What's the name of this algebraic property? [Complementary Subgroup Test] I'm looking for a name of a property of which I have a few examples:
$(1) \quad\color{green}{\text{even number}}+\color{red}{\text{odd number}}=\color{red}{\text{odd number}}$
$(2) \quad \color{green}{\text{rational number}}+\color{red}{\text{irrational number}}=\color{red}{\text{irrational number}}$
$(3) \quad\color{green}{\text{algebraic number}}+\color{red}{\text{transcendental number}}=\color{red}{\text{transcendental number}}$
$(4) \quad\color{green}{\text{real number}}+\color{red}{\text{non-real number}}=\color{red}{\text{non-real number}}$
If I were to generalise, this, I'd say that if we partition a set $X$ into two subsets $S$ and $S^c=X\setminus S$, then the sum of a member of $S$ and a member of $S^c$ is always in either $S^c$ or $S$.
My question is:
"Is there a name for this property (in these four cases) and is this property true in general?"
Also, does anyone have any more examples of this property?
 A: I call it the Complementary Subgroup Test, since the composition law arises via the  following complementary view of the Subgroup Test ($\rm\color{#c00}{ST} $), $ $  cf. below from one of my old sci.math posts.
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 $\rm\,H\!\!\overset{\ \large \color{#c00}{\rm ST}}\iff\! G\,$ is closed under subtraction, so, complementing
$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\  fails\\[.2em]
&\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\
&\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_i\in G,\ \ \bar g\in \bar G\\
&\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\  for\ some\ \ g_I\in G,\ \ \bar g\in \bar G\\[.2em]
&\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED}
\end{eqnarray}$ 
Instances of this law are ubiquitous in concrete number systems, e.g. below. For many further examples see some of my prior posts here (and also on sci.math).

A: I think this comes from the fact that if you have a group $G$ and $H$ a subgroup  of $G$ then if $h\in H$ and $x\not\in H$ we get $xh\not\in H$.
The proof is by contradiction, suppose $xh=l$ with $l\in H$. Then postmultiplying by $h^{-1}$ gives $x=lh^{-1}$ which is in $H$ since $H$ is a subgroup of $G$.
A: This is basically just "the complement" of the statement that the sets of numbers in green are subgroups of $(\Bbb R,+)$.
Given that for a subgroup $S<\Bbb R$ $a,b\in S$ implies $a-b\in S$, you can quickly compute that if $c\notin S$, $a+c=b\in S$ implies $a-b=c\in S$, a contradiction. Therefore $a+c\notin S$ for any $a\in S$, $c\notin S$.
You could also add to your list anything like "an integer plus a noninteger is a noninteger" and "for any subring $S\subseteq \Bbb R$, an element in $S$ plus an element outside of $S$ results in an element outside of $S$." Again, you don't even really need a subring: this is true for any proper subgroup.
A: Don't prime numbers offer a counter-example to the general truth of this property?
Prime $+$ not-prime $=$ not-prime $==> 17 + 4 = 21$
Prime $+$ not-prime $=$ prime $==> 7+4=11$
