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The question asks:

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

(f, g) ∈ R if and only if f(x) − g(x) ≥ 0 for all x ∈ 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) ≥ 0: which is true

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

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  • $\begingroup$ 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 at 3:46
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    $\begingroup$ Does the condition $f(x)-g(x)\ge0$ look "symmetric" in the functions $f$ and $g$? $\endgroup$ – Lord Shark the Unknown Apr 7 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 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 at 4:00
  • $\begingroup$ @ ph-quiett Consider $ x ^ 2 + 1 $ and $ x ^ 2 $ to disprove the symmetry $\endgroup$ – Minz Apr 7 at 4:02
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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.

Symmetric

$$f(x)-g(x)\geq 0 \forall x\in \mathbb{R}$$ $$g(x)-f(x)\leq 0 \forall x\in \mathbb{R}$$ 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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