Proof of Mantel's Theorem
$begingroup$
There is already a proof verification question on the site about Mantel's theorem, but the other proof looks very different to mine, uses Cauchy-Schwarz, etc.
First of all, the theorem:
Theorem (Mantel's Theorem). A graph $G$ is maximally triangle-free with respect to edges only if $$m = leftlfloorfrac{n^2}{4}rightrfloor.$$
Now my proof is by induction on the number of vertices. The idea is to take advantage of the fact that the desired graph is $K_{lfloor n/2rfloor,lceil n/2rceil}$, which we actually know from Turán's theorem.
Proof. Suppose $G$ is an $n$-vertex graph on $m$ edges that contains no triangles. Then any two adjacent vertices $u$ and $v$ cannot share a common neighbour, so $N(u) cap N(v) = varnothing$. Thus
begin{align*}
deg(u) + deg(v) = |N(u)| + |N(v)| &= |N(u) cup N(v)| - |N(u) cap N(v)|\
&leqslant |V(G)| - |varnothing| = n.
end{align*}
Now remove two adjacent vertices $u$ and $v$ from $G$ to get $G'$. By induction, this has $lfloor (n-2)^2/4rfloor$ edges. Thus
begin{align*}
m &= leftlfloorfrac{(n-2)^2}{4}rightrfloor + deg(u) + deg(v) - 1 qquadtext{(since $deg(u) + deg(v)$ counts ${u,v}$ twice)}\
&leqslant frac{(n-2)^2}{4} + n - 1 = frac{n^2}{4}.
end{align*}
It suffices to show that a triangle-free graph on $n$ vertices and $lfloor n^2/4rfloor$ edges always exists, since it must have maximal edges by the inequality $m leqslant n^2/4$ we have just obtained. Indeed, the complete bipartite graph $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no cycles and in particular no triangles, is on $lfloor n/2rfloor + lceil n/2rceil = n$ vertices and has
$$leftlfloorfrac{n}{2}rightrfloorleftlceilfrac{n}{2}rightrceil = begin{cases}
displaystyleleft(frac{n}{2}right)left(frac{n}{2}right) = frac{n^2}{4} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is even}\[15pt]
displaystyleleft(frac{n-1}{2}right)left(frac{n+1}{2}right) = frac{n^2 - 1}{2} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is odd}
end{cases} $$
edges, as required. $square$
I'd appreciate any feedback about this proof.
proof-verification graph-theory
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
$endgroup$
add a comment |
$begingroup$
There is already a proof verification question on the site about Mantel's theorem, but the other proof looks very different to mine, uses Cauchy-Schwarz, etc.
First of all, the theorem:
Theorem (Mantel's Theorem). A graph $G$ is maximally triangle-free with respect to edges only if $$m = leftlfloorfrac{n^2}{4}rightrfloor.$$
Now my proof is by induction on the number of vertices. The idea is to take advantage of the fact that the desired graph is $K_{lfloor n/2rfloor,lceil n/2rceil}$, which we actually know from Turán's theorem.
Proof. Suppose $G$ is an $n$-vertex graph on $m$ edges that contains no triangles. Then any two adjacent vertices $u$ and $v$ cannot share a common neighbour, so $N(u) cap N(v) = varnothing$. Thus
begin{align*}
deg(u) + deg(v) = |N(u)| + |N(v)| &= |N(u) cup N(v)| - |N(u) cap N(v)|\
&leqslant |V(G)| - |varnothing| = n.
end{align*}
Now remove two adjacent vertices $u$ and $v$ from $G$ to get $G'$. By induction, this has $lfloor (n-2)^2/4rfloor$ edges. Thus
begin{align*}
m &= leftlfloorfrac{(n-2)^2}{4}rightrfloor + deg(u) + deg(v) - 1 qquadtext{(since $deg(u) + deg(v)$ counts ${u,v}$ twice)}\
&leqslant frac{(n-2)^2}{4} + n - 1 = frac{n^2}{4}.
end{align*}
It suffices to show that a triangle-free graph on $n$ vertices and $lfloor n^2/4rfloor$ edges always exists, since it must have maximal edges by the inequality $m leqslant n^2/4$ we have just obtained. Indeed, the complete bipartite graph $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no cycles and in particular no triangles, is on $lfloor n/2rfloor + lceil n/2rceil = n$ vertices and has
$$leftlfloorfrac{n}{2}rightrfloorleftlceilfrac{n}{2}rightrceil = begin{cases}
displaystyleleft(frac{n}{2}right)left(frac{n}{2}right) = frac{n^2}{4} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is even}\[15pt]
displaystyleleft(frac{n-1}{2}right)left(frac{n+1}{2}right) = frac{n^2 - 1}{2} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is odd}
end{cases} $$
edges, as required. $square$
I'd appreciate any feedback about this proof.
proof-verification graph-theory
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
$endgroup$
$begingroup$
It's not immediately clear what "maximally triangle-free with respect to edges" means. From the context, I know you mean that the number of edges attains the maximum over all triangle-free graphs on the same number of vertices. Under another plausible interpretation of the words "maximally triangle-free", the graph $C_5$ is maximally triangle-free, since you can't add an edge to it without creating a triangle.
$endgroup$
– bof
Dec 4 '18 at 0:28
$begingroup$
By $|N(u)cup N(v)|-|N(u)cap N(v)|$ you mean $|N(u)cup N(v)|+|N(u)cap N(v)|$.
$endgroup$
– bof
Dec 4 '18 at 0:32
$begingroup$
You mean $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no odd cycles
$endgroup$
– hbm
Dec 4 '18 at 23:09
add a comment |
$begingroup$
There is already a proof verification question on the site about Mantel's theorem, but the other proof looks very different to mine, uses Cauchy-Schwarz, etc.
First of all, the theorem:
Theorem (Mantel's Theorem). A graph $G$ is maximally triangle-free with respect to edges only if $$m = leftlfloorfrac{n^2}{4}rightrfloor.$$
Now my proof is by induction on the number of vertices. The idea is to take advantage of the fact that the desired graph is $K_{lfloor n/2rfloor,lceil n/2rceil}$, which we actually know from Turán's theorem.
Proof. Suppose $G$ is an $n$-vertex graph on $m$ edges that contains no triangles. Then any two adjacent vertices $u$ and $v$ cannot share a common neighbour, so $N(u) cap N(v) = varnothing$. Thus
begin{align*}
deg(u) + deg(v) = |N(u)| + |N(v)| &= |N(u) cup N(v)| - |N(u) cap N(v)|\
&leqslant |V(G)| - |varnothing| = n.
end{align*}
Now remove two adjacent vertices $u$ and $v$ from $G$ to get $G'$. By induction, this has $lfloor (n-2)^2/4rfloor$ edges. Thus
begin{align*}
m &= leftlfloorfrac{(n-2)^2}{4}rightrfloor + deg(u) + deg(v) - 1 qquadtext{(since $deg(u) + deg(v)$ counts ${u,v}$ twice)}\
&leqslant frac{(n-2)^2}{4} + n - 1 = frac{n^2}{4}.
end{align*}
It suffices to show that a triangle-free graph on $n$ vertices and $lfloor n^2/4rfloor$ edges always exists, since it must have maximal edges by the inequality $m leqslant n^2/4$ we have just obtained. Indeed, the complete bipartite graph $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no cycles and in particular no triangles, is on $lfloor n/2rfloor + lceil n/2rceil = n$ vertices and has
$$leftlfloorfrac{n}{2}rightrfloorleftlceilfrac{n}{2}rightrceil = begin{cases}
displaystyleleft(frac{n}{2}right)left(frac{n}{2}right) = frac{n^2}{4} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is even}\[15pt]
displaystyleleft(frac{n-1}{2}right)left(frac{n+1}{2}right) = frac{n^2 - 1}{2} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is odd}
end{cases} $$
edges, as required. $square$
I'd appreciate any feedback about this proof.
proof-verification graph-theory
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
$endgroup$
There is already a proof verification question on the site about Mantel's theorem, but the other proof looks very different to mine, uses Cauchy-Schwarz, etc.
First of all, the theorem:
Theorem (Mantel's Theorem). A graph $G$ is maximally triangle-free with respect to edges only if $$m = leftlfloorfrac{n^2}{4}rightrfloor.$$
Now my proof is by induction on the number of vertices. The idea is to take advantage of the fact that the desired graph is $K_{lfloor n/2rfloor,lceil n/2rceil}$, which we actually know from Turán's theorem.
Proof. Suppose $G$ is an $n$-vertex graph on $m$ edges that contains no triangles. Then any two adjacent vertices $u$ and $v$ cannot share a common neighbour, so $N(u) cap N(v) = varnothing$. Thus
begin{align*}
deg(u) + deg(v) = |N(u)| + |N(v)| &= |N(u) cup N(v)| - |N(u) cap N(v)|\
&leqslant |V(G)| - |varnothing| = n.
end{align*}
Now remove two adjacent vertices $u$ and $v$ from $G$ to get $G'$. By induction, this has $lfloor (n-2)^2/4rfloor$ edges. Thus
begin{align*}
m &= leftlfloorfrac{(n-2)^2}{4}rightrfloor + deg(u) + deg(v) - 1 qquadtext{(since $deg(u) + deg(v)$ counts ${u,v}$ twice)}\
&leqslant frac{(n-2)^2}{4} + n - 1 = frac{n^2}{4}.
end{align*}
It suffices to show that a triangle-free graph on $n$ vertices and $lfloor n^2/4rfloor$ edges always exists, since it must have maximal edges by the inequality $m leqslant n^2/4$ we have just obtained. Indeed, the complete bipartite graph $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no cycles and in particular no triangles, is on $lfloor n/2rfloor + lceil n/2rceil = n$ vertices and has
$$leftlfloorfrac{n}{2}rightrfloorleftlceilfrac{n}{2}rightrceil = begin{cases}
displaystyleleft(frac{n}{2}right)left(frac{n}{2}right) = frac{n^2}{4} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is even}\[15pt]
displaystyleleft(frac{n-1}{2}right)left(frac{n+1}{2}right) = frac{n^2 - 1}{2} = leftlfloor frac{n^2}{4} rightrfloor & text{if $n$ is odd}
end{cases} $$
edges, as required. $square$
I'd appreciate any feedback about this proof.
proof-verification graph-theory
proof-verification graph-theory
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
asked Dec 3 '18 at 23:57
Luke CollinsLuke Collins
725417
725417
$begingroup$
It's not immediately clear what "maximally triangle-free with respect to edges" means. From the context, I know you mean that the number of edges attains the maximum over all triangle-free graphs on the same number of vertices. Under another plausible interpretation of the words "maximally triangle-free", the graph $C_5$ is maximally triangle-free, since you can't add an edge to it without creating a triangle.
$endgroup$
– bof
Dec 4 '18 at 0:28
$begingroup$
By $|N(u)cup N(v)|-|N(u)cap N(v)|$ you mean $|N(u)cup N(v)|+|N(u)cap N(v)|$.
$endgroup$
– bof
Dec 4 '18 at 0:32
$begingroup$
You mean $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no odd cycles
$endgroup$
– hbm
Dec 4 '18 at 23:09
add a comment |
$begingroup$
It's not immediately clear what "maximally triangle-free with respect to edges" means. From the context, I know you mean that the number of edges attains the maximum over all triangle-free graphs on the same number of vertices. Under another plausible interpretation of the words "maximally triangle-free", the graph $C_5$ is maximally triangle-free, since you can't add an edge to it without creating a triangle.
$endgroup$
– bof
Dec 4 '18 at 0:28
$begingroup$
By $|N(u)cup N(v)|-|N(u)cap N(v)|$ you mean $|N(u)cup N(v)|+|N(u)cap N(v)|$.
$endgroup$
– bof
Dec 4 '18 at 0:32
$begingroup$
You mean $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no odd cycles
$endgroup$
– hbm
Dec 4 '18 at 23:09
$begingroup$
It's not immediately clear what "maximally triangle-free with respect to edges" means. From the context, I know you mean that the number of edges attains the maximum over all triangle-free graphs on the same number of vertices. Under another plausible interpretation of the words "maximally triangle-free", the graph $C_5$ is maximally triangle-free, since you can't add an edge to it without creating a triangle.
$endgroup$
– bof
Dec 4 '18 at 0:28
$begingroup$
It's not immediately clear what "maximally triangle-free with respect to edges" means. From the context, I know you mean that the number of edges attains the maximum over all triangle-free graphs on the same number of vertices. Under another plausible interpretation of the words "maximally triangle-free", the graph $C_5$ is maximally triangle-free, since you can't add an edge to it without creating a triangle.
$endgroup$
– bof
Dec 4 '18 at 0:28
$begingroup$
By $|N(u)cup N(v)|-|N(u)cap N(v)|$ you mean $|N(u)cup N(v)|+|N(u)cap N(v)|$.
$endgroup$
– bof
Dec 4 '18 at 0:32
$begingroup$
By $|N(u)cup N(v)|-|N(u)cap N(v)|$ you mean $|N(u)cup N(v)|+|N(u)cap N(v)|$.
$endgroup$
– bof
Dec 4 '18 at 0:32
$begingroup$
You mean $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no odd cycles
$endgroup$
– hbm
Dec 4 '18 at 23:09
$begingroup$
You mean $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no odd cycles
$endgroup$
– hbm
Dec 4 '18 at 23:09
add a comment |
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$begingroup$
It's not immediately clear what "maximally triangle-free with respect to edges" means. From the context, I know you mean that the number of edges attains the maximum over all triangle-free graphs on the same number of vertices. Under another plausible interpretation of the words "maximally triangle-free", the graph $C_5$ is maximally triangle-free, since you can't add an edge to it without creating a triangle.
$endgroup$
– bof
Dec 4 '18 at 0:28
$begingroup$
By $|N(u)cup N(v)|-|N(u)cap N(v)|$ you mean $|N(u)cup N(v)|+|N(u)cap N(v)|$.
$endgroup$
– bof
Dec 4 '18 at 0:32
$begingroup$
You mean $K_{lfloor n/2rfloor, lceil n/2rceil}$ has no odd cycles
$endgroup$
– hbm
Dec 4 '18 at 23:09