Let Q be the field of rational numbers and let F Q(√2, √7). You may assume that √2, √7, √14 are irrational. (a) Show that √7 Q(√2). Hence obtain that [F : Q] = 4.
Let Q be the field of rational numbers and let F Q(√2, √7). You may assume that √2, √7, √14 are irrational. (a) Show that √7 Q(√2). Hence obtain that [F : Q] = 4.
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.1: Real Numbers
Problem 36E
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I don't understand 2 points in the solution for this problem. Can you please explain why? Thanks.
1. Why : {1, √2} is a basis for Q(√2) over Q then we conclude 2rs = 0 and r2 + 2s2 = 0 ???
2. Why: [F : Q(√2) ] = 2
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