Jutifying: irrational + rational = irrational I am not sure how to show this. This is what I tried. 
Proposition:
If x is rational number and y is an irrational number then x+y is an irrational number.
Logically speaking this is
P=x is rational
Q= Y is irrational
R= x+y is irrational
$P\wedge Q \rightarrow R$
Explanation:
What I did
$x=\frac{p}{q}$ P and Q integer
Y cannot be written as the a rational by definition
So therefore
$x+y=\frac{p}{q}+y$
Will not be a rational number as there cannot be a common denominator. 
 A: $\color{green}{x=\frac{a}{b}}$ (since $x$ is rational), for $a, b \in \mathbb{Z}$.
Assume, for a contradiction, that $x+y$ is rational.
Then $x+y=\frac{p}{q}$, for $p, q \in \mathbb{Z}$.
Re-arranging: $$\color{red}y=\frac{p}{q}-\color{green}x=\frac{p}{q}-\color{green}{\frac{a}{b}}=\color{red}{\frac{pb-aq}{qb}}\left(=\frac{p'}{q'}\right),$$  which is rational (it's the quotient of two integers)-- a contradiction (since $y$ was assumed to be irrational).
$\square$
...
PS: the following is not really necessary, but it's a useful question:
Why are $pb-aq$ and $qb$ integers? I'll leave you to answer that.
A: Hint : Assume that x+y is a rational. What can you say about y=(x+y)-x ?
A: Instances of this addition law are ubiquitous in concrete number systems, e.g. below.

A slight generalization reveals the group-theoretical essence of the matter:
Lemma $ $ Let $\,Q\,$ be a nonempty subset of the additive group $\,\Bbb R\,$ (or any abelian group). Let $\,\overline Q\,$ be the complement  of $\,Q\,\,$in $\,\Bbb R.\,$ Then $\,Q + \overline Q\,\subseteq\, \overline Q\!\iff\! Q - Q \subseteq Q\!\iff\! Q\,$ is a subgroup of $\,\Bbb R$.
So $\,\Bbb Q + \overline{\Bbb Q}\,\subseteq\, \overline{ \Bbb Q},\,$ i.e. rational + irrational = irrational, is simply a special case of the above, since $\,\Bbb Q\,$ is an additive subgroup of $\,\Bbb R,\,$ being closed under subtraction (so the Subgoup Test applies).
Therefore the above composition law is simply a complementary view of the Subgroup Test. See below for the simple proof

The irrationals are the complement $\,\overline{\Bbb Q}\,$ of the subgroup $\Bbb Q\subset \Bbb C$. But a complement of subgroup is not a subgroup since it does not contain the identity $\,0,\,$ nor is it closed under subtraction, not containing $\,\alpha -\alpha.$ However, we can do some group-like calculations with such complements, such as: rational + irrational = irrational. Such statements are a special case of the following complementary view of a subgroup.
Theorem $ $ Let $\rm\,G\,$ be a nonempty subset of an abelian group $\rm\,H,\,$ with complement set $\rm\,\bar G = H\backslash G.\,$
Then $\rm\,G\,$ is a subgroup of $\rm\,H\iff G + \bar G\, =\, \bar G. $
Proof $\ $ $\rm\,G\,$ is a subgroup of $\rm\,H\iff G\,$ is closed under subtraction, so, complementing
$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\  fails\\
&\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\
&\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\  for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED}
\end{eqnarray}$
