Rudin's proof that rationals have no GLB (or LUB) 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?
 A: 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 $q$. 
