chapter 2: relations chapter 2 - relationsultimate2.shoppepro.com/~ibidcoma/samples/maths hl option...

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Chapter 2: Relations

Chapter 2 - RelationsWe could use up two Eternities in learning all that is to be learned about our own world and the thousands of nations that have arisen and flourished and vanished from it. Mathematics alone would occupy me eight million years. Mark Twain

A binary relation, R, usually called a relation defined on set S1, is written aRb or (x, y) ∈ R, which is pronounced "a is related to b". A relation is a set of ordered pairs. If a is not in relation with b, we write aRb.

Examples of relations

a. x, y ∈ , |y| = |x|.

b. R = {(1, 2), (3, 4), (5, 6)}.

c. R is a relation on , in which a and b have the same remainder upon division by 2. So we have {(0, 2), (0, 4), …, (1, 3), (1, 5), …, (2, 0), …}.

d. R is a relation on the set of all polygons, aRb if a and b have the same number of sides.

e. If there are three persons {Bill, John, Mary} and three objects {ball, car, doll} and Bill owns the ball, Mary owns the doll, and John owns the car, then the binary relation, R “is owned by” is given as R = {(ball, Bill), (doll, Mary), (car, John)}.

f. D = {(1, 2), (3, 4), (4, 5), (6, 7)}, (c, d)R(a, b) if d + a = c + b, where (a, b), (c, d) ∈ D. It’s probably helpful to translate that algebra into words, “The sum of the two inner elements equals the sum of the two outer elements”. Notice that in this example we have an ordered pair of ordered pairs! The two members of R are ((1, 2), (6, 7)) and ((3, 4), (4, 5)), which can also be written as (1, 2)R(6, 7) and (3, 4)R(4, 5).

The word “binary” refers to the fact that two elements are related. We will only study binary relations.

The concept of a binary relation is exemplified by such ideas as “is greater than” or “is equal to” in arithmetic, “is congruent to” and “is similar to” in geometry, “is an element of ” or “is a subset of ” in set theory.

Sometimes relations can be graphed on the Cartesian plane, for example the preceding examples (a), (b), (c) and (f) and sometimes they cannot, for example the preceding examples (d) and (e). If the relation can be graphed on the Cartesian plane, doing so can help understand what is going on.

Definition

Two ordered pairs are equal if their corresponding components are equal.

The Cartesian product of two sets is a set of ordered pairs in which every member of the first set is paired with every member of the second set.

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Mathematics HL Topic 8: Sets, Relations and Groups Chapter 2: Relations

Example

Given set A is the numbers {1, 2} and set B is the letters {a, b, c},

A×B (pronounced “A cross B”), is {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}.

B×A is {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)}.

A×B and B×A are finite sets, because they have a finite number of elements.

The Cartesian product × (also written ℝ²) is the set of all points on the Cartesian plane. × is an infinite set, because it has a infinite number of elements. A Cartesian product of two finite sets can be represented by a table, with one set as the rows and the other as the columns. The ordered pairs are then the cells of the table. Any subset of the Cartesian product of two sets is a relation.

In general A×B ≠ B×A.

The number of elements in A×B is the number of elements in A times the number of elements in B." In symbols this is written n(A×B) = n(A) × n(B). Notice that, unfortunately, the symbol "×" has two completely different meanings here.

There are many ways that relations can be expressed.

1. The elements of the relation can simply be listed.

Example: R = {(6, 4), (-1, 0.7)} or R = {(1, 1), (1, 2), (1, 10), (2, 2), (2, 4), (5, 10)}.

2. A formal definition of a set might be used.

Example: R = {(x, y) | x ∈ ℤ, y ∈ ℤ, x is a factor of y where -2 < x < 5, -4 < y < 5}.

3. Relations can also be given with a set and the description of how the ordered pairs are formed.

Example: H = {students in your HL Maths group} and aRb means that a and b are of the same age where a, b ∈ H.

Definition

The domain of R consists of all possible first components of the ordered pairs of the relation. The range of R contains all possible second components.

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Mathematics HL Topic 8: Sets, Relations and Groups

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Exercise 2.1

1. B = {12, 3, 4, 5, 6, 1} and mRn if m is a factor of n, where m, n ∈ B.

a. Determine whether 12R3.

b. Determine whether 3R12.

c. Give an example for two integers from set B which are in relation and two integers which are not in relation.

2. C = {–6, –1, 0, 4, 12} and mRn if m + n < 10, where m, n ∈ C.

a. Determine whether 0R4


c. Give an example for two integers from set C which are in relation and two integers which are not in relation.

3. D = {(3, 4), (0, 4), (7, 23)}, (c, d)R(a, b) if d + a = c + b, where (a, b), (c, d) ∈ D.

a. Determine whether (3, 4)R(0, 4).

b. Determine whether (3, 4)R(3, 4).

c. Give an example for two elements of set D which are in relation and two elements which are not in relation. Note that an element can, but need not be, in relation with itself.

4. E = {1, 4, 3, 56, 8} and aRb means that a and b are co-primes, that is that their greatest common divisor is 1, where a, b ∈ E.

a. Determine whether 1R4.


c. Give an example for two elements of set E which are in relation and two elements which are not in relation.

5. R = {(1, 3), (4, 6), (9, 9)} is a relation. Give an example for two elements of set R which are in relation.

6. Let us define the following relation: xRy, if x2 + y2 = 1, x, y ∈ 핉. Give an example for two real numbers which are in relation and two real numbers which are not in relation.

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Properties of relationsReflexive property of relations

Definition

A reflexive relation is one in which every element relates to itself.

Let R be a non-empty relation in a set A.

R is reflexive means aRa for all a ∈ A.

Examples

The relation (x, y) ∈ R if x < y is not reflexive, because ‘<’ does not include ‘=’.

The relation (x, y) ∈ R if x ≤ y is reflexive, because ‘≤’ includes =.

The relation R = {(1, 1), (2, 2), ( 2, 1)} on set D = {1,2} is reflexive, because 1 and 2 are both related to themselves.

The relation R = {(1, 1), (2, 1)} on set D = {1,2 }is not reflexive, because 2 is not related to itself.

If a relation can be displayed on the Cartesian plane, the relation is reflexive if for all allowed values of x, the line y = x is on the graph.

Symmetric property of relations

Definition

Let R be a non-empty relation on a set A. If (a, b)∈R then (b, a) ∈ R as well for all a, b ∈ A. R is symmetric means that if aRb, then bRa for all a ∈ A.

If a relation can be graphed on the Cartesian plane, ‘the relation is symmetric’ means that all of the points (x, y) are symmetric about the line y = x.

Examples

{(1, 2), (2, 1), (3, 3)} is symmetric, since (1, 2) and (2, 1) are in the set.

{(1, 2), (2, 3), (3, 3)} is not symmetric, since (1, 2) is in the set an (2, 1) is not.

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Exercise 2.2

Are the following relations symmetric?

1. B = {12, 3, 4, 5, 6, 1} and mRn if m is a factor of n, where m, n ∈ B.

2. C = {-6, -1, 0, 4, 12} and mRn if m + n < 10, where m, n ∈ C.

3. D = {(3, 4), (0, 4), (7, 23)}, (c, d)R(a, b) if d + a = c + b, where (a, b), (c, d) ∈ D.

4. E = {1, 4, 3, 56, 8} and aRb means that a and b are co-primes, that is that their greatest common factor is 1, where a, b ∈ E.

Transitive property of relations

Definition

Let R be a non-empty relation on a set A. We say that :R is transitive if, given aRb and bRc, then aRc for all a, b, c ∈ A.

Examples

{(1, 2), (2, 3), (1, 3)} is transitive, {(1, 2), (2, 3), (1, 1)} is not transitive.

Exercise 2.3

Are the following relations transitive?

1. H = {students in your HL Maths group} and aRb means that a and b are of the same age where a, b ∈ H.

2. H = {students at your school} and aRb means that a is at least as tall as b.

3. H = {members of your extended family} and aRb means a and b live in the same city.

4. B = {12, 3, 4, 5, 6, 1} and mRn where m, n ∈ B if m is a factor of n.

5. C = {–6, –1, 0, 4, 12} and mRn if m, n ∈ C and m + n < 10.

6. D = {(3, 4), (0, 4), (7, 23) and (c, d)R(a, b) if (a, b), (c, d) ∈ D and d + a = c + b.

7. E = {1, 4, 3, 56, 8} and aRb means that a and b are co-primes.

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If you need to determine whether a relationship has any of these three properties, remember the simple rule of logic. Only one counter example is needed to show that the property does not hold. To show that the property does hold, you must find a way to show that it holds for all elements or equally for any arbitrary element of the given set.

SummaryBasic properties of binary relations

Property Definitionreflexive If aRa holds for every a ∈ A. symmetric If aRb then bRa for every a, b ∈ A.transitive If aRb and bRc, then aRc for every a, b, c ∈ A.

Equivalence Relations“Medicine makes people ill, mathematics makes them sad, and theology makes them sinful” - Martin Luther

Introduction

In primary school you learnt that in some way

...is equivalent to...

While in the same way

...is not equivalent to...

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Later you learnt that 2 __ 3 = 4 __ 6 = –12 ____ –18 , that is that these fractions are equal or equivalent.

You also learnt what it means for two triangles to be similar.

All three of these situations are examples of a special type of mathematical “relation” called an equivalence relation. The concept of an equivalence relation is a way of expressing “sameness”. In other words it is a generalization of the concept of equality.

An equivalence relation on a set partitions the set into disjoint subsets, called equivalence classes.

Here are some examples of equivalence classes.

If we consider integers, then the equivalence might be even and odd, which is the same as saying that the set of those integers divisible by 2 are equivalent classes and those that when divided by 2 have a remainder of 1 are equivalent.

Or if we consider integers, we might decide to consider that those integers divisible by 3 are an equivalence class, that those integers which when divided by 3 have a remainder of 1 are an equivalence class and that those integers which when divided by 3 have a remainder of 2 are an equivalence class.

If we consider the set {1, 2, 3, 4, 5, 6, 7, 8, 9}, we might decide to consider {2, 4, 6, 8} is an equivalence class, {1, 9} is an equivalence class and {3, 5, 7} is an equivalence class.

If we consider the points in ℤ+ × ℤ+, then we might decide to consider all points on the line y = k for a given k is an equivalence class. For example the points (1, 2), (2, 2), (3, 2), etc. would be equivalent (k = 2) and the points (1, 5), (2, 5), (3, 5), etc. would be equivalent (k = 5).

If we consider the points in ℝ × ℝ, we might consider all points such that n ≤ x < n + 1 for a given integer n are equivalent; that is the Cartesian plane would be divided into vertical bands one unit wide with all points in any one band equivalent to each other. See the figure.

1 2 3 4–1–2–3–4

1

2

3

4

–1

–2

–3

–4

x

y

1 2 3 4–1–2–3–4

1

2

3

4

–1

–2

–3

–4

x

y

Equivalence classes as vertical bands Equivalence classes as concentric circles

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If we consider the points in ℝ × ℝ, we might consider all points such that x² + y² = k for a given k ∈ ℝ+ ⋃ {0} are equivalent to each other; that is the Cartesian plane would be divided into equivalence classes consisting of the points on concentric circles about the origin.

The property of dividing sets into equivalence classes is a very useful feature; so we will study it in detail.

Definition

An equivalence relation is a relation that is reflexive, symmetric and transitive.

Equivalence1 can be designated by either the symbol ≅ or ≡. You will be asked to prove that relations are equivalence relations. To do so you will need to prove that these three properties hold, so it is necessary to memorize these three properties. We suggest the mnemonic ‘RST’.

Partition of a setDefinition

A partition of a set A is a collection of non-empty subsets of A (xi i = 1, 2, ..., n where n ∈ ℕ),such that they are pairwise disjoint2 and their union is A. In other words, every element in A is contained in exactly one of the partitions. We say that A has been partitioned.

Thus, the elements of a partition are like the pieces of a jigsaw puzzle:

x0x1

x2x3

xn

...

A

Examples for partition of a set

1. If set A contains all students at your school then a partition P is formed by the year levels at your school.

1 The first known mention of the concept of equivalence relations was by Euclid in his book The Elements. Common Notion 1 states: ‘Things which equal the same thing also equal one another’, which is a statement of the property transitivity. Unfortunately, he did not mention symmetry or reflexivity.

2 Sets A, B and C are said to be disjoint if A∩B∩C = ∅. Sets A, B and C are said to be pairwise disjoint if A∩B=∅, B∩C=∅ and A∩C=∅. For example the sets {a, b}, {b, c} and {c, a} are disjoint, because there is no member which is in all three sets, but not pairwise disjoint.

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2. The four suits (spades, hearts, diamonds, and clubs) are a partition of a deck of playing cards (not counting the Joker). Every card is in one of these suits, and no card is in more than one suit.

3. Suppose A = {1, 2, 3, 4, 5, 6}. The following is a partition of A:

12

34

5

6

4. Let 핑+ be the set of all positive integers, 핑– the set of all negative integers, and 핑0 the one-element set {0}. Then {핑+, 핑–, 핑0} is a partition of the set ℤ. It has three sets; the first two sets have an infinite number of elements. ℤ has many other partitions. If we put integers with the same modulus (absolute value) into the same class then we get another partition of 핑: P = {{0}, {-1, 1}, {2, -2}, etc.}. This partition has an infinite number of sets; each set contains two or one elements.

Fundamental theorem of equivalence relations3

Let A be a non-empty set and R an equivalence relation on A. The equivalence classes of R partition set A.

Theorem

If R is an equivalence relation on set A ≠ ∅, then R induces a partition of A.

Definition

The members of the partitions are called equivalence classes.

Proof

Step 1: Every element of A belongs to some equivalence class.

If R is an equivalence relation on A, every element xi of A is related to itself, because (xi, xi) is reflexive. Hence every element belongs to at least one equivalence class therefore every element of xi is contained in the union of the equivalence classes.

Step 2: We must show that the equivalence classes are disjoint: for any two different equivalence classes A and B, A is disjoint from B means that A and B have no common element. We will use an indirect proof. Let us assume that C and B have the element y in common. Let c be any element in C and b be any element in B.

3 The fundamental theorem of equivalence relations is also known as the Partition Theorem.

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Since c and y are both in C, cRy. Since b and y are both in B, yRb. Therefore cRb holds by the transitive property. Thus any element of C is in relation with b from B. Consequently, C and B are the same class. So if two equivalence classes have one common element, then they are the same equivalence class. This is a contradiction. So we conclude that our assumption was wrong and that any pair of different equivalence classes is disjoint.

To summarize what we have found:

Every equivalence relation on set A partitions A into disjoint subsets such that every element of A is in one subset, and two elements are in the same subset if they are in relation to one another. The subsets are called the equivalent classes defined by R.

The equivalence class containing element x is designated by [x]. In general the equivalence class representative is not unique. The element that represents an equivalence class does not matter. If set A has a natural order, for example if A is {1, 2, 3, 4}, we will adopt the convention of choosing the first element as the representative. So the equivalence classes of A might be [1] = {1, 4} and [2] = {2, 3}.

We will now give some examples of an equivalence relation inducing a partition and we will determine the equivalence classes.

Example

Let M be the set of polygons. Define R in M such as: aRb if a and b have the same number of sides where a, b ∈ M. Show that it is an equivalence relation.

Answer

a, R is reflexive since aRa always holds since every polygon has the same number of sides as itself.

b. R is symmetric since if a has the same number of sides as b then b also has the same number of sides as a.

c. R is transitive since if a has the same number of sides as b and b has the same number as c then a and c also have the same number of sides.

Thus R partitions M into an infinite number of equivalence classes (triangles, quadrilaterals, pentagons, hexagons etc.), each having infinite elements.

Exercise 2.4

1. A = {1, 2, 3, 4} and xRy if x + y is even.

a. Show that R is an equivalence relation.

b. Find the equivalence classes.

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2. Determine whether R is an equivalence relation on A where A and R defined as follows:

a. A = {a, b, c, d} and R = {(a, a), (b, a), (b, b), (c, c), (d, d), (d, c)}.

b. A = {1, 2, 3, 4, 5} and R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1), (2, 3), (3, 3), (4, 4), (3, 2), (5, 5)}.

3. Consider the relation R defined on the set of real numbers such that xRy if x > y for all x, y ∈ ℝ. Show that R is not an equivalence relation.

4. Consider the relation defined such that (x, y)R(a, b) if [x + y]= [a + b], where x, y, a, b ∈ ℝ. Here we mean [z] to be the floor function, the largest integer not exceeding z.

a. Show that R is an equivalence relation.

b. Find the equivalence classes

5. Define the following relation on 핑: aRb, if a – b is divisible by 2 where a, b ∈ 핑. Show that R is an equivalence relation and find the equivalence classes.

6. Define the following relation on 핑: xRy, if a – b is divisible by 3 where a, b ∈ 핑. Assuming that it R is an equivalence relation, find the equivalence classes.

7. A relation Rm

is defined on ℤ such that aRm

b if m is a divisor of a – b where 2 ≤ m and a, b ∈ 핑. Prove that R

m is an equivalence relation for every m ∈ ℕ, 2 ≤ m.

Congruence of integers modulo nMathematics was hard, dull work. Geography pleased me more. For dancing I was quite enthusiastic. - John James Audubon

Modular arithmetic is a system of arithmetic with integers in which numbers ‘wrap around’ after they reach a certain value, called the modulus. An example in everyday use is the 12-hour clock.4 Five hours after 3:00 is 8:00, but five hours after 9:00 is 2:00, not 14:00. Equally 30 hours after 3:00 is not 33:00, but 9:00. Since the time wraps around after 12 hours, this is an example of addition modulo 12.

In modular arithmetic the result is the remainder after division by the modulus. If two numbers x and y have the same remainder after division by the modulus, m, we say that “x is congruent to y modulo m”, written x ≡ y (mod m), x, y and m ∈ ℤ, m ≥ 2. 5

4 Therefore modular arithmetic is sometimes called “clock math”. For our purposes we ignore the 24-hour clock and the issue of am and pm.

5 The Swiss mathematician Leonhard Euler (pronounced “oiler”) (1707 – 1783) introduced the idea of congruence modulo n. Modular arithmetic was further advanced by Carl Friedrich Gauss (1777 – 1855).

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Example

34 ≡ 28 (mod 3), because 34 and 28 both have the same remainder 1 when divided by 3. That is 34 = 3·11 + 1, 28=3·9 + 1.

Definition

x ≡ y (mod m) means x and y have the same remainder on division by m.

If two integers x and y have the same remainder after division by the modulus, m, then x − y = k·m, k ∈ .

Examples

34 ≡ 28 (mod 3), because 34 – 28 = 6 = 3·2.

13 ≡ 49 (mod 9) because 49 – 13 = 36 = 9·4.

To express this we can write m | (x – y), pronounced “m divides x – y”.

m | (x – y) means x − y = k·m, k ∈ ℤ, which means x ≡ y (mod m)

To prove x ≡ y (mod m) one normally proves x − y = k·m, k ∈ .

Careful! A very similar notation is used to mean something different! When ‘=’ is used instead of ‘≡‘, we mean calculate the remainder.

Example

23 (mod 7) = 2 (2 only).

For example we might write (4 + 7) (mod 8) = 3. This can also be also written as 4 +8 7 = 3.

We showed in Exercise 7 in the preceding section that for a fixed integer m where 2 ≤ m congruence mod m is an equivalence relation in the set ℤ with m equivalence classes.

Definition

The equivalence class with respect to the relation of congruence of x mod m, is called the congruence class of x mod m, written [x], where x is the representative of the equivalence class. So [x] is the set of all integers y such that x≡y mod m. So [x] = {..., x – 2m, x – m, x, x + m, x + 2m, ...}.

We call the smallest non-negative integer in [x] r, the residue of x mod m. Since r must be one of the integers 0, 1, ..., m – 1, each [x] is one of the m classes [0], [1] , [2] , ..., [m – 1], which are called the residue classes modulo m. Thus there are exactly m different residue classes mod m and these form a partition of ℤ.

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Example

When m = 5, the equivalence classes (residue classes) under congruence mod 5 are integers which leave a remainder of 0, 1, 2, 3, or 4 upon division by 5. So the equivalence classes of x ≡ y (mod 5) are {..., –5, 0, 5, 10, 15, ...}, {..., –4, 1, 6, 11, 16, ...}, {..., -3, 2, 7, 12, 17, ...}, {..., –2, 3, 8, 13, 18, ...} and {..., –1, 4, 9, 14, 19, ...}.

More theoremsTheorem

R is an equivalence relation on A if and only if R is reflexive and the following property holds: if aRb and aRc, then bRc for every a, b, c ∈ A. (Note that this is not transitivity!)

Proof

Step 1 (“if ”): Let us assume that R is an equivalence relation. We have to show R is reflexive and if aRb and aRc, then bRc for every a, b, c ∈ A.

a. By definition R is reflexive.

b. We have to show that if aRb and aRc then bRc for every a, b, c ∈ A. For any a, b ∈ A if aRb then bRa since relations are symmetric. We now have bRa and aRc. By applying the transitive property, we conclude that bRc for every a, b, c ∈A.

Step 2 (“only if ”): Let us assume that R is reflexive and aRb and aRc implies bRc for every a, b, c ∈ A. We have to show that R is an equivalence relation.

a. Let us choose a, b ∈ A such that aRb. aRa holds for any a ∈ A since R is reflexive. We now have aRb and aRa. Using the given property and letting c→a, we know that bRa. We have shown that aRb implies bRa, so we have shown that the symmetric property holds.

b. If aRb and bRc then bRa and bRc since R is symmetric. If we apply the given property, we get aRc. Hence the transitive property holds as well. So we have shown that R is an equivalence relation.

Theorem

If R ⊆ A × A is reflexive and aRb and bRc implies cRa for every a, b, c ∈ A, then R is an equivalence relation.

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Proof

We have to show that R is symmetric and transitive.

a. Let us choose a, c ∈ A such that aRc. R is reflexive therefore aRa for every a ∈ A. If aRa and aRc, then cRa by using the given property. So R is symmetric.

b. If aRb and bRc then cRa by using the given property. Applying the symmetric property proved in part (a), aRc. Therefore R is transitive.

Exercise 2.5

Determine whether the following relations are reflexive, symmetric and transitive. If so, find the equivalence classes.

1. aRb if |a| = |b| where a, b ∈ ℝ.

2. mRn if and m, n ∈ + and 3 is a factor of m2 – n2. That is: 3|m2 – n2.

3. xRy if x, y ∈ and 2 is a factor of x2 + y2.

4. mRn if m, n ∈ ℕ+ and m×n is a perfect square.

5. aRb if a and b are co-primes (relative primes) a, b ∈ ℕ+. (a and b are co-primes if their greatest common divisor is 1, gcd(a, b) = 1.)

6. (x, y)R(a, b) if x2 + y2 = a2 + b2 (a, b), (x, y) ∈ 핉×핉.

7. aRb if |a| − |b| is even, where a, b ∈ ℝ.

8. (x, y)R(a, b) if x, y, a, b ∈ ℝ and x + y = a + b.

9. (a, b)R(c, d) if |a| +|b| = |c| + |d|, where (a, b) ∈ ²\{(0, 0)}.

10. (x, y)R(a, b) if x, y, a, b ∈ ℝ and [a] = [x], [b] = [y], where [z], the floor function.

IB Exam Type Problems1. The relation R is defined for a, b ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} by aRb if

a2≡b2 (mod n) where n is a fixed positive integer.

i. Show that R is an equivalence relation.

ii. Determine the equivalence classes when n = 6.

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2. The relation R is defined on ordered pairs by (a, b)R(c, d) if ad = bc where a, b, c, d ∈ ℝ+.


ii. Describe, geometrically, the equivalence classes.

3. The relation R is defined on ℝ by xRy if |x| ≤ |y|. Show that R is not an equivalence relation.

4. The relation R is defined on ℝ by xRy if x² – 4x = y² – 4y.

i. Prove that R is an equivalence relation.

ii. Determine the equivalence classes of R.

iii. Find the equivalence class containing only one element.

5. The relation R is defined for x, y ∈ ℤ+ such that xRy if 2x ≡ 2y (mod10).


ii. Identify all the equivalence classes.

6. Define the relation (x, y) R (a, b) if x² − y² = a² − b², where (x, y), (a, b) ∈ ℝ+ × ℝ+.

i. Prove that R is an equivalence relation on ℝ+ ×´ ℝ+.

ii. Describe geometrically the equivalence class of (0, 0).

7. Consider the relations R1, R2 and R3, represented by the following tables

R1 a b c R2 d e f R3 g h i

a 1 1 1 d 1 1 1 g 1 1

b 1 1 1 e 1 1 h 1

c 1 1 f 1 i 1 1

(A ‘1’ in the table means that the element in that row is related to the element in that column, e.g. in R2, d is related to f, but f is not related to d.)

i. For each relation, determine whether or not it is an equivalence relation. In each case, justify your answer.

ii. For those which are equivalence relations, describe the corresponding equivalence classes.

HL T8 Sets.indb 39 12/03/13 6:46 PM

chapter 2: relations chapter 2 - relationsultimate2.shoppepro.com/~ibidcoma/samples/maths hl option...

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