Understanding Rudin Theorem 1.20 (b): Integers and Real Numbers Explained

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SUMMARY

The discussion focuses on Rudin's Theorem 1.20 (b), specifically the proof involving the relationship between integers and real numbers. The key assertion is that for positive integers m1, m2, and n, and a real number x, there exists an integer m satisfying -m2 ≤ m ≤ m1 and m-1 ≤ nx < m. Participants confirm the logic behind selecting m as the least integer greater than nx, validating the proof's integrity. A demonstration was provided to clarify the proof's steps.

PREREQUISITES
  • Understanding of real analysis concepts, particularly the properties of integers and real numbers.
  • Familiarity with Rudin's "Principles of Mathematical Analysis".
  • Basic knowledge of inequalities and their manipulation.
  • Experience with mathematical proofs and logical reasoning.
NEXT STEPS
  • Study the implications of Rudin's Theorem 1.20 in real analysis.
  • Review examples of integer and real number relationships in mathematical proofs.
  • Explore additional proofs in Rudin's text to deepen understanding of analysis techniques.
  • Practice constructing proofs involving inequalities and integer bounds.
USEFUL FOR

Mathematics students, educators, and anyone interested in advanced real analysis concepts, particularly those studying Rudin's work.

Dschumanji
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I understand the proof except for the following:

Suppose that -m2 < nx < m1 for positive integers m1, m2, n, and real number x.

Then there is an integer m with -m2 ≤ m ≤ m1 such that m-1 ≤ nx < m.

It definitely sounds reasonable, but it seems like a big jump in logic.
 
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Let m be the least integer that is strictly greater than nx. It is a triviality to verify that this integer has the desired properties.
 
Simple proof

Hi there,
I have attached a simple demonstration of the bit you are asking.
Let me know if it is clear now.
I hope it helps
 

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