(i) **Every simple graph has at least two vertices of the same degree. --- TRUE**

proof by contradiction:-

**Every simple graph has none of the vertices of the same degree**.

let there are n vertices, in a simple graph maximum degree of a vertex = n-1.

therefore according to the statement, the degree sequence must be 0,1,2,3,.....,(n-1)

is this sequence is valid?

No, due to one vertex degree=0 ===> Maximum degree can exist = n-2, but in our sequence, a vertex with degree = n-1 is exist.

Therefore your assumption is wrong.

(ii) **If u is a vertex of odd degree in a graph, then there exists a path from u to another vertex v of the graph where v also has odd degree. ---- TRUE**

if you take connected graph, then it is trivial. ( i mean there exist a path every pair of vertices ), Take a disconnected graph ( therefore it have** more than one component.** )

Let take it have 2 components, first component have x vertices, second component have y vertices.

we know that every graph has even no.of odd degree vertices.

**think each component is separate graph**, therefore if the first (second) component have a vertex odd degree (let u), then it must have another vertex(let v) which is odd degree ===> x is a component therefore u and v should be connected, connected means there is a path between u and v

(iii) **If there are exactly two vertices a and b of odd degree, there is an Eulerian cycle in the graph --- False**

we all know that, A graph have Eulerian cycle iff every vertex has even degree

the correct statement is:-

If there are exactly two vertices a and b of odd degree, there is an **Eulerian path** in the graph