# homework_2 - Homework 2 2.4 Suppose for the sake of...

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Unformatted text preview: Homework 2 2.4. Suppose for the sake of contradiction that x = (5- √ 3) 1 / 3 represents a rational number. Then we can write x as p q where p and q are integers with no common factors. Also, we have that x satisfies ( x 3- 5) 2 = 3 x 6- 10 x 3 + 22 = 0 By the rational zeros theorem, this implies that x ∈ {± 1 , ± 2 , ± 11 , ± 22 } . Hence, | x 3- 5 | ≥ 3 for all of the rational candidates for x . Thus, ( x 3- 5) 2 ≥ 9 > 3 which contradicts our first equation. Hence x is irrational. 3.4. To prove that 0 < 1 from the axioms, consider ≤ 1 2 by (Theorem 3.2, iv) ≤ 1 by (M3) . If 1 = 0 then, a = a · 1 for all a by (M3) = a · since 1 = 0 = 0 by (Theorem 3.1, ii). Since this is true for all a , there is only one element in our field, but this contradicts the definition of a field. Hence, 0 < 1. To prove that 0 < a < b implies 0 < b- 1 < a- 1 from the axioms, observe that a > 0 and b > 0 so a- 1 > 0 and b- 1 > 0 by Theorem 3.2, (vi). By Theorem 3.2 (iii), we have a- 1 b- 1 ≥ 0. Then, a < b implies a ( a- 1 b- 1 ) < b ( a- 1 b- 1 ) by (O5) ( aa- 1 ) b- 1 < ( bb- 1 ) a- 1 by (M1) and (M2)...
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homework_2 - Homework 2 2.4 Suppose for the sake of...

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