Reputation
29,397
Next tag badge:
366/400 score
265/80 answers
Badges
3 38 88
Impact
~297k people reached

3h
revised Prove that the graph dual to Eulerian planar graph is bipartite.
added 215 characters in body
10h
comment Prove that the graph dual to Eulerian planar graph is bipartite.
$G''$ is the dual of the dual, it is isomorphic to $G$. It id the second property con the wikipedia page about the dual.en.wikipedia.org/wiki/Dual_graph#Properties
10h
answered Prove that the graph dual to Eulerian planar graph is bipartite.
11h
comment Prove that the graph dual to Eulerian planar graph is bipartite.
It's actually an if and only if. The other direction is simpler. If I recall correctly you need to use the dual of the dual of $G$ is isomorphic to $G$:
13h
comment How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
Yeah, $\lim_{n\to\infty}\frac{n}{n+1}=1$ :)
13h
revised How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
edited body
13h
comment How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
Oh ok I got it. it should be an $i$ instead of an $n$. Thanks
13h
comment How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
$1/2+1/3+1/4+1/5+1/6>1$, I don't see how that claim is possible.
15h
revised How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
added 53 characters in body
15h
revised How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
added 53 characters in body
16h
answered How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
16h
comment How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.
The series? so you are going to add them? or the sequence?
16h
comment How many surjective functions $f: X \to \{1,…,j\}$?
so $k!$ multiplied by the number of paritions of a set $X$ with $kj$ elements into $k$ equal parts
1d
answered What the difference between the smallest two numbers from these numbers?
2d
comment Find the smallest natural number $n$
Yeah, I got that now CiaPan, I think that's what it is.
2d
revised Find the smallest natural number $n$
added 32 characters in body
2d
revised Find the smallest natural number $n$
deleted 27 characters in body
2d
revised Find the smallest natural number $n$
edited body
2d
comment Does there exist a graph $G$ such that every edge is contained in a unique Hamiltonian circuit, that is not a cycle graph?
Now that I think about it if such a graph doesn't exist for two it doesn't exist for more so it is equivalent in that sense
2d
comment Does there exist a graph $G$ such that every edge is contained in a unique Hamiltonian circuit, that is not a cycle graph?
Henning, I don't see why there needs to be exactly two hamiltonian circuits, I do see however why all the hamiltonian circuits need to be disjoint and how the graph needs to be a disjoint union of hamiltonian circuits.