Proof that $(n+1)m!>(m+1)!−1$ does not hold for $m>n$
$begingroup$
I am currently doing a project that involves some work on Liouville's theorem for transcendental numbers and Liouville's constant. I have found a proof that Liouville's constant is transcendental however it come with this note:
"so that $(n+1)m!>(m+1)!−1$ for all sufficiently large m. But this is false for any value of m greater than n (the reader should give a detailed proof of this statement)" - Taken from What is Mathematics? R. Courant, H. Robbins, I. Stewart.
Since I want my project to be as thorough as possible I would like to include this proof however I don't know how to do it. I know that the starting point is $(m+1)!=(m+1)m!$ but not sure what my next steps should be.
Thanks in advance
proof-writing transcendental-numbers liouville-numbers
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
$endgroup$
add a comment |
$begingroup$
I am currently doing a project that involves some work on Liouville's theorem for transcendental numbers and Liouville's constant. I have found a proof that Liouville's constant is transcendental however it come with this note:
"so that $(n+1)m!>(m+1)!−1$ for all sufficiently large m. But this is false for any value of m greater than n (the reader should give a detailed proof of this statement)" - Taken from What is Mathematics? R. Courant, H. Robbins, I. Stewart.
Since I want my project to be as thorough as possible I would like to include this proof however I don't know how to do it. I know that the starting point is $(m+1)!=(m+1)m!$ but not sure what my next steps should be.
Thanks in advance
proof-writing transcendental-numbers liouville-numbers
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
$endgroup$
add a comment |
$begingroup$
I am currently doing a project that involves some work on Liouville's theorem for transcendental numbers and Liouville's constant. I have found a proof that Liouville's constant is transcendental however it come with this note:
"so that $(n+1)m!>(m+1)!−1$ for all sufficiently large m. But this is false for any value of m greater than n (the reader should give a detailed proof of this statement)" - Taken from What is Mathematics? R. Courant, H. Robbins, I. Stewart.
Since I want my project to be as thorough as possible I would like to include this proof however I don't know how to do it. I know that the starting point is $(m+1)!=(m+1)m!$ but not sure what my next steps should be.
Thanks in advance
proof-writing transcendental-numbers liouville-numbers
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
$endgroup$
I am currently doing a project that involves some work on Liouville's theorem for transcendental numbers and Liouville's constant. I have found a proof that Liouville's constant is transcendental however it come with this note:
"so that $(n+1)m!>(m+1)!−1$ for all sufficiently large m. But this is false for any value of m greater than n (the reader should give a detailed proof of this statement)" - Taken from What is Mathematics? R. Courant, H. Robbins, I. Stewart.
Since I want my project to be as thorough as possible I would like to include this proof however I don't know how to do it. I know that the starting point is $(m+1)!=(m+1)m!$ but not sure what my next steps should be.
Thanks in advance
proof-writing transcendental-numbers liouville-numbers
proof-writing transcendental-numbers liouville-numbers
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
838
838
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The inequality does hold for $m=1,n=0$ so you need to exclude that case. Otherwise $m! gt 1$ and we can say $(m+1)!-1=(m+1)m!-1=(n+1)m!+(m-n)m!-1 gt (n+1)m!$ because $m-n ge 1$
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
For $m=n+p>n$ we have
$$(n+1)m!=(n+1)(n+p)!<(n+p+1)(n+p)!-p(n+p)!=$$$$=(n+p+1)!-p(n+p)!=(m+1)!-p(n+p)!le(m+1)!−1$$
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
Hint: Factorize:
$$(n+1)m!>(m+1)!−1 iff \
m!(n+1-m-1)>-1 iff \
m!(n-m)>-1.$$
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
$endgroup$
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%2f3035434%2fproof-that-n1mm1%25e2%2588%25921-does-not-hold-for-mn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality does hold for $m=1,n=0$ so you need to exclude that case. Otherwise $m! gt 1$ and we can say $(m+1)!-1=(m+1)m!-1=(n+1)m!+(m-n)m!-1 gt (n+1)m!$ because $m-n ge 1$
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
The inequality does hold for $m=1,n=0$ so you need to exclude that case. Otherwise $m! gt 1$ and we can say $(m+1)!-1=(m+1)m!-1=(n+1)m!+(m-n)m!-1 gt (n+1)m!$ because $m-n ge 1$
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
$endgroup$
add a comment |
$begingroup$
The inequality does hold for $m=1,n=0$ so you need to exclude that case. Otherwise $m! gt 1$ and we can say $(m+1)!-1=(m+1)m!-1=(n+1)m!+(m-n)m!-1 gt (n+1)m!$ because $m-n ge 1$
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
$endgroup$
The inequality does hold for $m=1,n=0$ so you need to exclude that case. Otherwise $m! gt 1$ and we can say $(m+1)!-1=(m+1)m!-1=(n+1)m!+(m-n)m!-1 gt (n+1)m!$ because $m-n ge 1$
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
answered Dec 11 '18 at 16:05
Ross MillikanRoss Millikan
295k23198371
295k23198371
add a comment |
add a comment |
$begingroup$
For $m=n+p>n$ we have
$$(n+1)m!=(n+1)(n+p)!<(n+p+1)(n+p)!-p(n+p)!=$$$$=(n+p+1)!-p(n+p)!=(m+1)!-p(n+p)!le(m+1)!−1$$
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
For $m=n+p>n$ we have
$$(n+1)m!=(n+1)(n+p)!<(n+p+1)(n+p)!-p(n+p)!=$$$$=(n+p+1)!-p(n+p)!=(m+1)!-p(n+p)!le(m+1)!−1$$
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
$endgroup$
add a comment |
$begingroup$
For $m=n+p>n$ we have
$$(n+1)m!=(n+1)(n+p)!<(n+p+1)(n+p)!-p(n+p)!=$$$$=(n+p+1)!-p(n+p)!=(m+1)!-p(n+p)!le(m+1)!−1$$
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
$endgroup$
For $m=n+p>n$ we have
$$(n+1)m!=(n+1)(n+p)!<(n+p+1)(n+p)!-p(n+p)!=$$$$=(n+p+1)!-p(n+p)!=(m+1)!-p(n+p)!le(m+1)!−1$$
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
answered Dec 11 '18 at 16:02
gimusigimusi
92.8k84494
92.8k84494
add a comment |
add a comment |
$begingroup$
Hint: Factorize:
$$(n+1)m!>(m+1)!−1 iff \
m!(n+1-m-1)>-1 iff \
m!(n-m)>-1.$$
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
Hint: Factorize:
$$(n+1)m!>(m+1)!−1 iff \
m!(n+1-m-1)>-1 iff \
m!(n-m)>-1.$$
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
$endgroup$
add a comment |
$begingroup$
Hint: Factorize:
$$(n+1)m!>(m+1)!−1 iff \
m!(n+1-m-1)>-1 iff \
m!(n-m)>-1.$$
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
$endgroup$
Hint: Factorize:
$$(n+1)m!>(m+1)!−1 iff \
m!(n+1-m-1)>-1 iff \
m!(n-m)>-1.$$
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
answered Dec 11 '18 at 16:15
farruhotafarruhota
20.2k2738
20.2k2738
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%2f3035434%2fproof-that-n1mm1%25e2%2588%25921-does-not-hold-for-mn%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
