 The importance of proof in mathematics is often misunderstood. Proving something guarantees the truth of a statement, but you don't bother to prove statements you don't already believe true. Instead, proof does three things. First, it reviews what you should already know, it reveals new results, and it raises new questions to investigate. Let's see how that works. So we've been looking at ways to identify and eliminate bridges, so it helps to consider extreme cases. Suppose all edges in a graph with n vertices are bridges. Now one of the things we know is that in a connected graph if an edge is part of a cycle then it is not a bridge and conversely. So if everything is a bridge then no edge can be part of a cycle. But we also know that cycles can be unavoidable. A connected graph with n vertices and n edges must contain a cycle. So if there are n or more edges, there must be a cycle. Consequently, if g is a connected graph with n vertices where all edges are bridges then g has at most n-1 edges. Now suppose g is a connected graph with n vertices, what's the fewest number of edges it can have? A useful strategy in math and in life, if you build it, they will come. Suppose you have n vertices and zero edges. Obviously the graph isn't connected, so we'll add one edge. Since this edge joins two isolated vertices, the first edge we add leaves us n-2 isolated vertices. So we'll try and connect the other vertices by adding edges. But every edge we add will include at most one more isolated vertex. So we need to add at least n-2 more edges. And so we need our first edge, plus n-2 more, that's n-1 edges, to make a connected graph with n vertices. And this proves our theorem. Suppose g is a connected graph with n vertices, then g has at least n-1 edges. Now suppose we have a connected graph with n vertices and n-1 edges. Is there anything else we know? And a good way forward is, consider the corollaries. The easiest corollary to find is a contrapositive. So remember the contrapositive of the conditional if a then b is if not b then not a. The contrapositive and conditional are two for one deal, one for proven one, you get the other for free, so you might as well take it. To find the contrapositive, it helps to identify the antecedent, the if part, and the consequent, the then part. The consequent is that g has at least n-1 edges, so that's easy to identify. The antecedent is actually a conjunction consisting of three simple statements. g is a graph, g is connected, g has n vertices. This means we can split our premise and have a choice of antecedents. Since the result is meaningless unless g is a graph, then we really have three possibilities. We can take the compound statement, g is a connected graph with n vertices as our antecedent. We can split off g as a connected graph, and then the rest becomes our antecedent. Or we can take g as n vertices, and the rest as our antecedent. Which conditional should we find the contrapositive of? Clearly the one we should use is all of them. So if we use everything as our antecedent, our theorem is if g is a connected graph with n vertices, then g has at least n-1 edges. So we negate and switch the antecedent and consequent. The consequent, the then part, g has at least n-1 edges. Now at least means n-1 or more, so the negation would be less than n-1. So the negation of the consequent would be g has less than n-1 edges, and that becomes our new antecedent. The antecedent of our original theorem is g is a connected graph with n vertices. This is a compound statement, consisting of two simple statements. g is a connected graph, and g has n vertices. So remember the negation of a and b is not a or not b. So our negation will be g is not a connected graph, or g does not have n vertices. So the negation of the antecedent would be either g is not connected, or g does not have n vertices, and that will be our new consequent. Now remember we can split the premise so that we can make g a connected graph part of an overall premise and that our conditional becomes if g has n vertices, then g has at least n-1 edges. Now g is a connected graph as a separate premise. It's not part of the conditional, so we leave it. The consequent of our theorem is the same, so the negation and our new antecedent is still g has less than n-1 edges. And the consequent will be the negation of our original antecedent. So the negation of g has n vertices is g does not have n vertices, and that will be the consequent of our contrapositive giving us a theorem. We could also change our theorem to suppose g has n vertices if g is connected, then g has at least n-1 edges. So g having n vertices is a separate premise, so we'll copy it over into our contrapositive. And the consequent of our original statement is the same, so our antecedent of the contrapositive is still the same. The antecedent is g is connected, and so the negation will be g is not connected. And that gives us our new consequent and our new theorem. Now while we found three different contrapositives, they are not necessarily equally useful. And in particular, just knowing that g does not have n vertices doesn't tell us much. So these corollaries are not particularly insightful. However, g is not connected does give us a lot of information. So let's see how we can use it. Suppose g is a connected graph with n vertices and exactly n-1 edges. Using any edge gives it fewer than n-1 edges, and since g still has n vertices, it is no longer connected. Consequently, the removed edge is a bridge, giving us the result. Suppose g is a connected graph with n vertices and n-1 edges, then every edge is a bridge. Since our goal is to have no bridges, this is actually a promising result. In a connected graph, with n vertices and n-1 edges, every edge is a bridge. But if we add one edge, we have a cycle, and no edge on a cycle is a bridge. So adding one edge converts some of the bridges into non-bridges. Let's see where this takes us.