Step-by-step explanation:
Proof for i)
We will prove by mathematical induction that, for every natural [tex]n\geq 4[/tex], the number of diagonals of a convex polygon with n vertices is [tex]\frac{n(n-3)}{2}[/tex].
In this proof we will use the expression d(n) to denote the number of diagonals of a convex polygon with n vertices
Base case:
First, observe that:, for n=4, the number of diagonals is
[tex]2=\frac{n(n-3)}{2}[/tex]
Inductive hypothesis:
Given a natural [tex]n \geq 4[/tex],
[tex]d(n)=\frac{n(n-3)}{2}[/tex]
Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
Observe that, given a convex polygon with n vertices, wich we will denote by P(n), if we add a new vertix (transforming P(n) into a convex polygon with n+1 vertices, wich we will denote by P(n+1)) we have that:
Because of these observation we know that, for every [tex]n\geq 4[/tex],
[tex]d(n+1)=d(n)+1+(n-2)=d(n)+n-1[/tex]
Therefore:
[tex]d(n+1)=d(n)+n-1=\frac{n(n-3)}{2}+n-1=\frac{n^2-3n+2n-2}{2}=\frac{n^2-n-2}{2}=\frac{(n+1)(n-2)}{2}[/tex]
With this we have proved our statement to be true for n+1.
In conlusion, for every natural [tex]n \geq 4[/tex],
[tex]d(n)=\frac{n(n-3)}{2}[/tex]
Proof for ii)
Observe that:
Then, the statement is not true for n=1,2,3.
We will prove by mathematical induction that, for every natural [tex]n \geq 4[/tex],
[tex]2n<n![/tex].
Base case:
For n=4, [tex]2n=8<24=n![/tex]
Inductive hypothesis:
Given a natural [tex]n \geq 4[/tex], [tex]2n<n![/tex]
Now, we will assume the induction hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
Observe that,
[tex]n!+2\leq (n+1)! \iff n!+2\leq n!(n+1) \iff 1+\frac{2}{n!}\leq n+1 \iff 2\leq n*n![/tex]
wich is true as we are assuming [tex]n\geq 4[/tex]. Therefore:
[tex]2(n+1)=2n+2<n!+2\leq (n+1)![/tex]
With this we have proved our statement to be true for n+1.
In conlusion, for every natural [tex]n \geq 4[/tex],
[tex]2n<n![/tex]
