From: David Llewellyn-Jones <david@flypig.co.uk>
Date: Mon, 10 Nov 2014 04:00:39 +0000 (+0000)
Subject: Fixed the sum range variable subscipts.
X-Git-Url: https://www.flypig.org.uk/git/?p=binaryfield.git;a=commitdiff_plain;h=HEAD

Fixed the sum range variable subscipts.
---

diff --git a/main.tex b/main.tex
index d9af18f..c5931cd 100644
--- a/main.tex
+++ b/main.tex
@@ -121,7 +121,7 @@ $$
 Starting from $m = n - 1$ and decreasing to $m = 1$ we define functions $b_m : \GF(p) \to \{0, 1 \}$ inductively as follows.
 \begin{equation}
 \label{equation:getbit}
-b_m (x) = d_{m - 1}' \left( x - \vphantom{\sum} \smash{\sum_{m}^{n - 1}} 2^i b_i (x) \right).
+b_m (x) = d_{m - 1}' \left( x - \vphantom{\sum} \smash{\sum_{i = m}^{n - 1}} 2^i b_i (x) \right).
 \end{equation}
 \end{definition}
 We claim that the functions $b_m (x)$ for $1 \le m \le n$ will generate the binary bits of any number $x < 2^n$ where $b_n$ is the MSB and $b_1$ is the LSB.
@@ -133,11 +133,9 @@ We start by noting that for any number $x \le 2^i$, the $i$-th bit is 0 if and o
 
 We now prove the result by induction on $m$ starting at $m = n$ and working downwards. So for the base case take $m = n$. Then $b_n (x) = d_{n - 1}'(x)$. However, we also know that $x \le 2^n$ and that the $n$-th bit is zero iff $x \le 2^{n - 1}$. But this therefore follows directly from Lemma \ref{lemma:comparison} setting $m = 2^{n - 1}$.
 
-For the inductive step, assume the result holds for $m = a + 1$. Our inductive hypothesis is that $b_i$ will output the binary bits for $a + 1 \le i \le n$ and we want to show that $b_a$ will output the $a$-th bit.
-
-Set
+For the inductive step, assume the result holds for $m = a + 1$. Our inductive hypothesis is that $b_i$ will output the binary bits for $a + 1 \le i \le n$ and we want to show that $b_a$ will output the $a$-th bit. Set
 $$
-x' = x - \sum_{a + 1}^{n - 1} 2^i b_i (x).
+x' = x - \sum_{i = a + 1}^{n - 1} 2^i b_i (x).
 $$
 From the results noted at the start of the proof, and the inductive hypothesis, we can deduce that
 $$