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

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The discussion centers on the existence of a constant \( A \) such that for every positive \( a \), there exists a positive \( b \) satisfying the inequality \( |x - x_0| < b \) implies \( |\frac{1}{[x]} - A| < a \), where \([x]\) denotes the floor function of \( x \). Participants debate the role of \( x_0 \) and whether it is a fixed number or a variable that can take multiple values. The suggestion to set \( A = \frac{1}{[x_0]} \) is made, emphasizing that \( [x_0] \) must not equal zero. The discussion highlights the need for clarity on the nature of \( x_0 \) in the proof.

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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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