# Prove that a simple graph with 17 vertices and 73 edges cannot be bipartite

Prove that a simple graph with 17 vertices and 73 edges cannot be bipartite.

-I asked my professor for help on this and his hint was to break the graph up into two vertex sets and count the number of edges within each vertex set.

-We did several examples in class involving chromatic graphs that involved 6 vertices just to have basic examples, I do understand what bipartite means and how such a graph would look like although I don't understand how I would necessarily prove that the given graph isn't bipartite. Seems clear that I would use contradiction, just not sure how. Any help is appreciated.

• Hint: $1\times16=16,\ 2\times15=30,\ 3\times14=42,\ 4\times13=52,\ 5\times12=60,\ 6\times11=66,\ 7\times10=70,\ 8\times9=72,\ 9\times8=72,\ 10\times7=70,\ 11\times6=66,\ 12\times5=60,\ 13\times4=52,\ 14\times3=42,\ 15\times2=30,\ 16\times1=16.$
– bof
Commented Oct 23, 2015 at 1:24

If a bipartite graph has parts of size $m$ and $n$, it has at most $mn$ edges.
What are the possibilities if $m + n = 17$? In particular, how large can $mn$ be?
It can be found out that the maximum value of $$f(n)= n(17-n)$$ using calculus comes out to be $$72$$ which is less than $$73$$ i.e. the largest value of $$mn$$(when $$m+n=17$$) is $$72$$ and hence a simple graph on $$17$$ vertices and $$73$$ vertices cannot be bipartite.