Lecture 2 21 127 Concepts of Math 08 29 2012 Administrivia Homework 1 is posted Check it out It s due next Thursday September 6 I posted a page about using LATEX Check that out too It includes a template tex file for Homework 1 Office Hours Mine are posted Mon and Wed after class 2 30 3 30 and Tues 1 2 I m also reachable by email to set up an appointment and AMPDF is good too TAs office hours will be posted this week Axioms and Proofs I mentioned on Monday that we have to agree on some fundamental truths that we just believe are true These are called axioms Here are some fundamental facts we will assume are true for the purposes of this course Standard Numbers First we will assume some fundamental sets of numbers exist N is the set of natural numbers N 1 2 3 4 5 Note It s more natural to start with 1 instead of 0 Z is the set of integers Z 3 2 1 0 1 2 3 Note Zahlen is German for numbers Q is the set of rational numbers or quotients of integers o na a b Z and b 6 0 Q b R is the set of real numbers R This set is hard to define We think of it as a number line Properties of Numbers The set of real numbers using the operations and is a mathematical object known as a field It has several nice properties We will assume all of these hold true Some of these are technically field axioms some of them are basic results one can prove from the axioms Associativity x y z x y z and x y z x y z Commutativity x y y x and x y y x Additive identity x 0 x 1 Multiplicative identity x 1 x Additive inverses no matter what x is we can find a number y that has the property x y 0 Note we write y as x Multiplicative inverses no matter what x is as long as it is nonzero we can find a number y that has the property x y 1 Note we write y as x 1 or as x1 Distributivity x y z x y x z Adding to an equality assuming x y then we also know x z y z Multiplying an equality assuming x y and z 6 0 then we also know x z y z Adding to inequalities assuming x y then we also know x z y z Multiplying an inequality z 0 assuming x y and z 0 then we also know x z y z Multiplying an inequality z 0 assuming x y and z 0 then we also know x z y z Zero product law assuming x y 0 then we also know x 0 or y 0 or possibly both Are all of these facts still true when we are talking about just natural numbers Integers Rational numbers Think about it The real numbers also happen to be an ordered field This is where the idea of the number line comes into play The following are order axioms Reflexivity x x Anti symmetry Assuming x y and y x then we also know x y Transitivity Assumign x y and y z then we also know x z Trichotomy given x and y exactly one of the following holds true i x y ii x y iii x y Multiplying inequalities Assuming 0 x y and 0 u v then 0 xu yv Square law 0 x2 Are all of these facts still true when we are talking about just natural numbers Integers Rational numbers Think about it Notice that the last one multiplying inequalities needs the assumption that everything is non negative We can easily find a counterexample to this false claim x y and u v implies xu yv Notice that 1 1 and 2 1 but 2 6 1 Some examples of true facts that are not axioms and require proof are as follows x y x y and Assuming 0 x y then we can deduce 0 We will prove these facts soon 2 x y and 0 x2 y 2 Algebraic Techniques Based on these assumptions about the properties of numbers we will assume you know how to do some algebraic steps Multiplying polynomials Try an example 2x 4 x 3 2x2 4x 6x 12 2x2 2x 12 Completing the square Try an example x2 6x 8 2 where we want to involve x and to not involve x There is a neat geometric visual trick Start with a square of dimensions x x add two rectangles along the sides of dimensions 3 x and consider what square term we d have to add to make the whole figure a square again In this example we get x2 6x 8 x 3 2 1 Factoring polynomials Difference of squares is nice x2 y 2 x y x y Try an example x2 6x 8 x 4 x 2 How could we have factored this if we hadn t spotted those roots Do we know the Quadratic Formula Why does that work Quadratic Formula Theorem Suppose a b c are real numbers and a 6 0 Then the solutions to ax2 bx c 0 are of the form x b b2 4ac 2a How might we prove such a Theorem mathematically Let s motivate this and solve it in a bunch of steps then write up a proper proof How do we factor something like ax2 bx c If a 1 it s not so bad We would want real numbers r and s such that rs c and r s b This is what we did in the example above Q What if a 0 Another way to view that we are looking for the roots of x2 bx c 0 the values r and s that satisfy q r 0 and q s 0 since we would know q x x r x s x2 bx c Write p x ax2 bx c We want the first coefficient to be 1 so let s make that happen b c p x x2 x a a a Now notice that any root of p x is also a root of is essential 3 p x a and vice versa This back and forth property Try to find the roots by simplifying and solving for x c p x b x2 x a a a 2 b x 2a 2 b x 2a c a b 2a 2 c b2 2 a 4a Notice that we completed the square Set this equal to 0 and start to solve for x Notice also that p x 0 is equivalent to having since a 6 0 Thus we can just try to find the values of x that make the expression above 0 c b2 4a2 a x b 2a p x a 0 2 Now what Can we take the square root of both sides ONLY when the left hand side is non negative Thus we …