chivistaelmore

Answered

2022-07-18

Let G be a graph of two or more vertices. Prove by contradiction that
(All u,v vertices in G, $N\left(u\right)=N\left(v\right)$)
(All x vertex in G, $N\left(x\right)=$ {})
(Hint: N(u) denotes the neighbourhood of vertex u; i.e. N(u) is the set of every vertex v such that u and v are neighbours.)
Solution:
(All u,v vertices in G, $N\left(u\right)=N\left(v\right)$) and (Exist x, y vertices in G, $N\left(x\right)!=$ {} and so y is in N(x))
$=>\left\{N\left(x\right)=N\left(y\right)\right\}$
(Exist y vertex in G, y in N(y))
{from definition of neighbourhood: (All z vertex in G, not (z in N(z))}
F
I'm confused. It looks like the problem is telling me that each vertex in G shares the neighbourhood with each other vertex, that the graph is complete. From that I have to prove that for all vertexes in G, the neighbourhood is empty? I'm very confused, could someone explain the question better and walk me through the process?

Answer & Explanation

kitskjeja

Expert

2022-07-19Added 13 answers

Step 1
The crux of the contradiction is that N(x) does not contain x.
If x and y are neighbors, then $x\in N\left(y\right)$ and $y\in N\left(x\right)$. Therefore it is impossible for $N\left(x\right)=N\left(y\right)$ to hold, because x is N(y) but not in N(x).
Step 2
Phrased differently, if $N\left(u\right)=N\left(v\right)$ for any pair of vertices u,v, then no pair of vertices can be neighbors.

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