Prove $S$ is a subset of the irrational numbers Suppose $α$ is an irrational number. Prove that $$S = \left\{
\frac{\alpha}{n} + q \ \middle|\ n \in \mathbb{N},\text{ }q \in \mathbb{Q}\right\}$$
is a subset of the irrational numbers.
 A: HINT: $\frac{\alpha}{n}$ is an irrational number. The sum of an irrational and rational number is ...
A: Hint: To arrive at a contradiction, suppose that you can find a number $s$ in $S$ such that $s$ is rational. Since $s$ is a member of $S$, we can write $$s = \frac{\alpha}{n} + q$$ for some positive integer $n$ and rational number $q$. What can you say about $ns - nq$?
A: We prove $S \subset \Bbb R \setminus \Bbb Q$.  To do this, we show
$s \in S \Longrightarrow s \in \Bbb R \setminus \Bbb Q; \tag 0$
Let
$s \in S; \tag 1$
then
$s = \dfrac{\alpha}{m} + r, \tag 2$
for some $m \in \Bbb N$ and $r \in Q$; if
$s \in \Bbb Q, \tag 3$
then
$\dfrac{\alpha}{m} = s - r. \in \Bbb Q, \tag 4$
whence
$\alpha = m(s - r) \in \Bbb Q; \tag 5$
but this contradicts our hypothesis that $\alpha \notin \Bbb Q$; thus
$s \notin \Bbb Q, \tag 6$
that is
$s \in \Bbb R \setminus \Bbb Q, \tag 7$
the irrationals.  Therefore
$S \subset  \Bbb R \setminus \Bbb Q; \tag 8$
that is, $S$ is a subset of the irrational numbers.
Note Added in Edit, Sunday 17 September 2017 10:11 AM PST:  This in response to the comment of Yen Chuan Qi below.  We proved here that every element of $S$ is an element of $\Bbb R \setminus \Bbb Q$; this, by definition, implies $S \subset \Bbb R - \Bbb Q$; however, by far the majority of irrationals are not in $S$; this is seen since $S$ is clearly countable, whilst $\Bbb R - \Bbb Q$ is not, being all reals but for a countable set.  It is a relatively easy matter to find an irrational number not in $S$.  End of Note.
A: If $\dfrac\alpha n + q\vphantom{\frac 1{\displaystyle\int}}$ were rational, then you would have $\dfrac \alpha n + q = \dfrac a b$ for some $a,b\in\mathbb N$ and $q = \dfrac c d$ for some $c,d\in\mathbb N,$ so
\begin{align}
& \frac \alpha n + \frac c d = \frac a b \\[10pt]
& \alpha = \frac{nad - ncb}{bd} = \frac{\text{integer}}{\text{integer}},
\end{align}
so $\alpha$ would be rational.
