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 \(\displaystyle \mathbb{Z} \times \mathbb{Z}^*\) and then has proved the usual fundamental algebraic properties of the rationals ...Now ... we wish to prove that for \(\displaystyle r, s \in \mathbb{Q}\) where \(\displaystyle r \gt 0\) and \(\displaystyle s \gt 0\) that:

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

Solution Strategy

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

then

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

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

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

\(\displaystyle \Longrightarrow k^2 \gt \frac{1}{ s - r^2 }\)

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

(Note that I am assuming that for \(\displaystyle k \in \mathbb{N}\) that if we show that \(\displaystyle k^2\) exists, then we know that \(\displaystyle k\) 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 \(\displaystyle k \in \mathbb{N}\) such that \(\displaystyle ( r + \frac{1}{k} )^2 \lt s\) ...

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

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

I would prefer to see this as

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

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

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

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

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

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