Proving existence of k such that (r + 1/k)² < s for rationals

[haruspex]

I note that the terms $a^2dk^2$ and $cb^2k^2$ from (4) also appear in (1)

Yes, this is key.
It might help clarify matters if you put those terms on the left and all else on the right in each inequality.
Be careful to keep track of which inequality is known to be true and which you are trying to deduce.

 

[Math Amateur]

Yes, this is key.
It might help clarify matters if you put those terms on the left and all else on the right in each inequality.
Be careful to keep track of which inequality is known to be true and which you are trying to deduce.

Following your advice ... we have ...

$(1) \Longleftrightarrow a^2k^2d - cb^2k^2 \lt -2abdk - b^2d$ ... ... ... ... (5)

and

$(4) \Longleftrightarrow a^2k^2d - cb^2k^2 \le -k^2$ ... ... ... ... (6)(6) is true by assumption and we are trying to find a condition on $k \in \mathbb(N)$ from (5)BUT ... what step do I take now ... still perplexed as to how to proceed ...

Can you help ... ...?

Peter

 

[haruspex]

we are trying to find a condition on $k \in \mathbb(N)$ from (5)

You need to come up with a choice for k such that if (6) is true then (5) must be true.
Do you think that will be a large value of k or a small one?

 

[Math Amateur]

You need to come up with a choice for k such that if (6) is true then (5) must be true.
Do you think that will be a large value of k or a small one?

Well ... the size of k doesn't matter in (6) ... .. but in (5) it would be more likely true if k were large ,,, is that right?

Peter

 

[haruspex]

the size of k doesn't matter in (6)

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).

 

[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