$\mathbb{Q}$: Unique operation s.t. $1\star q=q$ and right distributivity hold? 
Given $\mathbb{Q}$ and the usual addition $+$ on it, do we have unicity of a binary operation $\star$ such that
  \begin{align*}
\tag{1}1\star q&=q\\
\tag{2}(q+r)\star s&=q\star s+r\star s
\end{align*}
  for all $q,r,s\in\mathbb{Q}$ ?

I know this is true for $\mathbb{N}$ and $\mathbb{Z}$ instead of $\mathbb{Q}$, so I suspect that the answer is yes. However, I don't see how to prove it.
For $\mathbb{N}$, existence and unicity are usually proved together in a Recursion Theorem.
For $\mathbb{Z}$, one can show that $\star$ is the usual multiplication by reducing to $\mathbb{N}$ using the partition
$$
\mathbb{Z}\times\mathbb{Z}=(\mathbb{N}\times\mathbb{N})\sqcup(-\mathbb{N}\times\mathbb{N})\sqcup(\mathbb{N}\times-\mathbb{N})\sqcup(-\mathbb{N}\times-\mathbb{N})
$$
(See here.)
I don't see a useful partition of $\mathbb{Q}$ in terms of $\mathbb{Z}$, $-\mathbb{Z}$ and $(\mathbb{Z}\backslash\{0\})^{-1}$.
 A: No reason to partition.
First, show $0\star q=0$.
Then show that $(-p)\star q=-(p\star q)$.
Then show that if $n$ is natural, then $n\star q=nq$, by induction.
Then show that if $r=\frac{m}{n}$ with $m$ an integer and $n$ natural, then $mq=n(r\star q)$.
A: Consider expressions like:
$$1=1\star 1=\left(\underbrace{\frac{1}n+\frac{1}n+\ldots+\frac{1}n+\frac{1}n}_{n\text{ times}}\right)\star 1=\underbrace{\frac{1}n\star 1+\frac{1}n\star 1+\ldots+\frac{1}n\star 1}_{n\text{ times}}=n\left(\frac{1}n\star 1\right)$$
This allows you to see that $\frac{1}n\star 1=\frac{1}n$. Essentially, all you can say about the relation of rational numbers to integers when we only have addition is that $$\underbrace{\frac{m}n+\frac{m}n+\ldots+\frac{m}n+\frac{m}n}_{n\text{ times}}=m.$$ See if you can't use relations like this to first show that if $x$ is rational and $y$ is an integer, then $x\star y = xy$. This is easy with the above setup - just replace the $1$'s with $y$'s. Then, using a similar extension, show this if $x$ and $y$ are both only assumed to be rational.
