 2.000overline{0}. $$
$endgroup$
– Xander Henderson
Dec 1 '18 at 20:42












$begingroup$
A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal.
$endgroup$
– Arthur
Dec 1 '18 at 20:43






$begingroup$
A normal number (even one which is only normal in base ten) must necessarily be irrational, yes. Most real numbers are normal, yet the only numbers we know to be normal are the ones we have explicitly constructed to be normal.
$endgroup$
– Arthur
Dec 1 '18 at 20:43






1




1




$begingroup$
Normal doesn't just refer to single digits. Yes, each digit should occur with probability $frac 1{10}$ but each pair of digits should occur with probability $frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational.
$endgroup$
– lulu
Dec 1 '18 at 20:46




$begingroup$
Normal doesn't just refer to single digits. Yes, each digit should occur with probability $frac 1{10}$ but each pair of digits should occur with probability $frac 1{100}$ and so on. See, e.g., this. But yes, normal implies irrational.
$endgroup$
– lulu
Dec 1 '18 at 20:46




1




1




$begingroup$
@Sasha Yes. But that does not imply irrational. The rational number $.overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property.
$endgroup$
– lulu
Dec 1 '18 at 20:57






$begingroup$
@Sasha Yes. But that does not imply irrational. The rational number $.overline {0123456789}$ is simply normal in that sense. Proper normality is a much more interesting property.
$endgroup$
– lulu
Dec 1 '18 at 20:57






1




1




$begingroup$
Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such.
$endgroup$
– lulu
Dec 1 '18 at 21:15




$begingroup$
Actually my proof should be modified slightly to take into account the fact that the period may not start right away. Easy to handle...suppose there are $Q$ terms before the period of length $P$ starts and replace the $2P$ in my expression by $2(P+Q)$ or such.
$endgroup$
– lulu
Dec 1 '18 at 21:15










1 Answer
1






active

oldest

votes


















2












$begingroup$

As you noted a normal number ( in some basis) is necessarily irrational. We know (it was proved by E. Borel) that ''almost all'' (in the sense of measure theory) real numbers are absolutely normal normal, that is normal in all basis, but the proof is not constructive, and it is not clear if there is some computable normal number. What we can say is that we dont know if nubers as $pi, e, sqrt{2}$ are normal also in base $10$ .



We know that the Champernowne number is normal in base $10$ but it is not normal in all basis.



You can also see here for the problem of construction of a normal number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:03










  • $begingroup$
    I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
    $endgroup$
    – Emilio Novati
    Dec 1 '18 at 21:20










  • $begingroup$
    Thank you. I'll enjoy reading this, I'm sure.
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:37











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1 Answer
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2












$begingroup$

As you noted a normal number ( in some basis) is necessarily irrational. We know (it was proved by E. Borel) that ''almost all'' (in the sense of measure theory) real numbers are absolutely normal normal, that is normal in all basis, but the proof is not constructive, and it is not clear if there is some computable normal number. What we can say is that we dont know if nubers as $pi, e, sqrt{2}$ are normal also in base $10$ .



We know that the Champernowne number is normal in base $10$ but it is not normal in all basis.



You can also see here for the problem of construction of a normal number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:03










  • $begingroup$
    I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
    $endgroup$
    – Emilio Novati
    Dec 1 '18 at 21:20










  • $begingroup$
    Thank you. I'll enjoy reading this, I'm sure.
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:37
















2












$begingroup$

As you noted a normal number ( in some basis) is necessarily irrational. We know (it was proved by E. Borel) that ''almost all'' (in the sense of measure theory) real numbers are absolutely normal normal, that is normal in all basis, but the proof is not constructive, and it is not clear if there is some computable normal number. What we can say is that we dont know if nubers as $pi, e, sqrt{2}$ are normal also in base $10$ .



We know that the Champernowne number is normal in base $10$ but it is not normal in all basis.



You can also see here for the problem of construction of a normal number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:03










  • $begingroup$
    I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
    $endgroup$
    – Emilio Novati
    Dec 1 '18 at 21:20










  • $begingroup$
    Thank you. I'll enjoy reading this, I'm sure.
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:37














2












2








2





$begingroup$

As you noted a normal number ( in some basis) is necessarily irrational. We know (it was proved by E. Borel) that ''almost all'' (in the sense of measure theory) real numbers are absolutely normal normal, that is normal in all basis, but the proof is not constructive, and it is not clear if there is some computable normal number. What we can say is that we dont know if nubers as $pi, e, sqrt{2}$ are normal also in base $10$ .



We know that the Champernowne number is normal in base $10$ but it is not normal in all basis.



You can also see here for the problem of construction of a normal number.






share|cite|improve this answer











$endgroup$



As you noted a normal number ( in some basis) is necessarily irrational. We know (it was proved by E. Borel) that ''almost all'' (in the sense of measure theory) real numbers are absolutely normal normal, that is normal in all basis, but the proof is not constructive, and it is not clear if there is some computable normal number. What we can say is that we dont know if nubers as $pi, e, sqrt{2}$ are normal also in base $10$ .



We know that the Champernowne number is normal in base $10$ but it is not normal in all basis.



You can also see here for the problem of construction of a normal number.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 1 '18 at 21:14

























answered Dec 1 '18 at 21:01









Emilio NovatiEmilio Novati

52k43474




52k43474












  • $begingroup$
    It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:03










  • $begingroup$
    I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
    $endgroup$
    – Emilio Novati
    Dec 1 '18 at 21:20










  • $begingroup$
    Thank you. I'll enjoy reading this, I'm sure.
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:37


















  • $begingroup$
    It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:03










  • $begingroup$
    I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
    $endgroup$
    – Emilio Novati
    Dec 1 '18 at 21:20










  • $begingroup$
    Thank you. I'll enjoy reading this, I'm sure.
    $endgroup$
    – saulspatz
    Dec 1 '18 at 21:37
















$begingroup$
It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
$endgroup$
– saulspatz
Dec 1 '18 at 21:03




$begingroup$
It actually known that it's not normal in all bases, or just unknown whether it is? I've always thought it was the latter, but I don't have any support for that. If the former, do you know of a reference?
$endgroup$
– saulspatz
Dec 1 '18 at 21:03












$begingroup$
I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
$endgroup$
– Emilio Novati
Dec 1 '18 at 21:20




$begingroup$
I did not verify the proof, but here (google.it/… ) the autors say that Champernowne’s number fails to be strongly normal.
$endgroup$
– Emilio Novati
Dec 1 '18 at 21:20












$begingroup$
Thank you. I'll enjoy reading this, I'm sure.
$endgroup$
– saulspatz
Dec 1 '18 at 21:37




$begingroup$
Thank you. I'll enjoy reading this, I'm sure.
$endgroup$
– saulspatz
Dec 1 '18 at 21:37


















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