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In my real analysis book, when they show that the rational numbers does not have the least upper bound property, they show that the set $\{r \in \mathbb Q : r>0 \text{ and } r^2 < 2\}$ is non-empty and bounded above, but does not have a least upper bound. I understand the argument fully, but I'm having trouble understanding where they came up with some of it. Their argument goes as follows

Define $B = \{r \in Q : r>0 \text{ and } r^2 < 2\}$. Suppose $p \in B$. Define the rational number $q$ by

$$q = p + \frac{2-p^2}{p+2} = \frac{2p+2}{p+2}$$

Then we have

$$q^2-2 = \frac{2(p^2-2)}{(p+2)^2}$$

I understand that this shows for any $p \in B$ there is a $q \in B$ such that $p<q$. My question is why did they decide to define $q$ this way? Where is the logic for choosing $q$ to be that expression?

Thank you.

Michael Hardy
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  • This is in the review queue as a duplicate. It looks like the above link does indeed answer the question, but it emphatically is not a duplicate – Sam OT Apr 29 '21 at 14:09
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    It looks like a duplicate to me. Both questions ask why (among the many possible choices) $q=(2p+2)/(p+2)$ is chosen as the rational number satisfying $p < q$ and $q^2 < 2$. – @SamOT: Can you elaborate why you think that is is not a duplicate? – Martin R May 02 '21 at 16:35
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    One asks for a least upper bound on a specific question. The other speaks very generally without even saying what the upper bound in question is. Sure, the OP could have searched for the question number. But that is meta information on the question. Merely searching for the question itself will not necessarily bring up this example. Moreover, the notation used is quite different---here $q = 2/(p+2)$ but there $q$ appears to be something else and $x = 2/(p+2)$. Knowing and understanding the answer, the questions are clearly the same. But "duplicate" doesn't mean "there exists an equivalence" :) – Sam OT May 03 '21 at 09:10
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    @SamOT: clearly the asker is really interested in the genesis of that expression for $q$ in terms of $p$. They understand rest of the proof dealing with upper bound property. The dupe target mentioned in comments by user "The Amplitwist" also deals with this expression for $q$. – Paramanand Singh May 03 '21 at 12:23
  • You seem to be missing the point. The questions certainly are fundmanetally the same! A perfectly suitable answer would be something along the lines of the following: "An equivalent question has been asked [HERE]. Try to solve your quesiton by transalting it into that set-up and using that answer. If you are still stuck, please update your original question with further details." A large part of maths is converting one set-up into another. This is not so difficult to do in this case once the original asker sees the new linked question :-) – Sam OT May 04 '21 at 08:43
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    I am not convinced. Here there question is why $q$ is defined as $q = p + \frac{2-p^2}{p+2}$. In the older thread, the question is why $q$ is defined as $q = p + (2-p^2)x$ with $x= \frac{1}{p+2}$. That makes it an exact duplicate to me. – Martin R May 04 '21 at 11:01

3 Answers3

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If I wanted to write down an example of this sort from scratch, I'd think this way:

Given $p \in B$, I want to get $q \in B$ s.t. $q>p$. To start let's find a rational number which is expressed as some positive (real) multiple of $\sqrt{2}-p$. An easy way to do that is to just consider $2-p^2=(\sqrt{2}+p)(\sqrt{2}-p)$. This is too big to add to $p$ without $p$ becoming too big; specifically it is too big by a factor of $p+\sqrt{2}$. So we divide $2-p^2$ by a rational number bigger than $p+\sqrt{2}$.

The "simplest" such rational number is $p+2$. This $p+2$ part is actually quite arbitrary, we could have chosen any rational number bigger than $p+\sqrt{2}$. Using $2-p^2$ is not really arbitrary, because $2-p^2$ is basically "the thing we can measure to know whether $|p|<\sqrt{2}$ or not", if we can only work with rational numbers.

This uses the reals for the scratchwork, even though you probably haven't constructed the reals yet...but that's OK, because $p+\frac{2-p^2}{p+2}$ is a bona fide rational number in $B$ bigger than $p$.

Ian
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Well, you want to add something positive to $p$ so that its square is still less than $2$.

Notice that $(p+h)^2=p^2+h^2+2ph$ so our $h$ must satisfy $h^2+2ph< 2-p^2$.

In fact we could finish the proof here, clearly if $h$ is small enough it will work.

But notice that we need $h(p+h)<2-p^2$ If we can find an $h<2$ such that $h(p+2)\leq 2-p^2$ we are done, since $h(p+h)<(p+2)$. So taking $h=\frac{2-p^2}{p+2}$ does the trick (clearly this number is positive and smaller than $2$).

Asinomás
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There are so many formulae $f(x)$ with $x<f(x)\in B$ when $x\in B$ that it's hard to say how it was found, except that $$(2p+2)/(p+2)>p\iff 2p+2>p^2+2p\iff 2>p^2,$$ and that if we assume the existence of the number $\sqrt 2$ then

$(2p+2)/(p+2)<\sqrt 2\iff$ $ 2p+2<\sqrt 2 (p+2)\iff$ $ p(2-\sqrt 2)<2(\sqrt 2 -1)\iff$

$ \sqrt 2 (p \sqrt 2 -1)<2(\sqrt 2 -1)\iff$ $p\sqrt 2<2\iff p<\sqrt 2.$

A method written by Hero (or Heron) of Alexandria about 19 centuries ago is that for $0<A$ and $0<x$ let $f(x)=(x+A/x)/2 .$ You can verify that $x^2>A\implies x^2>f(x)^2>A.$

Since $f(y)=f(A/y),$ if $0<y$ and $y^2<A$ then $$(A/y)^2>A \implies (A/y)^2>f(A/y)^2>A \implies y^2<(A/f(A/y))^2<A.$$ (Since $f(A/y)=f(y)$ we can write that last inequality as $y^2<(A/f(y))^2<A.$

DanielWainfleet
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