EXTENSION FIELDThe lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 }Can someone please explain the highlight parts in his works for me? I don't get it:- Why if F is a field then F is the smallest ?- Why √2 not in Q, then we have those stuff = 0 ?- Similarly, why √3 not in Q, then we have those stuff = 0. ?Thank you. QQ (√₂₁ √3)= { a+b√₂ +c√3 + d√6 | a, b, c, d = Q }F•F is cubring of R, and FCQ (√2, √3)• Is F a field?If yes, then F is smallest and F = Q(√2₁ √3)Check:1a+b√₂+c√3+d√61(a+b√₂) + √3 (c+d√₂)(a+b√z) - (c+d)√3(a+b√² ) ² = 3 [c+d√₂) ²(a+b√₂ = √3 (c+d√₂)).(a²-3c² + 26²-6d²-√₂ (2ab-6cd)) EF2(a²-3c² +26²-66²) ²EQ22 (2ab-6cd) ²1Hence+b√√₂ +c√3+d√√6the denominator is 0.(*)€ F aslongas Assume√₂ & QdenominatorHence= 0Qab-6cd = 0 ) ab-3cd=0a² +2b²-3c² - 6d² = 0√3 & Q = ab = cd = 0¢ =>a²+ 26² = 0 = c² + 2d²a = b = c = d

Answered: Image /qna-images/answer/f0770db0-5256-4ad7-aad6-e1c6edd3a5bc.jpg

EXTENSION FIELD The lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 } Can someone please explain the highlight parts in his works for me? I don't get it: - Why if F is a field then F is the smallest ? - Why √2 not in Q

EXTENSION FIELD The lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 } Can someone please explain the highlight parts in his works for me? I don't get it: - Why if F is a field then F is the smallest ? - Why √2 not in Q

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

EXTENSION FIELD

The lecturer proved that: Q( √2, √3) = { a + b √2 + c √3 + d √6 }

Can someone please explain the highlight parts in his works for me? I don't get it:

- Why if F is a field then F is the smallest ?

- Why √2 not in Q, then we have those stuff = 0 ?

- Similarly, why √3 not in Q, then we have those stuff = 0. ?

Thank you.

QQ (√₂₁ √3)= { a+b√₂ +c√3 + d√6 | a, b, c, d = Q }
F
•F is cubring of R, and FCQ (√2, √3)
• Is F a field?
If yes, then F is smallest and F = Q(√2₁ √3)
Check:
1
a+b√₂+c√3+d√6
1
(a+b√₂) + √3 (c+d√₂)
(a+b√z) - (c+d)√3
(a+b√² ) ² = 3 [c+d√₂) ²
(a+b√₂ = √3 (c+d√₂)).(a²-3c² + 26²-6d²-√₂ (2ab-6cd)) EF
2
(a²-3c² +26²-66²) ²
EQ
2
2 (2ab-6cd) ²
1
Hence
+b√√₂ +c√3+d√√6
the denominator is 0.
(*)
€ F as
long
as

Assume
√₂ & Q
denominator
Hence
= 0
Qab-6cd = 0 ) ab-3cd=0
a² +2b²-3c² - 6d² = 0
√3 & Q = ab = cd = 0
¢ =>
a²+ 26² = 0 = c² + 2d²
a = b = c = d
<contradiction,

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