2

For all $x, y \in \mathbb{R}$ define that $x \equiv y$ if $x^2 = y^2$. Then $\equiv$ is an equivalence relation on $\mathbb{R}$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.

Answer:

True. Reflectivity: To show that $\mathbb{R}$ is reflexive we need to show that $\forall x \in \mathbb{R} : x \equiv x$. Let $x \in \mathbb{R}$, $x \equiv x$ if $x^2 = x^2$, which is obvious.

Symmetry: Let $x,y \in \mathbb{R}$ with $x \equiv y$, so we have $x^2=y^2$ and trivially we get $y^2=x^2$, hence $y \equiv x.$

Transitivity: Let $x,y,z \in \mathbb{R}$ with $x \equiv y$ and $y \equiv z$. We now show that $x \equiv z$: Since $x^2=y^2$ and $y^2=z^2$ we get by transitivity of equality $x^2=z^2$ and therefore $x \equiv z.$

Is the answer to the question correct? And please tell me how I can prove that there are infinitely many equivalence classes one of them consists of one element and the rest consist of two elements. If it's not correct then please provide the correct answer.

David K
  • 98,388
user129834
  • 21

2 Answers2

1

Yes, your proof that $\equiv$ is an equivalence relation is fine. To consider equivalence classes, note that

$$x \equiv y \iff x = y \text{ or } x = -y$$

(do you see why?) Once you have this, study numbers which are zero and positive separately - you'll find one distinct equivalence class for every positive number.

0

It is right. And FYI, if you define $x \sim y \iff f(x) = f(y)$, for a given function $f$, it will always wield an equivalence relation. The equivalence classes will be the level surfaces of $f$. I think it's a good exercise to prove this.

Ivo Terek
  • 77,665