When $n$ is an integer such that $n>1$, $S$ is an arbitrary subset of $1,2,3,…,3n$ and the number of elements of $S$ is $n+2$, prove that The Next CEO of Stack OverflowConsider a set A of 100, 000 arbitrary integers. Prove that there is some subset of 22 integers that end in the same last three digits.An interesting problem using Pigeonhole principleif $|Scap 1,2,3,ldots,n|ge un$, etc. show that $Bbb Z_+subset S+T$number of round subset of $1,2,…,n$ has the same parity as nIf $Asubseteq 1, 2, … 2n $ has $n+1$ elements, then it contains $a$, $b$ such that $amid b$Let $T$ be any subset of $1,2,3,…,100$ with $69$ elements. Prove that one can find four distinct integers such that $a+b+c=d$.Find the maximal number of elements of such subset $S$Optimal number of elements in a subsetLet $S$ be a subset of positive integers. Show that no such set $S$ exist.

When you upcast Blindness/Deafness, do all targets suffer the same effect?

Why isn't the Mueller report being released completely and unredacted?

What does "Its cash flow is deeply negative" mean?

Grabbing quick drinks

Prepend last line of stdin to entire stdin

Is it professional to write unrelated content in an almost-empty email?

How I can get glyphs from a fraktur font and use them as identifiers?

Using Rolle's theorem to show an equation has only one real root

Writing differences on a blackboard

Reference request: Grassmannian and Plucker coordinates in type B, C, D

Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?

Poetry, calligrams and TikZ/PStricks challenge

Can MTA send mail via a relay without being told so?

Make solar eclipses exceedingly rare, but still have new moons

Why does the flight controls check come before arming the autobrake on the A320?

Bartok - Syncopation (1): Meaning of notes in between Grand Staff

Rotate a column

Recycling old answers

Is wanting to ask what to write an indication that you need to change your story?

Why is the US ranked as #45 in Press Freedom ratings, despite its extremely permissive free speech laws?

Yu-Gi-Oh cards in Python 3

Is the D&D universe the same as the Forgotten Realms universe?

Method for adding error messages to a dictionary given a key

What flight has the highest ratio of timezone difference to flight time?



When $n$ is an integer such that $n>1$, $S$ is an arbitrary subset of $1,2,3,…,3n$ and the number of elements of $S$ is $n+2$, prove that



The Next CEO of Stack OverflowConsider a set A of 100, 000 arbitrary integers. Prove that there is some subset of 22 integers that end in the same last three digits.An interesting problem using Pigeonhole principleif $|Scap 1,2,3,ldots,n|ge un$, etc. show that $Bbb Z_+subset S+T$number of round subset of $1,2,…,n$ has the same parity as nIf $Asubseteq 1, 2, … 2n $ has $n+1$ elements, then it contains $a$, $b$ such that $amid b$Let $T$ be any subset of $1,2,3,…,100$ with $69$ elements. Prove that one can find four distinct integers such that $a+b+c=d$.Find the maximal number of elements of such subset $S$Optimal number of elements in a subsetLet $S$ be a subset of positive integers. Show that no such set $S$ exist.










3












$begingroup$


When $n$ is an integer such that $n>1$, $S$ is an arbitrary subset of $1,2,3,...,3n$ and the number of elements of $S$ is $n+2$, prove that there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



more information)

with out loss of generality, there exists $3n$ in $S$ (I don't know the reason..)



when the intersection of $[n+1,2n-1] $ and $S$ is not empty,
there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



then, what about the intersection of $[n+1,2n-1] $ and $S$ is empty..?










share|cite|improve this question











$endgroup$
















    3












    $begingroup$


    When $n$ is an integer such that $n>1$, $S$ is an arbitrary subset of $1,2,3,...,3n$ and the number of elements of $S$ is $n+2$, prove that there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



    more information)

    with out loss of generality, there exists $3n$ in $S$ (I don't know the reason..)



    when the intersection of $[n+1,2n-1] $ and $S$ is not empty,
    there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



    then, what about the intersection of $[n+1,2n-1] $ and $S$ is empty..?










    share|cite|improve this question











    $endgroup$














      3












      3








      3





      $begingroup$


      When $n$ is an integer such that $n>1$, $S$ is an arbitrary subset of $1,2,3,...,3n$ and the number of elements of $S$ is $n+2$, prove that there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



      more information)

      with out loss of generality, there exists $3n$ in $S$ (I don't know the reason..)



      when the intersection of $[n+1,2n-1] $ and $S$ is not empty,
      there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



      then, what about the intersection of $[n+1,2n-1] $ and $S$ is empty..?










      share|cite|improve this question











      $endgroup$




      When $n$ is an integer such that $n>1$, $S$ is an arbitrary subset of $1,2,3,...,3n$ and the number of elements of $S$ is $n+2$, prove that there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



      more information)

      with out loss of generality, there exists $3n$ in $S$ (I don't know the reason..)



      when the intersection of $[n+1,2n-1] $ and $S$ is not empty,
      there exist integers $x, y$ in $S$ such that $n<y-x<2n$.



      then, what about the intersection of $[n+1,2n-1] $ and $S$ is empty..?







      combinatorics pigeonhole-principle






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 19 at 13:50







      magician

















      asked Mar 19 at 13:16









      magicianmagician

      216




      216




















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          $newcommandset[1]left#1right$ EDIT: To help make the argument clearer, here is a description of the argument. The claim is that for any subset $S$ of the $3n$ consecutive integers $1,ldots, 3n$ which contains $n+2$ elements, there is a pair of elements in the subset whose difference lies between $n$ and $2n$. For example, if $n=2$ and the set is $set1,2,3,4,5,6$, we want to show any set of $4$ elements has a pair whose difference is $3$. This is easy to see by checking all possibilities in this instance, but that is impractical for larger $n$, much less to prove the statement in general.



          Instead, we will use a divide and conquer approach. The first part of the argument will reduce the problem significantly, without loss of generality, by showing that the set $S$ can always be assumed to contain the largest element, and thus it suffices to prove the statement in the case that $S$ contains $3n$ and none of the elements in the middle third $n+1,ldots, 2n-1$. This allows us to apply the pidgeonhole principle by splitting the remaining possible elements of $S$ into pairs whose difference lies in the desired range, and showing that $S$ must contain one of those pairs.



          The Strategy



          First, note that this is really a question about the set of differences $delta S$ generated by the subset $S$. For example, when $n=2$ and $S=set1,2,4,6$, we have $2-1=1$, $4-2=6-4=2$, $4-1=3$ $6-2=4$, and $6-1=5$, hence the set of differences is $delta S=set1,2,3,4,5$. (Here, we exclude the negative differences as we will always have $x-y$ or $y-x$ in $delta S$ as defined.) The claim then is that one of these differences lies between $n$ and $2n$.



          To prove the claim, first we simplify by noting we can always assume $3nin S$. This is because the differences don't change if we shift every element of $S$ by the same amount; for example, if $S=set1,2,3,4$, and $S+2=set3,4,5,6$, the differences $delta S=set1,2,3=delta(S+2)$ are the same. In particular, if there were some $S$ such that the claim were false, there would be an $S$ containing $3n$ also proving the claim to be false.



          If $S$ contains $3n$ and any of the elements between $n$ and $2n$ steps away (namely, $n+1, n+2,ldots, 2n-1$), then we would be done. For instance, if $n=2$ and $3in S$, then $3=6-3in S$ proving the claim. So there is only more to argue if we assume those elements are not in $S$.



          In that case, we have that $S$ contains $3n$ and some other elements $S'$ which live in $1,2,ldots, n$ or $2n,2n+1,ldots, 3n-1$. (E.g. in the $n=2$ example,
          if $6in S$ and $3notin S$, then the remaining 3 elements of $S$ must be elements of $set1,2,4,5$.) We can pair these elements together as $set1,2n$, $set2,2n+1,ldots, setn,3n-1$ where the pairs are of the form $setk,2n-1+k$. (So in the $n=2$ case, the pairs are $set1,4$ and $set2,5$.)



          Now to prove the claim in this case, we use the pidgeonhole principle. To do so, we need our pidgeons and our holes. The pidgeons in this case are the elements of $S'$ (i.e. the elements of $S$ other than $3n$) and the holes are our pairs $setk,2n-1+k$. Each pidgeon lives in one of the holes, but there are $n+1$ pidgeons and $n$ holes, so two pidgeons must share a hole. (The formal way of putting this is defining a function and concluding its not one-to-one.)



          But if two distinct elements of $S$ belong to the same pair, their difference is $2n-1$ by construction, hence the claim follows. (Indeed, in our example, our subset $S$ must contain all but one of $1,2,4,5$, but if it contains both 1 and 4, or both 2 and 5, those elements have a difference in the required range.



          The proof:



          Suppose $S$ is a subset of $[1,3n]_mathbb Z=set1,ldots, 3n$ with $|S|=n+2$.
          Let $delta S=setx-y$. The claim can be formulated as $[n+1,2n-1]_mathbb Zcap delta Sneq emptyset$; if there is some $x,yin S$ such that $n<y-x<2n$, then $y-xin setn+1,n+2,ldots, 2n-1=[n+1,2n-1]_mathbb Z$, but also $y-xin delta S$ by definition, hence is an element of the intersection.



          Note that for any $ainmathbb Z$, if $S+a=sets+amid sin Ssubset [1,3n]_mathbb Z$, then $delta(S+a)=delta S$ (since for any $x,yin S$, $x+a,y+ain S+a$ and $|x-y|=|(x+a)-(y+a)|$, so we can freely translate $S$ without effecting the claim. In particular, we can assume $3nin S$; otherwise, there is some maximal $min S$, and we can consider the set $S+(3n-m)$.



          If $3nin S$, then the claim follows if $[n+1,2n-1]_mathbb Zcap Sneq emptyset$, as any $n+1leq xleq 2n-1$ satisfies $n+1leq 3n-xleq 2n-1$. Then we are reduced to considering the case that $[n+1,2n-1]_mathbb Zcap S= emptyset$. Let $S'=Ssetminus set3n$ and note that $S'subseteq set1,ldots, n, 2n,ldots, 3n-1$. Define a new set of sets $$mathcal P=setsetk,2n-1+kmid 1leq kleq n=setset1,2n,set2,2n+1,ldots,setn,3n-1.$$
          (In other words, we pair up the elements of $set1,ldots, n, 2n,ldots, 3n-1$ into subsets $setx,y$ so that $|x-y|=2n-1$.)
          Since any element $sin S'$ belongs to exactly one $Pinmathcal P$, we have a function $p:S'rightarrow mathcal P$ defined by $p(s)= P$ where $sin P$. Since $|S'|=|S|-1=n+1$ and $|mathcal P|=n$, the Pidgeonhole Principle implies $p$ is not one-to-one, hence there is some $xneq yin S'subset S$ such that $setx,yin mathcal P$, which implies $|x-y|=2n-1$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            what is δS ... ?
            $endgroup$
            – magician
            Mar 19 at 14:04










          • $begingroup$
            Just a notation for the set of differences of $S$; it's defined in the second sentence.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:04










          • $begingroup$
            Aha, thanks. Now I'm reading your answer. It will cost me much time.
            $endgroup$
            – magician
            Mar 19 at 14:05










          • $begingroup$
            The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
            $endgroup$
            – magician
            Mar 19 at 14:07










          • $begingroup$
            I added an explanation in the edit.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:11


















          1












          $begingroup$

          The proof does not explicitly require the pigeon hole principle.



          1) For the inequality $n<y-x$, one should choose as many elements of $S$ as possible which are at close proximity; this will keep $y-x$ small.



          More concretely, I choose $n$ consecutive elements in $S$, $$a<a+1<ldots<a+(n-1).$$
          For those elements, the difference is at most $n-1$. But you have to choose two additional elements $b,c$ with say $b<c$ outside the above chain. There are two cases: (1) If $b<c<a$, then $a+(n-1) - b > n$. (2) If $b<a$ and $a+(n-1)<c$, then $c-b>n$. Done.



          2) For the inequalty $y-x<2n$, one should choose as many elements of $S$ as possible which are very far apart; this will make $y-x$ large.



          More concretely, I choose $n$ elements in $S$ which can be divided into two groups and have a large gap in between:
          $$1<2<ldots<n-a ;mboxand; 3n-a+1<3n-a+2<ldots<3n.$$
          The size of the gap is $(3n-a+1)-(n-a) =2n+1$. Now you have to add two elements $b,c$ with say $b<c$. Since $b,c$ are inserted into the gap, $c-b<2n$. Done.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
            $endgroup$
            – magician
            Mar 19 at 13:54










          • $begingroup$
            If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
            $endgroup$
            – antkam
            Mar 22 at 21:13











          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
          );



          );













          draft saved

          draft discarded


















          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154039%2fwhen-n-is-an-integer-such-that-n1-s-is-an-arbitrary-subset-of-1-2-3%23new-answer', 'question_page');

          );

          Post as a guest















          Required, but never shown

























          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $newcommandset[1]left#1right$ EDIT: To help make the argument clearer, here is a description of the argument. The claim is that for any subset $S$ of the $3n$ consecutive integers $1,ldots, 3n$ which contains $n+2$ elements, there is a pair of elements in the subset whose difference lies between $n$ and $2n$. For example, if $n=2$ and the set is $set1,2,3,4,5,6$, we want to show any set of $4$ elements has a pair whose difference is $3$. This is easy to see by checking all possibilities in this instance, but that is impractical for larger $n$, much less to prove the statement in general.



          Instead, we will use a divide and conquer approach. The first part of the argument will reduce the problem significantly, without loss of generality, by showing that the set $S$ can always be assumed to contain the largest element, and thus it suffices to prove the statement in the case that $S$ contains $3n$ and none of the elements in the middle third $n+1,ldots, 2n-1$. This allows us to apply the pidgeonhole principle by splitting the remaining possible elements of $S$ into pairs whose difference lies in the desired range, and showing that $S$ must contain one of those pairs.



          The Strategy



          First, note that this is really a question about the set of differences $delta S$ generated by the subset $S$. For example, when $n=2$ and $S=set1,2,4,6$, we have $2-1=1$, $4-2=6-4=2$, $4-1=3$ $6-2=4$, and $6-1=5$, hence the set of differences is $delta S=set1,2,3,4,5$. (Here, we exclude the negative differences as we will always have $x-y$ or $y-x$ in $delta S$ as defined.) The claim then is that one of these differences lies between $n$ and $2n$.



          To prove the claim, first we simplify by noting we can always assume $3nin S$. This is because the differences don't change if we shift every element of $S$ by the same amount; for example, if $S=set1,2,3,4$, and $S+2=set3,4,5,6$, the differences $delta S=set1,2,3=delta(S+2)$ are the same. In particular, if there were some $S$ such that the claim were false, there would be an $S$ containing $3n$ also proving the claim to be false.



          If $S$ contains $3n$ and any of the elements between $n$ and $2n$ steps away (namely, $n+1, n+2,ldots, 2n-1$), then we would be done. For instance, if $n=2$ and $3in S$, then $3=6-3in S$ proving the claim. So there is only more to argue if we assume those elements are not in $S$.



          In that case, we have that $S$ contains $3n$ and some other elements $S'$ which live in $1,2,ldots, n$ or $2n,2n+1,ldots, 3n-1$. (E.g. in the $n=2$ example,
          if $6in S$ and $3notin S$, then the remaining 3 elements of $S$ must be elements of $set1,2,4,5$.) We can pair these elements together as $set1,2n$, $set2,2n+1,ldots, setn,3n-1$ where the pairs are of the form $setk,2n-1+k$. (So in the $n=2$ case, the pairs are $set1,4$ and $set2,5$.)



          Now to prove the claim in this case, we use the pidgeonhole principle. To do so, we need our pidgeons and our holes. The pidgeons in this case are the elements of $S'$ (i.e. the elements of $S$ other than $3n$) and the holes are our pairs $setk,2n-1+k$. Each pidgeon lives in one of the holes, but there are $n+1$ pidgeons and $n$ holes, so two pidgeons must share a hole. (The formal way of putting this is defining a function and concluding its not one-to-one.)



          But if two distinct elements of $S$ belong to the same pair, their difference is $2n-1$ by construction, hence the claim follows. (Indeed, in our example, our subset $S$ must contain all but one of $1,2,4,5$, but if it contains both 1 and 4, or both 2 and 5, those elements have a difference in the required range.



          The proof:



          Suppose $S$ is a subset of $[1,3n]_mathbb Z=set1,ldots, 3n$ with $|S|=n+2$.
          Let $delta S=setx-y$. The claim can be formulated as $[n+1,2n-1]_mathbb Zcap delta Sneq emptyset$; if there is some $x,yin S$ such that $n<y-x<2n$, then $y-xin setn+1,n+2,ldots, 2n-1=[n+1,2n-1]_mathbb Z$, but also $y-xin delta S$ by definition, hence is an element of the intersection.



          Note that for any $ainmathbb Z$, if $S+a=sets+amid sin Ssubset [1,3n]_mathbb Z$, then $delta(S+a)=delta S$ (since for any $x,yin S$, $x+a,y+ain S+a$ and $|x-y|=|(x+a)-(y+a)|$, so we can freely translate $S$ without effecting the claim. In particular, we can assume $3nin S$; otherwise, there is some maximal $min S$, and we can consider the set $S+(3n-m)$.



          If $3nin S$, then the claim follows if $[n+1,2n-1]_mathbb Zcap Sneq emptyset$, as any $n+1leq xleq 2n-1$ satisfies $n+1leq 3n-xleq 2n-1$. Then we are reduced to considering the case that $[n+1,2n-1]_mathbb Zcap S= emptyset$. Let $S'=Ssetminus set3n$ and note that $S'subseteq set1,ldots, n, 2n,ldots, 3n-1$. Define a new set of sets $$mathcal P=setsetk,2n-1+kmid 1leq kleq n=setset1,2n,set2,2n+1,ldots,setn,3n-1.$$
          (In other words, we pair up the elements of $set1,ldots, n, 2n,ldots, 3n-1$ into subsets $setx,y$ so that $|x-y|=2n-1$.)
          Since any element $sin S'$ belongs to exactly one $Pinmathcal P$, we have a function $p:S'rightarrow mathcal P$ defined by $p(s)= P$ where $sin P$. Since $|S'|=|S|-1=n+1$ and $|mathcal P|=n$, the Pidgeonhole Principle implies $p$ is not one-to-one, hence there is some $xneq yin S'subset S$ such that $setx,yin mathcal P$, which implies $|x-y|=2n-1$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            what is δS ... ?
            $endgroup$
            – magician
            Mar 19 at 14:04










          • $begingroup$
            Just a notation for the set of differences of $S$; it's defined in the second sentence.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:04










          • $begingroup$
            Aha, thanks. Now I'm reading your answer. It will cost me much time.
            $endgroup$
            – magician
            Mar 19 at 14:05










          • $begingroup$
            The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
            $endgroup$
            – magician
            Mar 19 at 14:07










          • $begingroup$
            I added an explanation in the edit.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:11















          3












          $begingroup$

          $newcommandset[1]left#1right$ EDIT: To help make the argument clearer, here is a description of the argument. The claim is that for any subset $S$ of the $3n$ consecutive integers $1,ldots, 3n$ which contains $n+2$ elements, there is a pair of elements in the subset whose difference lies between $n$ and $2n$. For example, if $n=2$ and the set is $set1,2,3,4,5,6$, we want to show any set of $4$ elements has a pair whose difference is $3$. This is easy to see by checking all possibilities in this instance, but that is impractical for larger $n$, much less to prove the statement in general.



          Instead, we will use a divide and conquer approach. The first part of the argument will reduce the problem significantly, without loss of generality, by showing that the set $S$ can always be assumed to contain the largest element, and thus it suffices to prove the statement in the case that $S$ contains $3n$ and none of the elements in the middle third $n+1,ldots, 2n-1$. This allows us to apply the pidgeonhole principle by splitting the remaining possible elements of $S$ into pairs whose difference lies in the desired range, and showing that $S$ must contain one of those pairs.



          The Strategy



          First, note that this is really a question about the set of differences $delta S$ generated by the subset $S$. For example, when $n=2$ and $S=set1,2,4,6$, we have $2-1=1$, $4-2=6-4=2$, $4-1=3$ $6-2=4$, and $6-1=5$, hence the set of differences is $delta S=set1,2,3,4,5$. (Here, we exclude the negative differences as we will always have $x-y$ or $y-x$ in $delta S$ as defined.) The claim then is that one of these differences lies between $n$ and $2n$.



          To prove the claim, first we simplify by noting we can always assume $3nin S$. This is because the differences don't change if we shift every element of $S$ by the same amount; for example, if $S=set1,2,3,4$, and $S+2=set3,4,5,6$, the differences $delta S=set1,2,3=delta(S+2)$ are the same. In particular, if there were some $S$ such that the claim were false, there would be an $S$ containing $3n$ also proving the claim to be false.



          If $S$ contains $3n$ and any of the elements between $n$ and $2n$ steps away (namely, $n+1, n+2,ldots, 2n-1$), then we would be done. For instance, if $n=2$ and $3in S$, then $3=6-3in S$ proving the claim. So there is only more to argue if we assume those elements are not in $S$.



          In that case, we have that $S$ contains $3n$ and some other elements $S'$ which live in $1,2,ldots, n$ or $2n,2n+1,ldots, 3n-1$. (E.g. in the $n=2$ example,
          if $6in S$ and $3notin S$, then the remaining 3 elements of $S$ must be elements of $set1,2,4,5$.) We can pair these elements together as $set1,2n$, $set2,2n+1,ldots, setn,3n-1$ where the pairs are of the form $setk,2n-1+k$. (So in the $n=2$ case, the pairs are $set1,4$ and $set2,5$.)



          Now to prove the claim in this case, we use the pidgeonhole principle. To do so, we need our pidgeons and our holes. The pidgeons in this case are the elements of $S'$ (i.e. the elements of $S$ other than $3n$) and the holes are our pairs $setk,2n-1+k$. Each pidgeon lives in one of the holes, but there are $n+1$ pidgeons and $n$ holes, so two pidgeons must share a hole. (The formal way of putting this is defining a function and concluding its not one-to-one.)



          But if two distinct elements of $S$ belong to the same pair, their difference is $2n-1$ by construction, hence the claim follows. (Indeed, in our example, our subset $S$ must contain all but one of $1,2,4,5$, but if it contains both 1 and 4, or both 2 and 5, those elements have a difference in the required range.



          The proof:



          Suppose $S$ is a subset of $[1,3n]_mathbb Z=set1,ldots, 3n$ with $|S|=n+2$.
          Let $delta S=setx-y$. The claim can be formulated as $[n+1,2n-1]_mathbb Zcap delta Sneq emptyset$; if there is some $x,yin S$ such that $n<y-x<2n$, then $y-xin setn+1,n+2,ldots, 2n-1=[n+1,2n-1]_mathbb Z$, but also $y-xin delta S$ by definition, hence is an element of the intersection.



          Note that for any $ainmathbb Z$, if $S+a=sets+amid sin Ssubset [1,3n]_mathbb Z$, then $delta(S+a)=delta S$ (since for any $x,yin S$, $x+a,y+ain S+a$ and $|x-y|=|(x+a)-(y+a)|$, so we can freely translate $S$ without effecting the claim. In particular, we can assume $3nin S$; otherwise, there is some maximal $min S$, and we can consider the set $S+(3n-m)$.



          If $3nin S$, then the claim follows if $[n+1,2n-1]_mathbb Zcap Sneq emptyset$, as any $n+1leq xleq 2n-1$ satisfies $n+1leq 3n-xleq 2n-1$. Then we are reduced to considering the case that $[n+1,2n-1]_mathbb Zcap S= emptyset$. Let $S'=Ssetminus set3n$ and note that $S'subseteq set1,ldots, n, 2n,ldots, 3n-1$. Define a new set of sets $$mathcal P=setsetk,2n-1+kmid 1leq kleq n=setset1,2n,set2,2n+1,ldots,setn,3n-1.$$
          (In other words, we pair up the elements of $set1,ldots, n, 2n,ldots, 3n-1$ into subsets $setx,y$ so that $|x-y|=2n-1$.)
          Since any element $sin S'$ belongs to exactly one $Pinmathcal P$, we have a function $p:S'rightarrow mathcal P$ defined by $p(s)= P$ where $sin P$. Since $|S'|=|S|-1=n+1$ and $|mathcal P|=n$, the Pidgeonhole Principle implies $p$ is not one-to-one, hence there is some $xneq yin S'subset S$ such that $setx,yin mathcal P$, which implies $|x-y|=2n-1$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            what is δS ... ?
            $endgroup$
            – magician
            Mar 19 at 14:04










          • $begingroup$
            Just a notation for the set of differences of $S$; it's defined in the second sentence.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:04










          • $begingroup$
            Aha, thanks. Now I'm reading your answer. It will cost me much time.
            $endgroup$
            – magician
            Mar 19 at 14:05










          • $begingroup$
            The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
            $endgroup$
            – magician
            Mar 19 at 14:07










          • $begingroup$
            I added an explanation in the edit.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:11













          3












          3








          3





          $begingroup$

          $newcommandset[1]left#1right$ EDIT: To help make the argument clearer, here is a description of the argument. The claim is that for any subset $S$ of the $3n$ consecutive integers $1,ldots, 3n$ which contains $n+2$ elements, there is a pair of elements in the subset whose difference lies between $n$ and $2n$. For example, if $n=2$ and the set is $set1,2,3,4,5,6$, we want to show any set of $4$ elements has a pair whose difference is $3$. This is easy to see by checking all possibilities in this instance, but that is impractical for larger $n$, much less to prove the statement in general.



          Instead, we will use a divide and conquer approach. The first part of the argument will reduce the problem significantly, without loss of generality, by showing that the set $S$ can always be assumed to contain the largest element, and thus it suffices to prove the statement in the case that $S$ contains $3n$ and none of the elements in the middle third $n+1,ldots, 2n-1$. This allows us to apply the pidgeonhole principle by splitting the remaining possible elements of $S$ into pairs whose difference lies in the desired range, and showing that $S$ must contain one of those pairs.



          The Strategy



          First, note that this is really a question about the set of differences $delta S$ generated by the subset $S$. For example, when $n=2$ and $S=set1,2,4,6$, we have $2-1=1$, $4-2=6-4=2$, $4-1=3$ $6-2=4$, and $6-1=5$, hence the set of differences is $delta S=set1,2,3,4,5$. (Here, we exclude the negative differences as we will always have $x-y$ or $y-x$ in $delta S$ as defined.) The claim then is that one of these differences lies between $n$ and $2n$.



          To prove the claim, first we simplify by noting we can always assume $3nin S$. This is because the differences don't change if we shift every element of $S$ by the same amount; for example, if $S=set1,2,3,4$, and $S+2=set3,4,5,6$, the differences $delta S=set1,2,3=delta(S+2)$ are the same. In particular, if there were some $S$ such that the claim were false, there would be an $S$ containing $3n$ also proving the claim to be false.



          If $S$ contains $3n$ and any of the elements between $n$ and $2n$ steps away (namely, $n+1, n+2,ldots, 2n-1$), then we would be done. For instance, if $n=2$ and $3in S$, then $3=6-3in S$ proving the claim. So there is only more to argue if we assume those elements are not in $S$.



          In that case, we have that $S$ contains $3n$ and some other elements $S'$ which live in $1,2,ldots, n$ or $2n,2n+1,ldots, 3n-1$. (E.g. in the $n=2$ example,
          if $6in S$ and $3notin S$, then the remaining 3 elements of $S$ must be elements of $set1,2,4,5$.) We can pair these elements together as $set1,2n$, $set2,2n+1,ldots, setn,3n-1$ where the pairs are of the form $setk,2n-1+k$. (So in the $n=2$ case, the pairs are $set1,4$ and $set2,5$.)



          Now to prove the claim in this case, we use the pidgeonhole principle. To do so, we need our pidgeons and our holes. The pidgeons in this case are the elements of $S'$ (i.e. the elements of $S$ other than $3n$) and the holes are our pairs $setk,2n-1+k$. Each pidgeon lives in one of the holes, but there are $n+1$ pidgeons and $n$ holes, so two pidgeons must share a hole. (The formal way of putting this is defining a function and concluding its not one-to-one.)



          But if two distinct elements of $S$ belong to the same pair, their difference is $2n-1$ by construction, hence the claim follows. (Indeed, in our example, our subset $S$ must contain all but one of $1,2,4,5$, but if it contains both 1 and 4, or both 2 and 5, those elements have a difference in the required range.



          The proof:



          Suppose $S$ is a subset of $[1,3n]_mathbb Z=set1,ldots, 3n$ with $|S|=n+2$.
          Let $delta S=setx-y$. The claim can be formulated as $[n+1,2n-1]_mathbb Zcap delta Sneq emptyset$; if there is some $x,yin S$ such that $n<y-x<2n$, then $y-xin setn+1,n+2,ldots, 2n-1=[n+1,2n-1]_mathbb Z$, but also $y-xin delta S$ by definition, hence is an element of the intersection.



          Note that for any $ainmathbb Z$, if $S+a=sets+amid sin Ssubset [1,3n]_mathbb Z$, then $delta(S+a)=delta S$ (since for any $x,yin S$, $x+a,y+ain S+a$ and $|x-y|=|(x+a)-(y+a)|$, so we can freely translate $S$ without effecting the claim. In particular, we can assume $3nin S$; otherwise, there is some maximal $min S$, and we can consider the set $S+(3n-m)$.



          If $3nin S$, then the claim follows if $[n+1,2n-1]_mathbb Zcap Sneq emptyset$, as any $n+1leq xleq 2n-1$ satisfies $n+1leq 3n-xleq 2n-1$. Then we are reduced to considering the case that $[n+1,2n-1]_mathbb Zcap S= emptyset$. Let $S'=Ssetminus set3n$ and note that $S'subseteq set1,ldots, n, 2n,ldots, 3n-1$. Define a new set of sets $$mathcal P=setsetk,2n-1+kmid 1leq kleq n=setset1,2n,set2,2n+1,ldots,setn,3n-1.$$
          (In other words, we pair up the elements of $set1,ldots, n, 2n,ldots, 3n-1$ into subsets $setx,y$ so that $|x-y|=2n-1$.)
          Since any element $sin S'$ belongs to exactly one $Pinmathcal P$, we have a function $p:S'rightarrow mathcal P$ defined by $p(s)= P$ where $sin P$. Since $|S'|=|S|-1=n+1$ and $|mathcal P|=n$, the Pidgeonhole Principle implies $p$ is not one-to-one, hence there is some $xneq yin S'subset S$ such that $setx,yin mathcal P$, which implies $|x-y|=2n-1$.






          share|cite|improve this answer











          $endgroup$



          $newcommandset[1]left#1right$ EDIT: To help make the argument clearer, here is a description of the argument. The claim is that for any subset $S$ of the $3n$ consecutive integers $1,ldots, 3n$ which contains $n+2$ elements, there is a pair of elements in the subset whose difference lies between $n$ and $2n$. For example, if $n=2$ and the set is $set1,2,3,4,5,6$, we want to show any set of $4$ elements has a pair whose difference is $3$. This is easy to see by checking all possibilities in this instance, but that is impractical for larger $n$, much less to prove the statement in general.



          Instead, we will use a divide and conquer approach. The first part of the argument will reduce the problem significantly, without loss of generality, by showing that the set $S$ can always be assumed to contain the largest element, and thus it suffices to prove the statement in the case that $S$ contains $3n$ and none of the elements in the middle third $n+1,ldots, 2n-1$. This allows us to apply the pidgeonhole principle by splitting the remaining possible elements of $S$ into pairs whose difference lies in the desired range, and showing that $S$ must contain one of those pairs.



          The Strategy



          First, note that this is really a question about the set of differences $delta S$ generated by the subset $S$. For example, when $n=2$ and $S=set1,2,4,6$, we have $2-1=1$, $4-2=6-4=2$, $4-1=3$ $6-2=4$, and $6-1=5$, hence the set of differences is $delta S=set1,2,3,4,5$. (Here, we exclude the negative differences as we will always have $x-y$ or $y-x$ in $delta S$ as defined.) The claim then is that one of these differences lies between $n$ and $2n$.



          To prove the claim, first we simplify by noting we can always assume $3nin S$. This is because the differences don't change if we shift every element of $S$ by the same amount; for example, if $S=set1,2,3,4$, and $S+2=set3,4,5,6$, the differences $delta S=set1,2,3=delta(S+2)$ are the same. In particular, if there were some $S$ such that the claim were false, there would be an $S$ containing $3n$ also proving the claim to be false.



          If $S$ contains $3n$ and any of the elements between $n$ and $2n$ steps away (namely, $n+1, n+2,ldots, 2n-1$), then we would be done. For instance, if $n=2$ and $3in S$, then $3=6-3in S$ proving the claim. So there is only more to argue if we assume those elements are not in $S$.



          In that case, we have that $S$ contains $3n$ and some other elements $S'$ which live in $1,2,ldots, n$ or $2n,2n+1,ldots, 3n-1$. (E.g. in the $n=2$ example,
          if $6in S$ and $3notin S$, then the remaining 3 elements of $S$ must be elements of $set1,2,4,5$.) We can pair these elements together as $set1,2n$, $set2,2n+1,ldots, setn,3n-1$ where the pairs are of the form $setk,2n-1+k$. (So in the $n=2$ case, the pairs are $set1,4$ and $set2,5$.)



          Now to prove the claim in this case, we use the pidgeonhole principle. To do so, we need our pidgeons and our holes. The pidgeons in this case are the elements of $S'$ (i.e. the elements of $S$ other than $3n$) and the holes are our pairs $setk,2n-1+k$. Each pidgeon lives in one of the holes, but there are $n+1$ pidgeons and $n$ holes, so two pidgeons must share a hole. (The formal way of putting this is defining a function and concluding its not one-to-one.)



          But if two distinct elements of $S$ belong to the same pair, their difference is $2n-1$ by construction, hence the claim follows. (Indeed, in our example, our subset $S$ must contain all but one of $1,2,4,5$, but if it contains both 1 and 4, or both 2 and 5, those elements have a difference in the required range.



          The proof:



          Suppose $S$ is a subset of $[1,3n]_mathbb Z=set1,ldots, 3n$ with $|S|=n+2$.
          Let $delta S=setx-y$. The claim can be formulated as $[n+1,2n-1]_mathbb Zcap delta Sneq emptyset$; if there is some $x,yin S$ such that $n<y-x<2n$, then $y-xin setn+1,n+2,ldots, 2n-1=[n+1,2n-1]_mathbb Z$, but also $y-xin delta S$ by definition, hence is an element of the intersection.



          Note that for any $ainmathbb Z$, if $S+a=sets+amid sin Ssubset [1,3n]_mathbb Z$, then $delta(S+a)=delta S$ (since for any $x,yin S$, $x+a,y+ain S+a$ and $|x-y|=|(x+a)-(y+a)|$, so we can freely translate $S$ without effecting the claim. In particular, we can assume $3nin S$; otherwise, there is some maximal $min S$, and we can consider the set $S+(3n-m)$.



          If $3nin S$, then the claim follows if $[n+1,2n-1]_mathbb Zcap Sneq emptyset$, as any $n+1leq xleq 2n-1$ satisfies $n+1leq 3n-xleq 2n-1$. Then we are reduced to considering the case that $[n+1,2n-1]_mathbb Zcap S= emptyset$. Let $S'=Ssetminus set3n$ and note that $S'subseteq set1,ldots, n, 2n,ldots, 3n-1$. Define a new set of sets $$mathcal P=setsetk,2n-1+kmid 1leq kleq n=setset1,2n,set2,2n+1,ldots,setn,3n-1.$$
          (In other words, we pair up the elements of $set1,ldots, n, 2n,ldots, 3n-1$ into subsets $setx,y$ so that $|x-y|=2n-1$.)
          Since any element $sin S'$ belongs to exactly one $Pinmathcal P$, we have a function $p:S'rightarrow mathcal P$ defined by $p(s)= P$ where $sin P$. Since $|S'|=|S|-1=n+1$ and $|mathcal P|=n$, the Pidgeonhole Principle implies $p$ is not one-to-one, hence there is some $xneq yin S'subset S$ such that $setx,yin mathcal P$, which implies $|x-y|=2n-1$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 19 at 14:49

























          answered Mar 19 at 14:02









          Sean ClarkSean Clark

          2,078813




          2,078813











          • $begingroup$
            what is δS ... ?
            $endgroup$
            – magician
            Mar 19 at 14:04










          • $begingroup$
            Just a notation for the set of differences of $S$; it's defined in the second sentence.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:04










          • $begingroup$
            Aha, thanks. Now I'm reading your answer. It will cost me much time.
            $endgroup$
            – magician
            Mar 19 at 14:05










          • $begingroup$
            The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
            $endgroup$
            – magician
            Mar 19 at 14:07










          • $begingroup$
            I added an explanation in the edit.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:11
















          • $begingroup$
            what is δS ... ?
            $endgroup$
            – magician
            Mar 19 at 14:04










          • $begingroup$
            Just a notation for the set of differences of $S$; it's defined in the second sentence.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:04










          • $begingroup$
            Aha, thanks. Now I'm reading your answer. It will cost me much time.
            $endgroup$
            – magician
            Mar 19 at 14:05










          • $begingroup$
            The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
            $endgroup$
            – magician
            Mar 19 at 14:07










          • $begingroup$
            I added an explanation in the edit.
            $endgroup$
            – Sean Clark
            Mar 19 at 14:11















          $begingroup$
          what is δS ... ?
          $endgroup$
          – magician
          Mar 19 at 14:04




          $begingroup$
          what is δS ... ?
          $endgroup$
          – magician
          Mar 19 at 14:04












          $begingroup$
          Just a notation for the set of differences of $S$; it's defined in the second sentence.
          $endgroup$
          – Sean Clark
          Mar 19 at 14:04




          $begingroup$
          Just a notation for the set of differences of $S$; it's defined in the second sentence.
          $endgroup$
          – Sean Clark
          Mar 19 at 14:04












          $begingroup$
          Aha, thanks. Now I'm reading your answer. It will cost me much time.
          $endgroup$
          – magician
          Mar 19 at 14:05




          $begingroup$
          Aha, thanks. Now I'm reading your answer. It will cost me much time.
          $endgroup$
          – magician
          Mar 19 at 14:05












          $begingroup$
          The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
          $endgroup$
          – magician
          Mar 19 at 14:07




          $begingroup$
          The claim can be formulated as [n+1,2n−1]Z∩δS≠∅. Can you tell me the reason..? So sorry
          $endgroup$
          – magician
          Mar 19 at 14:07












          $begingroup$
          I added an explanation in the edit.
          $endgroup$
          – Sean Clark
          Mar 19 at 14:11




          $begingroup$
          I added an explanation in the edit.
          $endgroup$
          – Sean Clark
          Mar 19 at 14:11











          1












          $begingroup$

          The proof does not explicitly require the pigeon hole principle.



          1) For the inequality $n<y-x$, one should choose as many elements of $S$ as possible which are at close proximity; this will keep $y-x$ small.



          More concretely, I choose $n$ consecutive elements in $S$, $$a<a+1<ldots<a+(n-1).$$
          For those elements, the difference is at most $n-1$. But you have to choose two additional elements $b,c$ with say $b<c$ outside the above chain. There are two cases: (1) If $b<c<a$, then $a+(n-1) - b > n$. (2) If $b<a$ and $a+(n-1)<c$, then $c-b>n$. Done.



          2) For the inequalty $y-x<2n$, one should choose as many elements of $S$ as possible which are very far apart; this will make $y-x$ large.



          More concretely, I choose $n$ elements in $S$ which can be divided into two groups and have a large gap in between:
          $$1<2<ldots<n-a ;mboxand; 3n-a+1<3n-a+2<ldots<3n.$$
          The size of the gap is $(3n-a+1)-(n-a) =2n+1$. Now you have to add two elements $b,c$ with say $b<c$. Since $b,c$ are inserted into the gap, $c-b<2n$. Done.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
            $endgroup$
            – magician
            Mar 19 at 13:54










          • $begingroup$
            If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
            $endgroup$
            – antkam
            Mar 22 at 21:13















          1












          $begingroup$

          The proof does not explicitly require the pigeon hole principle.



          1) For the inequality $n<y-x$, one should choose as many elements of $S$ as possible which are at close proximity; this will keep $y-x$ small.



          More concretely, I choose $n$ consecutive elements in $S$, $$a<a+1<ldots<a+(n-1).$$
          For those elements, the difference is at most $n-1$. But you have to choose two additional elements $b,c$ with say $b<c$ outside the above chain. There are two cases: (1) If $b<c<a$, then $a+(n-1) - b > n$. (2) If $b<a$ and $a+(n-1)<c$, then $c-b>n$. Done.



          2) For the inequalty $y-x<2n$, one should choose as many elements of $S$ as possible which are very far apart; this will make $y-x$ large.



          More concretely, I choose $n$ elements in $S$ which can be divided into two groups and have a large gap in between:
          $$1<2<ldots<n-a ;mboxand; 3n-a+1<3n-a+2<ldots<3n.$$
          The size of the gap is $(3n-a+1)-(n-a) =2n+1$. Now you have to add two elements $b,c$ with say $b<c$. Since $b,c$ are inserted into the gap, $c-b<2n$. Done.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
            $endgroup$
            – magician
            Mar 19 at 13:54










          • $begingroup$
            If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
            $endgroup$
            – antkam
            Mar 22 at 21:13













          1












          1








          1





          $begingroup$

          The proof does not explicitly require the pigeon hole principle.



          1) For the inequality $n<y-x$, one should choose as many elements of $S$ as possible which are at close proximity; this will keep $y-x$ small.



          More concretely, I choose $n$ consecutive elements in $S$, $$a<a+1<ldots<a+(n-1).$$
          For those elements, the difference is at most $n-1$. But you have to choose two additional elements $b,c$ with say $b<c$ outside the above chain. There are two cases: (1) If $b<c<a$, then $a+(n-1) - b > n$. (2) If $b<a$ and $a+(n-1)<c$, then $c-b>n$. Done.



          2) For the inequalty $y-x<2n$, one should choose as many elements of $S$ as possible which are very far apart; this will make $y-x$ large.



          More concretely, I choose $n$ elements in $S$ which can be divided into two groups and have a large gap in between:
          $$1<2<ldots<n-a ;mboxand; 3n-a+1<3n-a+2<ldots<3n.$$
          The size of the gap is $(3n-a+1)-(n-a) =2n+1$. Now you have to add two elements $b,c$ with say $b<c$. Since $b,c$ are inserted into the gap, $c-b<2n$. Done.






          share|cite|improve this answer











          $endgroup$



          The proof does not explicitly require the pigeon hole principle.



          1) For the inequality $n<y-x$, one should choose as many elements of $S$ as possible which are at close proximity; this will keep $y-x$ small.



          More concretely, I choose $n$ consecutive elements in $S$, $$a<a+1<ldots<a+(n-1).$$
          For those elements, the difference is at most $n-1$. But you have to choose two additional elements $b,c$ with say $b<c$ outside the above chain. There are two cases: (1) If $b<c<a$, then $a+(n-1) - b > n$. (2) If $b<a$ and $a+(n-1)<c$, then $c-b>n$. Done.



          2) For the inequalty $y-x<2n$, one should choose as many elements of $S$ as possible which are very far apart; this will make $y-x$ large.



          More concretely, I choose $n$ elements in $S$ which can be divided into two groups and have a large gap in between:
          $$1<2<ldots<n-a ;mboxand; 3n-a+1<3n-a+2<ldots<3n.$$
          The size of the gap is $(3n-a+1)-(n-a) =2n+1$. Now you have to add two elements $b,c$ with say $b<c$. Since $b,c$ are inserted into the gap, $c-b<2n$. Done.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Mar 20 at 8:51

























          answered Mar 19 at 13:22









          WuestenfuxWuestenfux

          5,3331513




          5,3331513











          • $begingroup$
            Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
            $endgroup$
            – magician
            Mar 19 at 13:54










          • $begingroup$
            If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
            $endgroup$
            – antkam
            Mar 22 at 21:13
















          • $begingroup$
            Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
            $endgroup$
            – magician
            Mar 19 at 13:54










          • $begingroup$
            If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
            $endgroup$
            – antkam
            Mar 22 at 21:13















          $begingroup$
          Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
          $endgroup$
          – magician
          Mar 19 at 13:54




          $begingroup$
          Can you tell me the details..? I can't understand your explanation that is so difficult for me Besides, please check the additional information in my question.
          $endgroup$
          – magician
          Mar 19 at 13:54












          $begingroup$
          If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
          $endgroup$
          – antkam
          Mar 22 at 21:13




          $begingroup$
          If I understand you correctly, part (1) of your proof shows that there exist two elements $y_1, x_1$ whose difference $y_1 - x_1 >n$, and part (2) shows there exist two elements $y_2, x_2$ whose difference $y_2 - x_2 <2n$. But you did not show they are the same two elements... The OP is asking for two elements $y, x$ to have a difference meeting both conditions simultaneously, $n < y-x < 2n$.
          $endgroup$
          – antkam
          Mar 22 at 21:13

















          draft saved

          draft discarded
















































          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.




          draft saved


          draft discarded














          StackExchange.ready(
          function ()
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3154039%2fwhen-n-is-an-integer-such-that-n1-s-is-an-arbitrary-subset-of-1-2-3%23new-answer', 'question_page');

          );

          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







          Popular posts from this blog

          How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer

          random experiment with two different functions on unit interval Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Random variable and probability space notionsRandom Walk with EdgesFinding functions where the increase over a random interval is Poisson distributedNumber of days until dayCan an observed event in fact be of zero probability?Unit random processmodels of coins and uniform distributionHow to get the number of successes given $n$ trials , probability $P$ and a random variable $X$Absorbing Markov chain in a computer. Is “almost every” turned into always convergence in computer executions?Stopped random walk is not uniformly integrable

          Lowndes Grove History Architecture References Navigation menu32°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661132°48′6″N 79°57′58″W / 32.80167°N 79.96611°W / 32.80167; -79.9661178002500"National Register Information System"Historic houses of South Carolina"Lowndes Grove""+32° 48' 6.00", −79° 57' 58.00""Lowndes Grove, Charleston County (260 St. Margaret St., Charleston)""Lowndes Grove"The Charleston ExpositionIt Happened in South Carolina"Lowndes Grove (House), Saint Margaret Street & Sixth Avenue, Charleston, Charleston County, SC(Photographs)"Plantations of the Carolina Low Countrye