Is infinity mathematically observable?
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
3 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
2 hours ago
3
$begingroup$
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
2 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
42 mins ago
add a comment |
$begingroup$
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 6 hours ago
Student
asked 6 hours ago
StudentStudent
6671418
6671418
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
3 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
2 hours ago
3
$begingroup$
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
2 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
42 mins ago
add a comment |
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
3 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
2 hours ago
3
$begingroup$
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
2 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
42 mins ago
1
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
3 hours ago
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
3 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
2 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
2 hours ago
3
3
$begingroup$
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
2 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
2 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
42 mins 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
42 mins ago
add a comment |
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 wan to check it out: Does Pi contain all possible number combinations? .
$endgroup$
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
$e$ 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
6 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
3 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins 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
6 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
6 hours ago
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162396%2fis-infinity-mathematically-observable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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 wan to check it out: Does Pi contain all possible number combinations? .
$endgroup$
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
$e$ 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
6 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
3 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins 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 wan to check it out: Does Pi contain all possible number combinations? .
$endgroup$
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
$e$ 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
6 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
3 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins 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 wan 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 wan to check it out: Does Pi contain all possible number combinations? .
edited 6 hours ago
answered 6 hours ago
SilSil
5,49021645
5,49021645
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
$e$ 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
6 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
3 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins ago
add a comment |
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
$e$ 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
6 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
3 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins ago
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
Well, for $e$ is it possible?
$endgroup$
– Student
6 hours ago
$begingroup$
$e$ 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
6 hours ago
$begingroup$
$e$ 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
6 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
3 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
3 hours ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins ago
$begingroup$
You said "you might wan to check it out", I think you mean "want" instead of "wan".
$endgroup$
– numbermaniac
55 mins 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
6 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
6 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
6 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
6 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 6 hours ago
Eevee TrainerEevee Trainer
8,51421439
8,51421439
$begingroup$
Is it known a non-normal number?
$endgroup$
– Student
6 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
6 hours ago
add a comment |
$begingroup$
Is it known a non-normal number?
$endgroup$
– Student
6 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
6 hours ago
$begingroup$
Is it known a non-normal number?
$endgroup$
– Student
6 hours ago
$begingroup$
Is it known a non-normal number?
$endgroup$
– Student
6 hours ago
1
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
6 hours ago
$begingroup$
Nope. After all, if it were known to be not normal, it wouldn't be conjectured to be normal. :p
$endgroup$
– Eevee Trainer
6 hours ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3162396%2fis-infinity-mathematically-observable%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Possible duplicate of Does Pi contain all possible number combinations?
$endgroup$
– Xander Henderson
3 hours ago
$begingroup$
arxiv.org/abs/math/0411418
$endgroup$
– Count Iblis
2 hours ago
3
$begingroup$
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
2 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
42 mins ago