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I know this question has already been asked, for example Choice of $q$ in Baby Rudin's Example 1.1, but I can't understand any of the methods. The possibilities seem to be:

  1. Quadratic formula
  2. Secant method
  3. Drawing a chord between $(p,p^2 - 2)$ and $(2, 2)$ and taking the intersection of that chord with the $x$-axis

or just saying we want $n$ so large that $(p + 1/n)^2 < 2$, which makes the most sense.

I was hoping someone could walk me through an intuitive explanation of why Rudin's choice of $q$ in proving analytic incompleteness of the rationals is "natural" and where I can learn more. I haven't learned much about where Newton's method and the secant method come from, so I don't have a good idea. of how those formulas could be used to write $q$ as a function of $p$.

Update: Using a helpful suggestion in the comments:

Say $p \in A$, so $p^2 - 2 < 0$. We seek to find an $h > 0$ so that $q = p + h > 0$ but $(p + h)^2 < 2$. Then $$ p^2 + 2ph + h^2 < 2, $$ so $$ h(2p + h) < 2 - p^2. $$ I can then get $$ h < \frac{2-p^2}{2p + h}, $$ but then I've written $h$ in terms of itself; I need to get rid of $h$ somehow.

JohnT
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  • Not an answer to your question. I think the way $h$ is chosen in 1.21 is very intuitive and you can choose $q$ in 1.1 in the same way. –  Mar 04 '22 at 06:29
  • @user912011 That proof made sense to me and I've edited my first post with an attempt. The only problem is getting rid of the remaining $h$ on the right-hand side. Do you have any hints for that? – JohnT Mar 04 '22 at 15:46
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    You want to find a positive $h$ which is so small that $(p+h)^2<2$. Hence you can restrict $h$ to $(0,1)$, or $(0,42)$. Then $p^2+2ph+h^2<p^2+(2p+1)h$ or $(2p+42)$. Hence $(p+h)^2<2$ if $h<\frac{2-p^2}{2p+1}$. Notice how the assumption $p^2<2$ is used. –  Mar 05 '22 at 22:28

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