Determining whether a relation is an equivalence relation

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.

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