Proof that no Eulerian Tour exists for graph with even number of vertices and odd number of edges

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How would you prove that for a connected graph with an even number of vertices and an odd number of edges, at least one of the vertices has an odd degree?



My first attempt at solving this has been to show that the smallest connected graph with an even number of vertices is a graph with 2 vertices and one edge connecting them. From here, to maintain a connected graph, even number of vertices and an odd number of edges we must add both the vertices and edges by even numbers, and when adding in even numbers the degree of a vertex will not change so it must be odd.



However, this does not seem strong, so could anyone inform me on how to start a proof by contradiction?



I understand the first line of the proof by contradiction would be to assume that every vertex has even degree, but where should I go from there?



Is this not a connected graph with an even number of vertices and an odd number of edges for which an Eulerian tour exists?






discrete-mathematics graph-theory




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    Yes, the graph you indicated in your edit is a good example of an Eulerian graph with an even number of vertices and an odd number of edges.
    – bof
    Nov 19 '14 at 1:48










  • For the question in the first sentence, not the question in the title: assume all the vertices have even degree, use the handshaking lemma, and check divisibility by $4$.
    – Rahul
    Nov 21 '14 at 17:24














up vote
0
down vote

favorite
1












How would you prove that for a connected graph with an even number of vertices and an odd number of edges, at least one of the vertices has an odd degree?



My first attempt at solving this has been to show that the smallest connected graph with an even number of vertices is a graph with 2 vertices and one edge connecting them. From here, to maintain a connected graph, even number of vertices and an odd number of edges we must add both the vertices and edges by even numbers, and when adding in even numbers the degree of a vertex will not change so it must be odd.



However, this does not seem strong, so could anyone inform me on how to start a proof by contradiction?



I understand the first line of the proof by contradiction would be to assume that every vertex has even degree, but where should I go from there?



Is this not a connected graph with an even number of vertices and an odd number of edges for which an Eulerian tour exists?






discrete-mathematics graph-theory




share|cite|improve this question


















  • 1




    Yes, the graph you indicated in your edit is a good example of an Eulerian graph with an even number of vertices and an odd number of edges.
    – bof
    Nov 19 '14 at 1:48










  • For the question in the first sentence, not the question in the title: assume all the vertices have even degree, use the handshaking lemma, and check divisibility by $4$.
    – Rahul
    Nov 21 '14 at 17:24












up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





How would you prove that for a connected graph with an even number of vertices and an odd number of edges, at least one of the vertices has an odd degree?



My first attempt at solving this has been to show that the smallest connected graph with an even number of vertices is a graph with 2 vertices and one edge connecting them. From here, to maintain a connected graph, even number of vertices and an odd number of edges we must add both the vertices and edges by even numbers, and when adding in even numbers the degree of a vertex will not change so it must be odd.



However, this does not seem strong, so could anyone inform me on how to start a proof by contradiction?



I understand the first line of the proof by contradiction would be to assume that every vertex has even degree, but where should I go from there?



Is this not a connected graph with an even number of vertices and an odd number of edges for which an Eulerian tour exists?






discrete-mathematics graph-theory




share|cite|improve this question














How would you prove that for a connected graph with an even number of vertices and an odd number of edges, at least one of the vertices has an odd degree?



My first attempt at solving this has been to show that the smallest connected graph with an even number of vertices is a graph with 2 vertices and one edge connecting them. From here, to maintain a connected graph, even number of vertices and an odd number of edges we must add both the vertices and edges by even numbers, and when adding in even numbers the degree of a vertex will not change so it must be odd.



However, this does not seem strong, so could anyone inform me on how to start a proof by contradiction?



I understand the first line of the proof by contradiction would be to assume that every vertex has even degree, but where should I go from there?



Is this not a connected graph with an even number of vertices and an odd number of edges for which an Eulerian tour exists?







discrete-mathematics graph-theory





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edited Jan 7 '17 at 15:24









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asked Nov 19 '14 at 0:40









user3746892

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  • 1




    Yes, the graph you indicated in your edit is a good example of an Eulerian graph with an even number of vertices and an odd number of edges.
    – bof
    Nov 19 '14 at 1:48










  • For the question in the first sentence, not the question in the title: assume all the vertices have even degree, use the handshaking lemma, and check divisibility by $4$.
    – Rahul
    Nov 21 '14 at 17:24












  • 1




    Yes, the graph you indicated in your edit is a good example of an Eulerian graph with an even number of vertices and an odd number of edges.
    – bof
    Nov 19 '14 at 1:48










  • For the question in the first sentence, not the question in the title: assume all the vertices have even degree, use the handshaking lemma, and check divisibility by $4$.
    – Rahul
    Nov 21 '14 at 17:24







1




1




Yes, the graph you indicated in your edit is a good example of an Eulerian graph with an even number of vertices and an odd number of edges.
– bof
Nov 19 '14 at 1:48




Yes, the graph you indicated in your edit is a good example of an Eulerian graph with an even number of vertices and an odd number of edges.
– bof
Nov 19 '14 at 1:48












For the question in the first sentence, not the question in the title: assume all the vertices have even degree, use the handshaking lemma, and check divisibility by $4$.
– Rahul
Nov 21 '14 at 17:24




For the question in the first sentence, not the question in the title: assume all the vertices have even degree, use the handshaking lemma, and check divisibility by $4$.
– Rahul
Nov 21 '14 at 17:24










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The claim you want to prove is false -- you've given a counterexample to it yourself.



So the answer to "How would you prove ..." is: That is impossible to prove, because it is not true.






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    1 Answer
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    up vote
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    The claim you want to prove is false -- you've given a counterexample to it yourself.



    So the answer to "How would you prove ..." is: That is impossible to prove, because it is not true.






    share|cite|improve this answer
























      up vote
      1
      down vote













      The claim you want to prove is false -- you've given a counterexample to it yourself.



      So the answer to "How would you prove ..." is: That is impossible to prove, because it is not true.






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        The claim you want to prove is false -- you've given a counterexample to it yourself.



        So the answer to "How would you prove ..." is: That is impossible to prove, because it is not true.






        share|cite|improve this answer












        The claim you want to prove is false -- you've given a counterexample to it yourself.



        So the answer to "How would you prove ..." is: That is impossible to prove, because it is not true.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 26 at 13:18









        Henning Makholm

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