Is the set $\{n: n=10k \text{ for some } k \in \Bbb Z\}$ closed under addition? is it associative? identity? Inverse?
I believe that it is closed under addition since the addition of factors of $10$ are also divisible by $10$.
I also believe it is to be associative.
I'm struggling on how to show that is has an identity and an inverse though
Best Answer
Let $G= \{10n~\vert~n \in \Bbb Z \}$ be the set of integer multiples of $10$.
Claim: $G$ is closed under addition, addition is associative in $G$, $G$ has an additive identity and $G$ is closed under additive inverses.
Proof: Assume $x, y \in G$. Then $\exists m, n \in \Bbb Z$ with $x=10m, y=10n$. Then $x+y=10m+10n=10(m+n)$, and $m, n \in \Bbb Z \Rightarrow m+n \in \Bbb Z \Rightarrow x+y \in G$, so $G$ is closed under addition.
Addition is associative in $G$ because it is associative in $\Bbb Z$, and $G \subseteq \Bbb Z$.
$0=10 \times 0 \in G. \forall x \in G~(x+0=0)$ so $G$ has an additive identity; namely, $0$.
Finally, assume $x =10m \in G$. Then $m \in \Bbb Z \Rightarrow -m \in \Bbb Z \Rightarrow 10(-m)=-10m = -x \in G$. Of course $x+(-x)=0$ so $G$ is closed under additive inverses.