Is a 𝑢−𝑣 path followed by a 𝑣−𝑤 path is a 𝑢−𝑤 path. [closed]

Can I say a 𝑢−𝑣 path followed by a 𝑣−𝑤 path is a 𝑢−𝑤 path.

Because if there is a path from 𝑢−𝑣 followed by 𝑣−𝑤 I believe there should be a 𝑢−𝑤 path.

I don't know how to prove it.

EDIT: A 𝑢−𝑣 path is a path between vertex 𝑢 and 𝑣 vertex. and 𝑣−𝑤 path is a path between vertex 𝑢 and 𝑤 vertex

A walk is a sequence of vertices and edges of a graph i.e. if we traverse a graph, then we get a walk.

A path is a walk with no repeated vertex. This directly implies that no edges will ever be repeated

• It's hard to address your question without knowing exactly how you define a "$u$-$v$ path". I suggest that you edit your question to provide the definition. Commented Feb 27, 2023 at 17:52
• When you write "a $u-v$ path", do you mean a path from some point $u$ to some other point $v$, or are you implying some sort of subtraction of $v$ from $u$? Your question does not make it clear. If it's the first meaning, then the answer by @MPW below answers you quite well. If it's the second, then we definitely need more details. Commented Feb 27, 2023 at 18:03
• @JonathanZsupportsMonicaC I have made an edit to the question Commented Feb 27, 2023 at 18:16
• Hey, i just noticed that you have the tag 'graph theory' on your post. Not everybody looks at the tags, so you might want to include the phrase "in a (directed) graph" in the body of your question. Me, I don't even know what formal definition of a path in a graph people use in graph theory. Commented Feb 27, 2023 at 18:34
• You also need to define path. If the definition prohibits repeated vertices, then... Commented Feb 27, 2023 at 18:42

Imagine that the $$vw$$ path contains the vertex $$u$$, the concatenation of $$uv$$ and $$vw$$ is not a path, but a walk. To be a path, you have to make sure it does not contains two identical vertices.
Hint: If $$f,g:[0,1]\to X$$ and $$f(1)=g(0)$$, then $$h:[0,1]\to X$$ and $$h(0)=f(0)$$ and $$h(1)=g(1)$$, where $$h(t) = \begin{cases} f(2t),& 0\leq t <\tfrac12\\ g(2t-1),& \tfrac12 \leq t \leq 1 \end{cases}$$