Given a book with $100$ pages and a $100$ lemmas, prove that there is some lemma written on the same page as...
up vote
15
down vote
favorite
A book consists of 100 pages and contains 100 lemmas and some images. Each lemma is at most one page long and can't be split into two pages (it has to fit in one page). The lemmas are numbered from 1 to 100 and are written in ascending order. Prove that there must be at least one lemma written on a page with the same number as the lemma's number.
If lemma no 1 is written on page no 1, then it is proved. Let's assume lemma nr 1 is written on page nr k, k>1. Then in at least one page there must be 2 lemmas. Let's assume that always in page k+i we have the lemma nr i+1 and so on. Then the last 100-k-i lemmas must fit in the last page, which means that there will be at least one lemma (number 100) in page 100.
But I don't know how to express it in a more mathematical way!
Any help?
combinatorics
add a comment |
up vote
15
down vote
favorite
A book consists of 100 pages and contains 100 lemmas and some images. Each lemma is at most one page long and can't be split into two pages (it has to fit in one page). The lemmas are numbered from 1 to 100 and are written in ascending order. Prove that there must be at least one lemma written on a page with the same number as the lemma's number.
If lemma no 1 is written on page no 1, then it is proved. Let's assume lemma nr 1 is written on page nr k, k>1. Then in at least one page there must be 2 lemmas. Let's assume that always in page k+i we have the lemma nr i+1 and so on. Then the last 100-k-i lemmas must fit in the last page, which means that there will be at least one lemma (number 100) in page 100.
But I don't know how to express it in a more mathematical way!
Any help?
combinatorics
1
Lemma 1 occurs on page 1.
– Chickenmancer
2 days ago
8
@Chickenmancer An image could be on the first page.
– Oldboy
2 days ago
I edited your tags from "number theory" and "elementary number theory" to "combinatorics", which I'm fairly confident more accurately reflects how other people would perceive the topic...
– paul garrett
2 days ago
1
Are the pages numbered (in ascending order)?
– Servaes
2 days ago
add a comment |
up vote
15
down vote
favorite
up vote
15
down vote
favorite
A book consists of 100 pages and contains 100 lemmas and some images. Each lemma is at most one page long and can't be split into two pages (it has to fit in one page). The lemmas are numbered from 1 to 100 and are written in ascending order. Prove that there must be at least one lemma written on a page with the same number as the lemma's number.
If lemma no 1 is written on page no 1, then it is proved. Let's assume lemma nr 1 is written on page nr k, k>1. Then in at least one page there must be 2 lemmas. Let's assume that always in page k+i we have the lemma nr i+1 and so on. Then the last 100-k-i lemmas must fit in the last page, which means that there will be at least one lemma (number 100) in page 100.
But I don't know how to express it in a more mathematical way!
Any help?
combinatorics
A book consists of 100 pages and contains 100 lemmas and some images. Each lemma is at most one page long and can't be split into two pages (it has to fit in one page). The lemmas are numbered from 1 to 100 and are written in ascending order. Prove that there must be at least one lemma written on a page with the same number as the lemma's number.
If lemma no 1 is written on page no 1, then it is proved. Let's assume lemma nr 1 is written on page nr k, k>1. Then in at least one page there must be 2 lemmas. Let's assume that always in page k+i we have the lemma nr i+1 and so on. Then the last 100-k-i lemmas must fit in the last page, which means that there will be at least one lemma (number 100) in page 100.
But I don't know how to express it in a more mathematical way!
Any help?
combinatorics
combinatorics
edited yesterday
Asaf Karagila♦
300k32421751
300k32421751
asked 2 days ago
Reyansh Laghari
1515
1515
1
Lemma 1 occurs on page 1.
– Chickenmancer
2 days ago
8
@Chickenmancer An image could be on the first page.
– Oldboy
2 days ago
I edited your tags from "number theory" and "elementary number theory" to "combinatorics", which I'm fairly confident more accurately reflects how other people would perceive the topic...
– paul garrett
2 days ago
1
Are the pages numbered (in ascending order)?
– Servaes
2 days ago
add a comment |
1
Lemma 1 occurs on page 1.
– Chickenmancer
2 days ago
8
@Chickenmancer An image could be on the first page.
– Oldboy
2 days ago
I edited your tags from "number theory" and "elementary number theory" to "combinatorics", which I'm fairly confident more accurately reflects how other people would perceive the topic...
– paul garrett
2 days ago
1
Are the pages numbered (in ascending order)?
– Servaes
2 days ago
1
1
Lemma 1 occurs on page 1.
– Chickenmancer
2 days ago
Lemma 1 occurs on page 1.
– Chickenmancer
2 days ago
8
8
@Chickenmancer An image could be on the first page.
– Oldboy
2 days ago
@Chickenmancer An image could be on the first page.
– Oldboy
2 days ago
I edited your tags from "number theory" and "elementary number theory" to "combinatorics", which I'm fairly confident more accurately reflects how other people would perceive the topic...
– paul garrett
2 days ago
I edited your tags from "number theory" and "elementary number theory" to "combinatorics", which I'm fairly confident more accurately reflects how other people would perceive the topic...
– paul garrett
2 days ago
1
1
Are the pages numbered (in ascending order)?
– Servaes
2 days ago
Are the pages numbered (in ascending order)?
– Servaes
2 days ago
add a comment |
5 Answers
5
active
oldest
votes
up vote
11
down vote
We claim more generally that a book of $n$ pages and $n$ lemmas numbered $1$ through $n$ has at least one lemma on a page matching its number.
Proof by induction on $n$: The case $n=1$ is obvious. Now suppose the statement is true for some $n$, and suppose we have a book of $n+1$ lemmas and $n+1$ pages. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ must be on pages $1$ through $n$, and there must be at least one lemma on a same-numbered page by the inductive hypothesis. If not, then lemma $n+1$ is on page $n+1$, and we're done.
+1 The way to go!
– Servaes
2 days ago
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
1
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
|
show 1 more comment
up vote
6
down vote
For each page $i$ assign a number $p(i)$ that defines the highest number of lemma printed on all pages starting from page 1 up to page $i$ (inclusive). So if we have lemmas 3, 4 and 5 printed on page 12, with page 13 having images only: $p(12)=p(13)=5$.
We have the following sequence:
$$p(1), p(2), dots, p(100)=100tag{1}$$
If $p(1)ge1$ it means that lemma 1 is printed on the first page and we are done.
Let us consider a case when $p(1)=0$ (which basically means that the first page is reserved for images only).
The sequence of page numbers $i$ is strictly increasing and the sequence of values $p(i)$ is non-decreasing. Both sequencies have the same number of items (100), with $p(1)<1$ and $p(100)=100$.
Because of that we must have a pair of consecutive pages $i,space j=i+1$ such that:
$$p(i)<i$$
$$p(j)ge j$$
This basically means that lemma $j$ is not printed on page $i$ or on any other page before it. It is actually printed on page $j$ and this completes the proof.
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
add a comment |
up vote
6
down vote
Slight rephrasing of Oldboy's argument:
Let $a_i = i - p_i$, where $p_i$ is the page number of lemma $i$.
Then $a_1 leq 0$, $a_{100} geq 0$, and $a_{i+1}-a_{i} leq 1$. Thus $a_i$ must be $0$ for some $i$.
New contributor
1
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
add a comment |
up vote
2
down vote
Consider the path of points $(L, p(L))$ where lemma $L$ is on page $p(L)$. Plot that path on the grid of lattice points $(x,y)$ for $1 le x, y le 100$. The path starts on or above the diagonal at point $(1, p(1))$ and ends on or below the diagonal at point $(100, p(100))$.
Following that path, you move to the right one step at a time as the lemma count increases. You may stay at the same horizontal level since many lemmas can appear on a page. Vertical steps can be longer, if pages of images intervene. Since you start out above the diagonal, you can't cross it for the first time on a vertical step. Since in order to end up on the other side of the diagonal you must cross it once, that must be a horizontal step, so you have landed on it on the way to the other side.
(This doesn't exactly answer your question, which calls "expressing [your proof] in a more mathematical way". I might be wrong, but I don't think that's possible.)
2
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
1
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
1
It’s ok now, +1.
– Oldboy
yesterday
add a comment |
up vote
1
down vote
Here's a slightly amended proof by induction, which I hope will make the induction step more obvious.
Let $P_n$ be the proposition that for a book with $n$ sequentially numbered pages and $n$ or more sequentially numbered lemmas, at least one lemma is on a page with the same number as the lemma.
Suppose $P_n$ is true for some $n=k$, and consider the case of $k+1$ pages and $k+1$ lemmas. To avoid being on its own-numbered page, lemma $k+1$ has to be somewhere in the first $k$ pages, and to keep the lemmas in sequence, no other lemma can go on page $k+1$.
We therefore have to fit all $k+1$ lemmas onto the first $k$ pages. By $P_k$, this puts at least one of them on its own-numbered page.
So wherever lemma $k+1$ is, at least one lemma is on its own-numbered page.
Clearly this remains true if we add more lemmas without adding more pages. That is, with $k+1$ pages and $ge k+1$ lemmas, at least one lemma is on its own-numbered page: ie $P_{k+1}$ holds. That is, if $P_n$ is true for $n=k$, it is also true for $n=k+1$.
Now consider the case of $1$ page and $ge 1$ lemmas. Clearly lemma 1 is on page 1, so $P_1$ is true. Therefore, by induction, $P_n$ is true for all $nge 1$.
Finally, putting $n=100$ proves the case in the original question.
New contributor
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
We claim more generally that a book of $n$ pages and $n$ lemmas numbered $1$ through $n$ has at least one lemma on a page matching its number.
Proof by induction on $n$: The case $n=1$ is obvious. Now suppose the statement is true for some $n$, and suppose we have a book of $n+1$ lemmas and $n+1$ pages. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ must be on pages $1$ through $n$, and there must be at least one lemma on a same-numbered page by the inductive hypothesis. If not, then lemma $n+1$ is on page $n+1$, and we're done.
+1 The way to go!
– Servaes
2 days ago
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
1
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
|
show 1 more comment
up vote
11
down vote
We claim more generally that a book of $n$ pages and $n$ lemmas numbered $1$ through $n$ has at least one lemma on a page matching its number.
Proof by induction on $n$: The case $n=1$ is obvious. Now suppose the statement is true for some $n$, and suppose we have a book of $n+1$ lemmas and $n+1$ pages. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ must be on pages $1$ through $n$, and there must be at least one lemma on a same-numbered page by the inductive hypothesis. If not, then lemma $n+1$ is on page $n+1$, and we're done.
+1 The way to go!
– Servaes
2 days ago
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
1
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
|
show 1 more comment
up vote
11
down vote
up vote
11
down vote
We claim more generally that a book of $n$ pages and $n$ lemmas numbered $1$ through $n$ has at least one lemma on a page matching its number.
Proof by induction on $n$: The case $n=1$ is obvious. Now suppose the statement is true for some $n$, and suppose we have a book of $n+1$ lemmas and $n+1$ pages. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ must be on pages $1$ through $n$, and there must be at least one lemma on a same-numbered page by the inductive hypothesis. If not, then lemma $n+1$ is on page $n+1$, and we're done.
We claim more generally that a book of $n$ pages and $n$ lemmas numbered $1$ through $n$ has at least one lemma on a page matching its number.
Proof by induction on $n$: The case $n=1$ is obvious. Now suppose the statement is true for some $n$, and suppose we have a book of $n+1$ lemmas and $n+1$ pages. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ must be on pages $1$ through $n$, and there must be at least one lemma on a same-numbered page by the inductive hypothesis. If not, then lemma $n+1$ is on page $n+1$, and we're done.
answered 2 days ago
awkward
5,70111022
5,70111022
+1 The way to go!
– Servaes
2 days ago
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
1
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
|
show 1 more comment
+1 The way to go!
– Servaes
2 days ago
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
1
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
+1 The way to go!
– Servaes
2 days ago
+1 The way to go!
– Servaes
2 days ago
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
That does not seem to work. If lemma $n+1$ is on a page numbered less than $n+1$, then lemmas $1$ through $n$ are spread over pages $1$ through $n+1$, not pages $1$ through $n$! So you can't apply the induction hypothesis.
– Stephan Kolassa
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
@StephanKolassa: Doesn't lemma $n$ have to be on the same page as, or an earlier page than, lemma $n+1$? In which case, if lemma $n+1$ is on a page numbered less than $n+1$ (thus less than or equal to $n$), lemma $n$ is too.
– Tim Pederick
yesterday
1
1
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@StephanKolassa The page numbers of the lemmas form a non-decreasing sequence. If the page number of lemma $n+1$ is less than $n+1$, then the page number of lemma $n$ is less than $n+1$, hence less than or equal to $n$.
– awkward
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
@awkward: thanks, that helps - I had overlooked that little detail. Better to read first, comment later.
– Stephan Kolassa
yesterday
|
show 1 more comment
up vote
6
down vote
For each page $i$ assign a number $p(i)$ that defines the highest number of lemma printed on all pages starting from page 1 up to page $i$ (inclusive). So if we have lemmas 3, 4 and 5 printed on page 12, with page 13 having images only: $p(12)=p(13)=5$.
We have the following sequence:
$$p(1), p(2), dots, p(100)=100tag{1}$$
If $p(1)ge1$ it means that lemma 1 is printed on the first page and we are done.
Let us consider a case when $p(1)=0$ (which basically means that the first page is reserved for images only).
The sequence of page numbers $i$ is strictly increasing and the sequence of values $p(i)$ is non-decreasing. Both sequencies have the same number of items (100), with $p(1)<1$ and $p(100)=100$.
Because of that we must have a pair of consecutive pages $i,space j=i+1$ such that:
$$p(i)<i$$
$$p(j)ge j$$
This basically means that lemma $j$ is not printed on page $i$ or on any other page before it. It is actually printed on page $j$ and this completes the proof.
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
add a comment |
up vote
6
down vote
For each page $i$ assign a number $p(i)$ that defines the highest number of lemma printed on all pages starting from page 1 up to page $i$ (inclusive). So if we have lemmas 3, 4 and 5 printed on page 12, with page 13 having images only: $p(12)=p(13)=5$.
We have the following sequence:
$$p(1), p(2), dots, p(100)=100tag{1}$$
If $p(1)ge1$ it means that lemma 1 is printed on the first page and we are done.
Let us consider a case when $p(1)=0$ (which basically means that the first page is reserved for images only).
The sequence of page numbers $i$ is strictly increasing and the sequence of values $p(i)$ is non-decreasing. Both sequencies have the same number of items (100), with $p(1)<1$ and $p(100)=100$.
Because of that we must have a pair of consecutive pages $i,space j=i+1$ such that:
$$p(i)<i$$
$$p(j)ge j$$
This basically means that lemma $j$ is not printed on page $i$ or on any other page before it. It is actually printed on page $j$ and this completes the proof.
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
add a comment |
up vote
6
down vote
up vote
6
down vote
For each page $i$ assign a number $p(i)$ that defines the highest number of lemma printed on all pages starting from page 1 up to page $i$ (inclusive). So if we have lemmas 3, 4 and 5 printed on page 12, with page 13 having images only: $p(12)=p(13)=5$.
We have the following sequence:
$$p(1), p(2), dots, p(100)=100tag{1}$$
If $p(1)ge1$ it means that lemma 1 is printed on the first page and we are done.
Let us consider a case when $p(1)=0$ (which basically means that the first page is reserved for images only).
The sequence of page numbers $i$ is strictly increasing and the sequence of values $p(i)$ is non-decreasing. Both sequencies have the same number of items (100), with $p(1)<1$ and $p(100)=100$.
Because of that we must have a pair of consecutive pages $i,space j=i+1$ such that:
$$p(i)<i$$
$$p(j)ge j$$
This basically means that lemma $j$ is not printed on page $i$ or on any other page before it. It is actually printed on page $j$ and this completes the proof.
For each page $i$ assign a number $p(i)$ that defines the highest number of lemma printed on all pages starting from page 1 up to page $i$ (inclusive). So if we have lemmas 3, 4 and 5 printed on page 12, with page 13 having images only: $p(12)=p(13)=5$.
We have the following sequence:
$$p(1), p(2), dots, p(100)=100tag{1}$$
If $p(1)ge1$ it means that lemma 1 is printed on the first page and we are done.
Let us consider a case when $p(1)=0$ (which basically means that the first page is reserved for images only).
The sequence of page numbers $i$ is strictly increasing and the sequence of values $p(i)$ is non-decreasing. Both sequencies have the same number of items (100), with $p(1)<1$ and $p(100)=100$.
Because of that we must have a pair of consecutive pages $i,space j=i+1$ such that:
$$p(i)<i$$
$$p(j)ge j$$
This basically means that lemma $j$ is not printed on page $i$ or on any other page before it. It is actually printed on page $j$ and this completes the proof.
edited 2 days ago
answered 2 days ago
Oldboy
5,7731628
5,7731628
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
add a comment |
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
+1 very nice. I wonder if my solution is essentially the same as yours.
– Ethan Bolker
2 days ago
add a comment |
up vote
6
down vote
Slight rephrasing of Oldboy's argument:
Let $a_i = i - p_i$, where $p_i$ is the page number of lemma $i$.
Then $a_1 leq 0$, $a_{100} geq 0$, and $a_{i+1}-a_{i} leq 1$. Thus $a_i$ must be $0$ for some $i$.
New contributor
1
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
add a comment |
up vote
6
down vote
Slight rephrasing of Oldboy's argument:
Let $a_i = i - p_i$, where $p_i$ is the page number of lemma $i$.
Then $a_1 leq 0$, $a_{100} geq 0$, and $a_{i+1}-a_{i} leq 1$. Thus $a_i$ must be $0$ for some $i$.
New contributor
1
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
add a comment |
up vote
6
down vote
up vote
6
down vote
Slight rephrasing of Oldboy's argument:
Let $a_i = i - p_i$, where $p_i$ is the page number of lemma $i$.
Then $a_1 leq 0$, $a_{100} geq 0$, and $a_{i+1}-a_{i} leq 1$. Thus $a_i$ must be $0$ for some $i$.
New contributor
Slight rephrasing of Oldboy's argument:
Let $a_i = i - p_i$, where $p_i$ is the page number of lemma $i$.
Then $a_1 leq 0$, $a_{100} geq 0$, and $a_{i+1}-a_{i} leq 1$. Thus $a_i$ must be $0$ for some $i$.
New contributor
New contributor
answered 2 days ago
user113102
612
612
New contributor
New contributor
1
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
add a comment |
1
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
1
1
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
I guess this is the same idea as both Ethan's and Oldboy's arguments, but feels cleaner to me.
– user113102
2 days ago
add a comment |
up vote
2
down vote
Consider the path of points $(L, p(L))$ where lemma $L$ is on page $p(L)$. Plot that path on the grid of lattice points $(x,y)$ for $1 le x, y le 100$. The path starts on or above the diagonal at point $(1, p(1))$ and ends on or below the diagonal at point $(100, p(100))$.
Following that path, you move to the right one step at a time as the lemma count increases. You may stay at the same horizontal level since many lemmas can appear on a page. Vertical steps can be longer, if pages of images intervene. Since you start out above the diagonal, you can't cross it for the first time on a vertical step. Since in order to end up on the other side of the diagonal you must cross it once, that must be a horizontal step, so you have landed on it on the way to the other side.
(This doesn't exactly answer your question, which calls "expressing [your proof] in a more mathematical way". I might be wrong, but I don't think that's possible.)
2
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
1
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
1
It’s ok now, +1.
– Oldboy
yesterday
add a comment |
up vote
2
down vote
Consider the path of points $(L, p(L))$ where lemma $L$ is on page $p(L)$. Plot that path on the grid of lattice points $(x,y)$ for $1 le x, y le 100$. The path starts on or above the diagonal at point $(1, p(1))$ and ends on or below the diagonal at point $(100, p(100))$.
Following that path, you move to the right one step at a time as the lemma count increases. You may stay at the same horizontal level since many lemmas can appear on a page. Vertical steps can be longer, if pages of images intervene. Since you start out above the diagonal, you can't cross it for the first time on a vertical step. Since in order to end up on the other side of the diagonal you must cross it once, that must be a horizontal step, so you have landed on it on the way to the other side.
(This doesn't exactly answer your question, which calls "expressing [your proof] in a more mathematical way". I might be wrong, but I don't think that's possible.)
2
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
1
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
1
It’s ok now, +1.
– Oldboy
yesterday
add a comment |
up vote
2
down vote
up vote
2
down vote
Consider the path of points $(L, p(L))$ where lemma $L$ is on page $p(L)$. Plot that path on the grid of lattice points $(x,y)$ for $1 le x, y le 100$. The path starts on or above the diagonal at point $(1, p(1))$ and ends on or below the diagonal at point $(100, p(100))$.
Following that path, you move to the right one step at a time as the lemma count increases. You may stay at the same horizontal level since many lemmas can appear on a page. Vertical steps can be longer, if pages of images intervene. Since you start out above the diagonal, you can't cross it for the first time on a vertical step. Since in order to end up on the other side of the diagonal you must cross it once, that must be a horizontal step, so you have landed on it on the way to the other side.
(This doesn't exactly answer your question, which calls "expressing [your proof] in a more mathematical way". I might be wrong, but I don't think that's possible.)
Consider the path of points $(L, p(L))$ where lemma $L$ is on page $p(L)$. Plot that path on the grid of lattice points $(x,y)$ for $1 le x, y le 100$. The path starts on or above the diagonal at point $(1, p(1))$ and ends on or below the diagonal at point $(100, p(100))$.
Following that path, you move to the right one step at a time as the lemma count increases. You may stay at the same horizontal level since many lemmas can appear on a page. Vertical steps can be longer, if pages of images intervene. Since you start out above the diagonal, you can't cross it for the first time on a vertical step. Since in order to end up on the other side of the diagonal you must cross it once, that must be a horizontal step, so you have landed on it on the way to the other side.
(This doesn't exactly answer your question, which calls "expressing [your proof] in a more mathematical way". I might be wrong, but I don't think that's possible.)
edited yesterday
answered 2 days ago
Ethan Bolker
39.7k543103
39.7k543103
2
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
1
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
1
It’s ok now, +1.
– Oldboy
yesterday
add a comment |
2
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
1
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
1
It’s ok now, +1.
– Oldboy
yesterday
2
2
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
"since at most one lemma fits on a page": This is not given.
– Oldboy
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
I believe that the path steps horizontally at most one but can step up many steps, right?
– Reyansh Laghari
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
@ReyanshLaghari I think my proof is fixable but I can't do it right now. You can edit it if you figure it out. If not I will return later today.
– Ethan Bolker
2 days ago
1
1
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
@Oldboy Fixed the argument thank you.
– Ethan Bolker
2 days ago
1
1
It’s ok now, +1.
– Oldboy
yesterday
It’s ok now, +1.
– Oldboy
yesterday
add a comment |
up vote
1
down vote
Here's a slightly amended proof by induction, which I hope will make the induction step more obvious.
Let $P_n$ be the proposition that for a book with $n$ sequentially numbered pages and $n$ or more sequentially numbered lemmas, at least one lemma is on a page with the same number as the lemma.
Suppose $P_n$ is true for some $n=k$, and consider the case of $k+1$ pages and $k+1$ lemmas. To avoid being on its own-numbered page, lemma $k+1$ has to be somewhere in the first $k$ pages, and to keep the lemmas in sequence, no other lemma can go on page $k+1$.
We therefore have to fit all $k+1$ lemmas onto the first $k$ pages. By $P_k$, this puts at least one of them on its own-numbered page.
So wherever lemma $k+1$ is, at least one lemma is on its own-numbered page.
Clearly this remains true if we add more lemmas without adding more pages. That is, with $k+1$ pages and $ge k+1$ lemmas, at least one lemma is on its own-numbered page: ie $P_{k+1}$ holds. That is, if $P_n$ is true for $n=k$, it is also true for $n=k+1$.
Now consider the case of $1$ page and $ge 1$ lemmas. Clearly lemma 1 is on page 1, so $P_1$ is true. Therefore, by induction, $P_n$ is true for all $nge 1$.
Finally, putting $n=100$ proves the case in the original question.
New contributor
add a comment |
up vote
1
down vote
Here's a slightly amended proof by induction, which I hope will make the induction step more obvious.
Let $P_n$ be the proposition that for a book with $n$ sequentially numbered pages and $n$ or more sequentially numbered lemmas, at least one lemma is on a page with the same number as the lemma.
Suppose $P_n$ is true for some $n=k$, and consider the case of $k+1$ pages and $k+1$ lemmas. To avoid being on its own-numbered page, lemma $k+1$ has to be somewhere in the first $k$ pages, and to keep the lemmas in sequence, no other lemma can go on page $k+1$.
We therefore have to fit all $k+1$ lemmas onto the first $k$ pages. By $P_k$, this puts at least one of them on its own-numbered page.
So wherever lemma $k+1$ is, at least one lemma is on its own-numbered page.
Clearly this remains true if we add more lemmas without adding more pages. That is, with $k+1$ pages and $ge k+1$ lemmas, at least one lemma is on its own-numbered page: ie $P_{k+1}$ holds. That is, if $P_n$ is true for $n=k$, it is also true for $n=k+1$.
Now consider the case of $1$ page and $ge 1$ lemmas. Clearly lemma 1 is on page 1, so $P_1$ is true. Therefore, by induction, $P_n$ is true for all $nge 1$.
Finally, putting $n=100$ proves the case in the original question.
New contributor
add a comment |
up vote
1
down vote
up vote
1
down vote
Here's a slightly amended proof by induction, which I hope will make the induction step more obvious.
Let $P_n$ be the proposition that for a book with $n$ sequentially numbered pages and $n$ or more sequentially numbered lemmas, at least one lemma is on a page with the same number as the lemma.
Suppose $P_n$ is true for some $n=k$, and consider the case of $k+1$ pages and $k+1$ lemmas. To avoid being on its own-numbered page, lemma $k+1$ has to be somewhere in the first $k$ pages, and to keep the lemmas in sequence, no other lemma can go on page $k+1$.
We therefore have to fit all $k+1$ lemmas onto the first $k$ pages. By $P_k$, this puts at least one of them on its own-numbered page.
So wherever lemma $k+1$ is, at least one lemma is on its own-numbered page.
Clearly this remains true if we add more lemmas without adding more pages. That is, with $k+1$ pages and $ge k+1$ lemmas, at least one lemma is on its own-numbered page: ie $P_{k+1}$ holds. That is, if $P_n$ is true for $n=k$, it is also true for $n=k+1$.
Now consider the case of $1$ page and $ge 1$ lemmas. Clearly lemma 1 is on page 1, so $P_1$ is true. Therefore, by induction, $P_n$ is true for all $nge 1$.
Finally, putting $n=100$ proves the case in the original question.
New contributor
Here's a slightly amended proof by induction, which I hope will make the induction step more obvious.
Let $P_n$ be the proposition that for a book with $n$ sequentially numbered pages and $n$ or more sequentially numbered lemmas, at least one lemma is on a page with the same number as the lemma.
Suppose $P_n$ is true for some $n=k$, and consider the case of $k+1$ pages and $k+1$ lemmas. To avoid being on its own-numbered page, lemma $k+1$ has to be somewhere in the first $k$ pages, and to keep the lemmas in sequence, no other lemma can go on page $k+1$.
We therefore have to fit all $k+1$ lemmas onto the first $k$ pages. By $P_k$, this puts at least one of them on its own-numbered page.
So wherever lemma $k+1$ is, at least one lemma is on its own-numbered page.
Clearly this remains true if we add more lemmas without adding more pages. That is, with $k+1$ pages and $ge k+1$ lemmas, at least one lemma is on its own-numbered page: ie $P_{k+1}$ holds. That is, if $P_n$ is true for $n=k$, it is also true for $n=k+1$.
Now consider the case of $1$ page and $ge 1$ lemmas. Clearly lemma 1 is on page 1, so $P_1$ is true. Therefore, by induction, $P_n$ is true for all $nge 1$.
Finally, putting $n=100$ proves the case in the original question.
New contributor
edited yesterday
New contributor
answered yesterday
timtfj
817
817
New contributor
New contributor
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3016149%2fgiven-a-book-with-100-pages-and-a-100-lemmas-prove-that-there-is-some-lemma%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
Lemma 1 occurs on page 1.
– Chickenmancer
2 days ago
8
@Chickenmancer An image could be on the first page.
– Oldboy
2 days ago
I edited your tags from "number theory" and "elementary number theory" to "combinatorics", which I'm fairly confident more accurately reflects how other people would perceive the topic...
– paul garrett
2 days ago
1
Are the pages numbered (in ascending order)?
– Servaes
2 days ago