# Graphs Problem 5

### Buggy Solution

Here is a partly-correct solution to the problem. It's on the right track but has a few bugs. See how it compares to your draft solution.

We must show that R is reflexive, symmetric, and transitive.

#### Reflexive

Let $$u \in V$$. Then uRu because there is a length zero path from u to u containing no edges.

#### Symmetric

Let $$u,v \in V$$. Suppose uRv, so there is a path P from u to v. To get a path from v to u we simply reverse the path P.

#### Transitive

Let $$u,v,w \in V$$. Suppose uRv and vRw. Then there is a path P from u to v, and another path Q from v to w. We join P and Q together at their common endpoint making a path from u to w.

Thus R is an equivalence relation.

The symmetric part has the right idea, but you could supply more specifics. Suppose the path from u to v has the edge sequence $$e_1,e_2,\ldots,e_n$$. Then what is the edge sequence for the path from v to u? This kind of detail isn't always necessary, but this is a good place to practice producing it when required.