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I am having trouble understanding the concepts of reflexivity, symmetric, and transitivity For the set of all real numbers, are the following two statements an equivalence relation:

x + y is an integer and

x-y is an irrational number

and whether x<=y and x + y = 0 are circular

Thanks

math12341123
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1 Answers1

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Here's a solved example for $x+y$ is an integer. Try to do the others along the same lines. Recall that then $x \sim y$ iff $x+y$ is an integer, $x, y \in \mathbb{R}$ (which is assumed).

(1) $x \sim x$ (reflexive): If $x \in \mathbb{R}$, then, usually, $x+x = 2x \notin \mathbb{Z}$. Example, $x = \frac{1}{3}$. As one counterexample is enough, it is not reflexive.

(2) $x \sim y \Rightarrow y \sim x$ (symmetric): If $x + y \in \mathbb{Z}$, then $y + x = x+y \in \mathbb{Z}$ as the reals commute, so it is symmetric

(3) $x \sim y, y \sim z \Rightarrow x \sim z$ (transitive): You are given $$x + y \in \mathbb{Z}, y + z \in \mathbb{Z}, \quad (1)$$ and want to conclude $$ x+z \in \mathbb{Z}. \quad (2)$$ Now this must hold for all reals that satisfy (1). So let $$x = \frac{1}{3} = z, y = \frac{2}{3},$$ which is easily seen to satisfy (1). But obviously $x+z = \frac{2}{3} \notin \mathbb{Z},$ so (2) is not satisfied. So you have found one example in which the relation would not be transitive, so it is not transitive, and so this is not an ER.

Edit: as to your comment whether $x \leq y$ is transitive, it is (by well-ordering), but you should try to show this yourself to get some exercise.

gnometorule
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  • I understood "For the set of real numbers" in the question to mean that we want to know whether "$x+y$ is an integer" defines an equivalence relation on the set of real numbers. With this meaning, reflexivity fails, since there are real numbers $x$ for which $x+x$ is not an integer. – Andreas Blass Mar 04 '13 at 01:01
  • @AndreasBlass: Good point, and shamed ducking. I will adjust the solution. Thanks for your vigilance. – gnometorule Mar 04 '13 at 01:02