If x is rational, $x\ne 0$, and $y$ irrational, prove $x+y, x-y, xy, x/y$ and $y/x$ are all irrational. I'm going through Apostol's Calculus. And I'm not sure how to tackle this. In this section he introduced the least upper bounds, and Archimedian properties of the real number system. Any hints would be appreciated.
 A: Hint:
Example: if $x + y$ was rational, then also
$$y = \underbrace{\underbrace{(x + y)}_{\text{rational}} - \underbrace{x}_{\text{rational}}}_{\text{rational}}.$$
Contradiction.
Use this idea for the other expressions.
A: Another example, multiplicatively:
Assume $x\in \mathbb Q,\; x\neq 0,\;$ and $\,y\in \mathbb R\setminus \mathbb Q$.
Note that $$x\neq 0, x\in \mathbb Q \iff \dfrac 1x \in \mathbb Q$$
Suppose for the sake of contradiction that $\;xy \in \mathbb Q.$
Then $$\dfrac 1x\cdot(xy) = \left(\dfrac 1x \cdot x\right)y = y \in \mathbb{Q}$$ since the set of rational numbers is closed under multiplication, and we have that $\dfrac 1x \in \mathbb Q$ and $xy \in \mathbb Q$. This contradicts the assumption that $y$ is not rational.
This can easily be modified and generalized for the cases $\dfrac xy$ and $\dfrac yx$.
A: It's true precisely because the rationals are a subset of reals closed under subtraction and division (by elements $\ne 0),$ i.e. they form additive and multiplicative subgroups of $\rm\,\Bbb R.\:$
If you know about groups then you can appreciate this from a more conceptual viewpoint. $ $ Let $\rm\:G\subset H\:$ be abelian groups, and $\rm\:\overline G = H\backslash G\:$ be the complement of $\rm\:G\:$ in $\rm\:H.\:$ Then 
Lemma $\rm\ \ \  g\in G,\ \ \bar g\in \overline G\ \Rightarrow\ g - \bar g\in \overline G\quad (Proof\!:\ else\ \ g - \bar g\, =\, g_2\!\!\in G\:\Rightarrow\: \bar g\, =\, g\!\!-g_2\in G)$
So $\rm\,\ x \in \Bbb Q,\ \ y \in \overline{\Bbb Q}\ \Rightarrow\: x\!-\!y \in \overline{\Bbb Q},\ $ so $\rm\ 2x \in \Bbb Q\: \Rightarrow\: 2x\!-\!(x\!-\!y) = x+y \in \overline{\Bbb Q}.\ $  In the same way:
So $\rm\ x\in \Bbb Q^*\!,\, \ y \in \overline{\Bbb Q^*}\Rightarrow\, x/y \in \overline{\Bbb Q^*},\:$ so $\rm\: x^2\in \Bbb Q^*\Rightarrow\: x^2/\,(x\,/\,y)\,  =\,  x\, *\, y \in \overline{\Bbb Q^*}$.   
Finally $\rm\ 1 \in \Bbb Q^*\Rightarrow\: 1/(x/y)\, =\, y/x \in \overline{\Bbb Q^*}.\ $ Note every inference is an instance of the Lemma.
Remark $\ $ This is a special case of the following complementary view of a subgroup.
Theorem $\ $ Let $\rm\,G\,$ be a nonempty subset of abelian group $\rm\,H,\,$ with complement set $\rm\,\overline G = H\backslash G.\,$
Then $\rm\,G\,$ is a subgroup of $\rm\,H\!\iff\! G - \overline G\, =\, \overline G. $ 
Proof  $\ $ Recall  that $\rm\,G\,$ is a subgroup of $\rm\,H\!\iff\! G\,$ is closed under subtraction (subgroup test). 
$\begin{eqnarray}\rm Complementing, & &\ \ \rm G\text{ is not a subgroup of }\, H\\
&\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ is\ false\\
&\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \overline G\\
&\iff&\ \rm\ g_1\, -\ \bar g\ \ =\,\ g_2\  for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \overline G\\
&\iff&\ \rm\ G\ -\ \overline G\ \subseteq\ \overline G\ \ is\ false\quad\ {\bf QED}
\end{eqnarray}$ 
Instances of this are ubiquitous in concrete number systems, e.g. below. For many further examples see various prior posts here.

