description/proof of that for field with characteristic non-\(2\) and nonzero element, \(b\), \(b\) plus \(r'\) divided by \(b\) plus \(r\) equals \(b\) minus \(r'\) divided by \(b\) minus \(r\) iff \(r'\) equals \(r\)
Topics
About: field
The table of contents of this article
Starting Context
- The reader knows a definition of field.
- The reader knows a definition of characteristic of ring.
Target Context
- The reader will have a description and a proof of the proposition that for any field with characteristic non-\(2\) and any nonzero element, \(b\), \(b\) plus \(r'\) divided by \(b\) plus \(r\) equals \(b\) minus \(r'\) divided by \(b\) minus \(r\) iff \(r'\) equals \(r\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\text{ the fields }\}\), such that \(Ch (F) \neq 2\)
\(b\): \(\in F\), \(\neq 0\)
\(r\): \(\in F\), such that \(b + r \neq 0 \land b - r \neq 0\)
\(r'\): \(\in F\)
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Statements:
\((b + r') / (b + r) = (b - r') / (b - r)\)
\(\iff\)
\(r' = r\)
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2: Note
As an example that this proposition does not hold when \(Ch (F) = 2\), let \(F = \mathbb{Z} / 2\), the integers modulo \(2\) field, and \(b = 1, r = 0, r' = 1\): \((b + r') / (b + r) = (1 + 1) / (1 + 0) = 0 / 1 = 0 = 0 / 1 = (1 - 1) / (1 - 0) = (b - r') / (b - r)\), but \(r' = 1 \neq 0 = r\).
3: Proof
Whole Strategy: Step 1: suppose that \(r' = r\), and see that \((b + r') / (b + r) = (b - r') / (b - r)\); Step 2: suppose that \((b + r') / (b + r) = (b - r') / (b - r)\), and see that \(r' = r\).
Step 1:
Let us suppose that \(r' = r\).
\((b + r') / (b + r) = (b + r) / (b + r) = 1 = (b - r) / (b - r) = (b - r') / (b - r)\).
Step 2:
Let us suppose that \((b + r') / (b + r) = (b - r') / (b - r)\).
\((b + r') (b - r) = (b - r') (b + r)\), so, \(r' (b - r) + r' (b + r) = b (b + r) - b (b - r)\), so, \(r' (b - r + b + r) = b (b + r - b + r)\), so, \(r' (b + b) = b (r + r) = b r + b r = r (b + b)\), so, \(r' b (1 + 1) = r b (1 + 1)\).
As \(b (1 + 1) \neq 0\) (because \(Ch (F) \neq 2\)), \(r' = r\).
