The question asks:

Define a relation $R$ on the set of functions from $R$ to $R$ as follows:

$$(f,g) \in R \text{ if and only if } f(x) − g(x) \geq 0 \text{ for all } x \in R$$

Is this relation reflexive? Symmetric? Transitive? Is it an equivalence relation? Explain.

So far I have that the relations is reflexive because $f(x)-f(x) \geq 0$, which is true.

But I'm not quite sure if the relation is symmetric or transitive as I am not quite familiar.

  • $\begingroup$ Welcome to MSE. $R$ is not symmetric, because $(f,g)\in R \not\implies (g,f)\in R$, so $R$ is not an equivalence relation $\endgroup$ – J. W. Tanner Apr 7 '19 at 3:46
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    $\begingroup$ Does the condition $f(x)-g(x)\ge0$ look "symmetric" in the functions $f$ and $g$? $\endgroup$ – Angina Seng Apr 7 '19 at 3:48
  • $\begingroup$ I don't quite understand why it is not symmetric. However, is it true that the relation is reflexive and transitive? $\endgroup$ – ph-quiett Apr 7 '19 at 3:51
  • $\begingroup$ For example, "$<$" is not symmetric on real numbers because $a<b$ does not imply $b<a$ $\endgroup$ – J. W. Tanner Apr 7 '19 at 4:00
  • $\begingroup$ @ ph-quiett Consider $ x ^ 2 + 1 $ and $ x ^ 2 $ to disprove the symmetry $\endgroup$ – Minz Apr 7 '19 at 4:02

Reflexive $$f(x)-f(x)\geq 0 \forall x\in \mathbb{R}$$ Yes it is reflexive.

Transitive $$f(x)-g(x)\geq 0 \forall x\in \mathbb{R}$$ $$g(x)-h(x)\geq 0 \forall x\in \mathbb{R}$$ Add above equations, $$\Longrightarrow f(x)-h(x)\geq 0 \forall x\in \mathbb{R}$$ Yes it is transitive.


$$f(x)-g(x)\geq 0 \forall x\in \mathbb{R}$$ $$g(x)-f(x)\leq 0 \forall x\in \mathbb{R}$$ Hence, $(f,g)\in R $ & $ (g,f)\in R$ iff $g=f$ Hence, this relation is not symmetric.

Hence, not equivalence relation.

Hope it helps:)


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