Thread: Simple Graphs and Paths

This might seem easy, but im having trouble putting my proofs into words.

Let G = (V,E) be simple graph,

I need to prove for all vertices (x,y,z) of G, the follwoing are true:

1) There is a path from x to x.
2) If there is a path from x to y then there is path y -x.
3)If there is a path from x- y, and y-z then there is path x-z.

Now im not sure whats expected when proofing this.

Do i just say a path from x to x exists since its a path to itself?
and...
if a path x-y exists then there is a path y- x because x and y are two connected vertices by a edge?
and..
if there is a path from x-y, and y-z then there is a path x-z since vertices x and y are connected by an edge, and y-z is connected by an edge, this provides a path from vertices x to z through vertices y.

Id like some feedback if possible thanks.

Just work from the definitions and find a path that satisfies the conditions.

1: You can take the vertice sequence {x} which starts and stops at x and exists by axiom.
2: if F: (a,b,...,z) is the path from x to y, then G: (z,...,b,a) is the path from y to x, which exists because every edge q in G exists by assumption since it exists in F.
3: if F: (a,b,...,m), G: (n,o...,z) are the respective paths from x to y and from y to z, then the path Ha,b,...,m,n,o,...,z) is the path from x to z, and all edges exist because they existed in F and G.

I couldn't be arsed with indices of the vertices, which you should use if you do this for homework, but using the alphabet is clearer.

General rule of thumb when solving these things: don't think too deeply and just construct the answers straight from the definition and question.