1. CSNB143 – Discrete Structure Topic 8 – Function

2. Learning Outcomes Students should be able to understand function and know how it maps. Students should be able to identify different type of function and solve it.

3. Topic 8 – Function Understanding function Function is a special type of relation A function f from A to B is written as we write a normal relation f : A  B The rule is for all a  Dom (f), f(a) contains just one element of B. If a is not in Dom (f), then f(a) = . The element a is called an argument of the function f and f(a) is called the value of the function for the argument a and is also referred to as the image of a under f

4. Topic 8 – Function Understanding function The rule for f : A  B is for all a  Dom (f), f(a) contains just one element of B. If a is not in Dom (f), then f(a) = . Example: Let A = {1, 2, 3, 4} and B = {a, b, c, d} and let f = {(1, a), (2, a), (3, d), (4, c)} We have f(1) = a f(2) = a f(3) = d f(4) = c Since each set f(n) is a single value, f is a function.

5. Topic 8 – Function Understanding function Another example: Let A = {1, 2, 3} and B = {x, y, z}. Consider the relations R = {(1, x), (2, x), (3, x)} S = {(1, x), (1, y), (2, z), (3, y)} T = {(1, z), (2, y)} Determine if the relations R, S and T are functions – Relation R is a function with Dom(R) = {1, 2, 3} and Ran(R) = {x}. – Relation S is not a function since S(1) = {x, y}. It shows that element 1 has two images under f. Note that if the elements have two or more images or value, it is not a function. – Relation T is not a function since 3 has no image under f.

6. Topic 8 – Function Special types of function Let f be a function from A to B, TypesCharacteristics Everywhere definedDom(f) = A OntoRan (f) = B One to oneIf f(a) = f(a’), then a = a’

7. Topic 8 – Function Special types of function Example: Let A = {1, 2, 3, 4} and B = {a, b, c, d} and let f = {(1, a), (2, a), (3, d), (4, c)}. Determine the type of function f is TypesCharacteristicsOutcomes Everywhere defined Dom(f) = ADom(f) = {1,2,3,4}. Hence Dom(f) = A Hence f is of type everywhere defined OntoRan (f) = BRan (f) = {a, d, c} Hence Dom(f)  B Hence f is not of type onto One to oneIf f(a) = f(a’), then a = a’Since f(1) = f(2) = a Hence f is not of type one to one

8. Topic 8 – Function Special types of functions Another example: Let A = {a 1, a 2, a 3 }; B = {b 1, b 2, b 3 }; C = {c 1, c 2 } and D = {d 1, d 2, d 3, d 4 }. Consider the following four functions, from A to B, A to D, B to C, and D to B respectively Identify the types of each function a)f = {(a 1, b 2 ), (a 2, b 3 ), (a 3, b 1 )} b)f = {(a 1, d 2 ), (a 2, d 1 ), (a 3, d 4 )} c)f = {(b 1, c 2 ), (b 2, c 2 ), (b 3, c 1 )} d)f = {(d 1, b 1 ), (d 2, b 2 ), (d 3, b 1 )}

9. Topic 8 – Function Invertible function A function f: A  B is said to be invertible if its reverse relation, f -1 is also a function Example: Consider f = {(1, a), (2, a), (3, d), (4, c)}. therefore, f -1 = {(a, 1), (a, 2), (d, 3), (c, 4)} We can clearly see that f -1 is not a function because f -1 (a) = {1, 2}. So f is not an invertible function. Note that if the elements have two or more images or value, it is not a function.

10. Topic 8 – Function Permutation Function This is a part in which set A have a relation to itself. Set A is finite set. In this case, a function must be onto and one-to-one. A bijection from a set A to itself is called a permutation of A. Example:

11. Topic 8 – Function Permutation Function Find a) P 4 -1 b) P 3  P 2 a) See that P 4 is a one-to-one function, we have P 4 = {(1, 3), (2, 1), (3, 2)} So, P 4 -1 = {(3, 1), (1, 2), (2, 3)} b)

12. Topic 8 – Function Cyclic Permutations The composition of two permutations is another permutation, usually referred to as the product of these permutations. If A = {a 1, a 2, …, a n } is a set contained n elements, so there will be n! = n.(n – 1)… 2.1 permutation for A. Let b 1, b 2, …. b r are r elements of set A = {a 1, a 2, ….. a n }. Permutation for P: A  A is given by: P(b 1 ) = b 2 P(b 2 ) = b 3. P(b r ) = b 1 P(x) = x if x  A, x  {b 1, b 2, … b r }

13. Topic 8 – Function Cyclic Permutations Example: Rule: P(b 1 ) = b 2 P(b 2 )= b 3 P(x) = x P(b r ) = b 1

14. Topic 8 – Function Cyclic Permutations Example: Let A = {1, 2, 3, 4, 5, 6}. Find: a)(4, 1, 3, 5)  (5, 6, 3) b)(5, 6, 3)  (4, 1, 3, 5)