Proof that $(n+1)m!>(m+1)!−1$ does not hold for $m>n$
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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
798
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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
798
$endgroup$
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
798
$endgroup$
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
798
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
798
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
798
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
798
asked Dec 11 '18 at 15:58
AntRobinsonAntRobinson
798
798
add a comment |
add a comment |
3 Answers
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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
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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
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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
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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active
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active
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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
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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
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$begingroup$
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 |
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