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For $n \geq 3$ proof that the number of diagonals of a polygon is $\frac{n(n-3)}{2} $ using induction.

I don't know how to start this problem, can you give me a hint?

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How many diagonals meet at each vertex? Hm, but that way doesn't use induction. – Daniel Fischer Feb 3 '14 at 21:03
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Let $d_n$ be the number of diagonals of a polygon.

$d_3=0$ (polygon is triangle) Assume that the number of diagonals of a $n$-polygon is $\frac{n(n−3)}{2}.$

Сonsider $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Under the assumption the number of diagonals of $n-$poligon $A_1 A_2 \ldots A_{n}$ is $\frac{n(n−3)}{2}.$ Besides there are $n-1$ diagonals $A_{n+1} A_2,\,\,$ $A_{n+1} A_3,\ldots , A_{n+1} A_{n-1},$ and $A_1A_n$ for $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Therefore the number of diagonals of a $n+1$-polygon $A_1 A_2 \ldots A_{n+1}$ equal to $d_{n+1}=\frac{n(n−3)}{2}+n-1=\frac{n^2−3n+2n-2}{2}=\frac{(n+1)((n+1)-3)}{2}.$ Sorry for my English.

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I don't know what you mean by induction. But it's simple: For each vertex (n), there is one diagonal linking for all vertex, except the same vertex and the two neighbours (n-3). Thus you have n(n-3), that you got to divide by two or else each diagonal will be counted going and coming back.

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"Induction" stands for a basic logical way of proving something. Basically it is used, when something need to be proven $\forall n \in \mathhbb{N}. So, e.g. here, where polygon can have arbitrary number of vertices, it is good to use induction. Not enough room for details, but basically: proof by induction is done in two steps, the first base, and then the induction step. For example look at my answer, for more, search the internet;) – quapka Apr 9 '14 at 6:08

Suppose you have an $n$-gon with vertices $V_1, \ldots, V_n$ which by induction hypothesis has $n (n - 3)/2$ diagonals. Now let us add another vertex $V_{n+1}$. How many new diagonals have appeared? Add that to the existing diagonals and verify that you get $(n + 1) (n + 1 - 3)/2$. (Do not forget that the side $V_n V_1$ becomes a diagonal when you add $V_{n+1}$ beween $V_n$ and $V_1$.)

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Well start with the first step: prove, that $\frac{n(n-3)}{2}$ is the number of diagonals for the smallest possible polygon. In this case it is $n$-polygon, where $n=3$.

In the second step assume, that $\frac{k(k-3)}{2}$ is the number of diagonals for $k$-polygon and you must show that $\frac{(k+1)(k+1-3)}{2}$ is the number of diagonals for $(k+1)$-polygon. It might be helpful to draw a simple picture with arbitrary number of vertices and add one vertex and find out, what it does with the number of diagonals.

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