Let ~ be an equivalence relation on the set $A = \{1,2,...,8\}$, and denote the equivalence class of $x ∈ A$ by $[x]$.
If $1 ∈ [3]$, $2 ∈ [4]$, and $1 ∈ [2]$, prove that $[4] = [3]$.
Work So Far
If $1 ∈ [2]$, $2 ∈ [4]$, then $1 ∈ [4]$
However, beyond this, I'm not sure where to go. I do have $1 ∈ [3]$ and $1 ∈ [4]$, but I don't see how to extend that into proving $[3] = [4].$ Would anyone be able to point me in the right direction?
In any equivalence relation, $a \in [b]$ if and only if $[a] = [b]$. So in your case, $1 \in [3]$ implies $[1] = [3]$, and $1 \in [4]$ implies $[1] = [4]$. So all together, $[3] = [4]$.