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I found this question in a problem set- Check whether the relation R in $\mathbb{R}$ defined by:

$$R=\{(a,b):a\leq b^3\}$$

is reflexive, symmetric or transitive I found an example to show that it is not transitive-

$$3<(\frac{3}{2})^3$$

$$\frac{3}{2}<(\frac{6}{5})^3$$

But:

$$3>(\frac{6}{5})^3$$

So it isn't transitive

But in many cases it becomes harder to find examples for instance where powers are different so is there a way to prove this in general for the given relation that it isn't transitive.

Ted
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    It can become arbitrarily hard to find counterexamples that disprove transitivity ... – Hagen von Eitzen Jul 16 '19 at 15:36
  • @HagenvonEitzen yes that's why I'm asking is there an analytical method to reach this conclusion? –  Jul 16 '19 at 15:36
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    No, because, depending on the relation at hand, it can become arbitrarily hard to prove $\exists x,y,z\colon xRy\land yRz\land \neg xRz$, whereas coming up with a witness is still the simplest method to prove such an existential statement – Hagen von Eitzen Jul 16 '19 at 15:41
  • @HagenvonEitzen I know it's arbitrarily hard but is there a way? Has there been a research paper or proof for the same? I know it's of no use especially at my level but I'm just curious for the answer –  Jul 16 '19 at 16:40

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Define the relation $R_n$ with $n\ge3$ by $$aR_nb\iff a\le b^n$$ This relation is not transitive because for $a\gt2$ we have $$aR(a-1)\iff a\le(a-1)^n\\(a-1)R(\sqrt[n]{a-1})^n)\iff(a-1)\le(\sqrt[n]{a-1})^n=(a-1)$$ But it is clear that $a$ is not in relation with $(\sqrt[n]{a-1})^n=(a-1)$ because $a\not\le(a-1)$.

Piquito
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