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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?

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Note that you're saying more than just set here: You're requiring that the objects of that set can be added. –  Semiclassical Jul 25 '14 at 16:50
@Semiclassical Yeah, thanks! What word should I use instead of set, then? –  alexqwx Jul 25 '14 at 16:51
This can be written as: $S$ is closed under subtraction. That is, if $a,b\in S$ and $a-b$ exists, then $a-b\in S$. –  Thomas Andrews Jul 25 '14 at 16:52
@Semiclassical I don't see how this contradicts $(2).$ –  alexqwx Jul 25 '14 at 16:54
@Semiclassical No, you misread my comment. The rationals are closed under subtraction. That is what the above says... –  Thomas Andrews Jul 25 '14 at 16:55

4 Answers 4

up vote 9 down vote accepted

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 $G$.

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Why have you used $xh$ instead of $x+h$ in your generalisation (does it matter?)? –  alexqwx Jul 25 '14 at 16:58
$xh$ just means the group operator acting on the pair $(x,h)$ if the operation in the particular group is addition then $xh$ is just $x+h$. the plus sign is normally used for abelian groups. But this result generalizes to any group. –  Modded Bear Jul 25 '14 at 17:03
Got it. Thanks! One final thing: why did you mention post-multiplying rather than just multiplying (since addition is commutative)? –  alexqwx Jul 25 '14 at 17:05
well, in general groups don't need to be commutative, however this assumption is not required for the result to be true. –  Modded Bear Jul 25 '14 at 17:22
Talking about this in terms of groups seems just a bit too strict. The first example given in the question works just as well, for example, if we were to only talk about positive even integers (no inverse or zero element under addition.) –  Semiclassical Jul 25 '14 at 17:51

I call this the complementary subgroup law, because 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

$\begin{eqnarray} & &\ \ \rm G\text{ is a subgroup of }\, H\ fails\\ &\iff&\ \rm\ G\ -\ G\ \subseteq\, G\,\ \ fails\\ &\iff&\ \rm\ g_1\, -\ g_2 =\,\ \bar g\ \ \ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ g_2\, +\ \bar g\ \ =\,\ g_1\ for\ some\ \ g_1,g_2\in G,\ \ \bar g\in \bar G\\ &\iff&\ \rm\ G\ +\ \bar G\ \subseteq\ \bar G\ \ fails\qquad\ {\bf QED} \end{eqnarray}$

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).

enter image description here

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Nice! ByteErrant, in the answers below, has shown that prime +non-prime can be either prime or non-prime itself, which seems to contradict the above theorem. Could you explain? –  alexqwx Jul 25 '14 at 21:36
@alex The set $\,\color{#0a0}G\,$ of $\rm\color{#0a0}{primes}$ is not a subgroup of $\,H =$ integers under addition, since $\,\color{#0a0}G\,$ is not closed under subtraction, e.g. $\,\color{#0e0}{11}-\color{#0a0}7 = \color{#c00}4\,$ is not $\rm\color{#0a0}{prime,\,}$ so $\, \color{#0a0}7+\color{#c00}4 = \color{#0e0}{11}\,\Rightarrow\, \color{#0a0}G+\color{#c00}{\bar G}\subseteq \color{#c00}{\bar G}\,$ fails. That agrees with the (negation of) the theorem, i.e. $\,G+\bar G\subseteq G\,$ fails $\iff$ $\,G\,$ is not a subgroup of $\,H\, $ $\iff$ $\,G\,$ is not closed under subtraction. –  Bill Dubuque Jul 25 '14 at 23:21

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.

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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 $==> 7+4=11$

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Ah. I hadn't thought about this. Thanks! Do you (or does anyone else) have any explanation for this? –  alexqwx Jul 25 '14 at 21:34
@alexqwx Primes don't form a group under addition. Evens do, rationals do, algebraics do, and reals do. –  Cory Jul 25 '14 at 21:41

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