I am not sure how this number is constructed.

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I know that $2>q^2>p^2$, I then also know that q>p, hence I could construct a q such that q>p. I therefore choose q=p+x, where x could be an element of of rationals and then given I know that $2>q^2$ then $2>(p+x)^2$, so I should be able to solve for x. But since I cant use square root( have not created the real number line yet) I need to figure out how to solve $2>p^2+2 p x+x^2$

How would I proceed?

Or am I completely off track?


  • $\begingroup$ What exactly do you want to do? I mean, for given $p$ he explicitly defined a $q$ which has the properties you want and you have $x = q - p$. $\endgroup$ – Paul K Jan 2 '17 at 11:41
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    $\begingroup$ @menag I think OP wants to know how does the author comes up with that formula for $q$. $\endgroup$ – user175968 Jan 2 '17 at 11:49
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    $\begingroup$ Yes, I would like to see the construction of q $\endgroup$ – ALEXANDER Jan 2 '17 at 11:50

This is what the OP says in a comment: Yes, I would like to see the construction of $\,q$ . If that's all, then we are lucky; I've recently posted a derivation of the formula as an answer to this question:

Contrary to a statement in an answer by Paramanand Singh, there is nothing mysterious / magical about the formula by Rudin. Seems that another author has been re-inventing the wheel, though :-(

Summary of the know how. To understand why the above link should be followed.
Essential ingredient is the mediant of two fractions. And how the Stern-Brocot tree is formed with help of these mediants. We initialize $p < \sqrt{2}$ and $q > \sqrt{2}$ as: $$ p = \frac{m}{n} \quad ; \quad q = \frac{2n}{m} $$ Where $m$ and $n$ are positive integers. Now form the mediant of $p$ and $q$ , two times: $$ q := \frac{m+2n}{n+m} \quad ; \quad q := \frac{m+(m+2n)}{n+(n+m)} = \frac{2m+2n}{m+2n} = \frac{2p+2}{p+2} $$ The rightmost formula is already Rudin's.


This is a serious problem with the way Rudin complicates a simple problem. In order to prove that the set $A$ has no largest element it is not really necessary to find an explicit formula for a member $q\in A$ with $q > p$ where $p$ is a given member of $A$. We only need to prove that if $p \in A$ then there is a $q \in A$ with $q > p$ and for that that we don't need any mysterious / magical formula for $q$ in terms of $p$.

Hardy in his textbook A Course of Pure Mathematics does it so much better and also teaches the way things work in real-analysis. The crux of the argument by Hardy is that the set of positive rationals is partitioned into $A, B$ such that $A \cup B = \mathbb{Q}^{+}, A\cap B = \emptyset$ and further that we can find a member of $A$ and a member of $B$ which are as close to each other as we please.

Clearly $1 \in A, 2 \in B$ and given any positive integer $n$ we have the following chain of increasing rational numbers $$1, 1 + \frac{1}{n}, 1 + \frac{2}{n}, \ldots, 1 + \frac{n}{n} = 2$$ such that successive numbers in the above chain differ by $1/n$. Since the first member of the chain lies in $A$ and the last member of the chain lies in $B$ it follows that there is a last number in the chain which lies in $A$ and the next one belongs to $B$. Thus we have found two rationals $q, r$ with $q \in A, r \in B$ such that $r - q = 1/n$. Thus we can find a member of $A$ and a member of $B$ which are as close to each other as we please. Moreover we can choose $q, r$ both less than 2.

Now suppose we have a $p \in A$ then by definition of $A$ we have $p^{2} < 2$ and hence $\epsilon = 2 - p^{2}$ is a positive rational number. We can now find a positive integer $n$ such that $1/n < \epsilon/4$. Then by argument in previous paragraph we can find $q, r$ with $q \in A, r \in B$ with $r - q = 1/n < \epsilon/4$. Also both $q, r$ can be chosen to be less than $2$ so $r + q < 4$ and hence $r^{2} - q^{2} = (r - q)(r + q) < \epsilon$. And this means that $$(r^{2} - 2) + (2 - q^{2}) < \epsilon$$ Since $q\in A, r \in B$ each expression in parentheses in above equation is positive and their sum is less than $\epsilon$ so that each expression itself is less than $\epsilon$. Therefore we have $$r^{2} - 2 < \epsilon, 2 - q^{2} < \epsilon$$ By definition of $\epsilon$ it now follows that $$2 - q^{2} < 2 - p^{2}$$ or $q > p$ and thus we have found $q > p, q \in A$.

Note that the most of the deep significant theorems in real-analysis are existential in nature where it is of utmost importance to focus on the existence of a quantity with certain specific desired properties rather than explicitly finding such a quantity and the above proof is a typical example of proofs seen in real-analysis where it shows the existence of an element $q \in A, q > p$ given an element $p \in A$ without explicitly giving a formula for $q$ in terms of $p$.

Rudin on the other hand uses some sort of numerical technique to have explicit formula for $q$ in terms of $p$. Hardy also gives this approach by asking his readers to prove the following simple theorem:

If $x$ is a positive approximation to $\sqrt{2}$ then $(x + 2)/(x + 1)$ is a better approximation to $\sqrt{2}$ but in a different direction so that if $x < \sqrt{2}$ then $(x + 2)/(x + 1) > \sqrt{2}$.

Applying this rule twice we get $$x \to \frac{x + 2}{x + 1}\to\dfrac{\dfrac{x + 2}{x + 1} + 2}{\dfrac{x + 2}{x + 1} + 1} = \frac{3x + 4}{2x + 3}$$ so we can choose $q = (3p + 4)/(2p + 3)$ also.

  • $\begingroup$ While I agree with you that Rudin's approach here is strange, we shouldn't forget that he certainly knows the way things are done in real analysis. Rudin was an top-rate mathematician. $\endgroup$ – zhw. Jan 3 '17 at 22:25
  • $\begingroup$ @zhw.: Rudin was a great fellow and the same is true of G. H. Hardy, but personally I feel that Rudin's book is so typical of American books with a highly formal and boring style. It is specifically designed for teaching undergraduates. But Hardy's book on the other hand is a classic and it is not exactly for teaching students at all. It is meant for self study and targeted at very young students who are totally new to calculus. But this is all a personal opinion and I respect your opinion very much. $\endgroup$ – Paramanand Singh Jan 4 '17 at 4:19
  • $\begingroup$ You make some good points. I too think there are problems w. Rudin's books. See math.stackexchange.com/questions/2077932/… $\endgroup$ – zhw. Jan 4 '17 at 6:58

I'd do it by trying to get $(p+1/n)^2 < 2$ for some $n\in \mathbb N.$ Note that for any $n\in \mathbb N,$

$$(p+1/n)^2 = p^2 +2p/n + 1/n^2 < p^2 +4/n + 1/n = p^2 + 5/n,$$

where we've used $p<2$ and $1/n^2 \le 1/n.$ So we'll be done if we can make $p^2 + 5/n < 2.$ Can we? Sure, it's the same as saying $n > 5/(2-p^2).$

  • $\begingroup$ Simple and nice. +1 I think OP is trying to solve $p^{2}+2px+x^{2}<2$ like a quadratic in-equation. This is a typical algebraic approach which does not give any dividends in analysis. The right way is to replace $x^{2}$ by $x$ and then we simply get $x < (2 - p^{2})/(2p +1)$. You make it even simpler replacing $2p+1$ by $5$. $\endgroup$ – Paramanand Singh Jan 4 '17 at 4:39

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