Some Properties of the Rationals .... Bloch Ex. 1.5.9 (3)

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In summary, the conversation discusses a problem in section 1.5 of Ethan D. Bloch's book "The Real Numbers and Real Analysis", specifically exercise 1.5.9. The problem involves proving that for rational numbers r and s, if r^2 < s then there is some natural number k such that (r + 1/k)^2 < s. The conversation includes a solution strategy and the attempt at a solution, as well as relevant equations and information from earlier parts of the book. The conversation also mentions two other related exercises (1.5.6 and 1.5.8) that may be useful in solving the problem.
  • #36
haruspex said:
It does. It might help to focus by calling the terms on the left of the inequalities just "LHS", as an opaque package.
As k increases, what does (6) tell you is happening to the LHS? Compare that with what happens as k increases in (5).
In (6) the LHS is decreasing as ##-k^2## ...

... but ... when k increases in (5) the LHS decreases (decreases because it's negative) at the lesser rate of k ... ( ! ignoring for the moment that the LHS contains ##k^2## terms )

EDIT I should be talking about the upper bound on the LHS ...

Peter
 
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  • #37
Math Amateur said:
In (6) the LHS is decreasing as ##-k^2## ...

... but ... when k increases in (5) the LHS decreases (decreases because it's negative) at the lesser rate of k ... ( ! ignoring for the moment that the LHS contains ##k^2## terms )

EDIT I should be talking about the upper bound on the LHS ...

Peter
Right. So for sufficiently large k ...?
 
  • #38
Q
haruspex said:
Right. So for sufficiently large k ...?
Hi haruspex ... still thinking about this ... ? ...

PeterEDIT: Can you give any further guidance?
 
  • #39
You need inequality (6) to imply inequality (5). What relationship between the two right-hand sides would lead to that implication?
 
  • #40
Well ... we require ##(6) \Longrightarrow (5)## so ... I THINK ...

we require that ##-2abdk - bd^2 \gt k^2##

Is that correct?

Peter
 
  • #41
Math Amateur said:
Well ... we require ##(6) \Longrightarrow (5)## so ... I THINK ...

we require that ##-2abdk - bd^2 \gt k^2##

Is that correct?

Peter
Sign error.
 
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  • #42
haruspex said:
Sign error.
Oh yes ... silly typo ... sorry

Should be ##-2abdk - b^2d \gt -k^2##

or could write it as

##2abdk + b^2d \lt k^2##

Peter
 
  • #43
Math Amateur said:
Oh yes ... silly typo ... sorry

Should be ##-2abdk - b^2d \gt -k^2##

or could write it as

##2abdk + b^2d \lt k^2##

Peter
Good.
Essentially,you can solve this as a quadratic in k.
 
  • #44
We have ##2abdk + b^2d \lt k^2##

##\Longleftrightarrow k^2 - 2abdk - b^2d \gt 0##

Assume for a moment that all variables represent real numbers ... and solve k in

##k^2 - 2abdk - b^2d = 0## ... ... where we take solution for ##k \gt 0##

We get ##k = \frac{ 2abd \pm \sqrt{ 4a^2b^2d^2 - 4(-b^2d) } }{2}##

So take ##k = abd + \sqrt{a^2b^2d^2 + b^2d} = \delta## (real number}

Take ##k## as the natural number just greater than or equal to ##\delta## ... ... (surely this number exists! since ##\delta## exists )

... and then

... we have ##k^2 - 2abdk - b^2d \gt 0## and problem is about finished ...

Is that correct ...?
 
  • #45
Math Amateur said:
finished
Indeed.
 
  • #46
Thank you for all your help

It is much appreciated ...

Peter
 

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