Union of sets proof Prove that $\{3t\}\cup\{3t+1\}\cup\{3t+2\}=\Bbb Z$, where $t$ is in the set of integers.
It makes sense that you can get every integer from this Union of sets but how would you prove something like that? 
 A: First show that $\{ 3 \mathbb{Z} \} \cup \{ 3\mathbb{Z} + 1 \} \cup \{ 3\mathbb{Z} +2 \} \subset \mathbb{Z}$
let $x \in \{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\}$. 
then $x = 3r$ for some $r$ in $\mathbb{Z}$, or  $x = 3s+1$ for some $s$ in $\mathbb{Z}$, $x = 3t+2$ for some $t$ in $\mathbb{Z}$
Since in any of these three cases $x$ is an integer, $x \in \mathbb{Z}$
So we have that if $x \in \{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\}$ then $x \in \mathbb{Z}$ which is the defenition of $\{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\} \subset \mathbb{Z}$.
Secondly show that  $\{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\} \supset \mathbb{Z}$
Let $x \in \mathbb{Z}$. Then consider the remainder $\frac{x}{3}$, it will be a remainder of either $1, 2$, or $0$. In the case where it is $1$, then $x = 3r+1$ for some $r \in \mathbb{Z}$, so $x \in \{3\mathbb{Z} + 1\}$ . If the remainder is $2$, then $x = 3r+2$ for some $r \in \mathbb{Z}$ and $x \in \{3\mathbb{Z} + 2\}$. If the remainder is $0$, then $x = 3r$ for some $r \in \mathbb{Z}$ and $x \in \{3\mathbb{Z} \}$. In any of the three cases, we have that $x in \{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\} $ which gives us that $\mathbb{Z} \subset \{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\}$.
Since $\mathbb{Z} \subset \{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\}$ and $\{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\} \subset \mathbb{Z}$, we have that $\{3\mathbb{Z}\}\cup \{3\mathbb{Z} + 1\} \cup \{3\mathbb{Z} +2\} = \mathbb{Z}$
A: Just apply the Euclidean division by $3$. For any integer $n$ there exist some $t\in\Bbb Z$ and some $r\in\{0,1,2\}$ such that $n=3t+r$. Furthermore, the pair $(t,r)$ with these properties is unique.
A: ajotatxe gave the easiest way to prove it. There's also another way of proving it. Define the relation $\sim$ on $\mathbb{Z}$ by:$$\forall a,b\in\mathbb{Z},\,a\sim b\Leftrightarrow3|a-b$$
$\sim$ is an equivalence relation on $\mathbb{Z}$ and you can easly check that $\{3t\}$, $\{3t+1\}$ and $\{3t+2\}$ are its equivalence classes. So they form a partition of $\mathbb{Z}$ and so $\mathbb{Z}=\{3t\}\cup\{3t+1\}\cup\{3t+2\}$.
You can also work on the quotient groups $\mathbb{Z}/3\mathbb{Z}$ if you have some knowledge on that, but that's part of group theory not elementary set-theory.
Edit: By the way, ajotatxe's answer is how you check that $\{3t\}$, $\{3t+1\}$ and $\{3t+2\}$ are $\sim$'s equivalence classes, so we're back to ajotatxe's answer.
