Lecture 12 21 127 Concepts of Math 09 24 2012 Implementing Proof Strategies Here are several more examples of good well written proofs that properly implement the strategies we went over on Friday Linear Diophantine Equations Proposition There is no integral solution to 3a 15b 100 Restate claim logically a b Z 3a 15b 6 100 Proof strategy Proof by contradiction Suppose we have a solution Find a contradiction Proof AFSOC a b Z 3a 15b 100 Let such a and b be given Notice that 3a 15b 3 a 5b 100 Dividing tells us 100 a 5b 3 Since a b Z we know a 5b Z However 100 3 Z so a 5b Z This is a contradiction We conclude that there is no such integral solution to the equation 1 Unique root This is meant to introduce the concept of uniqueness We say an object with a certain property is unique if it has the right property but no other object has that property That is we would say x is the unique element of S with property P x if and only if x S P x y S y 6 x P y Notice that this is logically equivalent to x S P x y S x P y Also we can write the contrapositive instead x S P x y S P y x y Proposition There is a unique natural root of the equation n3 n 6 0 Restate claim logically n N n3 n 6 0 m N m 6 n m3 m 6 6 0 Proof strategy First we will show such an n exists We need to find it and prove it has the right properties Second we need to show that n is unique We can do this in one of two ways Suppose that there is an m that solves the equation and deduce that m n Or suppose that m is arbitrary and m 6 n and show why m does not solve the euqation Proof Let s define n 2 Notice that 2 N and n3 n 6 8 2 6 0 Thus n 2 is a root Now we want to show n 2 is the unique root Suppose m N and m is a root we want to show m n This means m3 m 6 0 and n3 n 6 0 We can subtract the equations yielding 0 m3 m 6 n3 n 6 m3 n3 m n Factoring tells us 0 m n m2 n2 mn 1 We know that in general a product of two numbers is zero if and only if at least one of the numbers is 0 This gives us two cases Case 1 Suppose m n 0 Then m n Case 2 Suppose m2 n2 mn 1 0 We know m n N so m n 1 This means m2 n2 mn 1 and so m2 n2 mn 1 2 Thus this case is not possible Therefore m n 0 and so m n This shows that n 2 is the unique root of the equation Out of interest we ll show you another implementation of the uniqueness method Let s skip ahead and do just that part 2 Suppose m N and m is a root of the equation We can add 6 to both sides of the equation and factor yielding 6 m3 m m m2 1 m m 1 m 1 We know m N so we have three cases Case 1 Suppose m 1 Then m 1 0 so m m 1 m 1 0 6 6 Thus this case is not possible Case 2 Suppose m 2 Then m n 2 Case 3 Suppose m 3 Then m 1 2 and m 1 4 so m m 1 m 1 2 3 4 24 Thus m m 1 m 1 6 6 so this case is not possible We deduce that Case 2 must be true This shows n 2 is the only root 3 The rationals are dense in R Proposition Strictly between any two distinct rational numbers lies another rational number Restate claim logically a b Q a b c Q a c c b Note To have a and b distinct means a 6 b We may assume that one of them therefore is bigger and for our purposes we don t care which one is actually bigger so let s just say that a b Proof strategy Take two arbitrary rational numbers a and b Suppose they are distinct and that a b Construct a rational number c between a and b Show that c Q Show that a c b Proof Let a b Q be arbitrary and fixed Suppose further that a b By the definition of Q we know x y z w Z such that a x y and b z w Let such x y z w be given Note we also know that y 6 0 w 6 0 Define c a b 2 x y 2 z w xw zy yw 2 xw zy 2yw Notice that xw zy Z and 2yw Z since Z is closed under addition and multiplication Furthermore 2yw 6 0 because y 6 0 w 6 0 This shows c Q Next we need to show a c c b To do this it suffices to show that c a 0 b c 0 Observe that c a Since a b we know b a 0 Thus 1 2 a b a b 2a b a a 2 2 2 b a 0 as well Therefore c a 0 Also observe that b c b We know already that b a 2 a b 2b a b b a 2 2 2 0 and therefore b c 0 This shows that a c b and the proof is complete Summary We followed the quantifiers bringing in some objects whose existence was guaranteed then properly defining an object and proving it has the desired properties We also proved the conditional claim by supposing the hypothesis and satisfying the conclusion 4 Primality testing When checking whether a number p N is prime a computer algorithm can just verify whether or not all the numbers smaller than p are factors of p Conveniently the algorithm only needs to check those numbers up to p This saves some computational effort and time Why might this be the case Let s prove it Proposition Let p be a natural number that is at least 2 If none of the natural numbers between 2 and p inclusive divide p then p is prime Restate claim logically p N 1 m x N 2 x p m p p is prime Recall The formal definition of is Given a b Z we write a b if and only if k Z b ak Proof strategy We will prove this by contrapositive It does not seem helpful to assume that all of the possible divisors of p are not divisors how could we use this negative assumption to say something positive about p We ll supose p is not prime i e composite This …