Prove that if $g:\mathbb{N}\rightarrow \mathbb{N}$ and $\forall x,y\in \mathbb{N}, x<y\Rightarrow g(x)<g(y)$ then $n\leq g(n)\space\space\space \forall n\in \mathbb{N}$
My proof so far (induction): It's tru for 1 since 1 is less or equal any number in $\mathbb{N}$.
Let's suppose for $k$, $k\leq g(k)$, we got this:
$$k<k+1\Rightarrow g(k)<g(k+1)$$ so, $k\leq g(k)<g(k+1)$ And then I don't know what else to do. As far as I understand the operation "$-$" isn't defined yet for $\mathbb{N}$.
You're very close. Since $k < g(k+1)$, we must have $k+1 \leq g(k+1)$, since there are no numbers between $k$ and $k+1$. That finishes the induction step.
Together with my comment above that the problem needs to read "then $n \leq g(n)$" because of the function $g(n) = n$, the problem is now solved.