$A=B=\mathbb{Z}^+$.
Define a relation $R$ by $ a\;R; b \text{ iff } b = a \bmod 6.$
Please help me write the set relation. Will the set relation contain only the multiples of 6?
$A=B=\mathbb{Z}^+$.
Define a relation $R$ by $ a\;R; b \text{ iff } b = a \bmod 6.$
Please help me write the set relation. Will the set relation contain only the multiples of 6?
The elements of $R$ are not numbers, they are ordered pairs of numbers: an ordered pair $(x,y)$ will be in $R$ if and only if $x$ and $y$ are positive integers and $y=x\bmod 6$.
If $\bmod$ is the operation that returns the remainder modulo $6$, this being an integer betwen $0$ and $5$ (inclusive), then the pairs in your relation are exactly $\begin{align*} &(1,1), (7,1), (13,1), (19,1), \ldots\\ &(2,2), (8,2), (14,2), (20,2),\ldots\\ &(3,3), (9,3), (15,3), (21,3),\ldots\\ &(4,4), (10,4), (16,4), (22,4),\ldots\\ &(5,5), (11,5), (17,5), (23,5),\ldots \end{align*}$ and nothing else. If the $\bmod$ operator returns an integer between $1$ and $6$, then you get another list of infinitely many integer pairs.
If you mean that $aRb$ if and only if $b\equiv a\pmod{6}$ (if $a$ and $b$ have the same remainder modulo $6$), then there are lots of other pairs in your set, such as $(31,601)$, $(1000, 10)$, etc.
You take the union of all sets of this form $\{(m,m+6n): n \in \mathbb{N}\}$ and $\{(m+6n,m): n \in \mathbb{N}\}$ as $m$ ranges over $\mathbb{N}$. Convention is that $0$ is in $\mathbb{N}$.
No, because $1R7$. A relation is a set of pairs, not a set of numbers. $(1,7)$ is one of the elements, as is $(6,12)$. There are lots more.
You ask for help writing "the set," but you don't say which set. You ask whether the set will contain only multiples of 6, so it seems that the set you have in mind is a set of numbers. I thought the question was going to be about the relation $R$, which is not a set of numbers, but a set of pairs of numbers, namely, those pairs $(a,b)$ such that $b\equiv a\pmod6$. Clarify?