Can a Constant be Chosen to Satisfy an Inequality for All Real Numbers?

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Discussion Overview

The discussion revolves around the existence of a constant \( A \) that satisfies a specific inequality involving the floor function and a variable \( x_0 \). Participants explore the implications of this inequality for all real numbers, particularly focusing on the role of \( x_0 \) and its potential values.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that there exists a constant \( A \) such that for all \( a > 0 \), there exists \( b > 0 \) fulfilling the inequality involving \( |x - x_0| < b \) and \( |\frac{1}{[x]} - A| < a \).
  • Another participant questions the role of \( x_0 \) in the inequality, seeking clarification on whether it is a fixed number or if the result must hold for all possible values of \( x_0 \).
  • A different participant suggests setting \( A = \frac{1}{[x_0]} \), assuming \( [x_0] \) is not zero, as a potential approach to the problem.
  • Further clarification is requested regarding the status of \( x_0 \), with emphasis on whether it is predetermined or variable.
  • One participant notes that \( x_0 \) typically represents a constant that can take various non-integer values, providing examples such as 2.3, 4.1, and 12.9.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the nature of \( x_0 \) and whether the proposed constant \( A \) can be universally applied. There is no consensus on the implications of the inequality or the status of \( x_0 \.

Contextual Notes

The discussion highlights the ambiguity surrounding the definition and role of \( x_0 \), as well as the conditions under which the proposed inequality holds. The implications of the floor function and the choice of \( A \) remain unresolved.

solakis1
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Prove or disprove the following:
There exists $A$ such that for all $a>0$ there exists $b>0$ such that for all $ x$:

$|x-\ x_0|<b$ i mplies. $|\frac{1}{[x]}-A|<a$ where [x] is the floor value of x

Gvf
 
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solakis said:
Prove or disprove the following:
There exists $A$ such that for all $a>0$ there exists $b>0$ such that for all $ x$:

$|x-\ x_0|<b$ i mplies. $|\frac{1}{[x]}-A|<a$ where [x] is the floor value of x
How does $x_0$ come into this?
 
put A= $\frac{1}{[x_0]}$ and $[x_0]$ is not 0
 
You have not explained the status of $x_0$. Is it a fixed number given in advance, or are we supposed to prove that the result holds for all $x_0$?
 
With $x_0$ we usualy define a constant that can take several different values but not integers for example 2.3, 4.1,12.9 e.t.c. e. t. c
 

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