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0.2: Relations

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    131014
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    Definition: Binary Relation

    Let \(S\) be a nonempty set. Then any subset \(R\) of \(S \times S\) is said to be a relation over \(S\). In other words, a relation is a rule that is defined between two elements in \(S\). Intuitively, if \(R\) is a relation over \(S\), then the statement \(a R b\) is either true or false for all \(a,b\in S\).

    If the statement \(a R b\) is false, we denote this by \( a \not R b\).

    Example \(\PageIndex{1}\):

    Let \(S=\{1,2,3\}\). Define \(R\) by \(a R b\) if and only if \(a < b\), for \(a, b \in S\).

    Then \(1 R 2, 1 R 3, 2 R 3 \) and \( 2 \not R 1\).

    We can visualize the above binary relation as a graph, where the vertices are the elements of S, and there is an edge from \(a\) to \(b\) if and only if \(a R b\), for \(a b\in S\).

    sog8mVfmrjo7_-GoB-mBjEQ.png

    The following are some examples of relations defined on \(\mathbb{Z}\).

    Example \(\PageIndex{2}\):

    1. Define \(R\) by \(a R b\) if and only if \(a < b\), for \(a, b \in \mathbb{Z}\).
    2. Define \(R\) by \(a R b\) if and only if \(a >b\), for \(a, b \in \mathbb{Z}\).
    3. Define \(R\) by \(a R b\) if and only if \(a \leq b\), for \(a, b \in \mathbb{Z}\).
    4. Define \(R\) by \(a R b\) if and only if \(a \geq b\), for \(a, b \in \mathbb{Z}\).
    5. Define \(R\) by \(a R b\) if and only if \(a = b\), for \(a, b \in \mathbb{Z}\).

    Definition: Divisor/Divides

    Let \( a\) and \(b\) be integers. We say that \(a\) divides \(b\) is denoted \(a\mid b\), provided we have an integer \(m\) such that \(b=am\). In this case we can also say the following:

    • \(b\) is divisible by \(a\)
    • \(a\) is a factor of \(b\)
    • \(a\) is a divisor of \(b\)
    • \(b\) is a multiple of \(a\) 

    If \(a\)  does not divide \(b\), then it is  is denoted by \(a \not \mid b\).

    Properties of binary relation:

    Definition: Reflexive

    Let \(S\) be a set and \(R\) be a binary relation on \(S\). Then \(R\) is said to be reflexive if \( a R a, \forall a \in S.\)

     

    We will follow the template below to answer the question about reflexive.

    alt

    Example \(\PageIndex{7}\):

    Define \(R\) by \(a R b\) if and only if \(a < b\), for \(a, b \in \mathbb{Z}\). Is \(R\) reflexive?

    Counter Example:

    Choose \(a=2.\)

    Since \( 2 \not< 2\), \(R\) is not reflexive.

    Example \(\PageIndex{8}\):

    Define \(R\) by \(a R b\) if and only if \(a \mid b\), for \(a, b \in \mathbb{Z}\). Is \(R\) reflexive?

    Proof:

    Let \( a \in \mathbb{Z}\). Since \(a=(1) (a)\), \(a \mid a\).

    Thus \(R\) is reflexive. \( \Box\)

    Definition: Symmetric

    Let \(S\) be a set and \(R\) be a binary relation on \(S\). Then \(R\) is said to be symmetric if the following statement is true:

    \( \forall a,b \in S\), if \( a R b \) then \(b R a\), in other words, \( \forall a,b \in S, a R b \implies b R a.\)

    Example \(\PageIndex{8}\): Visually

    alt

    \(\forall a,b \in S, a R b \implies b R a.\) holds!

    We will follow the template below to answer the question about symmetric.

    stXO_6c-DcCh0t5duFY2Ejw.png

    Example \(\PageIndex{9}\):

    Define \(R\) by \(a R b\) if and only if \(a < b\), for \(a, b \in \mathbb{Z}\). Is \(R\) symmetric?

    Counter Example:

    \(1<2\) but \(2 \not < 1\).

    Example \(\PageIndex{10}\):

    Define \(R\) by \(a R b\) if and only if \(a \mid b\), for \(a, b \in \mathbb{Z}\). Is \(R\) symmetric?

    Counter Example:

    \(2 \mid 4\) but \(4 \not \mid 2\).

     

    Definition: Transitive

    Let \(S\) be a set and \(R\) be a binary relation on \(S\). Then \(R\) is said to be transitive if the following statement is true

    \( \forall a,b,c \in S,\) if \( a R b \) and \(b R c\), then \(a R c\).

    In other words, \( \forall a,b,c \in S\), \( a R b \wedge b R c \implies a R c\).

    Example \(\PageIndex{14}\): VISUALLY

    alt

    \( \forall a,b,c \in S\), \( a R b \wedge b R c \implies a R c\) holds!

    We will follow the template below to answer the question about transitive.

    ssUmTK4d9hHGQc-mZjO6p5g.png

    Example \(\PageIndex{15}\):

    Define \(R\) by \(a R b\) if and only if \(a < b\), for \(a, b \in \mathbb{Z}\). Is \(R\) transitive?

    Example \(\PageIndex{16}\):

    Define \(R\) by \(a R b\) if and only if \(a \mid b\), for \(a, b \in \mathbb{Z_+}\). Is \(R\) transitive?

     

    Definition: Equivalence Relation

    A binary relation is an equivalence relation on a nonempty set \(S\) if and only if the relation is reflexive(R), symmetric(S) and transitive(T).

    Definition: Equivalence Relation

    Example \(\PageIndex{1}\): \(=\)

     

    Let \(S=\mathbb{R}\) and \(R\) be =. Is the relation a) reflexive, b) symmetric, c) antisymmetric, d) transitive, e) an equivalence relation, f) a partial order.

    Solution:

    1. Yes \(R \) is reflexive.

      Proof:

      Let \(a \in \mathbb{R} \).

      Then \(a=a \).

      Hence \(R \) is reflexive. \(\blacksquare \)

       

      Yes \(R \) is reflexive.

      Proof:

      Let \((a,b) \in \mathbb{R} \).

      If \(a=b \), clearly \(b=a \).

      Hence \(R \) is symmetric. \(\blacksquare \)

      Yes \(R \) is antisymmetric.

      Proof:

      Let \(a,b \in \mathbb{R} \) s.t. \(a=b \) and \(b=a \) then clearly \(a=b \; \forall a,b \in \mathbb{R} \). \(blacksquare \)

       

      Yes \(R \) is transitive.

      Proof:

      Let \(a,b,c \in \mathbb{R} \) s.t. \(a=b \) and \(b=c \).

      We shall show that \(aRc \).

      Since \(a=b \) and \(b=c \) it follows that \(a=c \)

      Thus \(aRc \). \(blacksquare \)

       

      Yes \(R \) is an equivalence relation.

      Proof:

      Since \(R \) is reflexive, symmetric and transitive, it is an equivalence relation.

       

      Yes \(R \) is a partial order.

      Proof:

      Since \(R \) is a partial order since \(R \) is reflexive, antisymmetric and transitive.

    Example \(\PageIndex{2}\): Less than or equal to

    Let \(S=\mathbb{R}\) and \(R\) be \(\leq\). Is the relation a) reflexive, b) symmetric, c) antisymmetric, d) transitive, e) an equivalence relation, f) a partial order.

    Solution

    Proof:

    We will show that is true.

    Let that is .

    Since , is reflexive.◻

    Counterexample:

    Let and which are both .

    It is true that , but it is not true that .

    Thus is not symmetric.◻

    Proof:

    We will show that given and that .

    Since , s.t. .

    Further, since , , .

    Then .

    Thus, .

    Since, , thus .◻

    4. Yes, is transitive.

    Proof:

    We will show that given and that .

    Since , s.t. .

    Further, since , , .

    Then .

    Since, , thus .◻

    Definition: Equivalence Class

    Given an equivalence relation \( R \) over a set \( S, \) for any \(a \in S \) the equivalence class of a is the set \( [a]_R =\{ b \in S \mid a R b \} \), that is
    \([a]_R \) is the set of all elements of S that are related to \(a\).

    Example \(\PageIndex{3}\): Equivalence relation

    Define a relation that two shapes are related iff they are the same color. Is this relation an equivalence relation?

    Equivalence classes are:

    alt

    Example \(\PageIndex{4}\):

    Define a relation that two shapes are related iff they are similar. Is this relation an equivalence relation?

    Equivalence classes are:

    alt

    Definition: Partition

    Let \(A\) be a nonempty set. A partition of \(A\) is a set of nonempty pairwise disjoint sets whose union is A.

     

    Note

    For every equivalence relation over a nonempty set \(S\), \(S\) has a partition.

     

    Theorem \(\PageIndex{1}\)

    If \( \sim \) is an equivalence relation over a nonempty set \(S\). Then the set of all equivalence classes is denoted by \(\{[a]_{\sim}| a \in S\}\) forms a partition of \(S\).

    This means

    1. Either \([a] \cap [b] = \emptyset\) or \([a]=[b]\), for all \(a,b\in S\).

    2. \(S= \cup_{a \in S} [a]\).

    Proof

    Assume is an equivalence relation on a nonempty set .

    Let . If , then we are done. Otherwise,

    assume .

    Let be the common element between them.

    Let .

    Then and , which means that and .

    Since is an equivalence relation and .

    Since and (due to transitive property), .

    Thus and .

    Hence .

    Next, we will show that .

    First we shall show that .

    Let .

    Then exists and .

    Hence .

    Thus .

    Conversely, we shall show that .

    Let .

    Then for some .

    Thus .

    Thus .

    Since and , then .◻

     

     

    For the following examples, determine whether or not each of the following binary relations on the given set is reflexive, symmetric, antisymmetric, or transitive. If a relation has a certain property, prove this is so; otherwise, provide a counterexample to show that it does not. If is an equivalence relation, describe the equivalence classes of .

    Example \(\PageIndex{5}\):

    Let \(S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}\). Define a relation \(R\) on \(A = S \times S \) by \((a, b) R (c, d)\)  if and only if \(10a + b \leq 10c + d.\) 

     

    Solution

    1. is reflexive on .

    Proof:

    Let .

    We will show that .

    Since , then .

    Since is reflexive on .◻

    2. is not symmetric on .

    Counter Example:

    Let .

    Note , specifically, is true.

    However, , is false.

    Since , is not symmetric on .◻

    3. is antisymmetric on .

    Proof:

    Let s.t. and .

    Since and , .

    We will show that and .

    by definition.

    Since and , thus .

    Since , .

    Thus is antisymmetric on .◻​​​​

    5. is transitive on .

    Proof:

    Let s.t. and .

    We will show that .

    Since , .

    Thus is transitive on .◻

    6. is not an equivalence relation since it is not reflexive, symmetric, and transitive.

    Example \(\PageIndex{6}\):

    Let . Define a relation on , by if and only if

    Solution

    1. is reflexive on .

    Proof:

    Let we will show that .

    Clearly since and a negative integer multiplied by a negative integer is a positive integer in .

    Since , is reflexive on .◻​​​​​

     

    2. is symmetric on .

    Proof:

    We will show that if , then .

    Let s.t. , that is .

    Since , .

    Since , is symmetric on .◻

    3. is not antisymmetric on .

    Counter Example:

    Let and then and .

    Since , is not antisymmetric on .◻

    4. is transitive on .

    Proof:

    Let s.t and s.t. .

    There are two cases to be examined:

    Case 1: and .

    Since , thus .

    Case 2: and .

    Since , thus .

    Since in both possible cases is transitive on .◻

     

    5. Since is reflexive, symmetric and transitive, it is an equivalence relation. Equivalence classes are and .

    Let , then

    .

    Case 1: , then .

    .

    .

    Case 2: , then .

    .

    .

    Note this is a partition since or . So we have all the intersections are empty.

    .

    Further, we have . Note that is excluded from .

    Recall:

    Let \(m,n \in \mathbb{Z}\).

    We say that \(m\) divides \(n\) (written as \(m|n\)) if \(n=mk\) for some \(k\in \mathbb{Z}\).

     

    Example \(\PageIndex{1}\)

    Let \(S= \mathbb{Z}, m\in \mathbb{Z}_+\).  Define \(aRb\) if and only if \(m|(a-b), \forall a,b \in \mathbb{Z}\).

    Note that \(m\) is a positive integer.

    Let \(m=3\).  Choose \(a=1\) and \(b=2\).

    Are 1 and 2 related? i.e. does \(3|1-2\)?  Does \(1-2=3k, k\in \mathbb{Z}\)?  Since \(\notexists k\) s.t. \(-1 = 3k\), ∴ \(1 !R 2\).

     

    However, we can see that we have three groups as shown in the image below:

     

     Screen Shot 2023-06-28 at 2.41.58 PM.png

     

    Definition: Modulo

    Let \(m\) \(\in\) \(\mathbb{Z_+}\).

    \(a\) is congruent to \(b\) modulo \(m\) denoted as \( a \equiv b (mod \, n) \), if \(a\) and \(b\)  have the remainder when they are divided by \(n\), for  \(a, b \in \mathbb{Z}\).

    Properties

    Let \(n \in \mathbb{Z_+}\). Then

    Theorem 1 :

    Two integers \(a \) and \(b\) are said to be congruent modulo \( n\), \(a \equiv b (mod \, n)\), if all of the following are true:

    a) \(m\mid (a-b).\)

    b) both \(a\) and \(b \) have the same remainder when divided by \(n.\)

    c) \(a-b= kn\), for some \(k \in \mathbb{Z}\).

    NOTE: Possible remainders of \( n\) are \(0, ..., n-1.\)

    Reflexive Property

    Theorem 2:

    The relation " \(\equiv\) " over \(\mathbb{Z}\) is reflexive.

    Proof: Let \(a \in \mathbb{Z} \).

    Then \(a-a=0(n)\), and \( 0 \in \mathbb{Z}\).

    Hence \(a \equiv a (mod \, n)\).

    Thus congruence modulo n is Reflexive.

    Symmetric Property

    Theorem 3:

    The relation " \(\equiv\) " over \(\mathbb{Z}\) is symmetric.

    Proof: Let \(a, b \in \mathbb{Z} \) such that \(a \equiv b\) (mod n).

    Then \(a-b=kn, \) for some \(k \in \mathbb{Z}\).

    Now \( b-a= (-k)n \) and \(-k \in \mathbb{Z}\).

    Hence \(b \equiv a(mod \, n)\).

    Thus the relation is symmetric.

    Antisymmetric Property

    Is the relation " \(\equiv\) " over \(\mathbb{Z}\) antisymmetric?

    Counter Example: \(n\) is fixed

    choose: \(a= n+1, b= 2n+1\), then

    \(a \equiv b(mod \, n)\) and \( b \equiv a (mod \, n)\)

    but \( a \ne b.\)

    Thus the relation " \(\equiv\) "on \(\mathbb{Z}\) is not antisymmetric.

    Transitive Property

    Theorem 4 :

    The relation " \(\equiv\) " over \(\mathbb{Z}\) is transitive.

    Proof: Let \(a, b, c \in\) \(\mathbb{Z}\), such that \(a \equiv b (mod n)\) and \(b \equiv c (mod n).\)

    Then \(a=b+kn, k \in\) \(\mathbb{Z}\) and \(b=c+hn, h \in\) \(\mathbb{Z}\).

    We shall show that \(a \equiv c (mod \, n)\).

    Consider \(a=b+kn=(c+hn)+kn=c+(hn+kn)=c+(h+k)n, h+k \in\) \(\mathbb{Z}\).

    Hence \(a \equiv c (mod \, n)\).

    Thus congruence modulo \(n\) is transitive.

    Theorem 5:

    The relation " \(\equiv\) " over \(\mathbb{Z}\) is an equivalence relation.

    Modulo classes

    Let . The relation \( \equiv \) on by \( a \equiv b \) if and only if , is an equivalence relation. The Classes of have the following equivalence classes:

    .

     

    Example of writing equivalence classes:

    Example \(\PageIndex{3}\):

    The equivalence classes for \(( mod \, 3 )\) are (need to show steps):

    .

    .

    .

     Modulo Arithmetic

    In this section, we will explore arithmetic operations in a modulo world.

    Let \( n \in \mathbb{Z_+}\).

    Theorem 5 :

    Let \( a, b, c,d, \in \mathbb{Z}\) such that \(a \equiv b (mod\,n) \) and \(c \equiv d (mod\, n). \) Then \((a+c) \equiv (b+d)(mod\, n).\)

    Proof:

    Let \(a, b, c, d \in\mathbb{Z}\), such that \(a \equiv b (mod\, n) \) and \(c \equiv d (mod \,n). \)

    We shall show that \( (a+c) \equiv (b+d) (mod\, n).\)

    Since \(a \equiv b (mod\, n) \) and \(c \equiv d (mod\, n), n \mid (a-b)\) and \(n \mid (c-d)\)

    Thus \( a= b+nk, \) and \(c= d+nl,\) for \(k \) and \( l \in \mathbb{Z}\).

    Consider\( (a+c) -( b+d)= a-b+c-d=n(k+l), k+l \in \mathbb{Z}\).

    Hence \((a+c)\equiv (b+d) (mod \,n).\Box\)

    Theorem 6:

    Edit section

    Let \( a, b, c,d, \in \mathbb{Z}\) such that \(a \equiv b (mod \, n) \) and \(c \equiv d (mod \,n). \) Then \((ac) \equiv (bd) (mod \,n).\)

    Proof:

    Let \(a, b, c, d \in \mathbb{Z}\), such that \(a \equiv b (mod\, n) \) and \(c \equiv d (mod \, n). \)

    We shall show that \( (ac) \equiv (bd) (mod\, n).\)

    Since \(a \equiv b (mod \,n) \) and \(c \equiv d (mod\, n), n \mid (a-b)\) and \(n \mid (c-d)\)

    Thus \( a= b+nk, \) and \(c= d+nl,\) for \(k\) and \( l \in \mathbb{Z}\).

    Consider \( (ac) -( bd)= ( b+nk) ( d+nl)-bd= bnl+dnk+n^2lk=n (bl+dk+nlk), \) where \((bl+dk+nlk) \in \mathbb{Z}\).

    Hence \((ac) \equiv (bd) (mod \,n).\Box\)

    Theorem 7:

    Let \(a, b \in\) \(\mathbb{Z}\) such that \(a \equiv b (mod \, n) \). Then \(a^2 \equiv b^2 (mod \, n).\)

    Proof:

    Let \(a, b \in\) \(\mathbb{Z}\), and n \(\in\) \(\mathbb{Z_+}\), such that \(a \equiv b (mod \, n).\)

    We shall show that \(a^2 \equiv b^2 (mod \, n).\)

    Since \( a \equiv b (mod \, n), n\mid (a-b).\)

    Thus \( (a-b)= nx,\) where \(x \in\) \(\mathbb{Z}\).

    Consider \((a^2 - b^2) = (a+b)(a-b)=(a+b)(nx), = n(ax+bx), ax+bx \in \mathbb{Z}\).

    Hence \(n \mid a^2 - b^2,\) therefore \(a^2 \equiv b^2 (mod \, n).\) \(\Box\)

    Theorem 8:

    Let \(a, b \in\) \(\mathbb{Z}\) such that \(a \equiv b (mod \, n)\). Then \(a^m \equiv b^m (mod \, n)\), \(\forall \in\) \(\mathbb{Z}\).

    Proof:

    Exercise.

    By using the above result, we can solve many problems. Some of them are discussed below: 

    Example \(\PageIndex{1}\): \( mod\, 3\) Arithmetic

    Let \(n = 3\).

    Addition

    + [0] [1] [2]
    [0] [0] [1] [2]
    [1] [1] [2] [0]
    [2] [2] [0] [1]

    Multiplication

    x [0] [1] [2]
    [0] [0] [0] [0]
    [1] [0] [1] [2]
    [2] [0] [2] [1]

    Example \(\PageIndex{2}\): \( mod\, 4\) Arithmetic

    Let \(n=4\).

    x [0] [1] [2] [3]
    [0] [0] [0] [0] [0]
    [1] [0] [1] [2] [3]
    [2] [0] [2] [0] [2]
    [3] [0] [3] [2] [1]

    Example \(\PageIndex{3}\):

    Find the remainder when \((101)(103)(107)(109)\) is divided by \( 11.\)

    Answer

    \(101 \equiv 2 (mod \, 11)\)

    \(103 \equiv 4 (mod \, 11)\)

    \(107 \equiv 8 (mod \, 11)\)

    \(109 \equiv 10 (mod \,11)\).

    Therefore,

    \((101)(103)(107)(109) \equiv (2)(4)(8)(10) (mod \,11) \equiv 2 (mod \,11)\).

    Example \(\PageIndex{4}\):

    Find the remainder when\( 7^{1453}\) is divided by \( 8.\)

    Answer

    \(7^0 \equiv 1 (mod \, 8)\)

    \(7^1 \equiv 7 (mod \, 8)\)

    \(7^2 \equiv 1 (mod \, 8)\)

    \(7^3 \equiv 7 (mod \, 8)\),

    As there is a consistent pattern emerging and we know that \(1453\) is odd, then \(7^{1453} \equiv 7 (mod \, 8)\). Thus the remainder is \(7.\)

    Example \(\PageIndex{5}\):

    Find the remainder when \(7^{2020}\) is divided by \(18.\)

    Answer

    \(7^0 \equiv 1 (mod \, 18)\)

    \(7^1 \equiv 7 (mod \, 18)\)

    \(7^2 \equiv 13 (mod \, 18)\)

    \(7^3 \equiv 1 (mod \, 18)\),

    As there is a consistent pattern emerging and we know that \(2020=(673)3+1\), \(7^{2020}= 7^{(673)3+1}=\left( 7^3\right)^{673}7^1 \equiv 7 (mod \, 18).\) Thus the remainder is \(7.\)

    Example \(\PageIndex{6}\):

    Find the remainder when \( 26^{1453} \) is divided by \( 3.\)


    This page titled 0.2: Relations is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

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