degree of vertices in G [on hold]

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Prove there are two vertices of G with the same degree.



Can anyone help me with that? Or, can someone put the basis for the proof?



Any help is very much appreciated.










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put on hold as off-topic by Alexander Gruber Apr 11 at 16:23


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.




















    5












    $begingroup$


    Prove there are two vertices of G with the same degree.



    Can anyone help me with that? Or, can someone put the basis for the proof?



    Any help is very much appreciated.










    share|cite|improve this question











    $endgroup$



    put on hold as off-topic by Alexander Gruber Apr 11 at 16:23


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber
    If this question can be reworded to fit the rules in the help center, please edit the question.


















      5












      5








      5


      1



      $begingroup$


      Prove there are two vertices of G with the same degree.



      Can anyone help me with that? Or, can someone put the basis for the proof?



      Any help is very much appreciated.










      share|cite|improve this question











      $endgroup$




      Prove there are two vertices of G with the same degree.



      Can anyone help me with that? Or, can someone put the basis for the proof?



      Any help is very much appreciated.







      graph-theory






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      share|cite|improve this question













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      edited Apr 3 at 15:51







      Alex Freeman

















      asked Mar 16 at 14:51









      Alex FreemanAlex Freeman

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      155




      put on hold as off-topic by Alexander Gruber Apr 11 at 16:23


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber
      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by Alexander Gruber Apr 11 at 16:23


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Alexander Gruber
      If this question can be reworded to fit the rules in the help center, please edit the question.




















          2 Answers
          2






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          $begingroup$

          Hint: In a simple graph $G=(V, E)$ on $n$ vertices, the maximum degree of a vertex is $n-1$. In particular, the set of possible degrees is $A := 0, 1, ldots, n-1$, and has cardinality $n$. Now suppose that each vertex in your graph $G$ has a distinct degree. Then we can define an injection
          $$phi: V rightarrow A$$
          $$v mapsto deg(v).$$



          But $V$ and $A$ have the same cardinality, so injectivity actually implies ___________.



          Now note that the only way for a vertex to have degree $n-1$ is if it is adjacent to all other vertices in the graph. Can you find a contradiction?






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            $begingroup$

            Suppose all vertices have differnt degree. Then all the numebers $0,1,2,...,n-1$ apperas as degrees. But then there is a vertex (of degree $n-1$) which is connected to all vertices and including that one wit degree 0. A contradiction.






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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Hint: In a simple graph $G=(V, E)$ on $n$ vertices, the maximum degree of a vertex is $n-1$. In particular, the set of possible degrees is $A := 0, 1, ldots, n-1$, and has cardinality $n$. Now suppose that each vertex in your graph $G$ has a distinct degree. Then we can define an injection
              $$phi: V rightarrow A$$
              $$v mapsto deg(v).$$



              But $V$ and $A$ have the same cardinality, so injectivity actually implies ___________.



              Now note that the only way for a vertex to have degree $n-1$ is if it is adjacent to all other vertices in the graph. Can you find a contradiction?






              share|cite|improve this answer









              $endgroup$

















                3












                $begingroup$

                Hint: In a simple graph $G=(V, E)$ on $n$ vertices, the maximum degree of a vertex is $n-1$. In particular, the set of possible degrees is $A := 0, 1, ldots, n-1$, and has cardinality $n$. Now suppose that each vertex in your graph $G$ has a distinct degree. Then we can define an injection
                $$phi: V rightarrow A$$
                $$v mapsto deg(v).$$



                But $V$ and $A$ have the same cardinality, so injectivity actually implies ___________.



                Now note that the only way for a vertex to have degree $n-1$ is if it is adjacent to all other vertices in the graph. Can you find a contradiction?






                share|cite|improve this answer









                $endgroup$















                  3












                  3








                  3





                  $begingroup$

                  Hint: In a simple graph $G=(V, E)$ on $n$ vertices, the maximum degree of a vertex is $n-1$. In particular, the set of possible degrees is $A := 0, 1, ldots, n-1$, and has cardinality $n$. Now suppose that each vertex in your graph $G$ has a distinct degree. Then we can define an injection
                  $$phi: V rightarrow A$$
                  $$v mapsto deg(v).$$



                  But $V$ and $A$ have the same cardinality, so injectivity actually implies ___________.



                  Now note that the only way for a vertex to have degree $n-1$ is if it is adjacent to all other vertices in the graph. Can you find a contradiction?






                  share|cite|improve this answer









                  $endgroup$



                  Hint: In a simple graph $G=(V, E)$ on $n$ vertices, the maximum degree of a vertex is $n-1$. In particular, the set of possible degrees is $A := 0, 1, ldots, n-1$, and has cardinality $n$. Now suppose that each vertex in your graph $G$ has a distinct degree. Then we can define an injection
                  $$phi: V rightarrow A$$
                  $$v mapsto deg(v).$$



                  But $V$ and $A$ have the same cardinality, so injectivity actually implies ___________.



                  Now note that the only way for a vertex to have degree $n-1$ is if it is adjacent to all other vertices in the graph. Can you find a contradiction?







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 16 at 16:04









                  MauveMauve

                  1,6153516




                  1,6153516





















                      0












                      $begingroup$

                      Suppose all vertices have differnt degree. Then all the numebers $0,1,2,...,n-1$ apperas as degrees. But then there is a vertex (of degree $n-1$) which is connected to all vertices and including that one wit degree 0. A contradiction.






                      share|cite|improve this answer









                      $endgroup$

















                        0












                        $begingroup$

                        Suppose all vertices have differnt degree. Then all the numebers $0,1,2,...,n-1$ apperas as degrees. But then there is a vertex (of degree $n-1$) which is connected to all vertices and including that one wit degree 0. A contradiction.






                        share|cite|improve this answer









                        $endgroup$















                          0












                          0








                          0





                          $begingroup$

                          Suppose all vertices have differnt degree. Then all the numebers $0,1,2,...,n-1$ apperas as degrees. But then there is a vertex (of degree $n-1$) which is connected to all vertices and including that one wit degree 0. A contradiction.






                          share|cite|improve this answer









                          $endgroup$



                          Suppose all vertices have differnt degree. Then all the numebers $0,1,2,...,n-1$ apperas as degrees. But then there is a vertex (of degree $n-1$) which is connected to all vertices and including that one wit degree 0. A contradiction.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Mar 26 at 21:10









                          Maria MazurMaria Mazur

                          50k1361125




                          50k1361125












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