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The relation R is on the set of all integers, where (x,y) is an element of R if and only if x is a multiple of y.

Would this relation be reflexive, symmetric, antisymmetric, and/or transitive?

I have determined it to be reflexive, antisymmetric, and transitive... Apparently my textbook tells me that it is NOT antisymmetric. Yet I would think it would be because "x is a multiple of y" should mean that if (x,y) is an element and (y,x) is an element then x=y, because x can be simply (1*y).

ohbrobig
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    By "all integers" you mean all signed integers, I assume (because your book says $R$ is not antisymmetric). It seems you're confused about what antisymmetry means. It means: if $R(x, y)$ and $R(y,x)$ then $x = y$. $R$ is NOT antisymmetric if there are some $x,y$ such that $R(x, y)$ and $R(y,x)$ but $x \neq y$. You only have to find two numbers such $x,y$ to show $R$ not antisymmetric. – BrianO Nov 02 '15 at 00:18
  • All integers means all whole numbers (signed integers, so both positive and negative) including 0...I noticed I made a typo at first. I think I do for the most part get antisymmetry. – ohbrobig Nov 02 '15 at 00:22
  • The tag "equivalence relations" is misleading. This is clearly not an equivalence relation (it is not symmetric). Also, over all the integers, the relationship is indeed not antisymmetric, as $1$ is a multiple of $-1$ and vice versa. In proper jargon, $\mathbb{Z}$ has two different units (invertible elements), $1$ and $-1$, that's why it's not antisymmetric. It would be over the naturals though. – Fryie Nov 02 '15 at 01:12
  • That's a subtle but very important point in general if you start considering other structures e.g. rings of polynomials. Two objects that are multiples of each other are not necessarily the same - they are only the same up to a unit multiplier. For example, in $\mathbb{Q}$ every number is a multiple of every other number. That's because each number is invertible (i.e. a unit). Similarly, in the ring of polynomials over $\mathbb{Q}$, $5x$ and $2x$ are multiples of each other, but not the same. – Fryie Nov 02 '15 at 01:20

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Hint: $3$ is a multiple of $-3$, and $-3$ is a multiple of $3$.

Caleb Stanford
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