Exercise 1.1 · Q11: Points at the same distance from the origin

Basics you need

Solution

1. Translate the rule into an equation

   Write $O$ for the origin $(0,0)$⁠. By the distance formula (Basic 1), the rule of $R$ becomes

   $$(P,Q)\in R \iff |OP| = |OQ|.$$

   Try it on a few pairs, taking $P = (3,4)$⁠, which has $|OP| = 5$⁠:

   $Q$	$|OQ|$	Equal to $|OP| = 5$⁠?	$(P,Q)\in R$⁠?
   $(5, 0)$	$\sqrt{25+0} = 5$	yes	yes
   $(-4, 3)$	$\sqrt{16+9} = 5$	yes	yes
   $(0, -5)$	$\sqrt{0+25} = 5$	yes	yes
   $(1, 1)$	$\sqrt{1+1} = \sqrt2$	no	no

   The words "distance of $P$ from the origin is same as the distance of $Q$" are long. Replacing them by the equation $|OP| = |OQ|$ shows that the relation is nothing but equality of two real numbers. Every property we have to check then becomes a property of the symbol "$=$", which we know well.
   Each point has exactly one distance from $O$⁠, so $|OP|$ is a well-defined number for every $P\in A$⁠. This is what lets us compare the distances of any two points.

2. Reflexive: is every point related to itself?

   Take any point $P \in A$⁠. The number $|OP|$ is equal to itself:

   $$|OP| = |OP|.$$

   By Step 1 this says exactly $(P,P)\in R$⁠. Since $P$ was an arbitrary point, this holds for every point, and $R$ is reflexive.

   Reflexivity is a claim about every point, so we take an arbitrary one, a letter $P$ that could be any point of the plane. The argument uses nothing special about $P$⁠, so it works for all points at once. In words: every point is at the same distance from the origin as itself.
   Equality is reflexive: every number equals itself, $d = d$⁠.

3. Symmetric: if $P$ is related to $Q$⁠, is $Q$ related to $P$⁠?

   Suppose $(P,Q)\in R$⁠. Then $|OP| = |OQ|$⁠. An equation can be read from right to left, so

   $$|OQ| = |OP|.$$

   By Step 1 this means $(Q,P)\in R$⁠. Hence $R$ is symmetric.

   Symmetry is an "if … then" statement, so we start from the "if" part, $(P,Q)\in R$⁠, and use it. The rule "same distance" does not care which point is named first: if $P$ is as far from $O$ as $Q$ is, then $Q$ is as far from $O$ as $P$ is.
   Equality is symmetric: if $d_1 = d_2$ then $d_2 = d_1$⁠.

4. Transitive: does a chain of two links give a direct link?

   Suppose $(P,Q)\in R$ and $(Q,S)\in R$⁠, where $P, Q, S$ are any points. Then

   $$|OP| = |OQ| \qquad\text{and}\qquad |OQ| = |OS|.$$

   Both $|OP|$ and $|OS|$ are equal to the same number $|OQ|$⁠, so they are equal to each other:

   $$|OP| = |OQ| = |OS| \quad\Longrightarrow\quad |OP| = |OS|.$$

   By Step 1, $(P,S)\in R$⁠. Hence $R$ is transitive.

   The statement we want, $(P,S)\in R$⁠, does not mention the middle point $Q$⁠. The two given equations share the quantity $|OQ|$⁠, and chaining them through it removes $Q$⁠. In words: if $P$ and $Q$ are equally far from $O$⁠, and $Q$ and $S$ are equally far from $O$⁠, then $P$ and $S$ are equally far from $O$⁠.
   Equality is transitive: if $d_1 = d_2$ and $d_2 = d_3$ then $d_1 = d_3$ ("things equal to the same thing are equal to each other").

5. Conclude the first part

   $R$ is reflexive (Step 2), symmetric (Step 3) and transitive (Step 4). By definition, a relation with all three properties is an equivalence relation, so $R$ is an equivalence relation in $A$.

   The definition of an equivalence relation is exactly this list of three properties, so nothing more is needed. Notice the pattern: whenever a relation has the form "$P$ and $Q$ have the same value of some quantity", it is an equivalence relation, because equality is. Exercises 1.1 Q7, Q13 and Q14 follow the same pattern.

6. The points related to $P \ne O$ form the circle through $P$ centred at $O$

   Fix a point $P \ne (0,0)$ and put $r = |OP|$⁠. Because $P$ is not the origin, $r > 0$⁠. Let

   $$\begin{gathered} [P] = \{Q \in A : (P,Q)\in R\} \quad\text{and}\\ C = \{Q \in A : |OQ| = r\}. \end{gathered}$$

   $C$ is, by definition, the circle with centre $O$ and radius $r$⁠. We show $[P] = C$ by proving each set is contained in the other.

   • Every point related to $P$ lies on $C$⁠. Let $Q\in[P]$⁠. Then $(P,Q)\in R$⁠, so $|OQ| = |OP| = r$⁠. Hence $Q\in C$⁠.
   • Every point of $C$ is related to $P$⁠. Let $Q\in C$⁠. Then $|OQ| = r = |OP|$⁠, so $|OP| = |OQ|$⁠, that is $(P,Q)\in R$⁠. Hence $Q\in[P]$⁠.

   So $[P] = C$⁠. Finally, the circle passes through $P$⁠, because $|OP| = r$ means $P\in C$ (this is also reflexivity: $P$ is related to itself). In coordinates, if $P = (a,b)$⁠, the circle is $x^2 + y^2 = a^2 + b^2$⁠. So the set of all points related to $P$ is the circle through $P$ with centre at the origin.

   "The set of points related to $P$" and "the circle" are two descriptions of sets, and two sets are equal only when they have exactly the same elements. Showing just one inclusion is not enough: we must rule out both "a related point off the circle" and "a point of the circle that is not related". Each direction is one line, because both sets are defined by the same equation $|OQ| = |OP|$⁠.
   A circle is not just a curve that "looks round": it is defined as the set of all points at a fixed distance (the radius) from a fixed point (the centre). That definition is what makes the two inclusions immediate. Basics: equivalence classes

7. Why the question excludes $P = (0,0)$

   If $P = O$⁠, then $r = |OO| = 0$⁠, and the only point at distance 0 from $O$ is $O$ itself (because $\sqrt{x^2+y^2} = 0$ forces $x = y = 0$⁠). So the class of the origin is the single point $\{O\}$⁠, not a circle in the usual sense. That is why the question says $P \ne (0,0)$⁠.

   Putting everything together, $R$ sorts the plane into equivalence classes: the origin on its own, and one circle centred at $O$ for each radius $r > 0$⁠. Different radii give circles that do not meet, and every point of the plane lies on exactly one of them (or is $O$⁠).

   Understanding the excluded case shows the answer is complete: the classes are the origin and the circles about it, and together they cover the plane without overlap, exactly as the textbook says any equivalence relation does.
   $\sqrt{x^2+y^2} = 0$ means $x^2 + y^2 = 0$⁠. A sum of two squares of real numbers is 0 only when both are 0, because each square is $\ge 0$⁠.

Every fact used, and where it comes from