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objective: Show that IZ is irrational Proof by contradiction: Assume tz is rational: which means it can be expressed as to where b‡, and there is no common factors between and to So, √√√2 = 10 9 2/6²³-a³ Since a³ is equal to 26³, and 26³ must even integer, since any positive inteover musipried - 6Y 2 is even. a³ must be even, be an since a a*a*a, and a³ is even. a must be even as well. So a =2k, where KEZ => 26³ = (24) 263 = 8k³ 6²³ = 4K³ Since any integer, not 0, multiplied by an even number results in an even number. 16 must be even since both 6 and a are even Share band a a common factor of 2 Therefore, is irrational by contradiction GED

objective: Show that IZ is irrational Proof by contradiction: Assume tz is rational: which means it can be expressed as to where b‡, and there is no common factors between and to So, √√√2 = 10 9 2/6²³-a³ Since a³ is equal to 26³, and 26³ must even integer, since any positive inteover musipried - 6Y 2 is even. a³ must be even, be an since a a*a*a, and a³ is even. a must be even as well. So a =2k, where KEZ => 26³ = (24) 263 = 8k³ 6²³ = 4K³ Since any integer, not 0, multiplied by an even number results in an even number. 16 must be even since both 6 and a are even Share band a a common factor of 2 Therefore, is irrational by contradiction GED

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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objective: Show that IZ is irrational Proof by contradiction: Assume = is rational: which means it can be expressed as to where b#, and there is factors between and to no commen So, √√√2 = 16 120 -10³ 2/0²³-a³ Since a is equal to 26³, and 26³ must even integer, since any positive intover multiplied - 6Y 2 is even. a³ must be even, be an since d a*a*a, and a³ is even. a must be even as well. So a = 2K, where KEZ => 26³ = (24) 263 = 8k³ 6²³ = 4K³ Since any integer, not of multiplied by an even number results in an even number. 16 must be even since both 6 and a are even Share band a a common factor of 2 Therefore, is irrational by contradiction GED
Transcribed Image Text:objective: Show that IZ is irrational Proof by contradiction: Assume = is rational: which means it can be expressed as to where b#, and there is factors between and to no commen So, √√√2 = 16 120 -10³ 2/0²³-a³ Since a is equal to 26³, and 26³ must even integer, since any positive intover multiplied - 6Y 2 is even. a³ must be even, be an since d a*a*a, and a³ is even. a must be even as well. So a = 2K, where KEZ => 26³ = (24) 263 = 8k³ 6²³ = 4K³ Since any integer, not of multiplied by an even number results in an even number. 16 must be even since both 6 and a are even Share band a a common factor of 2 Therefore, is irrational by contradiction GED
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