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I've seen relations defined as functions between sets and as sets of ordered sets; however, I've never seen a relation defined as $3\mid(a+2b)$. What does this mean?

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I'll try and express my reasoning. Presuming I understand the relation as he expressed it, I presume it is symmetrical since if a=b then 3a is divisible by 3, and thus a multiple of it. For symmetry, it would not seem to matter whether one assigns value x to a and value y to b or vice versa, in either case the sum is the same, so they relate if 3 divides that sum. For transitivity, there is no third variable under consideration so I suppose it is vacuously true that they are transitive.

Martin Sleziak
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Hal
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    It means that $(a,b)\in R$ if and only if $a+2b$ is a multiple of $3$. Does that help you investigating reflexivity, symmetry and transitivity? – Daniel Fischer Nov 04 '14 at 14:22

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It means $aRb$ iff $a+2b$ is divisible by $3$. I'll check symmetry, leaving the rest to you: suppose $aRb$ so that $a+2b$ is divisible by $3$. Then, $$ b+2a=(3b+3a)-(a+2b) $$ is also divisible by $3$ so you also have $bRa$.

yurnero
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The relation means that $(a, b)\in R$, or $a \sim b$ ($a$ is related to $b$) if and only if $3$ divides $a + 2b$. So, for example, $(1, 1) \in R$, since $1 + 2(1) = 3$, which is certainly divisible by $3$.

You need to show whether the relation is reflexive, symmetric, and/or transitive.

E.g. Reflexivity: $\forall x \in \mathbb Z$, is $(x, x) \in R$? Well $x + 2x = 3x$, which is divisible by $3$. So the relation is reflexive.

Symmetry: Is it the case that for all $x, y \in \mathbb Z$, if $(x, y) \in R$, then $(y,x) \in R$? Suppose $(x, y) \in R$. Then $3\mid x + 2y$. This means that $x+2y = 3k$ for some integer $k$. Certainly, we also know that $3\mid 3(x+y)$. Now, $y + 2x = 3(x+y) - (x +2y)$ must be divisible by $3$ since $3(x+y)$ and $x+2y=(3k)$ is divisible by $3$. Hence, if $(x, y) \in R,$ then so too $(y, x) \in R$.

Transitivity: Is it the case that So for $x, y, z \in \mathbb Z$, IF $(x, y) \in R$ and $(y, z)\in R$, it follows that $(x, z) \in R$? I'll let you work at tackling this. Assume $(x, y)\in R,$ and $(y, z)\in R$. Now, unpack what this means, and from this assumption, check whether it then follows that $3\mid x + 2z$, i.e., if it then follows that $(x, z)\in R$.

amWhy
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  • I was going to try and post my reasoning but I don't feel I had the vocabulary needed to express it. But here goes: it is symmetrical since if a=b then 3a is divisible by 3, and thus a multiple of it. For symmetry, it would not seem to matter whether one assigns value x to a and value y to b or vice versa, in either case the sum is the same, so they relate if 3 divides that sum. For transitivity, there is no third variable under consideration so I suppose it is vacuously true that they are transitive. – Hal Nov 04 '14 at 14:34
  • @Hal What you say is actually no evidence for transitivity. In fact, rather tautologically, every binary relation, like the one you consider here, is defined for, well, two variables (up to parameters). For example, "a and b are related iff |a - b| < 3" is not transitive (since, e.g., 0 and 2 are related, 2 and 4 are related, but 0 and 4 are not). To examine transitivity, you have to assume that two arbitrary pairs (a,b) and (b,c) are in the relation, and then check if the relation "transits" the element b, so that (a,c) must also be in the relation. – Basil Nov 04 '14 at 15:11