Lecture 11 Enoch Cheung October 3 2013 Universal Claims Of the form x S P x Note Commonly P x is a conditional statement We ll revisit thse when we do the section on conditional claims Direct Proofs General idea Let x S be arbitrary Show that P x must hold Ex 1 Claim For all x y R x2 y 2 2xy Let x y R be arbitrary Consider x y 2 0 which we know to be non negative because squares of real numbers are non negative Through algebraic manipulation x2 y 2 2xy x y 2 0 so x2 y 2 2xy Ex 2 Claim The sum of any 2 rational numbers is a rational number We will write this out symbolically a b Q a b Q Let a b Q be arbitrary then by definition of rational numbers there exists s t u v Z where t 6 0 and v 6 0 such that s u a b t v Then it follows that s u sv tu a b Q t v tv because sv tu tv Z and tv 6 0 Indirect Proofs Observe that x S P x x S P x Proof strategy Assume for sake of contradiction that there exists an x S such that P x holds Arrive at a contradiction Ex 3 Claim m n Z 14m 21n 6 1 Assume for the sake of contradiction that m n Z are such that 14m 21n 1 Then 1 14m 21n 2 7m 3 7n 2m 3n 7 However this implies that 7 divides 1 which is a contradiction Ex 4 Claim 2 is irrational One way to write this out as a universal claim is x y Z xy 6 2 Assume for the sake of contradiction that there are such x y Z such that xy 2 Additionally we will require that the fraction xy is irreducible meaning that x y are coprime meaning that x and y are not both multiples of the same natural number a b c Z ac x bc y The reason why we can make this assumption is because given a fraction xy we can divide by whatever x and y are both multiples of to obtain smaller values x1 y1 such that x1 x and y1 y simplify the fraction We can iterate this process to eventually find xn yn such that xn yn is irreducible because if we cannot then we have an infinite sequence y y1 y2 such that y y1 y2 where each yi N This is impossible because there can be no infinte descending chain of natural numbers Therefore it is sufficient to arrive at a contradiction from assuming that there exists x y Z such that xy 2 and the fraction is irreducible 2 By assumption xy2 2 or equivalently x2 2y 2 This implies that x2 is even which means that x was even because the square of an odd number is odd so let x 2z for some x Z Substitution into our original equation gives 4z 2 2z 2 2y 2 so 2z 2 y 2 so y 2 is even as well which by the same argument means that y is also even However if x and y are both even then x y is not irreducible since we can divide both by 2 This gives a contradiction 1 Use the intermediate value theorem to come up with a non constructive direct proof of the following existential claim Claim x R x5 3x 1 0 Consider the continuous function f x x5 3x 1 Note that f 0 1 and f 1 1 Therefore by the intermediate value theorem since f 1 1 0 1 f 0 there exists some c R where 0 c 1 such that f c 0 It follows that f c c5 3c 1 0 so c is a real number satisfying our requirement 2