Classify the following relations as reflexive, irreflexive, symmetric, antisymmetric or transitive. Explain each property in the context of the question. “is greater than or equal to” on the set of real numbers.

My question is to classify the relations do I keep the values for $x, y$ and $z$ the same for each relation. For example say the set is $\{x,y,z\}$ which can be $\{1,2,3\}$ respectively? If so am I correct in the following?

Reflexive: $xRx$ eg. $1 \geq 1$, $2 \geq 2$ and since this is true, $R$ must be reflexive.


If $xRy$ then $yRx$. If $x = 1$ and $y = 2$ then $1>=2$ it is false so it is not symmetric Is this correct?

Antisymmetric if $xRy$ and $yRx$ then $x=y$

if $xRy$ and $yRx$ then $x=y$, this can only stand true on the condition that $x=y$ so on this condition it is Antisymmetric. Is this correct?

Transitive: If $xRy$ and $yRz$ then $xRz$ It is not transitive as $1 \geq 2 $ and $2 \geq 3$ and $1 \geq 3$ are all false.

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    $\begingroup$ You can use particular values of $x$, $y$, and $z$ to help you understand the relation, but your proof must be valid for any $x$, $y$, and $z$, not just the ones you happened to check. $\endgroup$ – Austin Mohr Apr 11 '13 at 5:57
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    $\begingroup$ You can also use particular values as counterexamples, as you did to violate symmetry. $\endgroup$ – Austin Mohr Apr 11 '13 at 6:06

To expand on my comment, let me show you how I might prove the reflexivity and antisymmetry of the $\leq$ relation. I invite you to try transitivity and ask questions in the comments.

To establish that $\leq$ is reflexive, we want to show $x \leq x$ for any (presumably real number) $x$. I would start with an absolutely true fact and work toward what I want:

\begin{align*} 0 &\leq 0\\ 0 + x &\leq 0 + x\\ x &\leq x \end{align*}

For antisymmetry, suppose any $x$ and $y$ are given. All we know about them is that $x \leq y$ and $y \leq x$. From the first fact, I can derive $x - y \leq 0$, and from the second I can derive $0 \leq x - y$. Putting them together, I have $0 \leq x - y \leq 0$, which means $x - y = 0$, which means $x = y$.

  • $\begingroup$ I don't see how your displayed equation proves anything. You can't prove $x\leq x$ without using the definition (as you did later for antisymmetry). $\endgroup$ – David Hill Feb 16 '16 at 15:08
  • $\begingroup$ @DavidHill I suppose it is a bit circuitous. $0 \leq 0$ is really no simpler than $x \leq x$, so I could have just cut to the chase. $\endgroup$ – Austin Mohr Feb 17 '16 at 16:14

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