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I have a question that I got wrong in my homework and I am having trouble understanding.

It says "Determine whether the relation R on the set of all real numbers is reflexive, sym- metric, antisymmetric, and/or transitive, where (x, y) ∈ R if and only if"

One of these are x=1.

This means the relation looks like this:

1,1 1,2 1,3 1,4

and so on. That means that X will never not be one. I determine this as reflexive because every x has a matching y, such as 1,1. However, if every Y also has to have a matching X, then I am wrong.

Is my latter guess correct? Does reflexive mean if x,y, every x must have a matching x,x or does it also apply that every y must have a y,y.

munchschair
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1 Answers1

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Reflexive means that for every $t$, $(t,t)$ is in the relation. The relation $x=1$ is not reflexive as $(2,2)$ is not in the relation.

please delete me
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  • It must go both ways then. If I have a 1,2, then a 2,2 must exist as well. – munchschair Dec 18 '14 at 17:39
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    @munchschair: no, for reflexive you just care about $(t,t)$. All pairs like that must be in the relation for the relation to be reflexive. Whether $(1,2)$ is in the relation is immaterial to whether the relation is reflexive-you can have all the extra pairs you want as long as you have the required ones. – Ross Millikan Dec 18 '14 at 17:50
  • So any number X in the relation must have a x,x pair in the relation? – munchschair Dec 18 '14 at 18:32
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    @munchschair: For each $x$ in the set on which the relation is based you must have $\langle x,x\rangle$ in the relation. In your problem that set is the set of real numbers, so $R$ would be reflexive only if $\langle x,x\rangle$ were in $R$ for every real number $x$: $\langle\pi,\pi,\rangle\in R,\langle 2,2\rangle\in R,\langle\sqrt2,\sqrt2\rangle\in R$, etc. – Brian M. Scott Dec 18 '14 at 21:40