Is infinity mathematically observable? [on hold]
$begingroup$
I have a little question. In fact, is too short.
Is infinity observable? (Can infinity be observed?)
I would like to explain it by example because the question seems unclear in this way.
A simple example:
$sqrt 2=1,41421356237309504880168872420969\807856967187537694807317667973799073247\846210703885038753432764157273501384623\091229702492483605585073721264412149709\993583141322266592750559275579995050115\278206057147010955997160597027453459686\201472851741864088 cdots$
Is it possible to prove that there is no combination of $left{0,0,0right}$, $left{1,1,1right}$ or $left{2,2,2right}$ in this writing?
By mathematical definition,
Let, $phi_{sqrt 2}(n)$ is n'th digit function of $sqrt 2.$
Question: Is there an exist such a $ninmathbb{Z^{+}}$, then $phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=0, phi_{sqrt 2}(n+2)=0$ ?
Or other combinations can be equal,
$$phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=1,phi_{sqrt 2}(n+2)=2, phi_{sqrt 2}(n+3)=3, phi_{sqrt 2}(n+4)=4, phi_{sqrt 2}(n+5)=5$$
Here, $sqrt 2$ is an only simple example. The question is not just
$sqrt 2$.
Generalization of the question is :
For function $phi _alpha (n)$, is it possible to find any integer sequence ? where $alpha$ is an any irrational number or constant ($e,picdots$ and etc).
I "think" , the answer is undecidability. Because, we can not observe infinity. Of course, I dont know the correct answer.
Sorry about the grammar and translation errors in my English.
Thank you very much.
algebra-precalculus soft-question math-history infinity irrational-numbers
$endgroup$
put on hold as unclear what you're asking by Xander Henderson, Tanner Swett, verret, Theo Bendit, Lord Shark the Unknown 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 3 more comments
$begingroup$
I have a little question. In fact, is too short.
Is infinity observable? (Can infinity be observed?)
I would like to explain it by example because the question seems unclear in this way.
A simple example:
$sqrt 2=1,41421356237309504880168872420969\807856967187537694807317667973799073247\846210703885038753432764157273501384623\091229702492483605585073721264412149709\993583141322266592750559275579995050115\278206057147010955997160597027453459686\201472851741864088 cdots$
Is it possible to prove that there is no combination of $left{0,0,0right}$, $left{1,1,1right}$ or $left{2,2,2right}$ in this writing?
By mathematical definition,
Let, $phi_{sqrt 2}(n)$ is n'th digit function of $sqrt 2.$
Question: Is there an exist such a $ninmathbb{Z^{+}}$, then $phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=0, phi_{sqrt 2}(n+2)=0$ ?
Or other combinations can be equal,
$$phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=1,phi_{sqrt 2}(n+2)=2, phi_{sqrt 2}(n+3)=3, phi_{sqrt 2}(n+4)=4, phi_{sqrt 2}(n+5)=5$$
Here, $sqrt 2$ is an only simple example. The question is not just
$sqrt 2$.
Generalization of the question is :
For function $phi _alpha (n)$, is it possible to find any integer sequence ? where $alpha$ is an any irrational number or constant ($e,picdots$ and etc).
I "think" , the answer is undecidability. Because, we can not observe infinity. Of course, I dont know the correct answer.
Sorry about the grammar and translation errors in my English.
Thank you very much.
algebra-precalculus soft-question math-history infinity irrational-numbers
$endgroup$
put on hold as unclear what you're asking by Xander Henderson, Tanner Swett, verret, Theo Bendit, Lord Shark the Unknown 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
20 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
20 hours ago
4
$begingroup$
I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask?
$endgroup$
– Tanner Swett
20 hours ago
1
$begingroup$
What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields.
$endgroup$
– Theo Bendit
18 hours ago
1
$begingroup$
There are certainly irrational numbers which do not contain000
anywhere in their decimal expansion. For example, the number whose digits begin0.10110111011110111110111111011111110111111110
and so on, with the length of the sequence of1
s increasing after each0
. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces000
as a sequence of digits.
$endgroup$
– Daniel Wagner
17 hours ago
|
show 3 more comments
$begingroup$
I have a little question. In fact, is too short.
Is infinity observable? (Can infinity be observed?)
I would like to explain it by example because the question seems unclear in this way.
A simple example:
$sqrt 2=1,41421356237309504880168872420969\807856967187537694807317667973799073247\846210703885038753432764157273501384623\091229702492483605585073721264412149709\993583141322266592750559275579995050115\278206057147010955997160597027453459686\201472851741864088 cdots$
Is it possible to prove that there is no combination of $left{0,0,0right}$, $left{1,1,1right}$ or $left{2,2,2right}$ in this writing?
By mathematical definition,
Let, $phi_{sqrt 2}(n)$ is n'th digit function of $sqrt 2.$
Question: Is there an exist such a $ninmathbb{Z^{+}}$, then $phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=0, phi_{sqrt 2}(n+2)=0$ ?
Or other combinations can be equal,
$$phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=1,phi_{sqrt 2}(n+2)=2, phi_{sqrt 2}(n+3)=3, phi_{sqrt 2}(n+4)=4, phi_{sqrt 2}(n+5)=5$$
Here, $sqrt 2$ is an only simple example. The question is not just
$sqrt 2$.
Generalization of the question is :
For function $phi _alpha (n)$, is it possible to find any integer sequence ? where $alpha$ is an any irrational number or constant ($e,picdots$ and etc).
I "think" , the answer is undecidability. Because, we can not observe infinity. Of course, I dont know the correct answer.
Sorry about the grammar and translation errors in my English.
Thank you very much.
algebra-precalculus soft-question math-history infinity irrational-numbers
$endgroup$
I have a little question. In fact, is too short.
Is infinity observable? (Can infinity be observed?)
I would like to explain it by example because the question seems unclear in this way.
A simple example:
$sqrt 2=1,41421356237309504880168872420969\807856967187537694807317667973799073247\846210703885038753432764157273501384623\091229702492483605585073721264412149709\993583141322266592750559275579995050115\278206057147010955997160597027453459686\201472851741864088 cdots$
Is it possible to prove that there is no combination of $left{0,0,0right}$, $left{1,1,1right}$ or $left{2,2,2right}$ in this writing?
By mathematical definition,
Let, $phi_{sqrt 2}(n)$ is n'th digit function of $sqrt 2.$
Question: Is there an exist such a $ninmathbb{Z^{+}}$, then $phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=0, phi_{sqrt 2}(n+2)=0$ ?
Or other combinations can be equal,
$$phi_{sqrt 2}(n)=0, phi_{sqrt 2}(n+1)=1,phi_{sqrt 2}(n+2)=2, phi_{sqrt 2}(n+3)=3, phi_{sqrt 2}(n+4)=4, phi_{sqrt 2}(n+5)=5$$
Here, $sqrt 2$ is an only simple example. The question is not just
$sqrt 2$.
Generalization of the question is :
For function $phi _alpha (n)$, is it possible to find any integer sequence ? where $alpha$ is an any irrational number or constant ($e,picdots$ and etc).
I "think" , the answer is undecidability. Because, we can not observe infinity. Of course, I dont know the correct answer.
Sorry about the grammar and translation errors in my English.
Thank you very much.
algebra-precalculus soft-question math-history infinity irrational-numbers
algebra-precalculus soft-question math-history infinity irrational-numbers
edited 23 hours ago
Student
asked yesterday
StudentStudent
6631418
6631418
put on hold as unclear what you're asking by Xander Henderson, Tanner Swett, verret, Theo Bendit, Lord Shark the Unknown 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Xander Henderson, Tanner Swett, verret, Theo Bendit, Lord Shark the Unknown 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
20 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
20 hours ago
4
$begingroup$
I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask?
$endgroup$
– Tanner Swett
20 hours ago
1
$begingroup$
What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields.
$endgroup$
– Theo Bendit
18 hours ago
1
$begingroup$
There are certainly irrational numbers which do not contain000
anywhere in their decimal expansion. For example, the number whose digits begin0.10110111011110111110111111011111110111111110
and so on, with the length of the sequence of1
s increasing after each0
. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces000
as a sequence of digits.
$endgroup$
– Daniel Wagner
17 hours ago
|
show 3 more comments
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
20 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
20 hours ago
4
$begingroup$
I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask?
$endgroup$
– Tanner Swett
20 hours ago
1
$begingroup$
What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields.
$endgroup$
– Theo Bendit
18 hours ago
1
$begingroup$
There are certainly irrational numbers which do not contain000
anywhere in their decimal expansion. For example, the number whose digits begin0.10110111011110111110111111011111110111111110
and so on, with the length of the sequence of1
s increasing after each0
. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces000
as a sequence of digits.
$endgroup$
– Daniel Wagner
17 hours ago
1
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
20 hours ago
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
20 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
20 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
20 hours ago
4
4
$begingroup$
I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask?
$endgroup$
– Tanner Swett
20 hours ago
$begingroup$
I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask?
$endgroup$
– Tanner Swett
20 hours ago
1
1
$begingroup$
What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields.
$endgroup$
– Theo Bendit
18 hours ago
$begingroup$
What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields.
$endgroup$
– Theo Bendit
18 hours ago
1
1
$begingroup$
There are certainly irrational numbers which do not contain
000
anywhere in their decimal expansion. For example, the number whose digits begin 0.10110111011110111110111111011111110111111110
and so on, with the length of the sequence of 1
s increasing after each 0
. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces 000
as a sequence of digits.$endgroup$
– Daniel Wagner
17 hours ago
$begingroup$
There are certainly irrational numbers which do not contain
000
anywhere in their decimal expansion. For example, the number whose digits begin 0.10110111011110111110111111011111110111111110
and so on, with the length of the sequence of 1
s increasing after each 0
. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces 000
as a sequence of digits.$endgroup$
– Daniel Wagner
17 hours ago
|
show 3 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Not sure why you multiplied it by $10$, but you can check $sqrt{2}$ written up to $1$ million digits for example here: https://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil . Full text search shows there are 899 occurences of $000$, 859 occurences of $111$ and 919 occurences of $222$. And that is "just" first one million of digits, that does not even come close to infinity...
Actually, there is possibility that $sqrt{2}$ is something called a normal number. If it is, it would mean it contains every finite combination of digits you can imagine. Unfortunately, it is currently unknown where it has this property. So in your second case, $012345$ would be there as well (although it already appears once in the first million digits referred above).
Also, there is one popular question here on MSE about whether $pi$ has this property, you might want to check it out: Does Pi contain all possible number combinations? .
$endgroup$
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
18 hours ago
$begingroup$
"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
$endgroup$
– Eevee Trainer
16 hours ago
add a comment |
$begingroup$
Less an answer than an extended comment:
This actually ties in quite nicely with the concept of a "normal" number. A number which is "normal" is one whose decimal expansion has any sequence of digits occurring equally as often as any other sequence, regardless of the base the number is in.
Of course, it is necessary for the number to be irrational for this to be achieved. "Almost every" real number is a normal number, in the sense that they have Lesbague measure $1$. Despite this, very few numbers are known to be normal, and most of those that are were artificially constructed for the purpose of showing them to be normal. For example, one such number is the concatenation of all the naturals in base $10$, which is known as Champernowne's constant:
$$0.12345678910111213141516171819202122232425...$$
It is suspected that many famous irrational constants - such as $e$, $pi$, and $sqrt 2$ - are indeed normal numbers. Thus, not only would these digit sequences you propose be in the expansion of $sqrt 2$, but every digit sequence would occur in every base - and equally often at that.
Of course, the proof for even $sqrt 2$ seems to elude us at this time. But I imagine that this is not conjectured without basis. As noted in Sil's answer, the three sequences you propose occur several times in just the first million digits. (I anecdotally played around and noticed the first few digits of $pi$ - $31415$ - occurred only once and no later sequences. But again, that's a finite truncation at like one million digits.)
$endgroup$
$begingroup$
Is it known a non-normal number?
$endgroup$
– Student
23 hours ago
1
$begingroup$
Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
$endgroup$
– Eevee Trainer
23 hours ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Not sure why you multiplied it by $10$, but you can check $sqrt{2}$ written up to $1$ million digits for example here: https://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil . Full text search shows there are 899 occurences of $000$, 859 occurences of $111$ and 919 occurences of $222$. And that is "just" first one million of digits, that does not even come close to infinity...
Actually, there is possibility that $sqrt{2}$ is something called a normal number. If it is, it would mean it contains every finite combination of digits you can imagine. Unfortunately, it is currently unknown where it has this property. So in your second case, $012345$ would be there as well (although it already appears once in the first million digits referred above).
Also, there is one popular question here on MSE about whether $pi$ has this property, you might want to check it out: Does Pi contain all possible number combinations? .
$endgroup$
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
18 hours ago
$begingroup$
"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
$endgroup$
– Eevee Trainer
16 hours ago
add a comment |
$begingroup$
Not sure why you multiplied it by $10$, but you can check $sqrt{2}$ written up to $1$ million digits for example here: https://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil . Full text search shows there are 899 occurences of $000$, 859 occurences of $111$ and 919 occurences of $222$. And that is "just" first one million of digits, that does not even come close to infinity...
Actually, there is possibility that $sqrt{2}$ is something called a normal number. If it is, it would mean it contains every finite combination of digits you can imagine. Unfortunately, it is currently unknown where it has this property. So in your second case, $012345$ would be there as well (although it already appears once in the first million digits referred above).
Also, there is one popular question here on MSE about whether $pi$ has this property, you might want to check it out: Does Pi contain all possible number combinations? .
$endgroup$
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
18 hours ago
$begingroup$
"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
$endgroup$
– Eevee Trainer
16 hours ago
add a comment |
$begingroup$
Not sure why you multiplied it by $10$, but you can check $sqrt{2}$ written up to $1$ million digits for example here: https://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil . Full text search shows there are 899 occurences of $000$, 859 occurences of $111$ and 919 occurences of $222$. And that is "just" first one million of digits, that does not even come close to infinity...
Actually, there is possibility that $sqrt{2}$ is something called a normal number. If it is, it would mean it contains every finite combination of digits you can imagine. Unfortunately, it is currently unknown where it has this property. So in your second case, $012345$ would be there as well (although it already appears once in the first million digits referred above).
Also, there is one popular question here on MSE about whether $pi$ has this property, you might want to check it out: Does Pi contain all possible number combinations? .
$endgroup$
Not sure why you multiplied it by $10$, but you can check $sqrt{2}$ written up to $1$ million digits for example here: https://apod.nasa.gov/htmltest/gifcity/sqrt2.1mil . Full text search shows there are 899 occurences of $000$, 859 occurences of $111$ and 919 occurences of $222$. And that is "just" first one million of digits, that does not even come close to infinity...
Actually, there is possibility that $sqrt{2}$ is something called a normal number. If it is, it would mean it contains every finite combination of digits you can imagine. Unfortunately, it is currently unknown where it has this property. So in your second case, $012345$ would be there as well (although it already appears once in the first million digits referred above).
Also, there is one popular question here on MSE about whether $pi$ has this property, you might want to check it out: Does Pi contain all possible number combinations? .
edited 15 hours ago
answered yesterday
SilSil
5,53021645
5,53021645
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
18 hours ago
$begingroup$
"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
$endgroup$
– Eevee Trainer
16 hours ago
add a comment |
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
18 hours ago
$begingroup$
"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
$endgroup$
– Eevee Trainer
16 hours ago
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
$e$ is not known to be normal, but (as pointed out in my pseudoanswer) it's conjectured to be. Pretty much all of the normal numbers aside from a few specific constants we know of were specifically constructed for the purpose of showing normal numbers exist.
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago.
$endgroup$
– Paul Sinclair
21 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
18 hours ago
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You said "you might wan to check it out", I think you mean "want" instead of "wan".
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– numbermaniac
18 hours ago
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"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
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– Eevee Trainer
16 hours ago
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"By curious coincidence, I just happened to watch this numberphile video that talks about normal numbers a couple hours ago" -- lol that was basically the inspiration behind my own post. I happened to recall that video when reading the OP and Sil's answer. Pretty much the main reason I know normal numbers exist. xD
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– Eevee Trainer
16 hours ago
add a comment |
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Less an answer than an extended comment:
This actually ties in quite nicely with the concept of a "normal" number. A number which is "normal" is one whose decimal expansion has any sequence of digits occurring equally as often as any other sequence, regardless of the base the number is in.
Of course, it is necessary for the number to be irrational for this to be achieved. "Almost every" real number is a normal number, in the sense that they have Lesbague measure $1$. Despite this, very few numbers are known to be normal, and most of those that are were artificially constructed for the purpose of showing them to be normal. For example, one such number is the concatenation of all the naturals in base $10$, which is known as Champernowne's constant:
$$0.12345678910111213141516171819202122232425...$$
It is suspected that many famous irrational constants - such as $e$, $pi$, and $sqrt 2$ - are indeed normal numbers. Thus, not only would these digit sequences you propose be in the expansion of $sqrt 2$, but every digit sequence would occur in every base - and equally often at that.
Of course, the proof for even $sqrt 2$ seems to elude us at this time. But I imagine that this is not conjectured without basis. As noted in Sil's answer, the three sequences you propose occur several times in just the first million digits. (I anecdotally played around and noticed the first few digits of $pi$ - $31415$ - occurred only once and no later sequences. But again, that's a finite truncation at like one million digits.)
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Is it known a non-normal number?
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– Student
23 hours ago
1
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Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
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– Eevee Trainer
23 hours ago
add a comment |
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Less an answer than an extended comment:
This actually ties in quite nicely with the concept of a "normal" number. A number which is "normal" is one whose decimal expansion has any sequence of digits occurring equally as often as any other sequence, regardless of the base the number is in.
Of course, it is necessary for the number to be irrational for this to be achieved. "Almost every" real number is a normal number, in the sense that they have Lesbague measure $1$. Despite this, very few numbers are known to be normal, and most of those that are were artificially constructed for the purpose of showing them to be normal. For example, one such number is the concatenation of all the naturals in base $10$, which is known as Champernowne's constant:
$$0.12345678910111213141516171819202122232425...$$
It is suspected that many famous irrational constants - such as $e$, $pi$, and $sqrt 2$ - are indeed normal numbers. Thus, not only would these digit sequences you propose be in the expansion of $sqrt 2$, but every digit sequence would occur in every base - and equally often at that.
Of course, the proof for even $sqrt 2$ seems to elude us at this time. But I imagine that this is not conjectured without basis. As noted in Sil's answer, the three sequences you propose occur several times in just the first million digits. (I anecdotally played around and noticed the first few digits of $pi$ - $31415$ - occurred only once and no later sequences. But again, that's a finite truncation at like one million digits.)
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Is it known a non-normal number?
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– Student
23 hours ago
1
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Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
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– Eevee Trainer
23 hours ago
add a comment |
$begingroup$
Less an answer than an extended comment:
This actually ties in quite nicely with the concept of a "normal" number. A number which is "normal" is one whose decimal expansion has any sequence of digits occurring equally as often as any other sequence, regardless of the base the number is in.
Of course, it is necessary for the number to be irrational for this to be achieved. "Almost every" real number is a normal number, in the sense that they have Lesbague measure $1$. Despite this, very few numbers are known to be normal, and most of those that are were artificially constructed for the purpose of showing them to be normal. For example, one such number is the concatenation of all the naturals in base $10$, which is known as Champernowne's constant:
$$0.12345678910111213141516171819202122232425...$$
It is suspected that many famous irrational constants - such as $e$, $pi$, and $sqrt 2$ - are indeed normal numbers. Thus, not only would these digit sequences you propose be in the expansion of $sqrt 2$, but every digit sequence would occur in every base - and equally often at that.
Of course, the proof for even $sqrt 2$ seems to elude us at this time. But I imagine that this is not conjectured without basis. As noted in Sil's answer, the three sequences you propose occur several times in just the first million digits. (I anecdotally played around and noticed the first few digits of $pi$ - $31415$ - occurred only once and no later sequences. But again, that's a finite truncation at like one million digits.)
$endgroup$
Less an answer than an extended comment:
This actually ties in quite nicely with the concept of a "normal" number. A number which is "normal" is one whose decimal expansion has any sequence of digits occurring equally as often as any other sequence, regardless of the base the number is in.
Of course, it is necessary for the number to be irrational for this to be achieved. "Almost every" real number is a normal number, in the sense that they have Lesbague measure $1$. Despite this, very few numbers are known to be normal, and most of those that are were artificially constructed for the purpose of showing them to be normal. For example, one such number is the concatenation of all the naturals in base $10$, which is known as Champernowne's constant:
$$0.12345678910111213141516171819202122232425...$$
It is suspected that many famous irrational constants - such as $e$, $pi$, and $sqrt 2$ - are indeed normal numbers. Thus, not only would these digit sequences you propose be in the expansion of $sqrt 2$, but every digit sequence would occur in every base - and equally often at that.
Of course, the proof for even $sqrt 2$ seems to elude us at this time. But I imagine that this is not conjectured without basis. As noted in Sil's answer, the three sequences you propose occur several times in just the first million digits. (I anecdotally played around and noticed the first few digits of $pi$ - $31415$ - occurred only once and no later sequences. But again, that's a finite truncation at like one million digits.)
answered yesterday
Eevee TrainerEevee Trainer
8,51321439
8,51321439
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Is it known a non-normal number?
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– Student
23 hours ago
1
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Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
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– Eevee Trainer
23 hours ago
add a comment |
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Is it known a non-normal number?
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– Student
23 hours ago
1
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Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
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– Eevee Trainer
23 hours ago
$begingroup$
Is it known a non-normal number?
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– Student
23 hours ago
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Is it known a non-normal number?
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– Student
23 hours ago
1
1
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Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
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– Eevee Trainer
23 hours ago
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Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
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– Eevee Trainer
23 hours ago
add a comment |
1
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Possible duplicate of Does Pi contain all possible number combinations?
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– Xander Henderson
20 hours ago
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arxiv.org/abs/math/0411418
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– Count Iblis
20 hours ago
4
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I see two questions here, and they are not the same question. The first question is, "Is infinity mathematically observable?" The second question is, "Do irrational numbers contain every possible sequence of digits?" Which of these two questions is the one you intend to ask?
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– Tanner Swett
20 hours ago
1
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What exactly do you mean by "observable" here? It's also not exactly clear what "infinity" means in this context either; it means many different things in many different mathematical fields.
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– Theo Bendit
18 hours ago
1
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There are certainly irrational numbers which do not contain
000
anywhere in their decimal expansion. For example, the number whose digits begin0.10110111011110111110111111011111110111111110
and so on, with the length of the sequence of1
s increasing after each0
. It is clearly not rational -- the decimal expansion never enters a repeating loop -- and also just as clearly never produces000
as a sequence of digits.$endgroup$
– Daniel Wagner
17 hours ago