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

  1. Jul 31, 2017 #1
    1. The problem statement, all variables and given/known data

    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:

    ?temp_hash=7ea11988c17c1b179b0660d32733467a.png



    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## ... ...


    2. Relevant equations ... and relevant information ...

    We 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 ...

    3. The attempt at a solution

    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

    ===========================================================================================

    ***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)

    ?temp_hash=7ea11988c17c1b179b0660d32733467a.png
    ?temp_hash=7ea11988c17c1b179b0660d32733467a.png
     

    Attached Files:

    Last edited: Jul 31, 2017
  2. jcsd
  3. Aug 1, 2017 #2

    andrewkirk

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    As is often the case in problems like this, one uses earlier parts to prove later parts. We can do that if we can turn the problem into one of searching for a ##n## that solves a problem of type (1) and a ##m## that solves a problem of type (2).

    Very broadly speaking, we want to convert it into a problem of showing that certain rational numbers are smaller than some other rational number.

    We have been asked to show there exists ##k## such that
    $$(r+\frac1k)^2<s$$
    One of the first things to try in any such situation is to expand a product. Doing that on the left allows us to restate the target inequality as
    $$r^2 +\frac {2r}k +\frac1{k^2}<s$$
    But we also need to use what we've been given, which is that ##r^2<s##. In other words ##h=s-r^2## is positive, and rational.

    We can use that to rewrite what we need to prove as (there exists ##k\in\mathbb N## such that)
    $$\frac {2r}k +\frac1{k^2}<h\quad\quad\quad\quad\quad\quad(\textrm{i})$$

    Rather than continue, and spoil your fun, can you think of a way to convert that into a problem of type (1) and another of type (2)? A minor obstacle is that we have only one inequality, and we need two, to get two problems. But we know in our guts that the inequality must be true and that we have lots of room to move, so we can adopt two inequalities that are sufficient but not necessary conditions for the above inequality to hold, and then use those two inequalities in parts (1) and (2).

    Can you think of two inequalities that, if both true, guarantee that (i) is true?

    EDIT: I just realised that, if you can prove that ##k\in\mathbb N\to \frac1k \leq1## then it can be solved in one part rather than two, using only (1).
     
    Last edited: Aug 1, 2017
  4. Aug 1, 2017 #3

    haruspex

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    If you are trying to prove something that depends on the variables being rational numbers then it is a good idea to invoke the definition of a rational number at some point. If r is a rational then there exist ....
     
  5. Aug 1, 2017 #4


    Hi Andrew ... thanks for the help!

    I take it that by problem of type (1) and type (2) you mean Exercise 1.5.9 (1) and Exercise 1.5.9 (2) ... ... I think you do mean this ... but just checking ...

    Peter
     
  6. Aug 1, 2017 #5
    Hi haruspex ... yes, indeed you are right ...

    Bloch, through a series of definitions and theorems shows us that we can legitimately think of rational numbers as expressions of the form ( "fractions" ) ##\frac{a}{b}## where ##a, b \in \mathbb{Z}## and where ##b \ne 0## ... ... and also proves that we may manipulate these entities as we did in high school ... I was taking this idea of "fractions" as my working definition of rationals ... but should have stated it ... mind you Bloch takes 5 pages of his text to show that this is legitimate ...

    Peter
     
  7. Aug 1, 2017 #6

    haruspex

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    Ok, so complete the sentence
     
  8. Aug 1, 2017 #7


    Thanks again, Andrew ...

    Think that this may be one way to go ...

    We have ##\frac{2r}{k} + \frac{1}{k^2} \lt h## where ##r, h \in \mathbb{Q}, \ k \in \mathbb{N}, \ r, h \gt 0##

    Now ...

    ##\frac{2r}{k} + \frac{1}{k^2} \lt h##

    ##\Longrightarrow 2rk + 1 \lt h k^2## ... (Multiply through by k^2 )

    ##\Longrightarrow 2rk \lt h k^2##

    ##\Longrightarrow 2 r \lt h k## ... ... (Divide through by k)

    Since ##2r \gt 0## and ##h \gt 0## we can use result in 1.5.9 (1) with ##2r = s## and ##k = n## and ##h = r## ...

    ... ... then we have that there exists a ##k## such that ##2r \lt k h## ...


    Is that correct?

    Peter


    Note: Andrew ... interested to know how you used both (1) and (2) ... I could not see how to do that ..
     
  9. Aug 1, 2017 #8
    To complete the sentence ...

    If ##r## is a rational then there exist ##a, b \in \mathbb{Z}## and ##\mathbb{Z}^*## respectively such that ##r = \frac{a}{b}## ...

    Hope that is correct ...

    Peter
     
  10. Aug 1, 2017 #9

    haruspex

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    Ok. So do the same with s and use these to replace r and s in the given relationship (r2<s).
     
  11. Aug 1, 2017 #10

    andrewkirk

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    Almost, but not quite. The arrows need to go the other way, because it is at the last bit that you use (1) to show that a ##k## exists that has certain properties, then, working backwards through the above (with reversed arrows) we prove that it also has the property of satisfying the original inequality.

    I usually do these problems the way you have, working forwards, but then checking that I can reverse it at the end, which relies on the arrows all being double-ended (##\Leftrightarrow##).

    Or we can do it in one pass, by requiring all the arrows to point in the other direction.

    There's probably often a better way, but I usually don't find it.

    Working from the bottom up, we start with
    $$\exists k\in \mathbb N:\ 2r<kh$$
    Given that ##k>0##, that is logically equivalent to (##\Leftrightarrow##)
    $$\exists k\in \mathbb N:\ 2rk<hk^2$$
    But then we strike a snag. Because we cannot reverse the next arrow. That is
    $$(\exists k\in \mathbb N:\ 2rk<hk^2)
    \not\Rightarrow
    (\exists k\in \mathbb N:\ 2rk+1<hk^2)$$

    Let's try to fix that:

    Taking ##\mathbb N## to exclude 0 so that ##k\in\mathbb N\Rightarrow k>0## we have
    \begin{align*}
    \exists k\in\mathbb N:\ k\in\mathbb N:\ \left(r+\frac1k\right)^2<s &\Leftrightarrow
    \exists k\in\mathbb N:\ r^2 + \frac{2r}k +\frac1{k^2}<s \\
    &\Leftrightarrow\ \
    \exists k\in\mathbb N:\ \frac{2r}{k} + \frac{1}{k^2} < h\\
    &\Leftrightarrow\ \
    \exists k\in\mathbb N:\ 2rk + 1 < hk^2
    \end{align*}

    That's where we previously hit the snag in reversing. We need to connect to a new inequality via a left-arrow ##\Leftarrow##, where the new inequality is something we can still achieve. Removing the ##1## won't do it. We need to make the LHS bigger. So instead let's replace ##1## by ##k##. Then our next step is
    $$\Leftarrow
    \exists k\in\mathbb N:\ 2rk + k < hk^2$$
    provided we have ##k\in\mathbb N\Rightarrow k\geq 1##. Do you have that given? If not we'll need to use a different trick.
    Assuming we have that, next we can divide by ##k## to get
    $$\ \ \Leftrightarrow\ \
    \exists k\in\mathbb N:\ 2r+1<kh$$
    and we can use (1) to show that last one is true.

    Then we can start with the last inequality, known to be true, and follow the left-pointing arrows to get to the original inequality that we wanted to prove.

    BTW, my approach using (1) and (2) was to go from
    \begin{align*}
    \exists k\in\mathbb N:\ k\in\mathbb N:\ \left(r+\frac1k\right)^2<s &\Leftrightarrow
    \exists k\in\mathbb N:\ r^2 + \frac{2r}k +\frac1{k^2}<s \\
    &\Leftrightarrow\ \
    \exists k\in\mathbb N:\ \frac{2r}{k} + \frac{1}{k^2} < h\\
    &\Leftarrow\ \
    \exists k\in\mathbb N:\bigg(\left(\frac{2r}k<\frac h2\right)\wedge
    \left(\exists k\in\mathbb N:\ \frac1{k^2}<\frac h2\right)\bigg)
    \end{align*}
    Then we use (1) to prove the first conjunct and (2) to prove the second, then take the greater of the two ##k##s from the two conjuncts.

    As the edit in my post shows, that is not necessary here, but it's a trick that is often useful in other analytical proofs - splitting a tolerance margin (##h## in this case) into sub-margins, proving a separate inequality for each one, then showing we can choose a number (##k## in this case) that satisfies them all simultaneously.
     
  12. Aug 2, 2017 #11

    To formulate the question as you wish we would proceed as follows:

    Let ##r = \frac{a}{b}## and let ##s = \frac{c}{d}## ...

    Then we have ##\frac{a}{b} \gt 0## and ##\frac{c}{d} \gt 0## where ##a,b,c,d \in \mathbb{Z}## ...

    and we can assume that ##a, b, c, d \gt 0## ... ( ... if ##r = \frac{a}{b}## where ##a, b## both negative then multiply ##a, b## by ##-1## and redefine ##a, b## ... )

    We assume ##r^2 \lt s## ... that is we assume ##\frac{a^2}{b^2} \lt \frac{c}{d}## ... ...

    We have to show that given the above there exists a ##k \in \mathbb{N}## such that:

    ##( \frac{a}{b} + \frac{1}{k} )^2 \lt \frac{c}{d}## ...

    If we follow the same strategy as Andrew followed then we would investigate the conditions for there to exist a ##k## such that ##( \frac{a}{b} + \frac{1}{k} )^2 \lt \frac{c}{d}## hoping to find a condition ... being careful to establish a chain of 'double' or two-way implications ... and then use the reverse implication chain to establish a proof ...

    Peter
     
  13. Aug 2, 2017 #12

    haruspex

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    Ok. Multiply that out. Does the resulting inequality suggest anything?
     
  14. Aug 2, 2017 #13
    ##\frac{a^2}{b^2} \lt \frac{c}{c} \ \Longleftrightarrow a^2 d \lt b^2 c## ... ...


    Now ... obviously from what you have said ... that should suggest something to me ... but ... no ideas ...

    Sorry ... :frown: ... can you help further ...

    Peter
     
  15. Aug 2, 2017 #14

    haruspex

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    What friendly sort of beast do you have on each side of ##a^2 d \lt b^2 c##? What species of mathematical animal were a, b, c and d defined to be?
     
  16. Aug 2, 2017 #15
    ... well ... both ##a^2 d## and ##b^2 c## are positive integers or natural numbers ... as are ##a, b, c## and ##d## ... ...

    Peter
     
  17. Aug 2, 2017 #16

    haruspex

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    Right. And if integer m < integer n, can you say something just a bit stronger about their relationship?
     
  18. Aug 2, 2017 #17
    ... For ##m, n \in \mathbb{Z}## then ... ... If ##m \lt n## then there exists ##p \in \mathbb{Z}## such that ##m + p = n## ... ...

    Peter
     
  19. Aug 2, 2017 #18

    haruspex

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    How near to being equal, while not being equal, can two integers be?
     
  20. Aug 2, 2017 #19

    If ##m,n## are as near equal as two integers can get then ##m + 1 = n## ... ...

    Peter
     
  21. Aug 2, 2017 #20

    haruspex

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    (Or m = n+1)
    Right, so what inequality can you write that is a bit stronger than just m < n?
     
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