Lecture 12 Enoch Cheung October 8 2013 1 Claim n N n X 1 i i2 i 1 We will prove this by induction P1 Base Case for n 1 i 1 1 i i2 1 Inductive case Assume for some n N we 1 n n n 1 2 1 1 2 2 Pn have i 1 1 i i2 n 1 X n X 1 i i2 1 n 1 n 1 2 i 1 i 1 1 i i2 1 n n n 1 2 then 1 n n n 1 1 n 1 n 1 2 by inductive hypothesis 2 1 n 1 n n 1 2 1 n 1 n 1 2 2 1 n 1 n 1 n 2n 2 2 1 n 1 n 1 n 2 2 n Pn so we have shown by induction that n N i 1 1 i i2 1 n n 1 2 2 a Prove that x R x2 6 1 x 6 1 We will prove this by contrapositive Let x R be arbitrary and x 1 then x2 1 Therefore x R x 1 x2 1 which is the contrapositive of the original statement b Prove that n N n 5 2n2 n 1 2 Suppose n N arbitrary n 5 Then 2n2 n2 n2 n2 4n n2 2n 1 n 1 2 where we made the substitution n 4 and 2n 1 c Let E x be the proposition x is even Prove that a b Z E a E b E a b E a b We first prove E a E b E a b E a b Consider a b Z arbitrary and E a E b so let a 2h and b 2k for h k Z Then a b 2h 2k 2 h k is even and a b 2h 2k 2 2hk is even so E a b E a b Now we prove E a E b E a b E a b by contraposition Consider a b Z arbitrary such that E a E b Equivalently E a E b Consider the case where E a so a is odd Then if a b is even then b is odd However this means that a b is odd which gives a contradiction Now consider the case where E b so b is odd Then if a b is even then a is odd so a b is odd which gives a contradiction 3 Claim There are no positive integer solutions to the equation x2 y 2 1 Symbolically this is x y N x2 y 2 6 1 Suppose for sake of contradiction that x y N such that x2 y 2 1 Then note that x y x y x2 y 2 1 so in particular x y divides 1 However since x y are positive integers x y 2 which is a contradiction because the only divisors of 1 are 1 and 1 1 4 Claim If a and b are both real numbers such that the product ab is irrational then either a or b must be irrational To write this out symbolically a b R ab Q a Q b Q Consider the contrapositive with De Morgan s law a b R a Q b Q ab Q Let a b R be arbitrary and suppose a b Q Then let a st b uv where s t u v Z and v t 6 0 Then ab su tv Q since uv 6 0 Therefore we have shown the contrapositive 5 Claim For any distinct integers p and q it is the case that p 1 is a multiple of q p if and only if q 1 is a multiple of q p To write this out symbolically p q Z p 6 q q p p 1 q p q 1 Consider p q Z arbitrary such that p 6 q Suppose q p p 1 then there is some n Z such that q p n p 1 It follows that q p n p 1 q p n p q p 1 p q q p n 1 q 1 so q p q 1 as desired On the other hand suppose q p q 1 then there is some n Z such that q p n q 1 then q p n q 1 q p n p q q 1 p q q p n 1 p 1 so q p p 1 as desired 2