Mixing
Relation between Q and R
$begingroup$
please how to prove these two properties:
$(1)quadforall xinmathbb{R}, forall varepsilon>0, exists rinmathbb{Q}, |x-r|leqvarepsilon$
$(2)quad forall xinmathbb{R}, exists (r_n)inmathbb{Q}, displaystylelim_{ntoinfty} r_n=x$
real-analysis
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
$endgroup$
add a comment |
$begingroup$
please how to prove these two properties:
$(1)quadforall xinmathbb{R}, forall varepsilon>0, exists rinmathbb{Q}, |x-r|leqvarepsilon$
$(2)quad forall xinmathbb{R}, exists (r_n)inmathbb{Q}, displaystylelim_{ntoinfty} r_n=x$
real-analysis
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
$endgroup$
1
$begingroup$
It refers to density.
$endgroup$
– Wuestenfux
Jan 11 at 9:37
2
$begingroup$
What are your definitions of $Bbb Q$ and of $Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another.
$endgroup$
– Arthur
Jan 11 at 9:37
$begingroup$
The second property is how my professor in mathematics defined $mathbb{R}$. And the first property is a consequence of the second one
$endgroup$
– Damien
Jan 11 at 10:27
add a comment |
$begingroup$
please how to prove these two properties:
$(1)quadforall xinmathbb{R}, forall varepsilon>0, exists rinmathbb{Q}, |x-r|leqvarepsilon$
$(2)quad forall xinmathbb{R}, exists (r_n)inmathbb{Q}, displaystylelim_{ntoinfty} r_n=x$
real-analysis
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
$endgroup$
please how to prove these two properties:
$(1)quadforall xinmathbb{R}, forall varepsilon>0, exists rinmathbb{Q}, |x-r|leqvarepsilon$
$(2)quad forall xinmathbb{R}, exists (r_n)inmathbb{Q}, displaystylelim_{ntoinfty} r_n=x$
real-analysis
real-analysis
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
asked Jan 11 at 9:36
Poline SandraPoline Sandra
897
897
1
$begingroup$
It refers to density.
$endgroup$
– Wuestenfux
Jan 11 at 9:37
2
$begingroup$
What are your definitions of $Bbb Q$ and of $Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another.
$endgroup$
– Arthur
Jan 11 at 9:37
$begingroup$
The second property is how my professor in mathematics defined $mathbb{R}$. And the first property is a consequence of the second one
$endgroup$
– Damien
Jan 11 at 10:27
add a comment |
1
$begingroup$
It refers to density.
$endgroup$
– Wuestenfux
Jan 11 at 9:37
2
$begingroup$
What are your definitions of $Bbb Q$ and of $Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another.
$endgroup$
– Arthur
Jan 11 at 9:37
$begingroup$
The second property is how my professor in mathematics defined $mathbb{R}$. And the first property is a consequence of the second one
$endgroup$
– Damien
Jan 11 at 10:27
1
1
$begingroup$
It refers to density.
$endgroup$
– Wuestenfux
Jan 11 at 9:37
$begingroup$
It refers to density.
$endgroup$
– Wuestenfux
Jan 11 at 9:37
2
2
$begingroup$
What are your definitions of $Bbb Q$ and of $Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another.
$endgroup$
– Arthur
Jan 11 at 9:37
$begingroup$
What are your definitions of $Bbb Q$ and of $Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another.
$endgroup$
– Arthur
Jan 11 at 9:37
$begingroup$
The second property is how my professor in mathematics defined $mathbb{R}$. And the first property is a consequence of the second one
$endgroup$
– Damien
Jan 11 at 10:27
$begingroup$
The second property is how my professor in mathematics defined $mathbb{R}$. And the first property is a consequence of the second one
$endgroup$
– Damien
Jan 11 at 10:27
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can show $(1)$ as follows. Let $N > frac{1}{varepsilon}$ be an integer. Then we observe that
$$Nx - left lfloor{Nx}right rfloor < 1 Leftrightarrow x - frac{left lfloor{Nx}right rfloor}{N} < frac{1}{N} < varepsilon$$
So taking $r = frac{left lfloor{Nx}right rfloor}{N}$, we are done.
$(2)$ follows from $(1)$ by considering a sequence $varepsilon_1,varepsilon_2,...$ with $varepsilon_n > 0$ and $lim_{n rightarrow infty}varepsilon_n = 0$ (for instance $varepsilon_n = frac{1}{n}$). Then we can find a sequence $r_n in mathbb{Q}$ such that $|x - r_n| < varepsilon_n$, from which $lim_{n rightarrow infty} r_n = x$ follows since $forall varepsilon > 0$ there is an $N$ such that $n > N$ implies
$$|x - r_n| < varepsilon_n < varepsilon$$
as required.
answered Jan 11 at 10:41
nesHannesHan
1735
$endgroup$
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%2f3069653%2frelation-between-q-and-r%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can show $(1)$ as follows. Let $N > frac{1}{varepsilon}$ be an integer. Then we observe that
$$Nx - left lfloor{Nx}right rfloor < 1 Leftrightarrow x - frac{left lfloor{Nx}right rfloor}{N} < frac{1}{N} < varepsilon$$
So taking $r = frac{left lfloor{Nx}right rfloor}{N}$, we are done.
$(2)$ follows from $(1)$ by considering a sequence $varepsilon_1,varepsilon_2,...$ with $varepsilon_n > 0$ and $lim_{n rightarrow infty}varepsilon_n = 0$ (for instance $varepsilon_n = frac{1}{n}$). Then we can find a sequence $r_n in mathbb{Q}$ such that $|x - r_n| < varepsilon_n$, from which $lim_{n rightarrow infty} r_n = x$ follows since $forall varepsilon > 0$ there is an $N$ such that $n > N$ implies
$$|x - r_n| < varepsilon_n < varepsilon$$
as required.
answered Jan 11 at 10:41
nesHannesHan
1735
$endgroup$
add a comment |
$begingroup$
We can show $(1)$ as follows. Let $N > frac{1}{varepsilon}$ be an integer. Then we observe that
$$Nx - left lfloor{Nx}right rfloor < 1 Leftrightarrow x - frac{left lfloor{Nx}right rfloor}{N} < frac{1}{N} < varepsilon$$
So taking $r = frac{left lfloor{Nx}right rfloor}{N}$, we are done.
$(2)$ follows from $(1)$ by considering a sequence $varepsilon_1,varepsilon_2,...$ with $varepsilon_n > 0$ and $lim_{n rightarrow infty}varepsilon_n = 0$ (for instance $varepsilon_n = frac{1}{n}$). Then we can find a sequence $r_n in mathbb{Q}$ such that $|x - r_n| < varepsilon_n$, from which $lim_{n rightarrow infty} r_n = x$ follows since $forall varepsilon > 0$ there is an $N$ such that $n > N$ implies
$$|x - r_n| < varepsilon_n < varepsilon$$
as required.
answered Jan 11 at 10:41
nesHannesHan
1735
$endgroup$
add a comment |
$begingroup$
We can show $(1)$ as follows. Let $N > frac{1}{varepsilon}$ be an integer. Then we observe that
$$Nx - left lfloor{Nx}right rfloor < 1 Leftrightarrow x - frac{left lfloor{Nx}right rfloor}{N} < frac{1}{N} < varepsilon$$
So taking $r = frac{left lfloor{Nx}right rfloor}{N}$, we are done.
$(2)$ follows from $(1)$ by considering a sequence $varepsilon_1,varepsilon_2,...$ with $varepsilon_n > 0$ and $lim_{n rightarrow infty}varepsilon_n = 0$ (for instance $varepsilon_n = frac{1}{n}$). Then we can find a sequence $r_n in mathbb{Q}$ such that $|x - r_n| < varepsilon_n$, from which $lim_{n rightarrow infty} r_n = x$ follows since $forall varepsilon > 0$ there is an $N$ such that $n > N$ implies
$$|x - r_n| < varepsilon_n < varepsilon$$
as required.
answered Jan 11 at 10:41
nesHannesHan
1735
$endgroup$
We can show $(1)$ as follows. Let $N > frac{1}{varepsilon}$ be an integer. Then we observe that
$$Nx - left lfloor{Nx}right rfloor < 1 Leftrightarrow x - frac{left lfloor{Nx}right rfloor}{N} < frac{1}{N} < varepsilon$$
So taking $r = frac{left lfloor{Nx}right rfloor}{N}$, we are done.
$(2)$ follows from $(1)$ by considering a sequence $varepsilon_1,varepsilon_2,...$ with $varepsilon_n > 0$ and $lim_{n rightarrow infty}varepsilon_n = 0$ (for instance $varepsilon_n = frac{1}{n}$). Then we can find a sequence $r_n in mathbb{Q}$ such that $|x - r_n| < varepsilon_n$, from which $lim_{n rightarrow infty} r_n = x$ follows since $forall varepsilon > 0$ there is an $N$ such that $n > N$ implies
$$|x - r_n| < varepsilon_n < varepsilon$$
as required.
answered Jan 11 at 10:41
nesHannesHan
1735
answered Jan 11 at 10:41
nesHannesHan
1735
answered Jan 11 at 10:41
nesHannesHan
1735
answered Jan 11 at 10:41
nesHannesHan
1735
1735
add a comment |
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%2f3069653%2frelation-between-q-and-r%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
$begingroup$
It refers to density.
$endgroup$
– Wuestenfux
Jan 11 at 9:37
2
$begingroup$
What are your definitions of $Bbb Q$ and of $Bbb R$? Because the details of the proof will depend on exactly how these two sets are related to one another.
$endgroup$
– Arthur
Jan 11 at 9:37
$begingroup$
The second property is how my professor in mathematics defined $mathbb{R}$. And the first property is a consequence of the second one
$endgroup$
– Damien
Jan 11 at 10:27