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The question in my book says:

Determine whether the relation defined on the set of positive integers is reflexive, symmetric, antisymmetric, transitive, and/or a partial order.

$x = y^2 \rightarrow (x,y) \in R$

I thought it was antisymmetric, but also transitive, and symmetric, and not reflexive. My reasoning was that $R = \{(1,1)\}$ because $1$ is the only positive integer that will equal its square. So it is trivially antisymmetric, transitive, and symmetric. While not being reflexive since $(2,2) \notin R$. Have I been mistaken? The answer in the back of my book says only:

Antisymmetric

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Leonardo
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  • It doesn't say $x=x^2$, which would not be a relationship between two variables, after all. It says $x=y^2$. – Thomas Andrews Feb 26 '13 at 06:16
  • What is the domain of this relationship - that is, what can $x,y$ be? If $x,y$ can be complex numbers, then it technically isn't even anti-symmetric, since if $z^3=1$ and $z\neq 1$ then $(z,z^2)\in R$ and $(z^2,z)\in R$ – Thomas Andrews Feb 26 '13 at 06:21
  • @ThomasAndrews It is defined on the positive integers. We can use $x = 1$ and $y = 1$ if we want. – Leonardo Feb 26 '13 at 06:22
  • Why do you insist that $(x,y)=(1,1)$ is the only element of $R$. $R$ contains lots more elements. – Thomas Andrews Feb 26 '13 at 06:25
  • @ThomasAndrews Wait, what elements? I need an example please. You are saying that there is a pair for $x, y \in \mathbb{Z}^+$ s.t. $x = y^2$ and $(x,y) \neq (1,1)$ ? – Leonardo Feb 26 '13 at 06:27
  • How about $(x,y)=(4,2)$? – Thomas Andrews Feb 26 '13 at 06:32
  • @ThomasAndrews Ah thanks, I don't know why I am having trouble thinking such a simple example! Probably just in the mind-set that $y = x$ is the law of the land for some reason. – Leonardo Feb 26 '13 at 06:34

1 Answers1

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Consider:

(Transitivity) If $x=y^2$, and $y=z^2$, does that imply that $x=z^2$?

(Symmetry) If $x=y^2$, does that imply $y=x^2$?

Zev Chonoles
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  • I think I see what you mean, because I was looking at it as 1 related to 1 as the only member in the relation. But you are looking at it from the perspective of the rule. Unfortunately in my opinion since 1 is the only number that I can think of besides 0 which equals its square, $x = y^2$ and $y = z^2$ does indeed imply that $x = z^2$ since 1 is still the only option for $x, y,$ and $z$. I can not think of a counter-example to transitivity or symmetry. – Leonardo Feb 26 '13 at 06:21
  • You're thinking of whether or not there exists some $x$ and $y$, or some $x$, $y$, and $z$, that make the statement true; what is necessary is that the statement be true for all $x$ and $y$, or all $x$, $y$, and $z$. Consider the counterexamples $$16=4^2,\quad 4=2^2,\quad 16\neq 2^2$$ and $$4=2^2,\qquad 2\neq 4^2$$ – Zev Chonoles Feb 26 '13 at 06:23
  • @Leonardo You need to read the definition. $(4,2)\in R$, for example. – Thomas Andrews Feb 26 '13 at 06:23
  • Ah I see where I have been mistaken now, it is for all $x,y \in X$ not $(x,y) \in R$, that was bad mistake! I understand it now. – Leonardo Feb 26 '13 at 06:25
  • @Leonardo: Glad I could help! – Zev Chonoles Feb 26 '13 at 06:26