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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Did I prove this well?
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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