Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.












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Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.




I attempted to prove this by contradiction but didn't make any progress.






graph-theory polyhedra







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  • $begingroup$
    Each edge borders exactly two faces.
    $endgroup$
    – Lord Shark the Unknown
    Nov 27 '18 at 3:35
















0












$begingroup$



Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.




I attempted to prove this by contradiction but didn't make any progress.






graph-theory polyhedra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Each edge borders exactly two faces.
    $endgroup$
    – Lord Shark the Unknown
    Nov 27 '18 at 3:35














0












0








0





$begingroup$



Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.




I attempted to prove this by contradiction but didn't make any progress.






graph-theory polyhedra







share|cite|improve this question









$endgroup$





Prove that in any convex polyhedron, the number of faces that have an odd number of edges is even.




I attempted to prove this by contradiction but didn't make any progress.






graph-theory polyhedra




graph-theory polyhedra






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asked Nov 27 '18 at 3:31









thetravellerthetraveller

1515




1515












  • $begingroup$
    Each edge borders exactly two faces.
    $endgroup$
    – Lord Shark the Unknown
    Nov 27 '18 at 3:35


















  • $begingroup$
    Each edge borders exactly two faces.
    $endgroup$
    – Lord Shark the Unknown
    Nov 27 '18 at 3:35
















$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35




$begingroup$
Each edge borders exactly two faces.
$endgroup$
– Lord Shark the Unknown
Nov 27 '18 at 3:35










3 Answers
3






active

oldest

votes


















1












$begingroup$

Create a graph in the following way:




  • Vertices are faces of the polyhedron

  • Two vertices are connected if their corresponding faces share an edge


Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
    where face $F_i$ has $n_i$ edges.
    This sum counts each edge twice: $S$ is even. So the number of $j$ for
    which $n_j$ is odd must be even.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
      $endgroup$
      – Dr. Richard Klitzing
      Nov 27 '18 at 19:47



















    1












    $begingroup$

    I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.




    Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$




    There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

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      active

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      active

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      1












      $begingroup$

      Create a graph in the following way:




      • Vertices are faces of the polyhedron

      • Two vertices are connected if their corresponding faces share an edge


      Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Create a graph in the following way:




        • Vertices are faces of the polyhedron

        • Two vertices are connected if their corresponding faces share an edge


        Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Create a graph in the following way:




          • Vertices are faces of the polyhedron

          • Two vertices are connected if their corresponding faces share an edge


          Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.






          share|cite|improve this answer









          $endgroup$



          Create a graph in the following way:




          • Vertices are faces of the polyhedron

          • Two vertices are connected if their corresponding faces share an edge


          Observe that the odd-edged faces are exactly the vertices of odd degree in our graph. We know that the number of vertices of odd degree in any graph $G$ is even.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 '18 at 3:47









          Santana AftonSantana Afton

          2,6192629




          2,6192629























              4












              $begingroup$

              Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
              where face $F_i$ has $n_i$ edges.
              This sum counts each edge twice: $S$ is even. So the number of $j$ for
              which $n_j$ is odd must be even.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
                $endgroup$
                – Dr. Richard Klitzing
                Nov 27 '18 at 19:47
















              4












              $begingroup$

              Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
              where face $F_i$ has $n_i$ edges.
              This sum counts each edge twice: $S$ is even. So the number of $j$ for
              which $n_j$ is odd must be even.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
                $endgroup$
                – Dr. Richard Klitzing
                Nov 27 '18 at 19:47














              4












              4








              4





              $begingroup$

              Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
              where face $F_i$ has $n_i$ edges.
              This sum counts each edge twice: $S$ is even. So the number of $j$ for
              which $n_j$ is odd must be even.






              share|cite|improve this answer









              $endgroup$



              Let the faces be $F_1,ldots,F_k$. Consider the sum $S=n_1+cdots+n_k$
              where face $F_i$ has $n_i$ edges.
              This sum counts each edge twice: $S$ is even. So the number of $j$ for
              which $n_j$ is odd must be even.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 27 '18 at 4:42









              Lord Shark the UnknownLord Shark the Unknown

              103k1160132




              103k1160132












              • $begingroup$
                well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
                $endgroup$
                – Dr. Richard Klitzing
                Nov 27 '18 at 19:47


















              • $begingroup$
                well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
                $endgroup$
                – Dr. Richard Klitzing
                Nov 27 '18 at 19:47
















              $begingroup$
              well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
              $endgroup$
              – Dr. Richard Klitzing
              Nov 27 '18 at 19:47




              $begingroup$
              well @Lord, this argument alone only holds for 1 type of faces, or for 1 odd-polygonal face type, while all others are even-polygonal. But why this rules out say 7 triangles plus 3 pentagons?
              $endgroup$
              – Dr. Richard Klitzing
              Nov 27 '18 at 19:47











              1












              $begingroup$

              I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.




              Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$




              There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.




                Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$




                There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.




                  Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$




                  There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.






                  share|cite|improve this answer









                  $endgroup$



                  I'll elaborate a bit more, which'll helpfully help anybody interested get off on the right foot.




                  Let $G$ be a graph depicting the planar representation of a convex polyhedron. Let $F in Bbb Z^{+}$ denote the total number of faces in $G$ and $f_{1},f_{2},...,f_{F}in Bbb Z^{+}$ denote the respective number of edges of each of the $F$ faces of $G$. Recall $$sum_{i=1}^F f_{i} =2e,$$ where $ein Bbb Z^{+}$ denotes the total of number of edges of $G$. Hence $$f_{1}+f_{2}+...+f_{F}=2e.$$




                  There we have it. You just use some basic properties of number theory/ arithmetic/ algebra to deduce that there must be an even number of odd $f$'s so that ultimately their sum is an even number. For example, note that $2+3+5=10$ is an even number, but $2+3+5+7=17$ is not.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 '18 at 3:38









                  greycatbirdgreycatbird

                  1377




                  1377






























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