How do I find the equivalence classes for $\vert x \vert = \vert y \vert$

Question

This was not mentioned in class. Isn't it the set of all real numbers because all real numbers satisfy the fact that $\vert x \vert = \vert y \vert$? Am I completely off? Am I supposed to to describe something about all three things, reflexive, transitive, symmetric(which I was able to show)?

Answer 1

Am I supposed to to describe something about all three things, reflexive, transitive, symmetric (which I was able to show)?

No, none of those ideas specify what the equivalence classes are.

This was not mentioned in class. Isn't it the set of all real numbers because all real numbers satisfy the fact that $\lvert x\rvert=\lvert y\rvert$?

The equivalence class of $x$ is, by definition the set of elements equivalent to $x$, that is $\mathrm{class}(x)=\{y\in \mathbb R\mid |y|=|x|\}$.

There are things that obviously have different absolute values, so your guess that everything is equivalent is not correct.

From the basic properties of the absolute value, you should be able to see why there is one equivalence class with a single element and the rest have two elements...

Another interesting version of this question would be to ask the same question with $\mathbb C$ instead of $\mathbb R$ and use the complex modulus instead of the real absolute value.

Answer 2

If you mean the equivalence relation given by $x \sim y \Leftrightarrow \vert x \vert = \vert y \vert$ then only those elements with the same absolute value are in an equivalence class. I sure do hope that $\vert 1 \vert = 1 \neq 0 = \vert 0 \vert$, so there cannot possibly be only one equivalence class...

Answer 3

The concepts may be abstract but:

Suppose $R$ is an equivelence relation on a set $X$.

Let $x \in X$. The equivalence class of $x$, which we can denote (not uniquely) as $[x]$ if want to, is $\{y \in X| y R x\}$.

So first question. If our equivalence relation is $|a| = |b|$. What is the equivalence class of $-27$? Of $\sqrt \pi$? Of $3$?

Thee equivalence class or $-27 =\{x\in \mathbb R: |x|=|27|\}=\{x\in\mathbb R: |x|=27\}=\{x\in \mathbb R: x = \pm 27\} = \{-27,27\}$ so the equivalence class of $27$ is $\{-27,27\}$. The equivalence class of $\sqrt \pi = \{x\in \mathbb R: |x| = |\sqrt{\pi}\} = \{-\sqrt{\pi}, \sqrt{\pi}\}$. Etc. What is the equivalence class of $0$?

The question is asking you to describe all the equivalence classes. That is the set of sets... of equivalence classes.

That is $\{[x]=\{y: y R x\}|x \in X\}$.

So what are the many different equivalence classes for the equivalence when $|x| = |y|$.

For any $x \in \mathbb R$ the set of all elements that are equivalent to $x$ is $\{-x, x\}$. So the set of all equivalence classes is $\{ \{-x, x\}|x\in \mathbb R\}$.

Answer 4

Many equivalence relations are given pretty much exactly in the form below:

Let $f : X \to Y$ be a function. Then the relation $R$ on $X$ given by $R := \{ (x, y) \in X \times X \mid f(x) = f(y) \}$ is an equivalence relation.

Furthermore, the equivalence classes of $R$ are exactly the sets of the form $f^{-1}(\{ y_0 \}) = \{ x \in X \mid f(x) = y_0 \}$ where $y_0$ is in the image (or range) of $f$. Also, the quotient set $X / R$ is in canonical bijection with the image of $f$.

In the given problem, $R$ is of the form above with $f : \mathbb{R} \to \mathbb{R}$ given by $f(x) := |x|$. So, to apply the above, you would need to ask: what is the range of $f$? For each $y_0$ in this range, what is $f^{-1}(\{ y_0 \}) = \{ x \in \mathbb{R} \mid |x| = y_0 \}$?

(If you have worked the problem directly for some sample functions $f$, it might also be instructive to see why the general statement above is true.)