From 3f5dcf3eb40978115716c3a78f61545e100efb5b Mon Sep 17 00:00:00 2001 From: Ingo Blechschmidt Date: Fri, 10 Nov 2017 17:59:21 +0100 Subject: [PATCH 1/2] very slightly sharpen the construction in the beta lemma --- .../incompleteness/representability-in-q/beta-function.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/content/incompleteness/representability-in-q/beta-function.tex b/content/incompleteness/representability-in-q/beta-function.tex index adf002ef..057b7bae 100644 --- a/content/incompleteness/representability-in-q/beta-function.tex +++ b/content/incompleteness/representability-in-q/beta-function.tex @@ -77,7 +77,7 @@ A couple of observations will help us in this regard. Given $y_0$, \dots,~$y_n$, let \[ -j = \max(n, y_0, \dots, y_n) + 1, +j = \max(n, y_0 + 1, \dots, y_n + 1), \] and let \begin{align*} @@ -101,7 +101,7 @@ Since $p$ divides $1 + (i+1)j\fac$, it can't divide $j\fac$ as well (otherwise, the first division would leave a remainder of~$1$). So $p$ divides $i-k$, since $p$ divides $(i-k)j\fac$. But $\left| i-k -\right|$ is at most~$n$, and we have chosen $j > n$, so this implies +\right|$ is at most~$n$, and we have chosen $j \geq n$, so this implies that $p \mid j\fac$, again a contradiction. So there is no prime number dividing both $x_i$ and $x_k$. Clause (2) is easy: we have $y_i < j < j\fac < x_i$. From 77d2675e0bd8c4952a592880724596ae68c86688 Mon Sep 17 00:00:00 2001 From: Ingo Blechschmidt Date: Fri, 10 Nov 2017 18:14:29 +0100 Subject: [PATCH 2/2] add explicit certificate of relative primality --- .../incompleteness/representability-in-q/beta-function.tex | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/content/incompleteness/representability-in-q/beta-function.tex b/content/incompleteness/representability-in-q/beta-function.tex index 057b7bae..b1a55df1 100644 --- a/content/incompleteness/representability-in-q/beta-function.tex +++ b/content/incompleteness/representability-in-q/beta-function.tex @@ -104,7 +104,10 @@ \right|$ is at most~$n$, and we have chosen $j \geq n$, so this implies that $p \mid j\fac$, again a contradiction. So there is no prime number dividing both $x_i$ and $x_k$. Clause (2) is easy: we have $y_i < -j < j\fac < x_i$. +j < j\fac < x_i$.\footnote{Using techniques of proof mining, it's possible to +extract from this argument an explicit certificate that~$x_i$ and~$x_k$ are +relatively prime: Writing $j! = (i-k)a$, we have $1 = (1 - (i+1)(i-k)a + +(i+1)^2 a) \cdot (1 + ij!) - (i+1)^2 a \cdot (1 + (k+1)j!)$.} Now let us prove the $\beta$ function lemma. Remember that we can use $0$, successor, plus, times, $\Char{=}$, projections, and any function