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I have a problem to do that is similar to this: $R_1$ is over the set of real numbers

(a) $(x, y) \in R_1$ if and only if $xy = 5$

decide whether it is reflexive, anti-reflexive, symmetric, anti-symmetric and transitive.

I'm confused, I know that reflexive means x=x and symmetric means that x,y implies y,x. I think it's the format of the question that is throwing me off. Help is very much appreciated.

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  • $\begingroup$ Hint: $R_1$ contains, among other pairs, $(1,5)$, $(5,1)$, and $(10,\frac{1}{2})$. $\endgroup$
    – vadim123
    Nov 4, 2015 at 0:53
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    $\begingroup$ Reflexive means (x ,x) in R $\endgroup$
    – Shailesh
    Nov 4, 2015 at 0:53
  • $\begingroup$ So, this would be that it is not reflexive., it is symmetric and it is transitive? If something is not reflexive is it safe to say it is anti-reflexive or does that have a unique definition, same quesiton applies to the other two relations. $\endgroup$
    – William
    Nov 4, 2015 at 0:55
  • $\begingroup$ Reflexive does not mean $x=x$; it means $(x,x)\in R_1$. Symmetric does not mean "$x,y$ implies $y,x$" (whatever that means), it means that $(x,y)\in R_1$ implies $(y,x)\in R_1$. $\endgroup$
    – bof
    Nov 4, 2015 at 0:55
  • $\begingroup$ Thank you bof, sorry, I'm really new to this. $\endgroup$
    – William
    Nov 4, 2015 at 0:56

1 Answer 1

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Is $R_1$ reflexive? That means: Does $xx=5$ hold for every real number $x$?

Is $R_1$ anti-reflexive? Does $xx\ne5$ hold for every real number $x$?

Is $R_1$ symmetric? Does $xy=5$ imply $yx=5$ for real numbers $x,y$?

Is $R_1$ transitive? Does $xy=5$ & $yz=5$ imply $xz=5$ for real numbers $x,y,z$?

First, $R_1$ is not reflexive, because $17$ is a real number and $17\cdot17\ne5$.

Next, $R_1$ is not anti-reflexive, because $\sqrt5$ is a real number and $\sqrt5\cdot\sqrt5=5$.

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    $\begingroup$ So, R1 is symmetric because for any x and y, xy = yx by commutative law R1 is not transitive because if 1x5 = 5 and 5x1 = 5 1x1 does not equal 5. $\endgroup$
    – William
    Nov 4, 2015 at 1:15
  • $\begingroup$ @William Right! $\endgroup$
    – bof
    Nov 4, 2015 at 1:21

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