Some Properties of the Rational Numbers .... Bloch Exercise 1.5.9 (3)

[Math Amateur]

I am reading Ethan D. Bloch's book: The Real Numbers and Real Analysis ...

I am currently focused on Section 1.5: Constructing the Rational Numbers ...

I need help with Exercise 1.5.9 (3) ...Exercise 1.5.9 reads as follows:
View attachment 7023We are at the point in Bloch's book where he has just defined/constructed the rational numbers, having previously defined/constructed the natural numbers and the integers ... so (I imagine) at this point we cannot assume the existence of the real numbers.

Basically Bloch has defined/constructed the rational numbers as a set of equivalence classes on $$\mathbb{Z} \times \mathbb{Z}^*$$ and then has proved the usual fundamental algebraic properties of the rationals ...Now ... we wish to prove that for $$r, s \in \mathbb{Q}$$ where $$r \gt 0$$ and $$s \gt 0$$ that:

If $$r^2 \lt s$$ then there is some $$k \in \mathbb{N}$$ such that $$( r + \frac{1}{k} )^2 \lt s$$ ... ...

Solution Strategy

Prove that there exists a $$k \in \mathbb{N}$$ such that $$( r + \frac{1}{k} )^2 \lt s$$ ... BUT ... without in the proof involving real numbers like $$\sqrt{2}$$ because we have only defined/constructed $$\mathbb{N}, \mathbb{Z}$$, and $$\mathbb{Q}$$ ... so I am assuming that we cannot take the square root of the relation $$( r + \frac{1}{k} )^2 \lt s$$ and start dealing with a quantity like $$\sqrt{s}$$ ... is this a sensible assumption ...?So ... assume $$( r + \frac{1}{k} )^2 \lt s$$ ..

then

$$( r + \frac{1}{k} )^2 \lt s$$

$$\Longrightarrow r^2 + \frac{2r}{k} + \frac{1}{k^2} \lt s$$

$$\Longrightarrow r^2 + \frac{1}{k^2} \lt s$$ ... ... since $$\frac{2r}{k} \gt 0$$ ... (but ... how do I justify this step?)

$$\Longrightarrow k^2 \gt \frac{1}{ s - r^2 }$$

But where do we go from here ... seems intuitively that such a $$k \in \mathbb{N}$$ exists ... but how do we prove it ...

(Note that I am assuming that for $$k \in \mathbb{N}$$ that if we show that $$k^2$$ exists, then we know that $$k$$ exists ... is that correct?Hope that someone can clarify the above ...

Help will be much appreciated ...

Peter

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***NOTE***

In Exercises 1.5.6 to 1.5.8 Bloch gives a series of relations/formulas that may be useful in proving Exercise 1.5.9 (indeed, 1.5.9 (1) and (2) may be useful as well) ... so I am providing Exercises 1.5.6 to 1.5.8 as follows: (for 1.5.9 (1) and (2) please see above)
https://www.physicsforums.com/attachments/7024
View attachment 7025

 

[Opalg]

Solution Strategy

Prove that there exists a $$k \in \mathbb{N}$$ such that $$( r + \frac{1}{k} )^2 \lt s$$ ...

.
.
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$$( r + \frac{1}{k} )^2 \lt s$$

$$\Longrightarrow r^2 + \frac{2r}{k} + \frac{1}{k^2} \lt s$$

I would prefer to see this as

$$( r + \frac{1}{k} )^2 \lt s$$

$${\color{red}\Longleftarrow} r^2 + \frac{2r}{k} + \frac{1}{k^2} \lt s$$,​

because that is the implication that you want to use!

In order to find $k$ such that $r^2 + \frac{2r}{k} + \frac{1}{k^2} \lt s$, let $t = s-r^2$. You are told (or at least, you can prove using Exercise 1.5.6) that $t>0$. Also, $\frac{2r}{k} + \frac{1}{k^2} \lt s - r^2 = t > 0.$ Use 1.5.6 again to write that as $tk\left(k - \frac{2r}t\right) + 1 >0$. That inequality will be satisfied whenever $k$ is an integer greater than $\frac{2r}t.$

Having found such a value for $k$, you can then reverse each step of the argument to show that $\left(r+\frac1k\right)^2 < s$.

 

[Math Amateur]

I would prefer to see this as

$$( r + \frac{1}{k} )^2 \lt s$$

$${\color{red}\Longleftarrow} r^2 + \frac{2r}{k} + \frac{1}{k^2} \lt s$$,​

because that is the implication that you want to use!

In order to find $k$ such that $r^2 + \frac{2r}{k} + \frac{1}{k^2} \lt s$, let $t = s-r^2$. You are told (or at least, you can prove using Exercise 1.5.6) that $t>0$. Also, $\frac{2r}{k} + \frac{1}{k^2} \lt s - r^2 = t > 0.$ Use 1.5.6 again to write that as $tk\left(k - \frac{2r}t\right) + 1 >0$. That inequality will be satisfied whenever $k$ is an integer greater than $\frac{2r}t.$

Having found such a value for $k$, you can then reverse each step of the argument to show that $\left(r+\frac1k\right)^2 < s$.

Thanks so much for the help Opalg ...

... in particular, thanks for the key point regarding the logic of the exercise ... I had not thought that out clearly ...

Peter