Functions. Prof. Susan Older. 20 October (CIS 375) Functions 20 Oct / 14

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1 Functions Prof. Susan Older 20 October 2016 Functions (CIS 375) Functions 20 Oct / 14 relation f is a function provided that for all a, b, c, if (a, b) f and (a, c) f, then b = c. (That is, no object a appears as the first component of two different pairs.) Some examples (and nonexamples): f 1 = {(1, 2), (2, 5), (4, 5), (10, 37)} is a function. f 2 = {(1, 2), (2, 5), (4, 5), (2, 37)} is not a function. g = {(Smith, ), (Jones, ), (Cook, D), (lake, )} is a function. h = {(x, x 2 ) : x N} and k = {(x, x 2 ) : x Z} are functions. l = {(x 2, x) : x Z} is not a function. When f is a function and (a, b) f : We often write: f (a) = b. We sometimes say: f maps a to b (CIS 375) Functions 20 Oct / 14

2 Domains and Images Let f be a function. The domain of f and the image of f are (respectively) the sets dom f = {x : y, (x, y) f } Recall from the previous slide: im f = {y : x, (x, y) f }. f 1 = {(1, 2), (2, 5), (4, 5), (10, 37)} g = {(Smith, ), (Jones, ), (Cook, D), (lake, )} h = {(x, x 2 ) : x N} dom f 1 = {1, 2, 4, 10} im f 1 = {2, 5, 37} dom g = {Smith, Jones, Cook, lake} im g = {,, D} dom h = N im h = {n 2 : n N} (CIS 375) Functions 20 Oct / 14 Functions from to Let f be a function, and let and be sets. Then f is a function from to (written f : ) provided that dom f = and im f. The domain of h is N and the image of h is {n 2 : n N}, so any of the following is permissible: h :N {n 2 : n N} h :N N h :N Z In contrast, the following are not correct: h : Z Z, h : N Z + g : {Smith, Jones, Cook, lake} {,, C, D, F } f 1 : {1, 2, 4, 10} {2, 5, 37} (CIS 375) Functions 20 Oct / 14

3 Functions from to : Second Look Let f be a function, and let and be sets. Then f is a function from to (written f : ) provided that dom f = and im f. Function from to Function (not from ) Not a function at all Each element of has exactly one arrow emerging from it. Domain is not : some element of has no arrows emerging. Some element of has multiple arrows emerging from it. (CIS 375) Functions 20 Oct / 14 What are the Domains and Images of These Functions? 1 The function that maps each nonnegative integer to its rightmost decimal digit (e.g., it would assign to the digit 7) Domain: N Image: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} 2 The function that maps each positive integer to its largest decimal digit (e.g., it would assign to the digit 8) Domain: Z + Image: {1, 2, 3, 4, 5, 6, 7, 8, 9} 3 The function that maps each real number to the largest integer that does not exceed that real number (i.e., the floor function) Domain: R Image: Z 4 The function that maps each pair of positive integers to the first integer of the pair Domain: Z + Z + Image: Z + (CIS 375) Functions 20 Oct / 14

4 Injections (1-1 Functions) function f : is injective (or one-to-one or 1-1) provided that for all x 1, x 2, if f (x 1 ) = f (x 2 ), then x 1 = x 2. That is, f maps distinct elements of to distinct elements of. Injective function Noninjective function Each element of has at most one arrow heading into it. Some element of has multiple arrows heading into it. (CIS 375) Functions 20 Oct / 14 Injections (1-1 Functions): Second Look function f : is injective (or one-to-one or 1-1) provided that for all x 1, x 2, if f (x 1 ) = f (x 2 ), then x 1 = x 2. n alternative definition (via the contrapositive): function f : is injective (or one-to-one or 1-1) provided that for all x 1, x 2, if x 1 x 2, then f (x 1 ) f (x 2 ). f 1 is not injective: f 1 (2) = f 1 (4) = 5. g is not injective: g(smith) = g(lake) =. h is injective: whenever h(x) = h(y), x = y. (CIS 375) Functions 20 Oct / 14

5 Is This Function Injective? Tips for determining whether or not a function is injective To trigger your intuition, consider sketching it (or a portion thereof). If you think it s not injective, try to identify a specific counterexample. If you think it is injective, try to prove it. If you absolutely can t decide, try proving it and see where you get. Two examples to try: h 1 : Z Z h 2 : Z Z h 1 (w) = w 2 h 2 (w) = 3w + 4 (CIS 375) Functions 20 Oct / 14 Proof about an Injection Claim Let h 2 : Z Z be the function defined as h 2 (w) = 3w + 4. Then h 2 is one-to-one. Proof (direct) Consider arbitrary a, b Z such that h 2 (a) = h 2 (b). [NTS: a = b.] y the definition of h 2, h 2 (a) = 3a + 4 and h 2 (b) = 3b + 4. ecause h 2 (a) = h 2 (b), we know that 3a + 4 = 3b + 4, and hence 3a = 3b. It follows that a = b. ecause a and b were arbitrary, h 2 is one-to-one. (CIS 375) Functions 20 Oct / 14

6 Surjections (onto functions) function f : is surjective (or onto) provided that for all y, there exists x such that f (x) = y. That is, im f = : each -element is mapped to from some -element. Surjective function Nonsurjective function Each element of has at least one arrow heading into it. Some element of has no arrows heading into it. (CIS 375) Functions 20 Oct / 14 Surjections (Onto Functions): Second Look function f : is surjective (or onto) provided that for all y, there exists x such that f (x) = y. That is, im f = : each -element is mapped to from some -element. Define f 3 : {a, b, c, d} {1, 2, 3}, f 4 : {a, b, c, d} {1, 2, 3, 4} as follows: f 3 (a) = 1 f 3 (b) = 3 f 4 (a) = 1 f 4 (b) = 3 f 3 (c) = 1 f 3 (d) = 2 f 4 (c) = 1 f 4 (d) = 2 f 3 is surjective: im f 3 = {1, 2, 3}. f 4 is not surjective: im f 4 {1, 2, 3, 4}. What about inc : N N defined as: inc(n) = n + 1? What about inc : Z Z defined as: inc (n) = n + 1? (CIS 375) Functions 20 Oct / 14

7 ijections function f : is a bijection provided that it is one-to-one and onto. Pigeonhole Principle Let, be finite sets, and suppose f :. Corollary If >, then f is not one-to-one. If <, then f is not onto. Let, be finite sets, and suppose f :. If f is a bijection, then =. (CIS 375) Functions 20 Oct / 14 s Time Permits Let s define some functions: function g 1 : N N that is not 1-1 and not onto function g 2 : N N that is 1-1 and not onto function g 3 : N N that is not 1-1 and onto function g 4 : N N that is 1-1 and onto (CIS 375) Functions 20 Oct / 14