If A is dense in Q, then it must be dense in R.Defining dense subsets of $mathbbR$Why aren't all dense subsets of $mathbbR$ uncountable?Show that $mathbbQ$ is dense in the real numbers. (Using Supremum)Characterization of dense open subsets of the real numbersProof of the infinitude of rational and irrational numbersMust a comeager set be dense?countable dense subset of R^kGiven that the rationals are countable and the denseness of $mathbbQ$ in $mathbbR$ how can $mathbbR$ be uncountable?Proof technique for between any two real numbers is an irrational numberprove that $mathbbQ^n$is dense subset of $mathbbR^n$
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If A is dense in Q, then it must be dense in R.
Defining dense subsets of $mathbbR$Why aren't all dense subsets of $mathbbR$ uncountable?Show that $mathbbQ$ is dense in the real numbers. (Using Supremum)Characterization of dense open subsets of the real numbersProof of the infinitude of rational and irrational numbersMust a comeager set be dense?countable dense subset of R^kGiven that the rationals are countable and the denseness of $mathbbQ$ in $mathbbR$ how can $mathbbR$ be uncountable?Proof technique for between any two real numbers is an irrational numberprove that $mathbbQ^n$is dense subset of $mathbbR^n$
$begingroup$
I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.
real-analysis real-numbers
edited 4 hours ago
Floris Claassens
1,04016
asked 4 hours ago
Priti DPriti D
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.
real-analysis real-numbers
edited 4 hours ago
Floris Claassens
1,04016
asked 4 hours ago
Priti DPriti D
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
4 hours ago
$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago
add a comment |
$begingroup$
I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.
real-analysis real-numbers
edited 4 hours ago
Floris Claassens
1,04016
asked 4 hours ago
Priti DPriti D
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have $A$ is a subset of $mathbbR$. If $A$ is dense in $mathbbQ$, then it must be dense in $mathbbR$. I am confused because $A$ is dense in $mathbbQ$. Does that imply that between any two rational numbers, there exists a real number? I understand for anything to be dense in R, there must exist something that lies between any two real numbers. However, how does knowing something is dense in $mathbbQ$ prove that it must be dense in the reals? Any help is appreciated.
real-analysis real-numbers
real-analysis real-numbers
edited 4 hours ago
Floris Claassens
1,04016
asked 4 hours ago
Priti DPriti D
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Floris Claassens
1,04016
asked 4 hours ago
Priti DPriti D
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 4 hours ago
Floris Claassens
1,04016
edited 4 hours ago
Floris Claassens
1,04016
edited 4 hours ago
Floris Claassens
1,04016
1,04016
asked 4 hours ago
Priti DPriti D
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Priti DPriti D
212
asked 4 hours ago
Priti DPriti D
212
212
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Priti D is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
4 hours ago
$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago
add a comment |
$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
4 hours ago
$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago
$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
4 hours ago
$begingroup$
Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
$endgroup$
– user334732
4 hours ago
$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago
$begingroup$
And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
$endgroup$
– user334732
4 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
$endgroup$
1
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
add a comment |
$begingroup$
$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.
We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$
answered 4 hours ago
MasonMason
1,7551630
$endgroup$
1
$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago
$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago
$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago
$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
$endgroup$
1
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
add a comment |
$begingroup$
Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
$endgroup$
1
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
add a comment |
$begingroup$
Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
$endgroup$
Since $A$ is dense in $Bbb Q$ so $overline A cap Bbb Q = Bbb Q subseteq overline A.$ So $Bbb R = overline Bbb Q subseteq overline A subseteq Bbb R.$ Therefore $overline A = Bbb R.$ This shows that $A$ is dense in $Bbb R.$
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
answered 4 hours ago
Dbchatto67Dbchatto67
1,965319
1,965319
1
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
add a comment |
1
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
1
1
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
Looks good to me. I would add the line: Where $A$ overbar refers to the closure of a set. It's unclear to me whether the OP is familiar with this expression so maybe a link to? mathworld.wolfram.com/SetClosure.html. +1
$endgroup$
– Mason
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
$begingroup$
This is true for any arbitrary metric space @Mason. Let $(X,d)$ be a metric space. Let $Y$ be metric subspace of $X.$ Let $A subseteq X.$ Then the closure of $A$ in $Y$ say $overline A^Y = overline A cap Y,$ where $overline A$ is the closure of $A$ in $X.$
$endgroup$
– Dbchatto67
4 hours ago
add a comment |
$begingroup$
$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.
We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$
answered 4 hours ago
MasonMason
1,7551630
$endgroup$
1
$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago
$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago
$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago
$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago
add a comment |
$begingroup$
$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.
We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$
answered 4 hours ago
MasonMason
1,7551630
$endgroup$
1
$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago
$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago
$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago
$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago
add a comment |
$begingroup$
$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.
We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$
answered 4 hours ago
MasonMason
1,7551630
$endgroup$
$A$ is dense in $mathbbQ$ if for any two rationals $q_1 < q_2$ there is some $ain A cap mathbbQ$ such that $q_1<a<q_2$. The dyadic rationals would be an example. Here is the way to think about the puzzle of nested dense sets. If you give me two reals $r_1$ and $r_2$ can I find a $q_1$ in between them? Yes. Why? because $mathbbQ$ is dense is $mathbbR$. Can I find two values? $q_1$ and $q_2$ in between $r_1$ and $r_2$? Because if I could find two... then I could exploit the density of $mathbbQ$ to finish the job.
We are given two reals and then we find $q_1,q_2$ inbetween the reals and then we find some $ain A$ inbetween these rationals. All told we have the following inequality: $$r_1<q_1<a<q_2<r_2$$
answered 4 hours ago
MasonMason
1,7551630
edited 4 hours ago
answered 4 hours ago
MasonMason
1,7551630
answered 4 hours ago
MasonMason
1,7551630
answered 4 hours ago
MasonMason
1,7551630
1,7551630
1
$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago
$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago
$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago
$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago
add a comment |
1
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Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
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– CiaPan
4 hours ago
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@CiaPan. I think you are right. This was written hastily.
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– Mason
4 hours ago
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:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
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– CiaPan
4 hours ago
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I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
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– Mason
4 hours ago
1
1
$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago
$begingroup$
Isn't it 'for any two $q_1, q_2 in mathbb Q$ there exists $ain Acolorredcapmathbb Q$ such that $q_1 < a < q_2$' ...? (Plus, of course, an assumption of $q_1<q_2.$)
$endgroup$
– CiaPan
4 hours ago
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@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago
$begingroup$
@CiaPan. I think you are right. This was written hastily.
$endgroup$
– Mason
4 hours ago
$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago
$begingroup$
:) I'm not familiar with the definition, but I thought about irrationals – they are certainly dense in reals, but I would hesitate to accept they are dense in rationals, which would fit your former formulation. Hence the red part.
$endgroup$
– CiaPan
4 hours ago
$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
$endgroup$
– Mason
4 hours ago
$begingroup$
I don't think that it would have fit because I defined it as a subset. So it would read "the irrational reals are dense in the rationals if (1) they are a subset of the rationals and (2) there is an irrational between any two rationals." So it wouldn't meet the (1)st criteria. So we might be dancing around equivalent or similar variations on the same concept.
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– Mason
4 hours ago
add a comment |
Priti D is a new contributor. Be nice, and check out our Code of Conduct.
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Priti D is a new contributor. Be nice, and check out our Code of Conduct.
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Not a proof, but it may help you to understand what's going on. For any pair of distinct real numbers, and any $a$ in-between, there is always a pair of rational numbers that also straddles $a$ and is also in-between the pair of real numbers. In fact there are infinitely many.
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– user334732
4 hours ago
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And yes, there is always an irrational real number between any pair of rational numbers. In fact there is an uncountable infinity of them.
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– user334732
4 hours ago