Dots and arrows. Is this true? Imagine a graph with $n$ dots and $2n$ connecting arrows that meet the following rules:
Rule 1) Every dot has at least one arrow going into it.
Rule 2) Every dot has two arrows going out, each going to another dot.
Rule 3) You can travel along the arrows from any dot to any other dot.
Then, is it true that if I add a dot that meets rules 1 & 2, the new graph will automatically follow rule 3, i.e. I can still travel from any dot to any other dot.
 A: Yes. We can show this as follows:

Let the new dot be called $d$, and let the new arrows be $c \to d, d \to e, d \to f$.
Then we can travel from $d$ to any dot through $e$ or $f$, and travel to $d$ from any dot through $c$.

However, the new graph is not of the same form as the original. Namely, to add $d$, we have to introduce three new arrows, one into $d$, and two out of $d$. Therefore, we have a graph with $n+1$ nodes and $2n+3$ edges.
Concretely, the node $c$ now has three arrows going out of it — two to nodes in the existing graph, and the new one to $d$. Therefore, the new graph as a whole does no longer satisfy rule 2.

In fact, the flaw is fundamental. For, the theorem:

Let $G$ be a graph on $n$ nodes such that:
  
  
*
  
*Every node has at least one arrow going into it;
  
*Every node has precisely two arrows going out of it
  
  
  Then there is a path between any two nodes of $G$.

for which this seems an attempt at a proof by induction, is false. Consider the following counterexample (created using Directed Graph Editor):

We see we cannot travel from $0$ to $4$.
A: now there is a system of $n$ dots , now we are adding one more which we are calling $x$ , following rule 1 and 2 . so there is one arrow going into it , lets say it comes from an arrow called $a$ and it has two arrows going out , lets say one of them is going to $b$ . so now we know in the prior system we could travel from any 1 dot to another . so now to go to $x$ from the system ,we can always go to $a$ and then through that to $x$ and to go to a random point from $x$ we can always go to $b$ from $x$ and then go to any desired point. thus we can say $x$ follows rule 3.
