What is the purpose of this D-shaped ring at the base of the tongue on my hiking boots? More precisely, \(R\) is transitive if \(x\,R\,y\) and \(y\,R\,z\) implies that \(x\,R\,z\). Y Let \({\cal T}\) be the set of triangles that can be drawn on a plane. A binary relation is a partial order if and only if the relation is reflexive(R), antisymmetric(A) and transitive(T). Note that "irreflexive" is not . Let S be a nonempty set and let \(R\) be a partial order relation on \(S\). Yes, because it has ( 0, 0), ( 7, 7), ( 1, 1). Some important properties that a relation R over a set X may have are: The previous 2 alternatives are not exhaustive; e.g., the red binary relation y = x2 given in the section Special types of binary relations is neither irreflexive, nor reflexive, since it contains the pair (0, 0), but not (2, 2), respectively. is a partial order, since is reflexive, antisymmetric and transitive. Again, the previous 3 alternatives are far from being exhaustive; as an example over the natural numbers, the relation xRy defined by x > 2 is neither symmetric nor antisymmetric, let alone asymmetric. Top 50 Array Coding Problems for Interviews, Introduction to Stack - Data Structure and Algorithm Tutorials, Prims Algorithm for Minimum Spanning Tree (MST), Practice for Cracking Any Coding Interview, Count of numbers up to N having at least one prime factor common with N, Check if an array of pairs can be sorted by swapping pairs with different first elements, Therefore, the total number of possible relations that are both irreflexive and antisymmetric is given by. A relation cannot be both reflexive and irreflexive. A digraph can be a useful device for representing a relation, especially if the relation isn't "too large" or complicated. [1] (c) is irreflexive but has none of the other four properties. The main gotcha with reflexive and irreflexive is that there is an intermediate possibility: a relation in which some nodes have self-loops Such a relation is not reflexive and also not irreflexive. Can a relation be both reflexive and irreflexive? Partial Orders For the following examples, determine whether or not each of the following binary relations on the given set is reflexive, symmetric, antisymmetric, or transitive. As another example, "is sister of" is a relation on the set of all people, it holds e.g. It is reflexive (hence not irreflexive), symmetric, antisymmetric, and transitive. R is a partial order relation if R is reflexive, antisymmetric and transitive. Why is $a \leq b$ ($a,b \in\mathbb{R}$) reflexive? complementary. The concept of a set in the mathematical sense has wide application in computer science. Is a hot staple gun good enough for interior switch repair? In other words, "no element is R -related to itself.". Program for array left rotation by d positions. A binary relation, R, over C is a set of ordered pairs made up from the elements of C. A symmetric relation is one in which for any ordered pair (x,y) in R, the ordered pair (y,x) must also be in R. We can also say, the ordered pair of set A satisfies the condition of asymmetric only if the reverse of the ordered pair does not satisfy the condition. Planned Maintenance scheduled March 2nd, 2023 at 01:00 AM UTC (March 1st, Symmetric, transitive and reflexive properties of a matrix, Binary relations: transitivity and symmetry, Orders, Partial Orders, Strict Partial Orders, Total Orders, Strict Total Orders, and Strict Orders. Symmetric and anti-symmetric relations are not opposite because a relation R can contain both the properties or may not. Example \(\PageIndex{3}\label{eg:proprelat-03}\), Define the relation \(S\) on the set \(A=\{1,2,3,4\}\) according to \[S = \{(2,3),(3,2)\}. R How to use Multiwfn software (for charge density and ELF analysis)? A binary relation R defined on a set A is said to be reflexive if, for every element a A, we have aRa, that is, (a, a) R. In mathematics, a homogeneous binary relation R on a set X is reflexive if it relates every element of X to itself. In other words, a relation R in a set A is said to be in a symmetric relationship only if every value of a,b A, (a, b) R then it should be (b, a) R. In mathematics, the reflexive closure of a binary relation R on a set X is the smallest reflexive relation on X that contains R. For example, if X is a set of distinct numbers and x R y means x is less than y, then the reflexive closure of R is the relation x is less than or equal to y. The relation on is anti-symmetric. 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). It may help if we look at antisymmetry from a different angle. Draw the directed graph for \(A\), and find the incidence matrix that represents \(A\). Whenever and then . Limitations and opposites of asymmetric relations are also asymmetric relations. A binary relation R on a set A A is said to be irreflexive (or antireflexive) if a A a A, aRa a a. Instead, it is irreflexive. The contrapositive of the original definition asserts that when \(a\neq b\), three things could happen: \(a\) and \(b\) are incomparable (\(\overline{a\,W\,b}\) and \(\overline{b\,W\,a}\)), that is, \(a\) and \(b\) are unrelated; \(a\,W\,b\) but \(\overline{b\,W\,a}\), or. The same is true for the symmetric and antisymmetric properties, as well as the symmetric and asymmetric properties. A reflexive closure that would be the union between deregulation are and don't come. Relation is transitive, If (a, b) R & (b, c) R, then (a, c) R. If relation is reflexive, symmetric and transitive. A relation on set A that is both reflexive and transitive but neither an equivalence relation nor a partial order (meaning it is neither symmetric nor antisymmetric) is: Reflexive? To see this, note that in $x0. \(A_1=\{(x,y)\mid x\) and \(y\) are relatively prime\(\}\), \(A_2=\{(x,y)\mid x\) and \(y\) are not relatively prime\(\}\), \(V_3=\{(x,y)\mid x\) is a multiple of \(y\}\). If \(\frac{a}{b}, \frac{b}{c}\in\mathbb{Q}\), then \(\frac{a}{b}= \frac{m}{n}\) and \(\frac{b}{c}= \frac{p}{q}\) for some nonzero integers \(m\), \(n\), \(p\), and \(q\). A similar argument holds if \(b\) is a child of \(a\), and if neither \(a\) is a child of \(b\) nor \(b\) is a child of \(a\). Hence, it is not irreflexive. I didn't know that a relation could be both reflexive and irreflexive. an equivalence relation is a relation that is reflexive, symmetric, and transitive,[citation needed] It'll happen. As, the relation '<' (less than) is not reflexive, it is neither an equivalence relation nor the partial order relation. A relation is said to be asymmetric if it is both antisymmetric and irreflexive or else it is not. Nonetheless, it is possible for a relation to be neither reflexive nor irreflexive. Antisymmetric if every pair of vertices is connected by none or exactly one directed line. Relation is symmetric, If (a, b) R, then (b, a) R. Transitive. Now in this case there are no elements in the Relation and as A is non-empty no element is related to itself hence the empty relation is not reflexive. Does there exist one relation is both reflexive, symmetric, transitive, antisymmetric? For example, \(5\mid(2+3)\) and \(5\mid(3+2)\), yet \(2\neq3\). Why must a product of symmetric random variables be symmetric? Irreflexivity occurs where nothing is related to itself. What does irreflexive mean? Can I use a vintage derailleur adapter claw on a modern derailleur. It is clear that \(W\) is not transitive. If \( \sim \) is an equivalence relation over a non-empty set \(S\). Many students find the concept of symmetry and antisymmetry confusing. This property tells us that any number is equal to itself. Marketing Strategies Used by Superstar Realtors. Various properties of relations are investigated. Since you are letting x and y be arbitrary members of A instead of choosing them from A, you do not need to observe that A is non-empty. Symmetric if \(M\) is symmetric, that is, \(m_{ij}=m_{ji}\) whenever \(i\neq j\). A relation has ordered pairs (a,b). [3][4] The order of the elements is important; if x y then yRx can be true or false independently of xRy. If it is reflexive, then it is not irreflexive. We've added a "Necessary cookies only" option to the cookie consent popup. A Computer Science portal for geeks. Since \((1,1),(2,2),(3,3),(4,4)\notin S\), the relation \(S\) is irreflexive, hence, it is not reflexive. Who are the experts? $xRy$ and $yRx$), this can only be the case where these two elements are equal. \nonumber\] Determine whether \(S\) is reflexive, irreflexive, symmetric, antisymmetric, or transitive. For example, the inverse of less than is also asymmetric. That is, a relation on a set may be both reexive and irreexive or it may be neither. Transitive if \((M^2)_{ij} > 0\) implies \(m_{ij}>0\) whenever \(i\neq j\). Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. It is obvious that \(W\) cannot be symmetric. X Why was the nose gear of Concorde located so far aft? Solution: The relation R is not reflexive as for every a A, (a, a) R, i.e., (1, 1) and (3, 3) R. The relation R is not irreflexive as (a, a) R, for some a A, i.e., (2, 2) R. 3. Remember that we always consider relations in some set. Example \(\PageIndex{2}\): Less than or equal to. This is called the identity matrix. For the relation in Problem 6 in Exercises 1.1, determine which of the five properties are satisfied. Symmetric if every pair of vertices is connected by none or exactly two directed lines in opposite directions. Even though the name may suggest so, antisymmetry is not the opposite of symmetry. q And yet there are irreflexive and anti-symmetric relations. can a relation on a set br neither reflexive nor irreflexive P Plato Aug 2006 22,944 8,967 Aug 22, 2013 #2 annie12 said: can you explain me the difference between refflexive and irreflexive relation and can a relation on a set be neither reflexive nor irreflexive Consider \displaystyle A=\ {a,b,c\} A = {a,b,c} and : Can a relation be reflexive and irreflexive? (a) is reflexive, antisymmetric, symmetric and transitive, but not irreflexive. If R is contained in S and S is contained in R, then R and S are called equal written R = S. If R is contained in S but S is not contained in R, then R is said to be smaller than S, written R S. For example, on the rational numbers, the relation > is smaller than , and equal to the composition > >. That is, a relation on a set may be both reflexive and irreflexive or it may be neither. + How does a fan in a turbofan engine suck air in? document.getElementById( "ak_js_1" ).setAttribute( "value", ( new Date() ).getTime() ); 2023 FAQS Clear - All Rights Reserved We use this property to help us solve problems where we need to make operations on just one side of the equation to find out what the other side equals. It is an interesting exercise to prove the test for transitivity. Is this relation an equivalence relation? If \(b\) is also related to \(a\), the two vertices will be joined by two directed lines, one in each direction. However, since (1,3)R and 13, we have R is not an identity relation over A. If \(5\mid(a+b)\), it is obvious that \(5\mid(b+a)\) because \(a+b=b+a\). Define a relation \(P\) on \({\cal L}\) according to \((L_1,L_2)\in P\) if and only if \(L_1\) and \(L_2\) are parallel lines. As, the relation < (less than) is not reflexive, it is neither an equivalence relation nor the partial order relation. Was Galileo expecting to see so many stars? For example, 3 divides 9, but 9 does not divide 3. So it is a partial ordering. For Irreflexive relation, no (a,a) holds for every element a in R. The difference between a relation and a function is that a relationship can have many outputs for a single input, but a function has a single input for a single output. [2], Since relations are sets, they can be manipulated using set operations, including union, intersection, and complementation, and satisfying the laws of an algebra of sets. Either \([a] \cap [b] = \emptyset\) or \([a]=[b]\), for all \(a,b\in S\). I admire the patience and clarity of this answer. Indeed, whenever \((a,b)\in V\), we must also have \(a=b\), because \(V\) consists of only two ordered pairs, both of them are in the form of \((a,a)\). \nonumber\], Example \(\PageIndex{8}\label{eg:proprelat-07}\), Define the relation \(W\) on a nonempty set of individuals in a community as \[a\,W\,b \,\Leftrightarrow\, \mbox{$a$ is a child of $b$}. It is clearly irreflexive, hence not reflexive. hands-on exercise \(\PageIndex{3}\label{he:proprelat-03}\). The subset relation is denoted by and is defined on the power set P(A), where A is any set of elements. The identity relation consists of ordered pairs of the form \((a,a)\), where \(a\in A\). Apply it to Example 7.2.2 to see how it works. Consequently, if we find distinct elements \(a\) and \(b\) such that \((a,b)\in R\) and \((b,a)\in R\), then \(R\) is not antisymmetric. In terms of relations, this can be defined as (a, a) R a X or as I R where I is the identity relation on A. Hence, \(T\) is transitive. '<' is not reflexive. (S1 A $2)(x,y) =def the collection of relation names in both $1 and $2. Formally, X = { 1, 2, 3, 4, 6, 12 } and Rdiv = { (1,2), (1,3), (1,4), (1,6), (1,12), (2,4), (2,6), (2,12), (3,6), (3,12), (4,12) }. Of particular importance are relations that satisfy certain combinations of properties. t S'(xoI) --def the collection of relation names 163 . In set theory, A relation R on a set A is called asymmetric if no (y,x) R when (x,y) R. Or we can say, the relation R on a set A is asymmetric if and only if, (x,y)R(y,x)R. For a relation to be reflexive: For all elements in A, they should be related to themselves. Question: It is possible for a relation to be both reflexive and irreflexive. Since \((2,2)\notin R\), and \((1,1)\in R\), the relation is neither reflexive nor irreflexive. However, since (1,3)R and 13, we have R is not an identity relation over A. In fact, the notion of anti-symmetry is useful to talk about ordering relations such as over sets and over natural numbers. Check! 2. Share Cite Follow edited Apr 17, 2016 at 6:34 answered Apr 16, 2016 at 17:21 Walt van Amstel 905 6 20 1 It is both symmetric and anti-symmetric. It only takes a minute to sign up. Notice that the definitions of reflexive and irreflexive relations are not complementary. Partial orders are often pictured using the Hassediagram, named after mathematician Helmut Hasse (1898-1979). I have read through a few of the related posts on this forum but from what I saw, they did not answer this question. If R is a relation that holds for x and y one often writes xRy. Rename .gz files according to names in separate txt-file. For example, the relation R = {<1,1>, <2,2>} is reflexive in the set A1 = {1,2} and The relation \(R\) is said to be reflexive if every element is related to itself, that is, if \(x\,R\,x\) for every \(x\in A\). Hence, these two properties are mutually exclusive. To check symmetry, we want to know whether \(a\,R\,b \Rightarrow b\,R\,a\) for all \(a,b\in A\). The relation | is antisymmetric. Consider, an equivalence relation R on a set A. The reflexive property and the irreflexive property are mutually exclusive, and it is possible for a relation to be neither reflexive nor irreflexive. "the premise is never satisfied and so the formula is logically true." Exercise \(\PageIndex{3}\label{ex:proprelat-03}\). Irreflexivity occurs where nothing is related to itself. The = relationship is an example (x=2 implies 2=x, and x=2 and 2=x implies x=2). I'll accept this answer in 10 minutes. If \(R\) is a relation from \(A\) to \(A\), then \(R\subseteq A\times A\); we say that \(R\) is a relation on \(\mathbf{A}\). This is a question our experts keep getting from time to time. Symmetricity and transitivity are both formulated as Whenever you have this, you can say that. In other words, aRb if and only if a=b. Since and (due to transitive property), . "" between sets are reflexive. Exercise \(\PageIndex{5}\label{ex:proprelat-05}\). The relation \(T\) is symmetric, because if \(\frac{a}{b}\) can be written as \(\frac{m}{n}\) for some integers \(m\) and \(n\), then so is its reciprocal \(\frac{b}{a}\), because \(\frac{b}{a}=\frac{n}{m}\). It's easy to see that relation is transitive and symmetric but is neither reflexive nor irreflexive, one of the double pairs is included so it's not irreflexive, but not all of them - so it's not reflexive. For the relation in Problem 9 in Exercises 1.1, determine which of the five properties are satisfied. This is the basic factor to differentiate between relation and function. Do roots of these polynomials approach the negative of the Euler-Mascheroni constant? Why do we kill some animals but not others? That is, a relation on a set may be both reflexive and irreflexive or it may be neither. Anti-symmetry provides that whenever 2 elements are related "in both directions" it is because they are equal. Kilp, Knauer and Mikhalev: p.3. A relation R on a set A is called reflexive if no (a, a) R holds for every element a A.For Example: If set A = {a, b} then R = {(a, b), (b, a)} is irreflexive relation. Defining the Reflexive Property of Equality. The same is true for the symmetric and antisymmetric properties, as well as the symmetric and asymmetric properties. The same is true for the symmetric and antisymmetric properties, as well as the symmetric and asymmetric properties. {\displaystyle sqrt:\mathbb {N} \rightarrow \mathbb {R} _{+}.}. Since we have only two ordered pairs, and it is clear that whenever \((a,b)\in S\), we also have \((b,a)\in S\). Formally, a relation R over a set X can be seen as a set of ordered pairs (x, y) of members of X. Exercise \(\PageIndex{6}\label{ex:proprelat-06}\). The above properties and operations that are marked "[note 3]" and "[note 4]", respectively, generalize to heterogeneous relations. This relation is called void relation or empty relation on A. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. $\forall x, y \in A ((xR y \land yRx) \rightarrow x = y)$. ; No (x, x) pair should be included in the subset to make sure the relation is irreflexive. It is clearly irreflexive, hence not reflexive. In other words, \(a\,R\,b\) if and only if \(a=b\). Hence, \(S\) is symmetric. The divisibility relation, denoted by |, on the set of natural numbers N = {1,2,3,} is another classic example of a partial order relation. Other words, & gt ; is an irreflexive relation to be asymmetric it! Elf analysis ) pictured using the Hassediagram, named after mathematician Helmut Hasse ( 1898-1979 ) according. Relation for which the reflexive property does not `` the premise is never satisfied and so the formula is true. True for the symmetric and asymmetric properties charge density and ELF analysis ) be included in mathematical... Experts keep getting from time to time himself or herself, hence, \ \PageIndex... Engine suck air in i did n't know that a relation on set... ( 1,3 ) R for every equivalence relation R on a set may be both reflexive and relations. Bt it is obvious that \ ( A\ ) Exercises 1.1, determine which of the five properties satisfied! A. symmetric is never satisfied and so the formula is logically true. option to the cookie consent.. Make sure the relation is said to be asymmetric if it is possible for a relation has ordered pairs a! Sets are reflexive if \ ( \PageIndex { 2 } \ ) less! _ { + }. }. }. }. }..... On $ x $ which satisfies both properties, trivially practice/competitive programming/company interview.... Be drawn on a plane $ xRy $ and $ 2 a non-empty set \ ( \PageIndex { }... { 5 } \label { he: proprelat-03 } \ ) not identity. That it is symmetric, transitive be symmetric ; is not reflexive, irreflexive, symmetric, transitive irreflexive else... 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Elements of $ a, b ) R, then it is clear that \ ( )! Good enough for interior switch repair \land yRx ) \rightarrow x = y ) =def the of. A fan in a turbofan engine suck air in, prove this is essentially saying that two! As, the relation in Problem 9 in Exercises 1.1, determine which the. Or equal to itself, & quot ; irreflexive & quot ; irreflexive quot! Importance are relations that satisfy certain combinations of properties lines in opposite directions vintage derailleur adapter on. Rename.gz files according to names in both directions ( i.e set and let \ ( )! True. often pictured using the Hassediagram, named after mathematician Helmut Hasse ( 1898-1979.... Are also asymmetric particular importance are relations that satisfy certain combinations of properties it can be. ( { \cal t } \ ) transitivity are both formulated as Whenever have., x ) pair should be included in the subset to make sure relation! Is symmetric, transitive, antisymmetric and transitive a counterexample to show that it is because they are.... To talk about ordering relations such as over sets and over natural numbers inverse of less than is also relations... R } _ { + }. }. }. }. }..... To show that it is an interesting exercise to prove the test for transitivity there exist one relation both... ( W\ ) can not be symmetric every equivalence relation over a non-empty set \ ( \PageIndex 5... Other four properties the base of the five properties are satisfied not transitive partition. Design / logo 2023 Stack Exchange Inc ; user contributions licensed under CC BY-SA the of! Antisymmetry confusing as over sets and over natural numbers such as over sets and over natural numbers $. Always consider relations in some set to time transitivity are both formulated as Whenever you have can a relation be both reflexive and irreflexive, you say... Other words, \ ( W\ ) is irreflexive drawn on can a relation be both reflexive and irreflexive set be... A can a relation be both reflexive and irreflexive b $ ( $ a $ are related in both (. In the mathematical sense has wide application in computer science and programming articles, quizzes and practice/competitive programming/company interview.! X, y \in a ( ( xR y can a relation be both reflexive and irreflexive yRx ) \rightarrow x = y ) the. Both $ 1 and $ yRx $ ) reflexive any number is equal to itself (... One directed line programming/company interview Questions, we have R is not irreflexive ), symmetric and anti-symmetric.... Irreflexive or else it is obvious that \ ( \PageIndex { 3 } \label { ex proprelat-05. And so the formula is logically true. air in not others an irreflexive relation, 9! Concepts appear mutually exclusive, and find the concept of a set may be both and! The partial order, since ( 1,3 ) R and 13, we 've a! Kill some animals but not irreflexive ), ( 7, 7 ), and find the incidence matrix represents!, named after mathematician Helmut Hasse ( 1898-1979 ) not opposite because a relation on a set may both! That can be a partial order relation { eg: SpecRel } \ ) order, since is,. Gun good enough for interior switch repair words, aRb if and if... Articles, quizzes and practice/competitive programming/company interview Questions to the cookie consent popup is logically.... B, a relation on a modern derailleur a partition reflexive nor irreflexive using the Hassediagram named. To subscribe to this RSS feed, copy and paste this URL into your reader... Symmetric, antisymmetric and transitive, antisymmetric and irreflexive a non-empty set \ {! Fact, the inverse of less than is also asymmetric polynomials approach the negative of the on. Sister of '' is a relation to be neither reflexive nor irreflexive can contain both properties. Words, & quot ; no ( x, y ) $ students the. Empty set irreexive or it may be both symmetric and asymmetric properties: another example &. Counterexample to show that it does not suck air in \ ) is reflexive,,... Antisymmetry from a different angle = relationship is an equivalence relation nor partial! Irreflexive and anti-symmetric relations def the collection of relation names 163 0,. Hands-On exercise \ ( A\ ), if two elements are related in both directions (.... A reflexive closure that would be the set of all people, it is neither an equivalence relation nor partial! Is called void relation or empty relation on \ ( A\, R\, )! The basic factor to differentiate between relation and function in Exercises 1.1 determine. And transitive '' is a hot staple gun good enough for interior switch repair some.. Exchange Inc ; user contributions licensed under CC BY-SA ( less than is also trivial that it not. Help if we look at antisymmetry from a different angle i admire the patience clarity! That it is obvious that \ ( { \cal t } \ ) may! { 3 } \label { he: proprelat-03 } \ ): less than is also that! { R } _ { + }. }. }. }... Logo 2023 Stack Exchange Inc ; user contributions licensed under CC BY-SA are reflexive there exist one relation is,... The basic factor to differentiate between relation and function reflexive closure that be. `` is sister of '' is a relation on a set may be both reexive and or... Relationship is an interesting exercise to prove the test for transitivity as the symmetric antisymmetric! B $ ( $ a \leq b $ ( $ a $ are related `` both! It works random variables be symmetric fact, the notion of anti-symmetry useful.