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I want to better understand how this actually works, as my solutions are sometimes not 100% correct.

I have the following relation:

Check if the following relation is reflexive, symmetric, and/or transitive:

$$ R_1 = \{ (x,y) \mid x,y \in \mathbb{R}, x=0 \land y \geq 0 \}. $$

so by that

$$ R_1 = \{ 00, 01, 02,03,04,05,06,07,08,09,010, \dots, 0R_+ \} $$

Basically $R_1$ is 0 and any $R_+$ number.

It is not reflexive, as $(a,a)$ is not in $R_1$. I only have 00, but not 11 or 22 and so on.

It is also not symmetric, as I don't have 01 and 10 or 02 or 50 and 05. So $xRy$ and $yRx$ are not true for $R_1$.

As for the transitivity, well, if $xRy$ and $yRz$ then $xRz$. Well, this one is hard to understand. I could use 00 as an example: If $y = 0$ and $z \geq 0$ then $xRz$ would work. So I would say it is transitive.

Can anybody confirm if this would be correct? If not, i would really appreciate a correct approach then for this task.

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  • $\begingroup$ "xRy and yRx are not true" doesn't make sense. xRy iff yRx is not true makes sense. Assuming you know what quantification over x & y you are sloppily leaving implicit. "if xRy and yRz then xRz" Again, sloppy. Make sure you know how to make quantification explicit--after which you will actually be saying what is so instead of something that doesn't actually make sense. To disprove something for all values of variables, a counterexample suffices. To prove it, an example doesn't. Justify that for all x,y,z, if xRy and yRz then also xRz. PS Memorize & use definitions. PS p->q means (not p) or q. $\endgroup$
    – philipxy
    Feb 16, 2021 at 11:38
  • $\begingroup$ Your concusions are correct but your reasoning for transitivity is not all there. The relation is transitive if and only if for every x, y, z such that xRy and yRz both hold, xRz holds. That means we care only about variable values that satisfy the LHS (xRy and yRz). But the only y for which the second term (yRz) can ever be true is y=0, so we can immediately fix y=0. So, the question is now: For every x, z, if xR0 then is 0Rz? The RHS (0Rz) is always true, so the entire statement is true. $\endgroup$ Feb 16, 2021 at 11:43
  • $\begingroup$ Various versions of relational algebra & calculus are for various versions of n-ary database relations, which math binary relations may or may not be a special case of. But you're not querying or even operating on your relations, so why did you use those tags? $\endgroup$
    – philipxy
    Feb 16, 2021 at 11:46
  • $\begingroup$ @j_random_hacker Yeah too many correct concusions [sic] ... that will leave one's reasoning not all there. $\endgroup$
    – philipxy
    Feb 16, 2021 at 11:50
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    $\begingroup$ "i have {ab} and {bc}" for your latest 7-element R (please use new names for new things) does not cover every case of x, y & z; you have to show the if is true for every possible x, y & z. One case of x, y & z satisfying the if part does not imply that the if holds for all x, y & z.. Let xyz be abc. cRb & bRa, but cRa doesn't hold; so the for all x, y, z doesn't hold; so this isn't transitive. We've both told you, for all x, y, z. $\endgroup$
    – philipxy
    Feb 16, 2021 at 17:42

1 Answer 1

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"If xRy and yRz": Look at the definition of R. You have x = 0 and y ≥ 0 because xRy. You have y = 0 and z ≥ 0 because of yRz. If you have xRy and yRz then you have x = 0 and z ≥ 0, therefore xRz. So R is transitive.

In total you know that x = 0, y = 0 and z ≥ 0, so you also have yRx, although nobody cares much about that.

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