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How could I induction-prove that

If a simple graph of $n$ vertices has at least $n$ edges, then the graph must have a circuit (=cycle, tour)

How should I do this proof? I would prefer simpler solutions, as I am a beginner at graph theory.

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    Also this hint can help you: http://math.stackexchange.com/a/2130833/100361 – 2017-02-20

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Induction: Base case is $n=3$, when the only possible simple graph with $3$ edges is the triangle which has a cycle.

Inductive hypothesis for $n>3$: The claim is true for all graphs with $v$ vertices, $3 \le v \le n{-}1$.

Consider a simple graph $G$ with $n$ vertices and $n$ edges. If $G$ is not connected, then there is some connected component of $G$ with $k

Otherwise for connected $G$, since the sum of vertex degrees is twice the number of edges, $\sum \delta_i = 2n$, we have two cases:

Case 1: All vertices are of degree $2$. Then $G$ is a cycle graph $C_n$ and has a cycle.

Case 2: There is some vertex with degree $\delta_i > 2$ and thus there is some other vertex $j$ with degree $\delta_j =1$. This degree-1 vertex and its associated edge can be removed without disconnecting $G$ or affecting any cycles in $G$ and produces a graph $G-j$ which has a cycle by the hypothesis.

For simple graphs with $n$ vertices and more than $n$ edges, we can remove edges to produce a graph for the above consideration and any cycles in the reduced graph will also be present in the original graph.

Thus by induction the claim is true.

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    I like that proof, but I don't entirely understand how it's a proof by induction. That's probably because I am not so familiar with proofs... But where do we prove that this 'property' is 'inherited'? So that if it is true for $n$, then it's also true for $n+1$? – 2017-02-23
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    I'm using strong induction, where I assume it's true for all numbers less than $n$, and show that it is also true for $n$. I use it twice (look for the phrase "by the hypothesis"); once to consider graphs that are not connected, and once after I remove a vertex of degree $1$. Cycle graphs (regular degree-2) cannot be reduced by vertex removal so they do not use induction. – 2017-02-23
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    I see, but could the proof be done with simple induction as well? Or would that be too complicated? – 2017-02-23
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    Just read $(n{+}1)$ where I wrote $n$, and $n$ where I wrote $n{-}1$. – 2017-02-23