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# INTR DISCRETE METH MTHSC 119

Clemson

GPA 3.76

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This 2 page Study Guide was uploaded by Jazmyn Braun on Saturday September 26, 2015. The Study Guide belongs to MTHSC 119 at Clemson University taught by Staff in Fall. Since its upload, it has received 72 views. For similar materials see /class/214290/mthsc-119-clemson-university in Mathematics (M) at Clemson University.

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Date Created: 09/26/15

MthSc 119 sections 16 and 17 161 Let the groups be B1B4 Then 1000 lBZ le 100 lBi BJ Bkl 10 and lBl Bg Bg B4l 1 Applying inclusion exclusion gives 31032033034 410007 g100 1071 400076004071 3439 165 We rst count the number of six digit numbers that are bad they have at least three consecutive digits the same Let B1 be the set of six digit numbers with digits 1 2 3 identical B2 be the set of six digit numbers with digits 234 identical B3 be the set of six digit numbers with digits 345 identical and B4 be the set of six digit numbers with digits 456 identical We apply inclusion exclusion to count the bad77 nurnbers Note that 39103 AlSO lBl Bgl lBg Bgl lBg B4l 103 lBl Bgl Bg B4 102 and lBl B4l Likewise Bl Bg Bgl lBg B3 B4l 102 and all other triple intersections have size 10 Finally lBl B2 B3 B4l gives lBl U B2 U B3 U B4l 4 39104 7 39 103 3 102 2 102 210 710 40000 7 3300 220 710 36910 Since there are 106 six digit numbers of which 36910 are bad there are 106 7 36910 963090 good77 numbers with no three consecutive digits the same 171 a If x2 is not odd then x is not odd or If 2 is even then x is even b If 2 7 2 is not divisible by p then p is not prirne c If 2 is not positive then x is zero e If the car does not start then the battery is not fully charged f If not C then not A and not B 172 Suppose the if then statement is A 7 B The contrapositive is therefore B 7 A The contrapositive of this latter statement is A 7 B which is logically equivalent to A 7 B the original staternent 173 To prove A if and only if B77 we need to show 1 A 7 B and 2 B 7 A However instead of 2 we can prove the equivalent contrapositive nA 7 nB 174 a Let A B and C be sets with A Q B and B Q 0 Suppose for sake of contradiction that A is not a subset of C b Let x and y be two negative integers Suppose for sake of contradiction 1 that z y 2 0 d Let p and q be primes with p q also a prime Suppose for sake of contradiction that p 31 2 and q 31 2 f Let 01 and Cg be distinct circles Suppose for sake of contradiction that 01 and Cg intersect in three or more points 175 Suppose for sake of contradiction that the integers z and x1 are both even Then there is an integer a with z 2a and an integer b with x1 2b Adding 1 to both sides of the equation x 2a we get z 1 2a 1 Since x1 2b it follows that 2b 2a 1 Subtracting 2a from both sides of the equation 2b 2a1 and then dividing both sides by 2 we get b7a But i is not an integer whereas b 7 a is an integer Therefore consecutive integers cannot both be even QED 177 Let p and q be prirnes for which p q is also prirne Suppose for sake of contradiction that neither p nor q is 2 Since p is prime p has no other positive divisors other than itself and 1 It follows that 2 X10 and hence that p is not even thus p must be odd Likewise q is odd Since the sum of two odd numbers is even it follows that 10 q is even Hence 2lpq But since p gt 2 and q gt 2 we know that p q gt 4 So 10 q has a positive divisor namely 2 that is equal to neither 1 nor 10 q Therefore if the sum of two prime numbers is also prirne one of prirnes must be 2 QED 178 Let A and B be sets Suppose for the sake of contradiction that A 7 B B 7 A 31 0 This means there is an x E A 7 B B 7 A Thus 6 A7B ande B7A Sinces A7BweknowthatisinAbut not in B However z E B 7 A implies that z E B Thus z Z B and z E B Therefore A 7 B B 7 A Q QED

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