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In page 2 of baby Rudin,

Rudin wants to prove that $ \forall p \in \mathbb{Q} $ such that $ p>0 $ and $p^2 < 2$ there exists a $q \in \mathbb{Q}$ such that $q>p$ and $q^2<2$

In his proof, he starts with let $q=p-(p^2-2)/(p+2)$

and then says $q^2-2 = 2(p^2-2)/(p+2)^2$

and he goes on to finish the proof.

His proof makes sense, my question is how am I supposed to think in order to solve such problems?

I mean how do you know what to set q equal to? How did he think about this problem to get the expressions?

For example, right now I am trying to find $X$ such that $ q=p-X $ different than the one Rudin set. But I can't think of anything.

Thank you!!

Edit: Yes, I apologize I did make a mistake, it's been fixed now

wwbb90
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  • If you take $p=2$, then necessarily $q^{2}>2$? – Lucien Sep 11 '13 at 14:55
  • and $p^2$ is alwasy greater than $0$ if $p>0$. Have you written the problem properly? I don't have Rudin atm to check it out – Jean-Sébastien Sep 11 '13 at 14:59
  • Presumably, you misquoted what you said Rudin wants to prove, because, as Lucien's example points out, what you wrote there is false. – Andreas Blass Sep 11 '13 at 15:00
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    Please, try to make the title of your question more informative. E.g., Why does $a\le b$ imply $a+c\le b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. – Lord_Farin Sep 11 '13 at 15:15
  • You're right thanks for pointing this out – wwbb90 Sep 11 '13 at 15:34
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    You have to understand that at that point in the book Rudin is merely interested in establishing the truth of the statement "${x \in \mathbb Q \mid x^2 < 2}$ has no least upper bound in the rationals." The definition of $q$ in this case happens to be derived from Newton's method for computing nth roots, but the why and how of this isn't relevant to what Rudin is trying to teach at that point, so he simply chooses the shortest/cleanest argument that yields what he wants. – kahen Sep 11 '13 at 16:11
  • Yes I understand that it's not relevant, I was just curious.. – wwbb90 Sep 11 '13 at 17:30

3 Answers3

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Suppose we just wanted to find any real $q$ which satisfied that. Well, all we'd have to do would be to choose some $q\in (p,\sqrt{2})$ and so the obvious choice would be to take the center of this interval and so set $q=\dfrac{\sqrt{2}-p}{2}$. The problem with this is that because $p$ is rational and $\sqrt{2}$ is irrational, we find that $q$ must be irrational.

So, we have to be a little more clever. We still want $q$ to be in the above stated range, because those values satisfy $q>p$ and $p^2<2$, but how do we guarantee that $q$ is also rational? Well, we need to find a rational number in the range $(0,\sqrt{2}-p)$ which we can add to $p$. Let's call that value $r$ for now. Well $r$ is going to be dependent on $p$ so lets write $r$ as a function $r(p)$ which we want $r(p)<\sqrt{2}-p$ in the interval $0<p<\sqrt{2}$. We want $r(p)$ to take rational values for all rational $p$ so we should write it as a rational function $$r(p)=\dfrac{s_1(p)}{s_2(p)}$$ where $s_i$ is a polynomial in $p$. Now, there are many polynomials we could take to be $s_1$ and $s_2$ so lets put in some simplifying conditions. We want $r(p)<\sqrt{2}-p$ and we note that the RHS has a root at $p=\sqrt{2}$ so lets make $r(p)$ also have a root there. We want $r(\sqrt{2})=0$ which happens precisely when $s_1(\sqrt{2})=0$. Probably the most obvious choice for $s_1(p)$ then would be $s_1(p)=2-p^2$.

Now we need to try and find an $s_2(p)$ which makes $r(p)$ small enough. As we're trying to make $r(p)$ smaller than a linear function, let's make $r(p)$ grow like a linear function away from its singularities, so let's make the denominator linear and set $s_2(p)=ap+b$. So, we want $\frac{-b}{a}$ to be outside of the interval $(0,\sqrt{2})$ and we also want $s_2(p)$ to be large enough to satisfy the condition we impose on $r(p)$. At this point, many values of $a$ and $b$ will work and I can only assume Rudin chooses $a=1$ and $b=2$ to make the algebra a little neater later on.

Given this, we get $$r(p)=\dfrac{2-p^2}{p+2}$$ and after this we need to check that that this choice of $r(p)$ really satisfies the conditions, which it does.

I'll admit, this 'method' of finding a good $q$ involves a lot of trial and error, and some arbitrary choices, but this is because there are many possible $q$ that we could have used and the dependance of $q$ on $p$ just makes things that little bit more complicated. I can offer no advice on how to 'get a feel' for making these choices other than practice and trying to make 'reasonable' choices along the way.

Dan Rust
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There were two cases.One with q such that q>p and q^2<2 and second with q< p and q^2>2. I think this X was chosen (in Rudin) so as to use it again in the case two.You can take any other X say X =(p-q)/2 in 1st case and X=(q-p)/2 in second.In the book it was chosen so as to avoid taking two different X's for the two cases.

O_huck
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I'm working through POMA too, and I had that same exact question when I read that section. Here's how I think you can derive it.

The easy way to create such a real $q$ is to have $q=p + \frac{\sqrt{2} -p}{k}$ for some $k>1$, but you want to avoid using $\sqrt{2}$ and irrational $k$ for $q$ to be rational. To get rid of the square root in the right term $$\frac{\sqrt{2} -p}{k}=\frac{2 -p^{2}}{k(\sqrt{2}+p)}$$

noting that $2+p > \sqrt{2}+p$ you get$$0<\frac{2 -p^{2}}{k(2+p)}<\frac{2 -p^{2}}{k(\sqrt{2}+p)}$$

where the middle function may be used as a replacement term for rational $k>1$ and, we see this inequality works for $k=1$ as well, which gives$$p<p+\frac{2 -p^{2}}{k(2+p)}<p+\frac{2 -p^{2}}{k(\sqrt{2}+p)}\ \leq \sqrt{2}$$

satisfying all of the conditions for q for rational $k \geq 1$. The case $k=1$ gives Rudin's Formula.

Also, something interesting I noticed when trying to figure this out was that if you view q and p recursively you can use Newton's formula for $f(x)=x^{2}-2$, which gives $q= p-\frac{p^{2}-2}{2p}$: a solution fairly close to the other answer.

Robearz
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  • Thank you! This is the clearest answer for me even though Daniel Rust's answer is really good too! – wwbb90 Sep 11 '13 at 17:31