1.  Let A and B be any sets A binary relation R from A to B is a subset of AxB Given an ordered pair (x, y), x is related to y by R iff (x, y) is in R. This is denoted If x is not related to y by R, we say  The term “binary” means that the relation is defined for two elements. Other types of relations, called n-ary relations, also exist. If you want, look them up. They’re very useful in CS

2.  Consider sets A={0,1,2} and B={1,2,3}. AxB is the set of all possible ordered pairs from A and B If element x from A is related to element y from B, we say xRy, “x is related to y”  If x<y, which elements in AxB are in set R?  Answer: {(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)}  Which elements are not in R?  Answer: {(1,1), (2,1), (2,2)}

3.  For any (x, y) in the real number plane (RxR), (x, y) are in relation C iff “x 2 + y 2 = 1” This is the equation for a circle This relation has infinitely many solutions because it is a subset of the cross product of two infinitely large sets (two sets of all real numbers) To check the relation, test some x and y values  (0,0) => 0 2 + 0 2 = 0 ≠ 1  (1,0) => 1 2 + 0 2 = 1  (2, 0) => 2 2 + 0 2 = 4 ≠ 1

4.  Given set A and relation S: S is reflexive iff for all x in A, (x, x) is in S  Informal: Each element is related to itself S is symmetric iff for all x and y in A, if (x, y) is in S, then (y, x) is in S  Informal: If any one element is related to any other element, then the second element is related to the first S is transitive iff for all x, y, and z in A, if (x, y) and (y, z) are in S, then (x, z) is in S  Informal: If any one element is related to a second, and that second is related to a third, then the first is related to the third  These definitions are each universal statements These properties can be proven true by proving that every element x, y, and z must satisfy the property Prove them false by finding a counter example

5.  S = {(0,0), (0,1), (0,3), (1,0), (1,1), (2,2), (3,0), (3,3)}  T ={(0,1), (2,3)}  W = {(0,0), (0,2), (0,3), (2,3)} Which properties do S, T, and W satisfy?  S is reflexive: Each element is related to itself – (0,0)…  S is symmetric: For every ordered pair in S, there is also a reverse pair – The reverse of (x, x) is (x, x)!!  S is not transitive: There is point (1,0) and point (0,3) but there is no point (1,3) – (x, y), (y, z), but not (x, z)  T is not reflexive: Not every point (x, x) is in T  T is not symmetric: Not every point (x, y) has a corresponding (y, x) point  T is transitive: It is not obvious why. Notice that there are no points (x,y) and (y, z) in T, so there is never an opportunity for transitivity to be false. We say that T is vacuously transitive, meaning the property is true by default  W is not reflexive: 1 is between 0 and 2, so is in set A on which W is a relation. The point (1,1), among other points like (2,2), is not in W  W is not symmetric: There is a (0,2) but not a (2,0)  W is transitive: There is a (0,2), (2,3) and a (0,3) – these are the only points of the form (x, y), (y, z), (x, z). There are no other (x, y), (y, z) points. It must be transitive  These properties extend readily to infinite relations like y = x

6.  We have a relation H that is not transitive, but we want it to be transitive The transitive closure of H, denoted H t, is the set that contains H but also contains all the points necessary to make H t a transitive relation H t satisfies the following properties  H t is transitive  H is a subset of or is equal to H t  If S is any other transitive relation that contains H, then it also contains H t

7.  A = {0,1,2,3} and R = {(0,1), (1,2), (2,3)}  What is the transitive closure of R? Every ordered pair in R is in R t Every grouping of points (x, y) and (y, z) must yield an (x, z) point  (0,1) + (1,2) → (0,2)  (1,2) + (2,3) → (1,3)  Since (0,2) is now in R t, (0,2) + (2,3) → (0,3) R t = {(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)}

8.  Consider the function F from a set A to a set B that satisfies the following two properties For every element x in A there is an element y in B such that (x, y) is in F. For all elements x in A and y, z in B  If (x, y) is in F and (x, z) is in F, then y = z If F is a function from A to B, we say y = F(x) iff (x, y) is in F

9.  Sets A = {2,4,6} and B = {1,3,5} are connected by two relations R = {(2,5), (4,1), (4,3), (6,5)} and S = “for all (x, y) in AxB, (x, y) is in S iff y = x + 1” Relation R is not a function because the x value 4 yields to separate y values 1 and 3  In a function, according to property 2, an x value (input) can yield only one unique y value (output) Relation S is not a function because there is no element y in B such that y = 6 + 1 = 7  Property 1 states that every element in A must yield an element in B

10.  If a function is defined from set X to set Y, we say that the first set, X, is the domain of the function and the second set, Y, is the range When you think of a function in terms of a machine, you can visualize the domain as the set of possible inputs and the range as the set of possible outputs If X is the set of all real numbers, there is no guarantee that Y will also contain all real numbers, as is demonstrated with the function y = x 2 The process of using a function to transform its domain into range values is known as mapping and consists of feeding in the inputs and recording the outputs  You will do a lot of relational mapping in computer science

11.  For a function to be injective, or one-to-one, every element in its domain must yield a unique value If F(x) = F(x[2]), then x = x[2]  F(2) can’t equal F(3). An example is the function F(x)=x, because F(x[2]) can only equal x[2], so if x[2] ≠ x, F(x[2]) won’t equal F(x)  For a function to be surjective, or onto, every element in the function’s domain must be matched with an element from its range A = {1,2,3}, B = {a,b,c,d}  F(A) = {a,b,c}  This function is not surjective because 1 yields a, 2 yields b, 3 yields c, but nothing yields d, which is in the range of F(x) because it is defined over the sets A and B Formally, this is described as: Given sets X and Y, F(X) is onto Y iff for every element y in Y, there exists an element x in X such that F(x) = y

12.  Is y= F(x) = x 2 for all real numbers x injective/surjective? F(x) is not injective  (-2) 2 = 2 2 = 4, therefore F(-2) = F(2) but -2 ≠ 2. F(x) is not surjective if Y (the range) is defined as all real numbers  The reason is straightforward: We have y = x 2. However, there is no negative real number y such that x 2 = y. EX: (-1/2) 2 = 1/4. All squared real numbers are positive, so none of the negative real values of y are matched with F(x) values  Y = G(x) = x 3 is both injective and surjective This is true because every real number raised to the third power has a unique value. G(-x) ≠ G(x), and it can be easily shown that every real value of y can be yielded from a particular value of x

13.  For a function to be defined over the real numbers, it must be true that for every real value of x, there is a real value y as the output  Also, if a relation is graphed on a RxR plane, it cannot be a function if a vertical line can be drawn through the graph and intersect two points  Both of these properties stem from properties 1 and 2 Notice that a circle is not a real-number function because it does not satisfy either property The second relation, however, is a real number function because it will continue to infinity in both directions without folding back into itself

14.  An inverse function is a function that “undoes” another function by mapping the range of a function back onto its domain In order for F(x) to have a single, all-encompassing inverse function F -1 (y), F(x) must be both injective and surjective An inverse function is defined as follows: F -1 (y) = x  y = F(x) If every range value, y, is produced by the function F(x), and no two domain values, x, yield the same y value, then F -1 (y) will yield every x value and no two y values will yield the same x value Most functions do not have a single, perfect inverse function. In these cases, the inverse relation is actually a system of inverse functions, each defined over a set domain and range  To obtain this set algebraically, swap the locations of all domain and range values and solve for the range values  EX: y = x 2 → x = y 2 → ±(x 1/2 ) = y.  Notice the ±. This means that the inverse of y = x 2 is actually a system of two functions, one having all positive and another having all negative values  As you would expect, these systems tend to be more complicated, and are a subject of higher mathematics

15.  If S is a relation from A to B, then its inverse S -1 is defined as follows S -1 = {(y, x) in BxA such that (x, y) is in S} In other words, (y, x) is in S -1 iff (x, y) is in S  To obtain an inverse relation, switch the x and y values for all of the ordered pairs To take the inverse of a real number graph, rotate the graph 90 degrees clockwise  Use the horizontal line test to determine whether a graph’s inverse is a function Both relations are functions, but only relation 2 has a single inverse function. To see this, you can perform the horizontal line test

16.  Now we get to some practical applications A hashing function is one which takes a variety of inputs (strings, large integers, binary data) and transforms those inputs into a distinct numerical output  Such functions are used to establish methods for sorting items, encrypting data, and improving the efficiency of storing data  Say you need a quick way to sort a data base of students based on social security numbers  These numbers are very large, so it would be impractical to sort data based on the numbers alone  Instead, we will use the hashing function H(n) = n mod 7, where mod is the modulo division operator defined by “n mod m = n-m*(n div m)”. Div is an operator which returns the integer component of division  This function yields a manageable way to sort data. However, sometimes the function will return the same value for different n values. This is known as a collision, and must be resolved using an algorithm

17.  A checksum function is a special class of hash functions that is designed to avoid and correct collision errors Essentially, you use a bit of programming magic to ensure that the hash is sustainably implemented As a conceptual example, consider the social security hash on the previous slide  If all of the numbers you entered into the system yielded the same hash value (yes this is possible), then the sort would be totally ineffective  To ensure redundancy, you would have to add other parameters to the sort, like last names, birthdays, or student IDs