Let $R$ be the equivalence relation on the real numbers given by $$R = \{(x, y) \in \Bbb R^2: (x−y)(x+y) = 0 \} $$ What are the equivalence classes of $R$?

So I wrote that, for every $x \in \Bbb R$, the corresponding equivalence class is $[x]_R = \{x, -x\}$.

Is my answer correct and complete or there is something I omitted? Thanks!

  • $\begingroup$ To clarify - by just $R$, you mean the relation, and by $R^2$ or R you're referring to $\mathbb{R}^2$ and $\Bbb R$ respectively, correct? $\endgroup$ – Eevee Trainer Feb 20 at 1:03
  • $\begingroup$ @EeveeTrainer exactly $\endgroup$ – JBuck Feb 20 at 1:08
  • $\begingroup$ I believe you are correct but I'd like to see a proof or an argument about why you believe this to be true. (A simple line that $ab = 0 \iff $ one of $a$ or $b$ is zero so....) Also it'd be good if you pointed out that the one exception is $[0]_R = \{0\}$ (although technically $\{0,-0\} = \{0\}$ so it is arguably correct. $\endgroup$ – fleablood Feb 20 at 1:09
  • $\begingroup$ I edited you question to try and clarify the ambiguities around $R$ and $\Bbb R$; I think I got it right. Did I? Cheers! $\endgroup$ – Robert Lewis Feb 20 at 1:10
  • $\begingroup$ @fleablood the justification is: (x+y)(x-y) = 0 ⟺ x = y or x = -y. Then the class of any $x ∈ R$ is defined: $[x]_R$ = {$y ∈ R: (x − y)(x + y) = 0$} = {x, -x}. $\endgroup$ – JBuck Feb 20 at 1:20

1) Make sure that your relation is indeed an equivalence relation, i.e. it is reflexive, symmetric and transitive. Indeed, it is (I just checked, but as an exercise, you should, too.)

2) Now find the ordered pairs , i.e. elements of R. For example, when x = y, or x = -y, R does indeed contain the elements (x, y), (x, -y). Does it contain (y, x), (-y, x)?

  • 1
    $\begingroup$ Equivalence is given in the exercise, also proving it wasn't a problem, so I did not include it in the question. Also, the ordered pairs you mention are included in R, by symmetry. $\endgroup$ – JBuck Feb 20 at 1:51

Yes, I believe you're correct.

We know

$$(x-y)(x+y) = x^2 - y^2$$

as it is the difference of two squares. From that,

$$(x-y)(x+y) = 0 \iff x^2 = y^2 \iff |x| = |y| \iff x = \pm y$$


$$(x,y) \in R \iff x = \pm y$$

Thus, any two real numbers $x,y$ are related if and only if one is the other, or its negative. Accordingly, for each real number, you have the equivalence class given by it and its negative.

As noted by fleablood in the comments it might not hurt to specifically set $0$ aside and thus let the equivalence classes be

$$[x]_R = \left\{\begin{matrix} \{ x,-x \} & x\neq 0\\ \{0\} & x = 0 \end{matrix}\right.$$

just to remove any potential ambiguity, but it's not like it's really wrong presented as-is since $0=-1 \cdot 0=-0$. Just depends on how paranoid you want to be about points on the assignment. :p

You can of course justify this by considering $x=y=0$ to show $(x,y)\in R$ and showing no other $x,y$ when one is $0$ can be in the same equivalence class. I maintain that, personally, it's fine as-is, but up to you.

Alternative Justification: (thanks to JMoravitz for pointing this out in the comments)

One recalls, when dealing with real numbers, this situation -- if $ab=0$, then what happens? At least one of $a,b$ are zero, i.e. $a=0,b=0$ or $a=b=0$.

This, as a bit of a tangent, comes as a consequence of the real numbers, when equipped with traditional addition and multiplication, forming what is known as a field. Fields are a special case of a certain, slightly more general algebraic structure known as an integral domain, itself a special case of another structure known as rings. I won't go into detail on these - the elaboration isn't necessary to understand for what follows, but it does provide a means of rigorously justifying this property. So some further reading if you're curious and have time to burn:

Anyhow, we essentially define an integral domain thusly: a commutative ring, in which there are no zero divisors. Equivalently, an integral domain $(R,\oplus,\otimes,0',1')$ is a commutative ring in which $a\otimes b=0'$ if and only if $a=0',b=0',$ or $a=b=0'$. Here, $0'$ denotes the "additive identity" for the ring $R$, the prime being used to distinguish it from the "actual" number zero. The notion of $0'$ being the additive identity simply means that $a\oplus 0' = 0' \oplus a = a$ for all $a \in R$.

In the case of the real numbers, $\Bbb R$, $0'$ is in fact the number zero, $1'$ the number one, $\oplus$ is the tradition addition operation $+$, and $\otimes$ is the traditional multiplication operation ($\cdot, \times$, however you choose to denote it).

Now we tie this back into where it becomes relevant. Or if you didn't care about the tangent into abstract algebra, I get back to the point. You're probably not going to be expected to justify all of this in your class since we just know from experience and such that the property holds for real numbers, but it's useful to know where it comes from, you know?


So, since $\Bbb R$ is a field, we know by definition it has the property that

$$ab = 0 \iff a=0, b=0, \; \text{or} \; a=b=0$$

Thus, since $xRy$ if and only if $(x-y)(x+y)=0$, we can immediately conclude by this property

$$x-y = 0 \;\;\; \text{and/or} \;\;\; x+y=0$$

If only the first holds, we know $xRy$ if and only if $x=y$.

If only the second holds, we know $xRy$ if and only if $x=-y$.

If both hold, we know $xRy$ if and only if $x=y=-y$, which only holds if $x=y=0$.

From there, it's a pretty clear exercise on how to construct the equivalence classes and so on, it more or less parallels the original approach.

  • $\begingroup$ I would skip the steps of expanding the product of $(x-y)(x+y)$ and simply cite or use the fact that $\Bbb R$ is an integral domain and so $(x-y)(x+y)=0$ directly implies (with no inbetween steps needed) that $x=y$ or $x=-y$. In doing so the proof could easily be extended to other integral domains like $\Bbb C$ where the implication $x^2=y^2\iff |x|=|y|$ might have been false. $\endgroup$ – JMoravitz Feb 20 at 1:25
  • $\begingroup$ @JMoravitz I haven't encountered integral domains yet, so I wouldn't know how to use them in my proof, but thanks anyway. $\endgroup$ – JBuck Feb 20 at 1:56
  • $\begingroup$ @JBuck Integral domains are a type of algebraic structure - a set equipped with a pair of operations in this case. You might study them later on. One property of integral domains is that if $a \cdot b = 0$, then one of $a,b$ are zero. (The $\cdot$ here does not necessarily mean "multiplication" as you know it, could be some other operation.) That's just glazing over the details. But basically the point JMoravitz is getting at is the fact that you know if $ab=0$ then $a=0, b=0,$ or $a=b=0$. [cont.] $\endgroup$ – Eevee Trainer Feb 20 at 2:16
  • $\begingroup$ (This is just an obvious fact when dealing with real numbers, rigorously justifiable if you want to; you don't need to appeal to integral domains since you're dealing with familiar territory, though.) In that sense, either $x-y=0$ or $x+y=0$ or both. The first yields $x=y$, the second $x=-y$, and the third yields $x=y=0$. That could be a very simple way to deal with this as opposed to expanding the product, so kudos to @JMoravitz for pointing it out. I'll edit it into my post in a sec. $\endgroup$ – Eevee Trainer Feb 20 at 2:17

The condition which defines $R$,

$(x - y)(x + y) = 0, \tag 1$

is clearly equivalent to

$x^2 - y^2 = 0 \equiv y^2 = x^2 \equiv y = \pm x. \tag 2$

Is $R$ and equivalence relation on $\Bbb R$? Well, writing in the usual way

$xRy \; \text{for} \; (x, y) \in R, \tag 3$

we check the usual conditions: reflexivity, symmetry, transitivity. Clearly

$x^2 = x^2 \equiv x = \pm x \equiv xRx; \tag 4$

$xRy \equiv x^2 = y^2 \Longleftrightarrow y^2 = x^2 \equiv yRx; \tag{5}$

so these two give us that $R$ is both reflexive and symmetric; that $R$ is transitive is just as easily seen:

$[xRy] \wedge [yRz] \equiv [x^2 = y^2] \wedge [y^2 = z^2] \Longrightarrow x^2 = z^2 \equiv xRz; \tag{6}$

(4)-(6) give us the reflexive, symmetric and transitive properties we need to accept that $R$ is in fact an equivalence relation on $\Bbb R$.

So, what are the equivalence classes determined by $R$? I think that's pretty easy to deduce at this point; since

$xRy \equiv x^2 = y^2 \equiv y = \pm x, \tag 7$

we have

$[x]_R = \{ x, -x \}, \; \forall x \in \Bbb R; \tag 8$

of course, this binds even in the event $x = 0$, there is nothing in the definition of equivalence classes which requires then all to be of the same cardinality.


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