Prove that there exists an isomorphism of monoids I have an exercise as preparation for an exam:
For $ n \in \mathbb{N} $, $ n \geq 1$ is defined the function $ f_n: \mathbb{R} \to \mathbb{R}$, $f_n(x) = \lfloor\frac{x}{n}\rfloor$, $\forall x \in \mathbb{R}$ and the set $M=\{f_n \mid n \in \mathbb{N}\}$
I have proven that $(M, \circ) $ is a commutative monoid, but I don't know how to solve the following 2 requirements:


*

*Prove that there exists an isomorphism of monoids 
$$
 (M, \circ) \simeq (\mathbb{N}, \cdot) 
$$

*Is $ M $ a submonoid in $(\mathbb{R}^\mathbb{R},\circ) $?

 A: To show it's an isomorphism (sending $f_n\to n$) we need to show that for all $x,m,n$ we have that
$$\left\lfloor\frac{\lfloor \frac xn\rfloor}m\right\rfloor=\left\lfloor \frac x{mn}\right\rfloor$$
Write
$$x=an+b$$
where $0\le b< n$ and $a$ is an integer. Then $\lfloor \frac xn\rfloor=a$. Write
$$a=cm+d$$
where $0\le d <m$ and $c$ is an integer. Then $\lfloor \frac am\rfloor=c$.
Now write
$$x=(cm+d)n+b=cmn+dn+b$$
We have that $d\le m-1$, and $b\le n-1+\epsilon$ where $0\leq \epsilon <1$ is such that $b-\epsilon$ is an integer. Then 
$$x-cmn\le (m-1)n+n-1+\epsilon=mn-1+\epsilon<mn$$
Thus
$$\left\lfloor \frac x{mn}\right\rfloor=c$$
and we are done.
This monoid is not a submonoid of the monoid of real valued functions because the identity elements are different.
A: The most obvious candidate to show that $M$ and $\mathbb{N}$ are isomorphic would be $\phi:M \to \mathbb{N}$ by $\phi(f_n)=n$. Are any of the isomorphism properties difficult to prove? 
As for the second question, what is your definition of submonoid and again are any of the properties difficult to prove?
