Morphism of monoids I am trying to solve an exercise about morphisms of monoids, but i am stuck at some point and not sure if my way idea for the proof is correct. I would be very grateful if somebody could help me or give me a hint how to move on. Here's the exercise:
Given is $f:\mathbb{R}\rightarrow \mathbb{R}$, assume that this map a morphism of monoids $f:(\mathbb{R},+)\rightarrow (\mathbb{R},+)$, from the additive group to itself and also a morphism of monoids $f:(\mathbb{R},\times )\rightarrow (\mathbb{R},\times )$, from the multiplicative monoid to itself. Show the following:
a) If $x \leq y \Rightarrow  f(x)\leq f(y)$.
b) $f(r)=r$ for all $r\in \mathbb{Q}$
c) $f=Id_{\mathbb{R}}$
About morphisms of monoids i know that for $f:(\mathbb{R},+)\rightarrow (\mathbb{R},+)$,  $f(a+b)=f(a)+f(b)$ and $f(0)=0$ for $0$, identity element of the additive group. For $f:(\mathbb{R},\times )\rightarrow (\mathbb{R},\times )$ i know that $f(ab)=f(a)f(b)$ and $f(1)=1$, $1$ acts as identity element for the multiplicative monoid.
a) Can i say, that if $x \leq y$, i multiply both sides by $x^{-1}$, then i get $1\leq yx^{-1}$. We have $f(1)=1$, so i apply $f$ on the both sides and use the properties of the monoids above, $f(1)\leq f(yx^{-1})\Leftrightarrow f(xx^{-1})\leq f(y)f(x^{-1})\Leftrightarrow f(x)f(x^{-1})\leq f(y)f(x^{-1})\Leftrightarrow f(x)\leq f(y)$. So this equality holds. 
I am stuck in part b) and c)...Can anybody help me? 
Thank you in advance!
P.S: I think my approach to part a) is wrong. Today i tried to solve it this way:
$x\leq y\Leftrightarrow z=y-x\geq 0=f(0)$. To show is: $f(z)\geq 0$.
$f(y-x)=f(y)-f(x)\geq f(0)=f(x+(-x))=f(x)+f(-x)\Leftrightarrow f(y)-f(x)\geq 0\Leftrightarrow f(y-x)=f(z)\geq 0$. Is this correct? Thank you very much!

Here's a hint for part $b$: $$f(2)f(1/2)=f(2\cdot 1/2)=f(1/2+1/2)=f(1/2)+f(1/2)=2\cdot f(1/2)$$ and so $f(2)=(f(2)\cdot f(1/2)) f(1/2)^{-1}=(2\cdot f(1/2))\cdot f(1/2)^{-1}=2$. Can you generalise this to all rationals?
For part $c$ note that part $a$ implies that $f$ is a non-decreasing function and part $b$ implies that $f$ is the identity on a dense subset of $\mathbb{R}$. Can you show from this that $f$ must be the identity everywhere (Suppose $f(s)>s$ for some irrational $s$ and choose some $r\in\mathbb{Q}$ such that $s<r<f(s)$)?
