Consider the “function” 1 f(x) = . x − 4 In fact, this is not a function until I have told you it’s domain and codomain. More accurately we have to talk about the function 1 f : R \{4} → R, f(x) = . x − 4 It’s important to get in the habit of specifying the domain and codomain. In fact the definition of function f : A → B specifies it as a subset of the cartesian produce A × B. From this point of view, our function is the set {(x, 1/(x − 4) : x ∈ R \{4}} ⊂ R \{4} × R 1 Note: my definition of function is different from that of the book. Book does not require that the function be defined everywhere in A. Range The range of a function f : A → B is the subset of B consisting of values of the function. A function is surjective (or onto) if the range equals the codomain. The function f : R \{4} → R given by f(x) = 1/(x − 4) is not surjective because it misses the value 0. That is, the equation g(x) = 0 has no solution in R \{4}. 2 However, if we define a new function g : R \{4} → R \{0}, using the same formula g(x) = 1/(x−4), then g is a surjective function. Note that although the formula is the same, this is a different function from f because we have changed the codomain. To prove g is surjective, we must show that if y ∈ R \{0}, then the equation g(x) = y has a solution x ∈ R\{4}. This is a matter of solving the equation 1 = y x − 4 for x, to get 1 x = + 4. y 3 The fact that you can solve the equation is what makes f surjec- tive. A proof of surjectivity does not need to contain the solving procedure, however, merely the solution itself: Proof that g is surjective: Let y ∈ R, y 6= 0. then since y 6= 0, then I can define 1 x = + 4 y Then g(x) = 1/(x − 4) does indeed equal y. 4 Here is the formal definition of surjectivity. Definition A function f : A → B is surjective (also called onto) if for all y ∈ B, there exists an x ∈ A such that f(x) = y. Another important notion is the notion of an injective or one- to-one function. Definition A function f : A → B is injective if, for all x1 and x2 in A, if x1 6= x2, then f(x1) 6= f(x2). 2 Example: g(x) = x is not injective. For example, x1 = −1 and 2 2 x2 = 1. Then x1 6= x2, but g(x1) = (−1) = 1 = 1 = g(x2). However, the function f : R \{4} → R, f(x) = 1/(x − 4) is injective. 5 How do we prove a function is injective? This is a case where it is easier to use the contrapositive. Rather than showing x1 6= x2 ⇒ f(x1) 6= f(x2) we show f(x1) = f(x2) ⇒ x1 = x2. Proof that f is injective: Suppose that 1 1 = . x1 − 4 x2 − 4 Multiply both sides by (x1 − 4)(x2 − 4), to get x2 − 4 = x1 − 4 add 4 to both sides to get x2 = x1. 6.