If graph $G$ is connected and has at least two vertices, prove that there exist vertices $u$ and $v$ so...
2
$begingroup$
I would like to know if proof below is correct for this problem.
Let $n$ be the number of vertices of the graph $G$.
As $G$ is connected graph, that means there are at least $n-1$ edges in the graph.
- If $G$ has exactly $n-1$ edges and $n$ vertices, it is a tree. Tree has at least two leaves. Let $u$ and $v$ be those leaves. By removing a vertex of degree 1 from a connected graph, it stays connected. That means $G-u$ and $G-v$ are connected too.
- If $G$ has more than $n-1$ edges, and it is connected, we can construct a spanning tree for all the vertices of the graph $G$. That spanning tree also has at least two leaves. Let $u$ and $v$ be those leaves. These vertices are either leaves in $G$ or are connected with some another vertex in $G$ so they make a cycle. Either way, $G-u$ is connected, so is $G-v$.
Thanks :)
graph-theory
asked Jan 23 at 20:33
HarisHaris
263
$endgroup$
add a comment |
2
$begingroup$
I would like to know if proof below is correct for this problem.
Let $n$ be the number of vertices of the graph $G$.
As $G$ is connected graph, that means there are at least $n-1$ edges in the graph.
- If $G$ has exactly $n-1$ edges and $n$ vertices, it is a tree. Tree has at least two leaves. Let $u$ and $v$ be those leaves. By removing a vertex of degree 1 from a connected graph, it stays connected. That means $G-u$ and $G-v$ are connected too.
- If $G$ has more than $n-1$ edges, and it is connected, we can construct a spanning tree for all the vertices of the graph $G$. That spanning tree also has at least two leaves. Let $u$ and $v$ be those leaves. These vertices are either leaves in $G$ or are connected with some another vertex in $G$ so they make a cycle. Either way, $G-u$ is connected, so is $G-v$.
Thanks :)
graph-theory
asked Jan 23 at 20:33
HarisHaris
263
$endgroup$
$begingroup$
If $V(G)$ is finite, then you can use induction on the number of vertices. Since the set is finite, there is also a "longest path" which you can use to base your argument. The base case should begin with $|V(G)|=2,$ and $G$ is connected, then pick $u$ and $v$ as the only two vertices. Then proceed by an argument about the longest path.
$endgroup$
– Chickenmancer
Jan 23 at 20:42
$begingroup$
Yeah. That seems logical, but it was not obvious to me while I was trying to prove this statement. Thank You! Could you also please say if this proof may be considered correct or not (the proof I posted above)?
$endgroup$
– Haris
Jan 23 at 20:50
$begingroup$
I would just try to explain the final point better.... Notice that it does not matter whether $u$ and $v$ are leaves or not in the original graph. The important part is to say that since $u$ is a leaf in the chosen spanning tree, removing $u$ will not disconect the tree and then this new tree that was formed is a spanning tree of $G-u$, implying it is connected.
$endgroup$
– Daniel
Jan 23 at 20:50
$begingroup$
Oh.. That looks better indeed. Thanks
$endgroup$
– Haris
Jan 23 at 20:53
$begingroup$
Your proof is correct! And you don't have to divide in cases, just take a spanning tree from the beginning (this tree will be everything in the first case).
$endgroup$
– yamete kudasai
Jan 23 at 21:08
add a comment |
2
2
2
$begingroup$
I would like to know if proof below is correct for this problem.
Let $n$ be the number of vertices of the graph $G$.
As $G$ is connected graph, that means there are at least $n-1$ edges in the graph.
- If $G$ has exactly $n-1$ edges and $n$ vertices, it is a tree. Tree has at least two leaves. Let $u$ and $v$ be those leaves. By removing a vertex of degree 1 from a connected graph, it stays connected. That means $G-u$ and $G-v$ are connected too.
- If $G$ has more than $n-1$ edges, and it is connected, we can construct a spanning tree for all the vertices of the graph $G$. That spanning tree also has at least two leaves. Let $u$ and $v$ be those leaves. These vertices are either leaves in $G$ or are connected with some another vertex in $G$ so they make a cycle. Either way, $G-u$ is connected, so is $G-v$.
Thanks :)
graph-theory
asked Jan 23 at 20:33
HarisHaris
263
$endgroup$
I would like to know if proof below is correct for this problem.
Let $n$ be the number of vertices of the graph $G$.
As $G$ is connected graph, that means there are at least $n-1$ edges in the graph.
- If $G$ has exactly $n-1$ edges and $n$ vertices, it is a tree. Tree has at least two leaves. Let $u$ and $v$ be those leaves. By removing a vertex of degree 1 from a connected graph, it stays connected. That means $G-u$ and $G-v$ are connected too.
- If $G$ has more than $n-1$ edges, and it is connected, we can construct a spanning tree for all the vertices of the graph $G$. That spanning tree also has at least two leaves. Let $u$ and $v$ be those leaves. These vertices are either leaves in $G$ or are connected with some another vertex in $G$ so they make a cycle. Either way, $G-u$ is connected, so is $G-v$.
Thanks :)
graph-theory
graph-theory
asked Jan 23 at 20:33
HarisHaris
263
asked Jan 23 at 20:33
HarisHaris
263
edited Jan 23 at 20:39
Haris
asked Jan 23 at 20:33
HarisHaris
263
asked Jan 23 at 20:33
HarisHaris
263
asked Jan 23 at 20:33
HarisHaris
263
263
$begingroup$
If $V(G)$ is finite, then you can use induction on the number of vertices. Since the set is finite, there is also a "longest path" which you can use to base your argument. The base case should begin with $|V(G)|=2,$ and $G$ is connected, then pick $u$ and $v$ as the only two vertices. Then proceed by an argument about the longest path.
$endgroup$
– Chickenmancer
Jan 23 at 20:42
$begingroup$
Yeah. That seems logical, but it was not obvious to me while I was trying to prove this statement. Thank You! Could you also please say if this proof may be considered correct or not (the proof I posted above)?
$endgroup$
– Haris
Jan 23 at 20:50
$begingroup$
I would just try to explain the final point better.... Notice that it does not matter whether $u$ and $v$ are leaves or not in the original graph. The important part is to say that since $u$ is a leaf in the chosen spanning tree, removing $u$ will not disconect the tree and then this new tree that was formed is a spanning tree of $G-u$, implying it is connected.
$endgroup$
– Daniel
Jan 23 at 20:50
$begingroup$
Oh.. That looks better indeed. Thanks
$endgroup$
– Haris
Jan 23 at 20:53
$begingroup$
Your proof is correct! And you don't have to divide in cases, just take a spanning tree from the beginning (this tree will be everything in the first case).
$endgroup$
– yamete kudasai
Jan 23 at 21:08
add a comment |
$begingroup$
If $V(G)$ is finite, then you can use induction on the number of vertices. Since the set is finite, there is also a "longest path" which you can use to base your argument. The base case should begin with $|V(G)|=2,$ and $G$ is connected, then pick $u$ and $v$ as the only two vertices. Then proceed by an argument about the longest path.
$endgroup$
– Chickenmancer
Jan 23 at 20:42
$begingroup$
Yeah. That seems logical, but it was not obvious to me while I was trying to prove this statement. Thank You! Could you also please say if this proof may be considered correct or not (the proof I posted above)?
$endgroup$
– Haris
Jan 23 at 20:50
$begingroup$
I would just try to explain the final point better.... Notice that it does not matter whether $u$ and $v$ are leaves or not in the original graph. The important part is to say that since $u$ is a leaf in the chosen spanning tree, removing $u$ will not disconect the tree and then this new tree that was formed is a spanning tree of $G-u$, implying it is connected.
$endgroup$
– Daniel
Jan 23 at 20:50
$begingroup$
Oh.. That looks better indeed. Thanks
$endgroup$
– Haris
Jan 23 at 20:53
$begingroup$
Your proof is correct! And you don't have to divide in cases, just take a spanning tree from the beginning (this tree will be everything in the first case).
$endgroup$
– yamete kudasai
Jan 23 at 21:08
$begingroup$
If $V(G)$ is finite, then you can use induction on the number of vertices. Since the set is finite, there is also a "longest path" which you can use to base your argument. The base case should begin with $|V(G)|=2,$ and $G$ is connected, then pick $u$ and $v$ as the only two vertices. Then proceed by an argument about the longest path.
$endgroup$
– Chickenmancer
Jan 23 at 20:42
$begingroup$
If $V(G)$ is finite, then you can use induction on the number of vertices. Since the set is finite, there is also a "longest path" which you can use to base your argument. The base case should begin with $|V(G)|=2,$ and $G$ is connected, then pick $u$ and $v$ as the only two vertices. Then proceed by an argument about the longest path.
$endgroup$
– Chickenmancer
Jan 23 at 20:42
$begingroup$
Yeah. That seems logical, but it was not obvious to me while I was trying to prove this statement. Thank You! Could you also please say if this proof may be considered correct or not (the proof I posted above)?
$endgroup$
– Haris
Jan 23 at 20:50
$begingroup$
Yeah. That seems logical, but it was not obvious to me while I was trying to prove this statement. Thank You! Could you also please say if this proof may be considered correct or not (the proof I posted above)?
$endgroup$
– Haris
Jan 23 at 20:50
$begingroup$
I would just try to explain the final point better.... Notice that it does not matter whether $u$ and $v$ are leaves or not in the original graph. The important part is to say that since $u$ is a leaf in the chosen spanning tree, removing $u$ will not disconect the tree and then this new tree that was formed is a spanning tree of $G-u$, implying it is connected.
$endgroup$
– Daniel
Jan 23 at 20:50
$begingroup$
I would just try to explain the final point better.... Notice that it does not matter whether $u$ and $v$ are leaves or not in the original graph. The important part is to say that since $u$ is a leaf in the chosen spanning tree, removing $u$ will not disconect the tree and then this new tree that was formed is a spanning tree of $G-u$, implying it is connected.
$endgroup$
– Daniel
Jan 23 at 20:50
$begingroup$
Oh.. That looks better indeed. Thanks
$endgroup$
– Haris
Jan 23 at 20:53
$begingroup$
Oh.. That looks better indeed. Thanks
$endgroup$
– Haris
Jan 23 at 20:53
$begingroup$
Your proof is correct! And you don't have to divide in cases, just take a spanning tree from the beginning (this tree will be everything in the first case).
$endgroup$
– yamete kudasai
Jan 23 at 21:08
$begingroup$
Your proof is correct! And you don't have to divide in cases, just take a spanning tree from the beginning (this tree will be everything in the first case).
$endgroup$
– yamete kudasai
Jan 23 at 21:08
add a comment |
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$begingroup$
If $V(G)$ is finite, then you can use induction on the number of vertices. Since the set is finite, there is also a "longest path" which you can use to base your argument. The base case should begin with $|V(G)|=2,$ and $G$ is connected, then pick $u$ and $v$ as the only two vertices. Then proceed by an argument about the longest path.
$endgroup$
– Chickenmancer
Jan 23 at 20:42
$begingroup$
Yeah. That seems logical, but it was not obvious to me while I was trying to prove this statement. Thank You! Could you also please say if this proof may be considered correct or not (the proof I posted above)?
$endgroup$
– Haris
Jan 23 at 20:50
$begingroup$
I would just try to explain the final point better.... Notice that it does not matter whether $u$ and $v$ are leaves or not in the original graph. The important part is to say that since $u$ is a leaf in the chosen spanning tree, removing $u$ will not disconect the tree and then this new tree that was formed is a spanning tree of $G-u$, implying it is connected.
$endgroup$
– Daniel
Jan 23 at 20:50
$begingroup$
Oh.. That looks better indeed. Thanks
$endgroup$
– Haris
Jan 23 at 20:53
$begingroup$
Your proof is correct! And you don't have to divide in cases, just take a spanning tree from the beginning (this tree will be everything in the first case).
$endgroup$
– yamete kudasai
Jan 23 at 21:08