Prove that $R$ is an equivalence relation on $F$.

Question

Question: Let $S = \{1, 2, 3, 4\}$. Let $F$ be the set of all functions $f: S \to S$. Let $R$ be the relation on $F$ defined by

For any $f, g \in F$, $fRg$ if and only if $f (1) + f (2) = g (1) + g (2)$.

Prove that $R$ is an equivalence relation on $F$.

I understand that to do this we must prove that $R$ is reflexive, symmetric, and transitive. I'm just having trouble using the definitions of these 3 properties to make an actual proof.

Answer 1

I'll try and get you started:

Reflexivity:

Let $f\in F$. Then $f(1)+f(2)=\dots$.

You need $f$ in $g$'s place.

Symmetry:

Let $f,g\in F$. Then we have

$$\begin{align} fRg &\iff f(1)+f(2)=g(1)+g(2) \\ &\iff g(1)+g(2)=f(1)+f(2)\quad\text{ (by symmetry of equality)} \\ &\iff \dots \end{align}$$

You need to conclude $gRf$ (preferably using "if and only if" statements, although implication is sufficient).

Transitivity:

Let $f, g,h\in F$ with $fRg$ and $gRh$. Then, by definition of $R$, we have $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=\dots$

You need to conclude that $fRh$.

Answer 2

Reflexivity: For all $f \in F$, we have $f(1)+f(2)=f(1)+f(2)$ so $f R f$

symmetric: Let $f, g \in F$ and $fRg$, then $f(1)+f(2)=g(1)+g(2) \Rightarrow g(1)+g(2)=f(1)+f(2) \Rightarrow g R f$

Transitivity: Let $f,g,h \in F$ and $f R g$ and $g R h$ then $f(1)+f(2)=g(1)+g(2)$ and $g(1)+g(2)=h(1)+h(2)$ $\Rightarrow$ $f(1)+f(2)=h(1)+h(2) \Rightarrow f R h.$

So $R$ is an equivalence relation on $F$.