The problem I am stuck on is this one: If $n>1$ is a natural number then $n-1$ is also a natural number. I am told to use induction.
Normally I would just do the following:
My statement is $P(n)=n-1$ such that $n-1$ is a natural number.
Base case. Since $n>1$, I use the base case of $n=2$. Thus $P(2)=1$ and this is a natural number. Now I choose some $k \in \mathbb{N}$ such that $k > 2$. So $P(k)=k-1 \in \mathbb{N}$.
Past this point I get speculative.
Now for my inductive step. $P(k+1)=(k+1)-1=k$. I don't know if this is allowed, but I noticed that by rearranging $P(k+1)=(k-1)+1 \Rightarrow P(k+1)=P(k)+1$.