Result If r is a nonzero real number such that r + is an integer, then r" + is an integer for every positive integer n.
Result If r is a nonzero real number such that r + is an integer, then r" + is an integer for every positive integer n.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8e237d3f-b8e6-4775-a6f9-5671b153aef2%2F7281adc9-8d2e-40f8-9ff6-a05975f6c893%2Fb071ata_processed.png&w=3840&q=75)
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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