In the proof of "the global properties of the Hilbert symbol", there are telling: This section makes use of the embedding of $\mathbb{Q}$ into $\mathbb{Q}_v$, where $v$ is a prime or $\infty$, adopting the convention that $\mathbb{Q}_{\infty}$ is equal to $\mathbb{R}$. For $a$,$b \in \mathbb{Q}^*$ we let $(a,b)_v$ denote the Hilbert Symbol on the images of $a$,$b$ in $\mathbb{Q}_v$ or $\mathbb{R}$.
I just don't understand why the images are in $\mathbb{Q}_v$ or $\mathbb{R}$ when the Hilbert symbols results are only $\{+1,-1\}$. And why given with the facts before can we conclude, that $\mathbb{Q}$ is dense in $\mathbb{Q}_v$ (or $\mathbb{R}$)? I don't understand this connection.
As stated, the Hilbert symbols are maps $\Bbb Q_p^*\times\Bbb Q_p^*\to\{\pm1\}$. To define it on $\Bbb Q^*\times\Bbb Q^*$ we take the inclusion map $i_p:\Bbb Q^*\to\Bbb Q_p^*$ which extends to an inclusion $i_p\times i_p:\Bbb Q^*\times \Bbb Q^*\to\Bbb Q_p^*\times\Bbb Q_p^*$ and compose with the Hilbert symbol $\Bbb Q_p^*\times\Bbb Q_p^*\to\{\pm1\}$ which gives a Hilbert symbol $\Bbb Q^*\times\Bbb Q^*\to\{\pm1\}$.
More pedantically, $(a,b)_p$ for $a$, $b\in\Bbb Q^*$ is defined as $(i_p(a),i_p(b))_p$ where $i_p(a)$ is the image of $a$ in $\Bbb Q_p^*$ (under the natural inclusion map $i_p$).