Help create this relation

$A=B=\mathbb{Z}^+$.

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?

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Is it really obvious what B is supposed to be? I am a bit confused. – Rankeya Nov 4 '11 at 21:19
@Rankeya: Sorry, I fixed it. It was a typo error. – Fahad Uddin Nov 4 '11 at 21:22
@Akito: The relation doesn't contain integers; it contains pairs of integers. So it cannot contain multiples of 6, and cannot contain only multiples of 6. Every element of the relation is an ordered pair of integers. – Arturo Magidin Nov 4 '11 at 22:01

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.

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So, the way in which you have written this list, isn't $(10,4)$ not an element of your list above? But it is an element in the relation right as defined by Akiro right? – Rankeya Nov 4 '11 at 21:57
@Rankeya: Oops; all my pairs are reversed, aren't they? ... Thanks! – Arturo Magidin Nov 4 '11 at 21:58
@Akito: You still have not clarified if you are using $\bmod$ as a binary operator (given positive integers $a$ and $b$, $b\bmod a$ is the unique nonnegative integer $r$, $0\leq r\lt a$, such that $b-r$ is a multiple of $a$); or as the indicator of the congruence relation. – Arturo Magidin Nov 5 '11 at 20:45
@ArturoMagidin: It is the modulo operator. – Fahad Uddin Nov 6 '11 at 5:55

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.

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I actually deleted my answer before I saw your comment because I realized my error. I think I fixed it now. – Rankeya Nov 4 '11 at 21:44

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?

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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 $\mathbb{N}$.

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I hope you include $0$ in $\mathbb{Z}^+$. – Rankeya Nov 4 '11 at 21:35