Proof that ${n\choose r}$ always yields a natural number proof by induction

Question

(https://i.stack.imgur.com/vfy4d.jpg) Hi I'm trying to prove that the ${n\choose r}$ formula gives a natural number by using proof by induction and I'm stuck on the last stage (in the picture above). Have I messed up earlier on and if not then how can I show that $\frac{n+1}{n-r+1}$ is a natural number. I know there are other articles in the stack exchange but I can't seem to find an answer to this. Cheers

Answer 1

I don't know if this falls under category of "induction", but if you knew about Pascal's triangle you could guess the following:

$$ {n\choose r}+{n\choose r+1}={n+1\choose r+1} $$

this can be verified by algebraic computation, and you only need to worry about cases when $r=0$ or $r=n+1$, which are simple enough to directly compute ($=1$)

rest of the terms are guranteed to be integers assuming the result for $n$