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# math 103 week 3 notes Math 103

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This 3 page Class Notes was uploaded by Ariel Harris on Monday February 8, 2016. The Class Notes belongs to Math 103 at James Madison University taught by Debra Warne in Spring 2016. Since its upload, it has received 12 views. For similar materials see THE NATURE OF MATHEMATICS in General at James Madison University.

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Date Created: 02/08/16

CLASS NOTES January 27th 2016 When mathematicians notice a pattern they propose a law Examples do NOT prove a rule Our de nitions for using natural numbers for even and odd was n is even iffn2k for some k E N n is odd iff n2k1 for some k E N what about the number one in N gut instinct 1 is odd but we can39t represent 1 as odd with our de nition because there is amp k E N 1 2K1 we would need k0 but 0 N We can x this problem by changing our de nition slightly weneedOto paykENU 0 With this change we can mathematically justify that 1 is odd because 1201 What about 0 Gut instinct 0 is neither but with a changed de nition 0 is even even n2k for some k E N U 0 with this de nition zero is now even zero is even because 020 and 0 E N U 0 Is it possible for 0 to be odd Is there a whole number k such that st 0 2k1 k would need to be 5 no 0 is not odd Negative Natural Numbers EXAMPLE if you have sheep then all your sheep catch a disease and die you then have zero sheep But you still need to feed your family you ask your neighbor to borrow four sheep After borrowing four sheep you now have 4 sheep because at some point you need to return the sheep you borrowed So now we move on to enlarge our number system again to allow us to talk about the concept of owing etc being below 0 ie negative 3210123 The set of integers there are equally many notation Z The countability of this set also never ending 11 correspondence can be found How could we match up integers Z and natural numbersN Our new number system Z aka the integers gives us further generalization for the concepts of even and odd n is even iffn2k for some kE Z n is odd iffn2k1 for some k E Z 4 is even because 4 22 where 2 E Z 9 is odd because 9 2 5 1 where 5 E Z CLASS NOTES January 29th 2016 Z 3210123 Considering the general property of odd odd even can we come up with an example where all three integers are negative 3 1 4 so our though progression from what we39ve been doing since the beginning of the semester and through 5 2 and 3 on January 27th homework leads us to suspect these properties hold true with all integers 1 odd odd even 2 even even even 3 even odd even One of the goals is to PROVE these really are true for all integers BUT FIRST 4 OF THE HOMEWORK How to create a 11 correspondence between Z and N U 0 possibility 1 match the zeroes in each group the match the odd N to negative Z Z 321Q l23 NUO Q123456 N U 0 1357 U 246 A positive integer n is matched with what whole number 2n A negative integer n is matched with what whole number ON TO MORE STRUCTURED MATHEMATICAL PROOF Let39s look at our property which thus far we39ve stated as oddoddeven we can more carefully state that as PROVE The sum of ANY two odd integers is even OR if two integers are each odd then their sum is even lfl know this If two integers are each odd Then I can prove this Then their sum is even Prove the sum of any two odd integers is even Proof Let m and n be odd integers so m and n must have the form m2k1 and n2j1 where k andj E Z adding m and n we have mn 2k1 2j1 2k12ji k g 2 we can now factor out a 2 of each 2kj12bwherebkj1

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