Exercise 1.1 · Q13: Polygons with the same number of sides

Basics you need

Solution

1. Translate the rule into an equation

   With $n(P)$ the number of sides of the polygon $P$ (Basic 1),

   $$(P_1,P_2)\in R \iff n(P_1) = n(P_2).$$

   For example, a square and a rectangle are related ($n = 4$ for both), and a square and a regular pentagon are not ($4 \ne 5$⁠).

   "Have same number of sides" is a statement in words. Writing it as an equation between two whole numbers makes clear that the relation is just equality of one number attached to each polygon, so all three properties will follow from the properties of "$=$".
   Each polygon has exactly one number of sides, so $n(P)$ is a well-defined whole number for every $P\in A$⁠.

2. Reflexive: is every polygon related to itself?

   Take any polygon $P$⁠. Then $n(P) = n(P)$⁠: a polygon has the same number of sides as itself. So $(P,P)\in R$⁠. Since $P$ was arbitrary, $R$ is reflexive.

   The claim is about every polygon, so we argue with an arbitrary one. The argument does not depend on which polygon $P$ is.
   Equality is reflexive: every number equals itself.

3. Symmetric: if $P_1$ is related to $P_2$⁠, is $P_2$ related to $P_1$⁠?

   Suppose $(P_1,P_2)\in R$⁠, so $n(P_1) = n(P_2)$⁠. Reading the equation the other way, $n(P_2) = n(P_1)$⁠. So $(P_2,P_1)\in R$⁠, and $R$ is symmetric.

   We start from the "if" part and use it. "$P_1$ and $P_2$ have the same number of sides" is a statement that does not depend on the order in which the two polygons are named.
   Equality is symmetric: $a = b \Rightarrow b = a$⁠.

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

   Suppose $(P_1,P_2)\in R$ and $(P_2,P_3)\in R$⁠. Then

   $$\begin{gathered} n(P_1) = n(P_2) \qquad\text{and}\qquad n(P_2) = n(P_3),\\ \text{so}\qquad n(P_1) = n(P_3). \end{gathered}$$

   Hence $(P_1,P_3)\in R$⁠, and $R$ is transitive.

   The middle polygon $P_2$ does not appear in the conclusion, so we remove it: both $n(P_1)$ and $n(P_3)$ equal the same number $n(P_2)$⁠, so they equal each other. For example, if $P_2$ is a hexagon, the first link says $P_1$ has 6 sides and the second says $P_3$ has 6 sides, so $P_1$ and $P_3$ are related.
   Equality is transitive: $a = b$ and $b = c$ give $a = c$⁠.

5. Conclude the first part

   $R$ is reflexive, symmetric and transitive (Steps 2 to 4), so by definition $R$ is an equivalence relation in $A$.

   An equivalence relation is defined as one with exactly these three properties. The proof has the same shape as Exercise 1.1 Q7 (books with the same number of pages): "same value of a quantity" is always an equivalence relation.

6. The polygons related to $T$

   $T$ is a triangle, so $n(T) = 3$⁠. (Its sides 3, 4, 5 satisfy $3^2 + 4^2 = 25 = 5^2$⁠, which is why it is a right triangle, but that fact is not needed here.) Let

   $$[T] = \{P \in A : (T,P)\in R\} = \{P \in A : n(P) = 3\}.$$

   We show $[T]$ is the set of all triangles, proving both inclusions.

   • Every polygon related to $T$ is a triangle. If $(T,P)\in R$⁠, then $n(P) = n(T) = 3$⁠. A polygon with 3 sides is a triangle.
   • Every triangle is related to $T$⁠. If $P$ is any triangle, then $n(P) = 3 = n(T)$⁠, so $(T,P)\in R$⁠.

   So the set of all elements of $A$ related to $T$ is the set of all triangles. It includes right, acute and obtuse triangles of every size, and $T$ itself.

   The question deliberately gives extra information (the side lengths and the right angle) to test whether we read the relation correctly. $R$ only looks at the number of sides, so every triangle has the same "value" as $T$⁠, whatever its shape or size. Both inclusions are needed for the two sets to be equal: the first shows nothing else sneaks in, the second shows no triangle is left out.
   "Related to $T$" means $(T,P)\in R$⁠. Since $R$ is symmetric, it makes no difference whether we write $(T,P)$ or $(P,T)$⁠. Basics: equivalence classes

7. The whole picture: one class for each number of sides

   The same argument works for any polygon: the class of a polygon with $n$ sides is the set of all $n$-sided polygons. So $R$ splits $A$ into the classes

   $$\{\text{triangles}\},\ \{\text{quadrilaterals}\},\ \{\text{pentagons}\},\ \{\text{hexagons}\},\ \dots$$

   one for each whole number $n \ge 3$⁠. Every polygon lies in exactly one class, because it has exactly one number of sides.

   This is what an equivalence relation does: it splits a set into non-overlapping groups of related elements. Seeing all the classes at once confirms that the answer to the second part is one of them, the class of triangles.
   The equivalence classes of any equivalence relation are disjoint and their union is the whole set (Section 1.2, after Example 5).

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