2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Hint: $R_1$ contains, among other pairs, $(1,5)$, $(5,1)$, and $(10,\frac{1}{2})$. $\endgroup$ – vadim123 Nov 4 '15 at 0:53
  • 1
    $\begingroup$ Reflexive means (x ,x) in R $\endgroup$ – Shailesh Nov 4 '15 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 '15 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 '15 at 0:55
  • $\begingroup$ Thank you bof, sorry, I'm really new to this. $\endgroup$ – William Nov 4 '15 at 0:56
3
$\begingroup$

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 $\sqrt5\cdot\sqrt5=5$.

$\endgroup$
  • 2
    $\begingroup$ 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 '15 at 1:15
  • $\begingroup$ @William Right! $\endgroup$ – bof Nov 4 '15 at 1:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.