$\{a_1,...,a_n,b\}$ is discrete if and only if $t_j \in \mathbb{Q}$ 
Let $G= \mathbb{R}^{n}$ with the usual topology. If $\{a_1,...,a_n\} \subset \mathbb{R}^n$ is linearly independent and $b=\sum_{j=1}^n t_ja_j$ show that the subgroup generated by $\{a_1,...,a_n,b\}$ is discrete if and only if $t_j \in \mathbb{Q}$ for each $j=1,2,...,n$.

Let $G$ be a topological group and $ \Gamma $ a subgroup of $ G $. $ \Gamma $ is a discrete space if and only if $ \Gamma $ has an isolated point.
Let $\Gamma$ be the subspace generated by $\{a_1,...,a_n,b\}$ and suppose that $t_j \in \mathbb{Q}$, my idea is to show that $ b $ is an isolated point  and conclude with the previous result, it is possible to show that $ b $ is an isolated point?
Any help, I would appreciate it.
 A: Here's 95% of the details. You can check the other 5% yourself.
First, take the linear transformation $T$ which maps $a_i$ to the $i$-th standard basis vector. Then $T$ is a homeomorphism, and from that it follows that $\langle a_1, \dots, a_n, b \rangle$ is discrete, if and only if $\langle e_1, \dots, e_n, Tb \rangle$ is discrete.
Now, assume $Tb = (t_1, \dots, t_n) \in \mathbb{Q}^n$. If we write the numerators and denominators as $t_i = p_i/q_i$ then
$$\langle e_1, \dots, e_n, Tb \rangle \subseteq \left\langle \frac{e_1}{q_1},\dots,\frac{e_n}{q_n}, Tb \right\rangle \cong \langle e_1,\dots,e_n,(p_1,\dots,p_n)\rangle = \langle e_1,\dots,e_n\rangle = \mathbb{Z}^n.$$
Where for the isomorphism, we map $e_i \mapsto q_ie_i$.
Conversely, suppose that some $t_j \notin \mathbb{Q}$. Actually, for convenience's sake, let's suppose that $t_1,\dots,t_m \notin \mathbb{Q}$ and $t_{m + 1} + \dots + t_n \in \mathbb{Q}$. Since
$$\mathbb{Z}^n = \langle e_1,\dots,e_n \rangle \subset \langle e_1, \dots, e_n, Tb \rangle,$$
we can speak of the fractional part of a vector in this group. If $x$ is in this group, let us denote its fractional part by $\{x\}$.
Consider the sequence $x_k = \{k (Tb)\}$. In the $m+1$-st through $n$-th coordinates, $x_k$ is periodic. In the $1$-st through $m$-th coordinates, $x_k$ is aperiodic. Therefore the sequence $x_k$ represents infinitely many points in the unit cube $[0,1]^n$.
So now if we cover $[0,1]^n$ by finitely many balls of radii $\varepsilon$ then by the pigeonhole principle, one of these balls must contain at least two elements of our sequence. Thus $0 < |x_k - x_l| < \varepsilon$ and it follows that $\vec{0}$ is not an isolated point.
See also:

*

*Why is this quotient space not Hausdorff?

*Multiples of an irrational number forming a dense subset
