For all $k$, $n\in{\Bbb{N_0}}$ such that $k ≤ n$ we define:
$\binom{n}{k}:=\frac{n!}{k!(n-k)!}\in{\Bbb{Q}}$
I am trying to do a proof by induction for this question.
(a) Show that $\binom{n}{k} = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$ for all $1 \leq k \leq n - 1$.
(b) Show that $\binom{n}{k} \in \mathbb{N}$ for all $0 \leq k \leq n$.
Any help is welcome as I am clueless!
$\binom{n-1}{k-1}+ \binom{n-1}{k}$
$=\frac{(n-1)!}{(k-1)!(n-k)!} + \frac{(n-1)!}{(k)!(n-k-1)!}$
$=\frac{k(n-1)!}{(k)!(n-k)!} + \frac{(n-k)(n-1)!}{(k)!(n-k)!}$
$=\frac{n(n-1)!}{(k)!(n-k)!}$
$=\frac{n!}{(k)!(n-k)!}$
$=\binom{n}{k}$
Now for the second part, first notice that $\binom{n}{1} = \frac{n!}{(1)!(n-1)!} = n$ is always an integer. This is going to be part of our base case.
We are going to use strong induction from now on.
Inductive step: assuming that $\binom{i}{j}$ is an positive integer for all $\binom{i}{j}$ that is both defined and satisfy $i \leq n$ and $j\leq n$, $j \leq i$.
now consider $\binom{n+1}{j}$. for j = n+1, this is 1 and we are done. if $j \le n+1$, then
$\binom{n+1}{j}$ = $\binom{n}{j-1}+ \binom{n}{j}$ by part (a). Then by inductive assumption, $\binom{n}{j-1}$ and $\binom{n}{j}$ are both integer so $\binom{n+1}{j}$ must be an integer
therefore $\binom{n+1}{j}$ is always an integer for all $j \leq n+1$
By induction, we can extend this too all $\binom{i}{j}$ that is defined. therefore, $\binom{n}{k}\in \mathbb{N} $ for all $0\leq k \leq n$
