Apostol Calculus Vol.1 Exercise 2 Help Requested

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Discussion Overview

The discussion revolves around an exercise from Apostol's Calculus Vol. 1, specifically Exercise 2 on page 28, which asks participants to prove that for any arbitrary real number x, there exist positive integers m and n such that m < x < n. The scope includes conceptual clarification and mathematical reasoning related to the exercise.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant requests clarification on the problem statement, indicating it is challenging.
  • Another participant suggests that the problem must adhere to Apostol's axioms, implying that the solution may be constrained by these foundational principles.
  • A participant interprets the problem as stating that for any real number x, integers m and n can be found such that m < x < n, arguing that the problem does not specify that m and n must be positive.
  • There is a question raised about the implications of x being equal to 0 and how that affects the existence of integers m and n.
  • A later reply discusses the negation of the statement, questioning whether it is true that there exists a real number x such that for all integers m and n, either m > x or x > n, suggesting a consideration of boundedness of integers.

Areas of Agreement / Disagreement

Participants express differing interpretations of the problem, particularly regarding the necessity for m and n to be positive integers and the implications of x being 0. The discussion remains unresolved with multiple competing views on the interpretation of the exercise.

Contextual Notes

There are limitations in the discussion regarding assumptions about the nature of integers and the definitions of boundedness, which are not fully explored or resolved.

danne89
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Hello! Anyone read Apostol's Calculus vol. 1. On p. 28 the exercises feels very hard. Can somebody help me with nr. 2?
 
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can u post the prob. here?
 
I thought it would be useless because the answer must build on his axioms...
But here it comes:
If x is an arbitrary real number, prove that there exist positive integers such as m<x<n.
 
do you mean m < abs(x) < n?
 
And what if x = 0? danne89, what is the problem, word-for-word?
 
danne89 said:
I thought it would be useless because the answer must build on his axioms...
But here it comes:
If x is an arbitrary real number, prove that there exist positive integers such as m<x<n.

The problem states that if x is an arbitary real number, then there exist integers m and n such that m < x < n. The problem makes no reference to positive or otherwise. No wonder you're having such a hard time with the problem.
 
The negation says that there exists a real number x such that, for all integers m and n, (m > x) or (x > n). IOW, that the set of integers is bounded below or bounded above or both. Is that true?
 

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