Homomorphism from $\mathbb{Q}$ to an ordered field F I know that there exists a unique injective function $\gamma : \mathbb Q →F$ for any ordered field F.
I don't understand why 'Prove $\gamma(r) = r•1_F$ for every $r\in \mathbb Q$' is an exercise.. Don't we just see $\gamma(r)$ as an element of $\mathbb Q$, hence $\gamma(r)=r$? Am i missing something?

If I am interpreting this question correctly: 
$\gamma : \mathbb{Q} \rightarrow F$ is probably defined by mapping $0 \mapsto 0_F$ and $1 \mapsto 1_F$. This extends to an injective ring homomorphism into $F$. 
Let $\times$ denote the multiplication on $F$. I will define $\cdot$. I think the notation $r \cdot 1_F$ is defined to be by if $r \in \mathbb{Z}$ and $r \geq 0$, then $r\cdot 1_F = 1_F + ... + 1_F$, $r$-times. If $r \in \mathbb{Z}$ and $r < 0$, then $r \cdot 1_F = (-1_F) + ... + (-1_F)$. In general, if $r = \frac{p}{q}$, then $r \cdot 1_F = (p \cdot 1_F)\times (q \cdot 1_F)^{-1}$, recall that $\times$ is the operation on $F$. 
Now I think the question is that $\gamma(r) = r \cdot 1_F$ where $\cdot$ is not the multiplication operation on $F$ but the thing I defined above. Now you have a actually question to answer, but it is still pretty easy. 
