How to prove that a connected graph with $|V| -1= |E|$ is a tree?












2














I could neither show myself nor find a proof of the following result: if $G=(V,E)$ is a connected graph with $|E|=|V|-1$ then $G$ is a tree.



Could somebody please provide an argument to establish this.






combinatorics graph-theory trees







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    I could neither show myself nor find a proof of the following result: if $G=(V,E)$ is a connected graph with $|E|=|V|-1$ then $G$ is a tree.



    Could somebody please provide an argument to establish this.






    combinatorics graph-theory trees







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      2












      2








      2







      I could neither show myself nor find a proof of the following result: if $G=(V,E)$ is a connected graph with $|E|=|V|-1$ then $G$ is a tree.



      Could somebody please provide an argument to establish this.






      combinatorics graph-theory trees







      share|cite|improve this question















      I could neither show myself nor find a proof of the following result: if $G=(V,E)$ is a connected graph with $|E|=|V|-1$ then $G$ is a tree.



      Could somebody please provide an argument to establish this.






      combinatorics graph-theory trees




      combinatorics graph-theory trees






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      edited Nov 22 '15 at 20:11









      quid♦

      36.9k95093




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      asked Nov 22 '15 at 19:47









      Serkan KlvzSerkan Klvz

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          An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).



          Suppose a connected graph on $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.






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            Suppose $|V| ge 2$. Recall that the sum of the degrees of all vertices is $2|E|$. Since The graph is connected there cannot be a vertex of degree $0$.



            Thus as $2|V|> 2|E|$ there is a vertex of degree one. Removing that vertex and the adjacent edge does not change that the graph is connected. And if the graph after removal is a tree, so is the graph itself.



            Using this starting idea you can set up a proof by induction.






            share|cite|improve this answer





























              -1














              Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
              Therefore the graph must not contain cycles, and then it is a tree by definition.






              share|cite|improve this answer























              • Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                – platty
                Nov 30 '18 at 2:33










              • @platty: Thank you for informing. I have already fixed my mathematical notations.
                – Quan Nguyen
                Dec 4 '18 at 10:01











              Your Answer





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              3 Answers
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              3 Answers
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              active

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              active

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              3














              An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).



              Suppose a connected graph on $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.






              share|cite|improve this answer


























                3














                An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).



                Suppose a connected graph on $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.






                share|cite|improve this answer
























                  3












                  3








                  3






                  An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).



                  Suppose a connected graph on $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.






                  share|cite|improve this answer












                  An empty graph on $n$ vertices has $n$ connected components, Suppose you have a graph and add an edge, then the number of connected components is reduced by at most one ( since this edge touches at most two connected components). Therefore a connected graph on $n$ vertices has at least $n-1$ edges).



                  Suppose a connected graph on $n$ vertices has $n-1$ edges, we must prove no cycle exists, suppose it does, if we remove an edge from the cycle we get a connected graph (because if we have a path that uses this edge, instead of using the edge we can go around the remaining part of the cycle). This is a contradiction, since the graph would have $n-2$ edges, and would hence not be connected.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 22 '15 at 21:17









                  Jorge FernándezJorge Fernández

                  75.1k1190191




                  75.1k1190191























                      2














                      Suppose $|V| ge 2$. Recall that the sum of the degrees of all vertices is $2|E|$. Since The graph is connected there cannot be a vertex of degree $0$.



                      Thus as $2|V|> 2|E|$ there is a vertex of degree one. Removing that vertex and the adjacent edge does not change that the graph is connected. And if the graph after removal is a tree, so is the graph itself.



                      Using this starting idea you can set up a proof by induction.






                      share|cite|improve this answer


























                        2














                        Suppose $|V| ge 2$. Recall that the sum of the degrees of all vertices is $2|E|$. Since The graph is connected there cannot be a vertex of degree $0$.



                        Thus as $2|V|> 2|E|$ there is a vertex of degree one. Removing that vertex and the adjacent edge does not change that the graph is connected. And if the graph after removal is a tree, so is the graph itself.



                        Using this starting idea you can set up a proof by induction.






                        share|cite|improve this answer
























                          2












                          2








                          2






                          Suppose $|V| ge 2$. Recall that the sum of the degrees of all vertices is $2|E|$. Since The graph is connected there cannot be a vertex of degree $0$.



                          Thus as $2|V|> 2|E|$ there is a vertex of degree one. Removing that vertex and the adjacent edge does not change that the graph is connected. And if the graph after removal is a tree, so is the graph itself.



                          Using this starting idea you can set up a proof by induction.






                          share|cite|improve this answer












                          Suppose $|V| ge 2$. Recall that the sum of the degrees of all vertices is $2|E|$. Since The graph is connected there cannot be a vertex of degree $0$.



                          Thus as $2|V|> 2|E|$ there is a vertex of degree one. Removing that vertex and the adjacent edge does not change that the graph is connected. And if the graph after removal is a tree, so is the graph itself.



                          Using this starting idea you can set up a proof by induction.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 22 '15 at 20:08









                          quid♦quid

                          36.9k95093




                          36.9k95093























                              -1














                              Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
                              Therefore the graph must not contain cycles, and then it is a tree by definition.






                              share|cite|improve this answer























                              • Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                                – platty
                                Nov 30 '18 at 2:33










                              • @platty: Thank you for informing. I have already fixed my mathematical notations.
                                – Quan Nguyen
                                Dec 4 '18 at 10:01
















                              -1














                              Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
                              Therefore the graph must not contain cycles, and then it is a tree by definition.






                              share|cite|improve this answer























                              • Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                                – platty
                                Nov 30 '18 at 2:33










                              • @platty: Thank you for informing. I have already fixed my mathematical notations.
                                – Quan Nguyen
                                Dec 4 '18 at 10:01














                              -1












                              -1








                              -1






                              Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
                              Therefore the graph must not contain cycles, and then it is a tree by definition.






                              share|cite|improve this answer














                              Suppose the graph contains cycles, then we can remove an edge without removing a vertex and the graph is still connected. After removing, it will has $n$ vertices and $n - 2$ edges. Continue to do so until we reach a tree (because there is no cycle) with $n$ vertices and $n - k$ edges $(k > 1)$. However it contradicts with the fact that a tree with $n$ vertices must has $n - 1$ edges.
                              Therefore the graph must not contain cycles, and then it is a tree by definition.







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited Dec 4 '18 at 10:00

























                              answered Nov 30 '18 at 1:52









                              Quan NguyenQuan Nguyen

                              11




                              11












                              • Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                                – platty
                                Nov 30 '18 at 2:33










                              • @platty: Thank you for informing. I have already fixed my mathematical notations.
                                – Quan Nguyen
                                Dec 4 '18 at 10:01


















                              • Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                                – platty
                                Nov 30 '18 at 2:33










                              • @platty: Thank you for informing. I have already fixed my mathematical notations.
                                – Quan Nguyen
                                Dec 4 '18 at 10:01
















                              Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                              – platty
                              Nov 30 '18 at 2:33




                              Welcome to MSE! Please format questions and answers using MathJax for mathematical notation. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
                              – platty
                              Nov 30 '18 at 2:33












                              @platty: Thank you for informing. I have already fixed my mathematical notations.
                              – Quan Nguyen
                              Dec 4 '18 at 10:01




                              @platty: Thank you for informing. I have already fixed my mathematical notations.
                              – Quan Nguyen
                              Dec 4 '18 at 10:01


















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