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

[Math Amateur]

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

 

[haruspex]

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 ...?

 

[Math Amateur]

Q

Right. So for sufficiently large k ...?

Hi haruspex ... still thinking about this ... ? ...

PeterEDIT: Can you give any further guidance?

 

[haruspex]

You need inequality (6) to imply inequality (5). What relationship between the two right-hand sides would lead to that implication?

 

[Math Amateur]

Well ... we require $(6) \Longrightarrow (5)$ so ... I THINK ...

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

Is that correct?

Peter

 

[haruspex]

Well ... we require $(6) \Longrightarrow (5)$ so ... I THINK ...

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

Is that correct?

Peter

Sign error.

 

[Math Amateur]

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

 

[haruspex]

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.

 

[Math Amateur]

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 ...?

 

[haruspex]

finished

Indeed.

 

[Math Amateur]

Thank you for all your help

It is much appreciated ...

Peter