# Check if a relation is reflexive, symmetric and transitive

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.

• "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. Feb 16, 2021 at 11:38
• 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. Feb 16, 2021 at 11:43
• 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? Feb 16, 2021 at 11:46
• @j_random_hacker Yeah too many correct concusions [sic] ... that will leave one's reasoning not all there. Feb 16, 2021 at 11:50
• "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. Feb 16, 2021 at 17:42