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by: Skylar Evans

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# Math statements and proofs chapter 1 Math 307

Marketplace > University of Oregon > Math > Math 307 > Math statements and proofs chapter 1
Skylar Evans
UO
GPA 4.0
Introduction to Proof

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Detailed first day of class notes
COURSE
Introduction to Proof
PROF.
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Math

This 2 page Class Notes was uploaded by Skylar Evans on Thursday October 1, 2015. The Class Notes belongs to Math 307 at University of Oregon taught by in Summer 2015. Since its upload, it has received 11 views. For similar materials see Introduction to Proof in Math at University of Oregon.

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Date Created: 10/01/15
Math statements and ProofsProof by Contradiction EX De ne x is odd or even It depends on the nature of x Case 1 Is an integer x is called even if x has the form x2k for some integer of k x is called odd if x2k1 for some integer Case 2 X is a function X is called even for every t in the domain of x t is also in the domain of x and x t xt EX Xttquot2 X is also called if for every t in the domain of x t is also in the domain of x and x txt Problem Show that the square root of 2 is a irrational not rational number In other words proof show that the square root of two cannot be written in the form square root of two mn where m and n are natural numbers By contradiction suppose it is possible to represent the square root of 2 in the form the square root of 2 mn where m and n are natural numbers we can assume that m and n have NO common factors other 1 Then 2mquot1nquot2 or 2nquot2mquot2 so mquot2 is even because it is equal to 2 times something so mquot2 is even Explain if mquot2 is even then m is even By contradiction assume mquot2 is even and m is not even m is odd So m2k1 for some integer k Then mquot22k1quot2 4kquot22k1 22kquot2tk1kquot1 So mquot2 is odd This contradicts the assumption mquot2 is even Thus m is even So m2k for some integer k Hence 2kquot2 ie 4kquot22nquot2 Hence nquot22kquot2 Thus nquot2 is even and so n is also even Therefore 2 is a common factor to m and n This contradicts the fact that m and n have no common factors other than 1 Remark Does the technique used above work for the square root of 3 and the cube root of two

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