Determine if the following relationships are equivalent. If they are, determine the equivalence classes, give a set of indices and the quotient set. Sketch out the graph of each relationship (whether or not it is an equivalence one): In $\mathbb{R}$, we define the relationship “∼” as $x ∼ y$ ⇐⇒ $x^2$ − 2x + 1 = $y^2$ + 4y + 4$. I have a question about transitivity, how can i prove it? i need to show that x∼z I tried to do this:
Suppose $x∼y$ and $y∼z$ thus: $x^2$ − 2x + 1 = $y^2$ + 4y + 4 and $y^2$ − 2y + 1 = $z^2$ + 4z + 4.
so we equal to 0 $x^2$ − 2x + 1- ($y^2$ + 4y + 4)=0 and $y^2$ − 2y + 1-($z^2$ + 4z + 4)=0
then we equal the equations: $x^2$ − 2x + 1- ($y^2$ + 4y + 4)=$y^2$ − 2y + 1-($z^2$ + 4z + 4) we resolve and we have :
(( $x^2$ − 2x + 1)+1)+(-$y^2$ + y)+7.5=((-1)($z^2$ + 4z + 4))+(-$y^2$ -y)-7.5
So i think this relation is transitive, but i dont know if i`m good, and i dont have any idea how to make the equivalence classes or set of indices
This relation is not transitive. A single counterexample suffices to prove that.
$x\sim y\iff (x-1)^2=(y+2)^2$.
Can you show $5\sim2$ and $2\sim -3$ but not $5\sim-3$?