Rudin's proof that rationals have no GLB (or LUB) [duplicate]

In Rudin's 3e of Principles of Mathematical analysis, he associates $q = (2p+1)/(p+1)$. How does he create this number? What is the method for discovering this answer?

marked as duplicate by A.Γ., Mark Bennet, Trevor Gunn, José Carlos Santos, ziggurismNov 19 '17 at 17:46

• The number he creates is $q=(2p+2)/(p+2)$. It is a very elegant construction. The attached link has an explanation math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_exs.pdf See page 3. – student Nov 19 '17 at 5:00
• Can you please provide a reference to the page number where you came across it? – Shatabdi Sinha Nov 19 '17 at 5:18
• page 2 of the book. Example 1.1. I think the person is asking for an explanation of Rudin's proof that the set of rational numbers less than $\sqrt{2}$ has no least upper bound within the set of rational numbers, and that the set rational numbers greater than $\sqrt{2}$ has no greatest lower bound within the set of rational numbers. – student Nov 19 '17 at 5:33

There is no need to rely on such non-obvious tricks. Given a positive rational $p$ with $p^2<2$ one can show that there exists rational $q>p$ with $q^2<2$. Let $q=p+h$ where $h$ is a positive rational to be chosen suitably. We want $q^{2}<2$ ie $(p+h) ^2<2$ or $$h(2p+h)<2-p^{2}$$ or $$h<\frac{2-p^{2}}{2p+h}\tag{1}$$ If we choose $0<h<1$ then clearly $$\frac{2-p^{2}}{2p+1}<\frac{2-p^{2}}{2p+h}\tag{2}$$ From $(1),(2)$ it is obvious that we can choose $h$ to be any positive rational number less than $\min(1,(2-p^{2})/(2p+1))$ and $q=p+h$ will do our job. Also the number $1$ in $0<h<1$ is arbitrary and we could equally well have $0<h<n$ and then we just need to have $h$ less than $\min(n, (2-p^{2})/(2p+n))$.
In fact if we choose $n=1.5$ then clearly $(2-p^2)/(2p+1.5)<2/1.5<1.5$ and hence we just need to have $h<(2-p^2)/(2p+1.5)$ and we can have $h=(2-p^2)/(2p+2)$ so that $q=(2+p^{2}+2p)/(2p+2)$ works fine.
The above type of reasoning is used routinely in various proofs in real analysis and it is much better to focus on such ideas rather than coming up with a non-obvious formula for $q$.
• Isn't it a bit silly to have $h = 2$? – Jossie Calderon Nov 20 '17 at 2:43
• @JossieCalderon: I have nowhere put $h=2$. – Paramanand Singh Nov 20 '17 at 4:29
• Sorry, I meant $n = 2$. – Jossie Calderon Nov 24 '17 at 7:24
• I still don't get why is it silly to put $n=2$. We can put $n=1000$ or $n=10^{10}$ if we wish and still get an expression for $q$ which works. – Paramanand Singh Nov 24 '17 at 12:59
• Yes, but it doesn't tell us much. For example, if $n = 5$, then the $min$ isn't necessary. – Jossie Calderon Nov 26 '17 at 8:53