 So, I'll probably get my mathematician card revoked for saying this, but don't learn mathematics. Instead, learn how to create mathematics. It's important to remember, it's the journey, not the destination. To learn mathematics, but why would you want to? Skip ahead to 7 minutes 27 seconds. To learn how to create mathematics, play on. So remember that if a connected graph has a bridge, it can be disconnected by the removal of an edge. We might want to prevent this, so let's see how we can avoid having bridges. And so a useful strategy is, book both ways. If we want to avoid having bridges, let's try to make as many bridges as possible. Since no edge on a cycle can be a bridge, how can we avoid having cycles? So let's consider a graph with n vertices. How can we avoid having cycles? You might begin with n edges that form the obvious cycle. To break the cycle, we need to remove an edge. But here's another useful strategy. Making the problem harder sometimes makes it easier. Breaking a cycle by removing an edge is easy. Instead, what if we move an edge so it joins different vertices? Notice that if we move an edge, we get a smaller cycle, and this suggests a connected graph with n vertices and n edges must contain a cycle. So how can we prove this? The key concept is that a bridge breaks a connected graph into smaller graphs. So if we can prove a statement true for small n, we can conclude it's true for larger n. And this suggests that many proofs about bridges can be done using induction. We'll prove it's true for a first case, and then prove that if it's true for some case it's also true for the next larger case. And so we'll use induction. There's one problem. There's no graph with one vertex and one edge, and in fact there's no graph with two vertices and two edges. We have to start with a graph with three vertices and three edges. That's not a real problem, but it will require us to do some additional steps later on. So there's only one graph with three vertices and three edges, and it contains a cycle. Suppose the statement is true for all graphs with k vertices and k edges. Let's consider a graph with k plus one vertices and k plus one edges. And for convenience let's consider the edge between vertices one and two. If this edge is not a bridge, then our graph has to have a cycle since there must be an alternate route between vertices one and two that does not use this edge. If the edge is a bridge, then the graph splits into two subgraphs, g1 and g2. So here's why starting our induction at n equals 3 requires us to take some additional steps. It's possible that one of our graphs has one or two vertices, so we need to consider these separately because we started at n equals 3. So if one of these subgraphs has one vertex, then the remainder has to have k vertices and k edges, and so by our induction hypothesis it has a cycle. If one of these has two vertices, they must be connected since the original graph was connected, so this graph has two vertices and one edge. Consequently, the remaining graph has k minus one vertices and k minus one edges, and so by our induction hypothesis we, uh oh, our induction hypothesis only says what happens if we have k vertices and k edges. Fortunately, that's easy enough to fix, we'll modify it and use strong induction. So remember the idea behind strong induction is we assume our statement is true for everything from some starting point up to our value. So now if the removal of an edge produces a graph with two vertices and one edge, the other graph has k minus one vertices, k minus one edges, and since our induction began with k equals three, we know the k is greater than three, and so k minus one is greater than two, and so k minus one is greater than or equal to three, and obviously k minus one is less than k. So our strong induction hypothesis holds and the graph has a cycle. So we've disposed to the case where one of our graphs has one or two vertices, so otherwise let the graphs have p and q vertices and p and q edges. We don't know how the edges will split among the graphs. So to solve the problem we'll employ a useful strategy in math. Not so much in life, change the game. In this case let's modify what we're trying to prove and prove that in any connected graph with n vertices and n or more edges contains a cycle. Now this doesn't change our analysis for one, two, or three vertices. It changes our induction assumption, but not by much. In fact the only real change is to reflect the theorem we're trying to prove. So I've disposed the removal of the edge, splits the graph into a graph with p vertices and the graph with q vertices where remember these are all of our original vertices so p plus q is equal to k plus one. At least one of the following must be true. The graph with p vertices has at least p edges or the graph with q vertices has at least q edges. If not the two graphs have at most p minus one edges and q minus one edges for a total of at most p plus q minus two edges. But since we obtained this by removing one edge from the original graph then the original graph would have had at most k edges but it had k plus one edges and so at least one of these two graphs will meet the conditions of our induction hypothesis so at least one of the graphs has a cycle and this gives us our theorem. Now if you want to learn how mathematicians think about mathematics don't read the textbook. The textbook has a cleaned up version of the proof of the cleaned up version of the theorem. That being said once we found the proof we should at least clean it up so let's do that. So let's prove the theorem any connected graph with n vertices and n or more edges contains a cycle. We'll use strong induction. Clearly the statement is true for graphs with n equals three vertices and n equals three or more edges. Suppose the statement is true for all graphs with between three and k vertices and at least as many edges and let g be a graph with k plus one vertices and at least k plus one edges. Now consider any edge. If the edge is not on a bridge then it is part of a cycle so g has a cycle. If the edge is a bridge the graph either splits into an isolated point at a graph with k vertices and at least k edges which by our induction hypothesis has a cycle or two points with an edge between them leaving a graph with k minus one vertices and at least k minus one edges which by our induction hypothesis has a cycle or two graphs with p and q vertices where p plus q is k plus one and so if the removal of the edge separates the graph into two components with p and q vertices then either the first graph has at least p edges or the second graph has at least q edges. In either case our induction hypothesis is met and so the graph has a cycle and that proves our theorem.