Question

Use strong induction to show that when a simple polygon P with consecutive vertices ${\mathbf{v}}_{\mathbf{1}}\mathbf{,}{\mathbf{v}}_{\mathbf{2}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{v}}_{\mathbf{n}}$is triangulated into n-2 triangles, the n-2 triangles can be numbered$\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathbf{n}\mathbf{-}\mathbf{2}$ so that${\mathbf{v}}_{\mathbf{i}}$ is a vertex of triangle i for $\mathbf{i}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}\mathbf{n}\mathbf{-}\mathbf{2}$.

Short answer

The statement “$n-2$ triangles can be numbered$1,2,....,n-2$ so that$v_i$ is a vertex of triangle i for $i=1,2,....,n-2$” is true for$n=3$ and it is also true for by$n=k+1$ assuming that it is true for $n=1,2,...,k$.

Step-by-step solution

Significance of the strong induction

The strong induction is referred to as a particular type of process that mainly proves a statement which holds true for the natural numbers. The simple and the strong induction are very closely related.

Prove of the question statement

Let$P\left(n\right)$ be the problem statement that “the$n-2$ triangles can be numbered$1,2,....,n-2$ so that is a vertex of triangle i for $i=1,2,....,n-2$”.

In the basic step, let$n=3$, in this, a polygon P having three vertices $v_1,v_2,v_3$is described as a triangle. Let the entire triangle be $v_1$. Hence,$P\left(3\right)$ holds true.

In the inductive step, it has been assumed that all are true in the case of $P\left(4\right),P\left(5\right)....P\left(k\right)$.

Let, the vertices of the polygon P arerole="math" localid="1668577300545" $v_1,v_2,...,v_k,v_{k+1}$ and the vertices are consecutive. Moreover, it has been assumed that the diagonal has triangulated that polygon P into$n-1$ number of triangles. It is safer to assume that the diagonal lies between the vertices$v_i$ to $v_i$having $i\in 3,4,...,k$.

The polygon $P^{\text{'}}$which is convex is nature consists consecutive number of vertices that$v_1,v_2,...,v_i$ are has been triangulated into$i-2$ triangles and the$i-2$ triangles can be labelled as$1,2,....,i-2$ as the triangle J consists of a vertex $v_j$. Hence,$P\left(i\right)$ holds true.

The polygon $P^{\text{'}\text{'}}$which is convex is nature consists consecutive number of vertices thatrole="math" localid="1668577434145" $v_1,v_{i+1},...,v_1$ are has been triangulated into$(n+1-i)-2$ triangles and the$(n+1-i)-2$ triangles can be labelled as$i,i+1,....,n-1$ as the triangle J consists of a vertex . Hence, holds true.

Hence, the above equations form the triangulation of the polygon P in which the$n-2$ triangles is being numbered hence the triangle consists of a vertex for which $j=1,2,....,n-2$. Hence,$P(n+1-i)$ is true.

Thus, the statement “$n-2$triangles can be numbered $1,2,....,n-2$so that $v_i$is a vertex of triangle i for $i=1,2,....,n-2$” is true for $n=3$and it is also true for $n=1,2,...,kn=1,2,...,k$by assuming that it is true for $n=1,2,...,k$.