Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists …
$begingroup$
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that 
for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
$endgroup$
add a comment |
$begingroup$
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
$endgroup$
$begingroup$
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
$endgroup$
– Andrés E. Caicedo
Dec 28 '18 at 14:07
add a comment |
$begingroup$
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
$endgroup$
Prove that if $nm ≤ nx < nm + n $, where $ n,m ∈ ℤ $ and $ x ∈ ℝ$, then there exists $j$ such that $j ∈ ℤ$ and $0≤ j <n$ for which $ nm+j≤ nx <nm+j+1 $.
I'm trying to prove Hermite's identity, which states that for every real number $x$, and every positive integer $n$.
Part of the proof states what is underlined in red:
I don't understand why that is true; I've tried to prove it, but have failed to do so. Any ideas on why this happens?
algebra-precalculus functions discrete-mathematics floor-function
algebra-precalculus functions discrete-mathematics floor-function
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
edited Dec 28 '18 at 14:05
Andrés E. Caicedo
65.5k8159250
65.5k8159250
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
asked Dec 28 '18 at 4:48
Daniel Bonilla JaramilloDaniel Bonilla Jaramillo
464310
464310
$begingroup$
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
$endgroup$
– Andrés E. Caicedo
Dec 28 '18 at 14:07
add a comment |
$begingroup$
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
$endgroup$
– Andrés E. Caicedo
Dec 28 '18 at 14:07
$begingroup$
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
$endgroup$
– Andrés E. Caicedo
Dec 28 '18 at 14:07
$begingroup$
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
$endgroup$
– Andrés E. Caicedo
Dec 28 '18 at 14:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
1
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
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%2f3054574%2fprove-that-if-nm-%25e2%2589%25a4-nx-nm-n-where-n-m-%25e2%2588%2588-%25e2%2584%25a4-and-x-%25e2%2588%2588-%25e2%2584%259d-then-there-ex%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
$begingroup$
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
1
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
$begingroup$
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
1
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
$begingroup$
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
$endgroup$
Let $ j = [nx] - nm$ and then the magic happens. Think that the inequality is a fancy way to find the remainder of the floor modulo $n$.
Your inequalities still hold when applying floor, so $nm leq [nx] < nm + n$. But, by definition, $[nx] = nm + j$ (such $j$ must exist because of $[nx]$ bounds) and you'll get $[nx] = mn + j leq [nx] < mn + j + 1 = [nx]+1$.
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
edited Dec 28 '18 at 13:55
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
answered Dec 28 '18 at 4:55
Lucas HenriqueLucas Henrique
1,026414
1,026414
1
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
add a comment |
1
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
1
1
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
$begingroup$
Can the downvoter please explain?
$endgroup$
– Lucas Henrique
Dec 28 '18 at 4:58
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%2f3054574%2fprove-that-if-nm-%25e2%2589%25a4-nx-nm-n-where-n-m-%25e2%2588%2588-%25e2%2584%25a4-and-x-%25e2%2588%2588-%25e2%2584%259d-then-there-ex%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
$begingroup$
What did you try? What do you know about the floor function? Have you tried particular cases? Do you understand what is happening, but do not see how to make it rigorous, or do not have an intuitive understanding yet?
$endgroup$
– Andrés E. Caicedo
Dec 28 '18 at 14:07