Result If r is a nonzero real number such that r + is an integer, then r" + is an integer for every positive integer n.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Please explain this proof in more detail step by step, if able give why every step is taken, I don't understand any of it, I understand The strong induction just this example puzzles me, with the fraction and making it longer. Thank you in advance

Result
Proof
If r is a nonzero real number such that r + ½ is an integer, then r¹ + is an integer for
every positive integer n.
1
We use the Strong Principle of Mathematical Induction. Let r be a nonzero real number
such that r + is an integer. Since r + = r¹+ is an integer, the statement is true
for n = 1. Assume for an arbitrary integer k ≥ 1 that m; = = pi + is an integer for every
integer i with 1 ≤ i ≤ k. We show that r³+1+is an integer. Observe that
1
1
p²+¹ + 2 + ₁ = (x^² + — ^) (r + - ) − (−¹+²₁)
1
pk +1
= mkm₁ — mk-1 € Z.
By the Strong Principle of Mathematical Induction, if r is a nonzero real number such
that r + € Z, then r + EZ for every positive integer n.
r
Transcribed Image Text:Result Proof If r is a nonzero real number such that r + ½ is an integer, then r¹ + is an integer for every positive integer n. 1 We use the Strong Principle of Mathematical Induction. Let r be a nonzero real number such that r + is an integer. Since r + = r¹+ is an integer, the statement is true for n = 1. Assume for an arbitrary integer k ≥ 1 that m; = = pi + is an integer for every integer i with 1 ≤ i ≤ k. We show that r³+1+is an integer. Observe that 1 1 p²+¹ + 2 + ₁ = (x^² + — ^) (r + - ) − (−¹+²₁) 1 pk +1 = mkm₁ — mk-1 € Z. By the Strong Principle of Mathematical Induction, if r is a nonzero real number such that r + € Z, then r + EZ for every positive integer n. r
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