I was studying binary relations and, while solving some exercises, I got stuck in a question.

I am sharing the question and my thoughts on solving it, and I am looking for some advice and comments about my attempt (what is wrong or what should I do to improve it).

Question. Let $A = \mathbb{N} \setminus \{1\}$ and consider the following binary relations on $A.$ $$R_1 = \{(x,X) : X \in \mathcal{P}(A) \wedge x \in X\}, \quad R_2 = \{(x,y) \in A^2 : x < y\}, \quad \quad R_3 = \{(x,y) \in A^2 : y > x^2\}.$$

Determine, justifying, if each of the above relations are reflexive, symmetric, transitive or antisymmetric.

Sketch. Of course, to solve this problem, one must understand what does it mean for a binary relation to be reflexive, symmetric, transitive or antisymmetric. So, let’s, first, recall the definition of each concept.

Let $A$ be a set $R \subseteq A^2$ a binary relation on $A.$ The binary relation $R$ is

$1.\quad$ reflexive, if $\quad \forall a \in A, aRa$;

$2.\quad$ symmetric, if $ \quad \forall a,b \in A, aRb \implies bRa$;

$3.\quad$ transitive, if $ \quad \forall a, b, c \in A, aRb \wedge bRc \implies aRc$;

$4.\quad$ antisymmetric, if $\quad \forall a,b \in A, aRb \wedge bRa \implies a = b.$

Hence, we must check if these conditions are satisfied for each of the above relations.


$\qquad 1. \quad R_1.$

This relation was include in this exercise, but I don’t agree with this. Because, $R_1 \subseteq \mathcal{P}(A) \times A,$ and the question states that the relations that we are working on are relation on $A.$

But forgetting this for a moment, those properties were only defined for binary relations on a set $A$ and not for a binary relation from $A$ to $B.$ Therefore, it makes no sense in talking about those properties in this example.

$\qquad 2. \quad R_2.$

Let $n \in A.$ The proposition $n < n$ is false, hence $(n,n) \notin R_2.$ Therefore, $R_2$ is not reflexive.

Let $m, n \in A.$ Suppose that $(m,n) \in R_2.$ Then, by definition of $R_2$ we have that $m < n.$ Then, it is not true that $n < m.$ So, $(n,m) \notin R_2.$ Therefore, $R_2$ is not symmetric.

Let $m, n, p \in A.$ Suppose that $m R_2 n$ and $n R_2 p.$ Then, $m < n$ and $n < p.$ Since $<$ is transitive, then $m < p$ and so $m R_2 p.$ Therefore, $R_2$ is transitive.

Let $m, n \in A.$ Suppose that $m R_2 n$ and $n R_2 m.$ Hence, we have that $m < n$ and $n < m$ which is a contradiction and so $$\forall a,b \in A, aRb \wedge bRa \implies a = b$$ is vacuously true, Therefore, $R_2$ is antisymmetric.

$\qquad 3. \quad R_3.$

Let $n \in A.$ Since $n \geq 2,$ then $n^2 > n.$ So, it is not true, that $n > n^2.$ Hence, $(n,n) \notin R_3.$ Therefore, $R_3$ is not reflexive.

Let’s $m, n \in A.$ Suppose that $m R_3 n.$ Then, $n > m^2.$ It follows that $n^2 > m^4$ and $m^4 > m.$ Hence, $n^2 > m.$ Therefore, $R_3$ is symmetric.

Let’s $m, n, p\in A.$ Suppose that $mR_3n$ and $nR_3p.$ Then, $n > m^2$ and $p > n^2.$ Because $n^2 > n,$ then $p > m^2.$ Therefore, $R_3$ is transitive.

Let’s $m,n \in A.$ Suppose that $mR_3n$ and $nR_3m.$ Then $n > m^2$ and $m > n^2.$ Since, $m^2 > m$ then $n > m.$ So $n \neq m.$ Therefore, $R_3$ is not antisymmetric.

My biggest doubt is definitely on $R_3.$ I don’t know why, but that looks a bit suspicious to me. Although I have no clue of what is wrong. Can you help me?

Thank you in advance for your attention.

  • 1
    $\begingroup$ Your suspicion for $R_3$ is right, there's an issue with one of the proofs. Maybe try checking each property with an example like $(2,5)$. $\endgroup$
    – Dan Simon
    Jan 3, 2021 at 19:45
  • $\begingroup$ @DanSimon it is clear that $(5,2) \notin R_3$ and for that $R_3$ can’t be symmetric... but what was the error with my argument? $\endgroup$
    – Air Mike
    Jan 3, 2021 at 20:05
  • $\begingroup$ "It follows that $n^2>m^4$ and $m^4>m$. Hence, $n^2>m$." This is technically a true statement, but it's not showing symmetry for $R_3$. Just pay really close attention to what you're actually saying vs what you need to prove. $\endgroup$
    – Dan Simon
    Jan 3, 2021 at 21:28

1 Answer 1


You’re right about $R_1$, except that it’s a subset of $X\times\wp(A)$, not of $\wp(A)\times A)$.

Your analysis of $R_2$ is correct.

Note that $R_3$ would not be reflexive even if $1$ were in $A$: as long as there is at least one $a\in A$ such that $\langle a,a\rangle\notin R_3$, $R_3$ is not reflexive.

$R_3$ is not symmetric: if $\langle n,m\rangle,\langle m,n\rangle\in R_3$, then $m>n^2$ and $n>m^2$, so


and hence $m>m^2$, which is false for every $m\in A$. Thus, not only is $R_3$ not symmetric, it is asymmetric: if $m\mathrel{R_3}n$, then $n\not\mathrel{R_3}m$. It is true that if $n>m^2$, then $n^2>m^4>m$, so $n^2>m$, but that actually implies that $n\not\mathrel{R_3}m$: $n\mathrel{R_3}m$ means that $m>n^2$.

Your argument for transitivity of $R_3$ is correct. But as I showed above, $R_3$ is asymmetric, so it, like $R_2$, is vacuously antisymmetric.

  • $\begingroup$ Thank you so much for the answer. At first I didn’t understood why $R_1$ was not a subset of $A \times \mathcal{P}(A)$ but now it is all clear in my mind. Also, in fact, there was a mistake that I did (it was required to prove that $m > n^2$ and not $n^2 > m$). $\endgroup$
    – Air Mike
    Jan 4, 2021 at 15:19
  • $\begingroup$ Once again, thank you for the answer. Just one short question. I would really like to know more about binary relations. Is there any books or texts that you would recommend as a good introduction to the study of binary relations? $\endgroup$
    – Air Mike
    Jan 4, 2021 at 15:21
  • 1
    $\begingroup$ @AirMike: You’re welcome. Most of the common sophomore-level discrete math texts have basic coverage, some more than others; I’ve been retired long enough that I no longer have a good picture of what’s available, but I seem to remember that the chapter on relations in the text by Kolman, Busby, and Ross had a bit more than some others that I used over the years. $\endgroup$ Jan 4, 2021 at 18:14
  • 1
    $\begingroup$ Once again, thank you, i really appreciate it. I will take a look at those texts :) $\endgroup$
    – Air Mike
    Jan 5, 2021 at 14:53

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