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Functions Definition a Function F from a to B Is a Relation from a to B Such That

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Functions Definition a Function F from a to B Is a Relation from a to B Such That Functions Definition A function f from A to B is a relation from A to B such that: (i) Dom(f) = A, and (ii) If (x,y) ∈ f and (x,z) ∈ f then y = z. If A = B, we say that f is a function on A. In terms of ordered pairs, (i) and (ii) say that every element of A appears as a first coordinate in one and only one ordered pair. If f is a function and (x,y) ∈ f then we will write y = f(x). Since f is the name of a relation we could write x f y, but absolutely no one ever writes that. More Terminology Functions are also called mappings. We usually write f: A → B read as "f maps A to B" The set B is called the codomain of f. Recall that the range of f, Rng(f) is the set of second coordinates of the relation f. So, Rng(f) ⊆ B, but they need not be equal. Example: Consider f : {1,2,3} → {a,b,c,d,e} given by f = {(1,a), (2,b), (3,c)} f is a function, Rng(f) = {a,b,c}, Codomain of f = {a,b,c,d,e}, Dom (f) = {1,2,3}, f(1) = a, f(2) = b and f(3) = c. More Examples Consider: f : {1,2,3,4} → {a,b,c,d,e} given by f = {(1,a), (2,b), (3,c)} This is not a function since Dom(f) ≠ A. Consider: f : {1,2,3} → {a,b,c,d,e} given by f = {(1,a), (2,b), (3,c), (1,e)} This is not a function since 1 appears as a first coordinate in more than one ordered pair. More Terminology If f: A → B is a function and f(x) = y we say that: y is the value of f at x; y is the image of x under f; x is the pre-image of y under f; x is an argument of f. Notice that f(x) is the image of x under f, it is NOT the name of the function, f is the name of the function. It is an abuse of the language to call f(x) a function, although many people do. Functions Notice that each argument of a function has exactly one image, but an element of the codomain may have several pre-images or none at all. Given a function f: A → B we may write f = {(x, f(x)) | x ∈ A }. This underscores the notion that a function is a set ... however, we don't often think of functions that way. Typically, we specify a function by giving a "rule" for computing the image of an arbitrary argument such as: f(x) = 2x -3 where x ∈ ℤ. Note that giving the "rule" by itself does not define a function, you must also specify the domain ( x ∈ ℤ in the above example). Real-Valued Functions A real-valued function is a function whose codomain is ℝ. A real-valued function of a real variable is a function whose codomain is R and whose domain is a subset of ℝ. Real-valued functions of a real variable are so common in some branches of mathematics that we have a special convention concerning them: If the domain of a function is not specified, it is assumed to be a real-valued function of a real variable whose domain is the largest set of real numbers for which the "rule" produces a real value. Real-Valued Functions Examples: f(x) = 5x2 -2x + ½ by convention is a function f : ℝ ℝ. x2−2 x3 f x= x1 x−5 by convention is a function f :ℝ−{−1,5}ℝ. f x= x2−4 by convention is a function f :ℝ−{−2,2}ℝ. Equality of Functions Since functions are sets of ordered pairs, we already know what it means for two functions to be equal, since we know what it means for two sets to be equal and we know what it means for two ordered pairs to be equal. But we shall restate this in the function terminology: Two functions f and g are equal iff (a) Dom(f) = Dom(g), and (b) ∀ x ∈ Dom(f), f(x) = g(x). Examples x2−1 for x≠1 f x={ x−1 } and g x=x1 2 for x=1 are the same function, but f(x) = 2x and g(x) = 2x for x ∈ ℤ are not the same function. Inverses and Compositions Since functions are special types of relations, the definitions of inverses and compositions of relations applies to them. However, the definitions of inverses and compositions do not guarantee that the inverse or composition of functions will be a function. For instance, the function given by f(x) = x2 contains the ordered pairs (1, 1), (-1, 1), (2, 4), (-2, 4) , ... etc. The inverse would contain (1,1), (1,-1), (4,2), (4,-2), ... etc. and it is not a function (but it still is a relation). Only under special conditions, which we shall see later, will the inverse of a function be a function. Composition of Functions Recall that if f: A → B and g: B → C are functions then g⊙ f = {(x,z) | ∃ y ∈ B so that (x,y) ∈ f and (y,z) ∈ g }. Using our functional notation we can rewrite this as: g⊙ f = {(x,z) | ∃ y ∈ B so that f(x) = y and g(y) = z }. Now, because f is a function, for each x ∈ A, there is a y ∈ B (and only one) so that f(x) = y and because g is a function, there will be a z ∈ C (again, only one) so that g(y) = z. So, we can go further and write: g⊙ f (x) = z = g(y), where y = f(x) or even more compactly: g⊙ f (x) = g(f(x)) . The formula above is the reason for choosing the "backwards" notation for composition. Composition of Functions Theorem: If f: A → B and g: B → C are functions then g⊙ f : A → C is a function. Pf: We already know that g⊙ f is a relation between A and C. Suppose that (x,y) ∈ g⊙ f and (x,z) ∈ g⊙ f . We then have y = g⊙ f (x) = g(f(x)) and z = g⊙ f (x) = g(f(x)). Since g is a function y = z. Dom(g⊙ f ) ⊆ A by the definition of composition. Suppose a ∈ A. Since f is a function, Dom(f) = A and f(a) ∈ B. Because g is a function and Dom(g) = B, g(f(a)) exists in C. Thus, a ∈ Dom(g⊙ f ) so A ⊆ Dom(g⊙ f ). We now have Dom(g⊙ f ) = A, so g⊙ f is a function. Restrictions If f: A → B is a function and D ⊆ A, we can define a new function, the restriction of f to D, denoted by f | , by keeping only the D ordered pairs of f which have their first coordinate in D. The domain of the restriction of f to D is D. It is clear that the restriction of a function is also a function. Surjections We know that for a function f: A → B, the range of f need not be the same as the codomain. Definition: A function f: A → B is onto B iff Rng(f) = B. Such functions are referred to as onto functions or surjections. Note that the common English word "onto" has a technical mathematical meaning. One should be careful when using this word in any mathematical context. Also, realize that you can not tell if a function is onto unless you know the codomain. The range can be calculated, but generally you have to be told what the codomain is. Examples Let f: ℤℤ be given by f(x) = x2. This function is not onto since, for instance, 6 in the codomain is not in the range of f. For this function, the Rng(f) consists only of squares, and the codomain ℤ also contains non-squares. Let f: ℝ [0, ∞) be given by f(x) = x2. This function is a surjection because Rng(f) = [0, ∞). That is, every non-negative real number is the square of a real number. More precisely, if real number c ≥ 0 is the square of √c, which is a real number, i.e., f(√c) = c. Proving a Function is Onto Suppose that f: A → B is a function that we wish to prove is a surjection. Since we always have Rng(f) ⊆ B, we need to prove B ⊆ Rng(f). That is we need to show that (∀ b) ( b ∈ B ⇒ b ∈ Rng(f)). To show that b ∈ Rng(f), we have to find an x in Dom(f) = A so that f(x) = b. Example: Show that the function given by f(x) = 2x + 3 is onto. By convention, the codomain of this function is ℝ and the domain is also ℝ. Let z ∈ ℝ be an arbitrary element of the codomain. We seek an x in the domain, so that f(x) = z, i.e., 2x + 3 = z. Solving for x we get x = ½(z – 3). Now this x is a real number if z is a real number, so is in the domain of f and f(½(z – 3)) = z. So f is onto. Proving a Function is Not Onto Suppose that f: A → B is a function that we wish to prove is not a surjection. We usually show that a function is not onto by providing a counterexample, i.e., an element of the codomain which is not in the range of the function.
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