AMB111F Notes 1: Sets and Real Numbers

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1 AMB111F Notes 1: Sets and Real Numbers A set is a collection of clearly defined objects called elements (members) of the set. Traditionally we use upper case letters to denote sets. For example the set of all natural numbers is denoted by N. We also list elements of the set inside braces separated by commas (or semicolons). For example the set N = {1, 2, 3, }. We may also use the set-builder notation to describe a set. For example the set N = {x x is a natural number }, read as : the set of all x such that x is a natural number. Or N = {x : x is a natural number }. Ifx is a natural number, we write x N, read as: x is an element of N. Equality of Sets: Two sets A and B are equal and we write A = B iff each element of A is in B and each element of B is in A. So in order to prove that two sets A and B are equal, we show that if x A then x B, and if x B then x A. Subsets : A is a subset of B and we write A B if whenever x A then x B. So two sets A and B are equal iff A B and B A. Now let Z = {, 3, 2, 1, 0, 1, 2, 3, } be the set of all integers, then N Z and N Z, i.e. N is not equal to Z. In this case N is a proper subset of Z and we may write N Z. Intervals : Let R denote the set of all real numbers. Then we also write R as the interval (, ) which is the set {x <x< }. An interval is a set of real numbers denoted by a pair of brackets (square or round). E.g. The interval [a, b]={x R a x b}. Its number line is as follows: a b A number line is a (horizontal) line used to denote some sets of real numbers. The interval [a, b)={x R a x<b}. Its number line is as follows: a b The interval (a, b]={x R a<x b}. Its number line is as follows: a b The interval (a, b)={x R a<x<b}. Its number line is as follows: a b The interval [a, ) ={x R a x< }. Its number line is as follows: 1

2 a Similarly for other intervals such as (a, ) = {x R a < x < }; (a, ] ={x R a<x }, etc. Union and Interesection: Let A and B be sets. The union of A and B is the set A B = {x : x A or x B}. Clearly A B = B A. Examples : (1) Let A = {1, 2, 3, 4} and B = {2, 4, 6, 7}. Then A B = {1, 2, 3, 4, 6, 7}. (2) Let A =( 1, 3) and B =[2, 4). Find A B. Soln From the above number line we see that A B =( 1, 4). The intersection of A and B is the set A B = {x : x A and x B}. Clearly A B = B A. In examples (1) and (2) above, A B = {2, 4} and [2, 3) respectively. Venn Diagrams. Let U be a universal (large) set. Let A and B be subsets of U. A Venn diagram for the sets is shown below, assuming A and B have common elements. U A B Difference of two sets: The difference A B = {x A : x B}. It is also called the relative complement. The complement of A is A c = R A. Example: (1) Let A = {1, 2, 3, 4} and B = {2, 4, 6, 7}. Then A B = {1, 3} and B A = {6, 7}. (2) Let A = ( 1, 3) and B = [2, 4). Then A B = ( 1, 2). A c = (, 1] [3, ). See the number line in the previous example. 2

3 Symmetric Difference. Let A and B be sets. The symmetric difference of A and B is the set A B =(A B) (A B). A B From the diagram (see shaded area), it is easy to see that A B =(A B) (B A). Example. Let A =[ 2, 5] and B =(0, 10]. Find (i) A B; (ii) B A; (iii) A B; (iv) A B and (v) A B. Soln. A B (i) A B =[ 2, 0]; (ii) B A =(5, 10]; (iii) A B =(A B) (B A) = [ 2, 0] (5, 10]; (iv) A B =(0, 5]; (v) A B =[ 2, 10]. Cartesian Product. Let A and B be non-empty sets. The cartesian product of A and B is A B = {(a, b) :a A and b B}, where (a, b) is an ordered pair. Example. Let A = {1, 2, 3} and B = {2, 4, 6}. Then A B = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6) }. Let A denote the number of elements of A if A is a finite set. A is also called the cardinality (or cardinal number) of A. If A = m and B = n then A B = mn. So the above example A B = 3 3=9. Functions. A function (or map or mapping) f from a set A to a set B, written f : A B, is a rule that assigns to each a A a unique element b B. 3

4 A f B a b. b is the image of a and we write b = f(a). f(a), called the image of A, also written as Im (f), is the set {f(a) :a A}. Sof(A) B. A is the domain of f and f(a) is the range of f. Iff(A) =B, then f is onto or surjective. Thus f is onto if and only if for each b B there exists a A such that b = f(a). f is one-to-one or injective if whenever f(a) =f(b) then a = b. For example f : Z Z given by f(n) = n is injective and surjective: Proof. Suppose f(m) =f(n), then m = n, hence m = n. Therefore f is injective. Let m Z, then m = ( m) =f( m). Hence f is surjective. (Note: Z is the set of all integers). Equality of functions. Let f and g be functions on the same domain D. Then f = g if f(x) =g(x) for all x D. A one-to-one and onto function is called a bijection or a one-to-one correspondence. Inverse image or Pre-image of f. Let f : A B and B 1 B. The pre-image of B 1 under f is f 1 (B 1 )={a A : f(a) B 1 }. Thus f 1 (B 1 ) consists of precisely all the elements of A that are mapped to elements of B 1 by f. Example. Let A = {a, b, c, d, e} and B = {1, 2, 3, 4, 5, 6}. Suppose f is defined by a 1, b 2, c 2, d 4, e 4. Then f 1 ({1, 2}) ={a, b, c}. f 1 ({3}) =. f 1 (4) = {d, e}. Inverse of a function. A bijective map f : A B has an inverse f 1 : B A which is also bijective, given by f 1 (b) =a f(a) =b. Thus f(f 1 (b)) = f(a) =b and f 1 (f(a)) = f 1 (b) =a. A map f : A A such that f(a) =a for all a A is called an identity map on A. Composition of Functions. Let f : A B and g : B C. Then the composite function gof : A C is given by (gof)(a) =g(f(a)) for all a A. Let f :(0, ) R and g :[1, ) R be given by f(x) =x and g(x) = x 1. Find (gof)(x) for any x R. 4

5 Soln. (gof)(x) =g(f(x)) = g(x 2 +1)= x =x. The Real Number System. Z = {, 3; 2; 1; 0; 1; 2; 3; } is the set of all integers. N = {1; 2; 3; } is the set of all natural numbers or positive integers. A rational number is a number that can be expressed in the form p/q where p and q are integers, with q 0. The set of all rational numbers is denoted by Q. A real number which is not a rational number is called an irrational number. The set of all irrational numbers is denoted by Q c. Examples of rational numbers: 3/2; -5/6; -8; 9; 0; 3 27; 3 1. Examples of irrational numbers: 3; 3 4; π; 27; 3 3. Numbers such as 4 and 4 16 are called imaginary numbers because they do not exist in the set of real numbers. Domain and Range of a function. Let f : B be a function from A into B. Then A is called the domain of f. f(a) ={f(a) a A} is called the range of f. B is the co-domain of f. If a domain of f is not specified, then take it to be largest set on which f can be defined or exists. Examples : (1) Let f(x) = x. Find the domain and the range of f. Soln. dom(f) ={x R x 0} =[0, ). Range is R(f) ={ x x 0} =[0, ). (2) Let f(x) = 2 x. Find the domain and the range of f. Soln. dom(f) ={x R 2 x 0} = {x R x 2} =(, 2]. Range is R(f) ={ 2 x x 2} =[0, ) (3) Let f(x) =1/x. Find the domain and the range of f. Soln. dom(f) ={x R x 0} =(, 0) (0, ). Range is R(f) ={1/x x 0} =(, 0) (0, ) The Remainder Theorem A polynomial is an expression f(x) =a n x n + + a 1 x + a 0. If a polynomial f(x) is divided by x k until the remainder is free of x, then the remainder is f(k). An example of long division and remainder: Divide x 2 5x +6 by x +2 and note the remainder. Soln. 5

6 divisor dividend output Explanation x +2 x 2 5x +6 x first divide x 2 by x x 2 +2x multiply x +2byx 7x + 6 subtract from dividend 7 divide the previous expression by x + 2 thus divide 7x by x 7x 14 multiply x +2by 7 20 subtract bottom expression from the previous one Thus when dividing x 2 5x+6 by x+2, the answer is x 7 and the remainder is 20. The next two examples show an application of the Remainder Theorem. Example 22(a): Let f(x) =3x 4 +2x 3 6x 4. Find the remainder when f(x) is divided by x 2. Soln. By the Remainder Theorem, the required remainder is f(2) = = 48. (Check this by using long division) Example 22(b) : When f(x) is divided by x + 1, the remainder is 10. If f(x) =x 3 + ax 2 5x + 4 find a. Soln. By the Rem Theorem, the remainder is 10 = f( 1) = a + 8, thus a =2. F actors: If f(x) =(x a)(x b)(x c) then x a, x b and x c are factors of f(x). The Factor Theorem If f(b) = 0, then x b is a factor of f(x). Example 23 : (a) Prove that x 2 is a factor of x 3 +2x 2 5x 6 and then factorise the expression into linear factors. (b) Solve x 3 +2x 2 5x 6=0 Soln. (a) f(2) = 0, hence x 2 is a factor of f(x). To find other factors, divide f(x) byx 2 until the remainder is 0. 6

7 divisor dividend output Explanation x 2 x 3 +2x 2 5x 6 x 2 first divide x 3 by x x 3 2x 2 multiply x 2byx 2 4x 2 5x 6 subtract from dividend 4x divide the previous expression by x 2 thus divide 4x 2 by x 4x 2 8x 3x 6 multiply x 2by4x subtract from previous expression 3 divide the previous expression by x 2 thus divide 3x by x 3x 6 multiply x 2by3 0 subtract bottom expression from the previous one Thus f(x) =(x 2)(x 2 +4x +3) =(x 2)(x + 1)(x + 3). We have now factorised f(x) into linear factors. Observe that x 2 +4x +3=(x + 1)(x +3) and (x + 1)(x +3)=x(x + 3) + 1(x +3)=x 2 +3x + x +3=x 2 +4x +3. (b) From (a) we see that x =2orx = 3 orx = 1. Example 24 : If x 2 and x + 1 are factors of x 3 + ax 2 + bx + 8, calculate a and b. Soln. f(2) = 0 and f( 1) = 0 since x 2 and x + 1 are factors. Substitute into the cubic expression; thus 4a +2b = 16 and a b = 7 implying 2a 2b = 14. Adding the 1st and the 3rd eqns we have 6a = 30, thus a = 5. Substitute into the 2nd eqn: therefore b =2. Completing the Square Let f(x) =ax 2 + bx + c. Iff(x) =m(x a) 2 + b, then the rhs is a complete square. The expression ax 2 + bx + c can be expressed as a complete square as follows: f(x) =ax 2 +bx+c = a[x 2 +(b/a)x+(b/2a) 2 (b/2a) 2 +c/a] =a[(x+ b/2a) 2 +c/a (b/2a) 2 ]. Let d = a[c/a (b/2a) 2 ], then f(x) =a(x+b/2a) 2 +d. Thus f(x) is expressed in the form of a complete square. Completing the square helps us to solve quadratic equations. Example: Solve the equation x 2 +5x +6=1. Soln. We complete the square as follows: x 2 +5x+6 = 1 implies x 2 +5x+5 = 0. Therefore x 2 +5x +(5/2) 2 (5/2) = 0, thus (x +5/2) 2 +( 5/4) = 0. So (x +5/2) 2 =5/4. Taking square roots, we have x +5/2 =± 5/4, implying 7

8 x = 5/2 ± 5/4 = 5± 5 2. Now back to f(x) =a(x + b/2a) 2 + d: If f(x) = 0, then (x + b/2a) 2 = d/a, implying (x + b/2a) = ± d/a. Thus x = b/2a ± d/a = b/2a ± a[c/a (b/2a) 2 ]/a = b/2a ± [c/a (b/2a) 2 ] = b/2a ± [c/a b2 /4a 2 ]= b/2a ± [b 2 4ac]/4a 2 = b± b 2 4ac. This gives us a 2a shorter way to solve a quadratic equation. For example in solving x 2 +5x + 6 = 1, we have x 2 +5x + 5 = 0, thus x = b± b 2 4ac 2a = 5± 5 2 4(1)(5) 2(1) = 5±

Sets and Functions. MATH 464/506, Real Analysis. J. Robert Buchanan. Summer Department of Mathematics. J. Robert Buchanan Sets and Functions

Sets and Functions. MATH 464/506, Real Analysis. J. Robert Buchanan. Summer Department of Mathematics. J. Robert Buchanan Sets and Functions Sets and Functions MATH 464/506, Real Analysis J. Robert Buchanan Department of Mathematics Summer 2007 Notation x A means that element x is a member of set A. x / A means that x is not a member of A.