diff --git a/blueprint/src/content.tex b/blueprint/src/content.tex index ad6509c..9bc0b41 100644 --- a/blueprint/src/content.tex +++ b/blueprint/src/content.tex @@ -185,4 +185,53 @@ \section{Content} \begin{theorem} An ordered semigroup without anomalous pairs is an Archimedean commutative semigroup. -\end{theorem} \ No newline at end of file +\end{theorem} +\begin{proof} +Let $a$ and $b$ be elements of an ordered semigroup $S$. +If either $a$ or $b$ is zero, then they commute. + +We begin with the case that $a$ and $b$ are positive. +If $a + b < b + a$, then for all $n\in \mathbb{N}^+$, we have that +\begin{align} +(n+1)(a+b) &= a + n(b+a) + b \\ +&> n(b+a) + b \\ +&> n(b+a) \\ +&> n(a+b) +\end{align} +And so $a+b$ and $b+a$ form an anomalous pair. + +The same for the case that $b + a < a + b$. +Therefore, we must have that $a+b = b+a$. + +The case where $a$ and $b$ are negative follows similarly. + +We now look at the case where $a$ is negative and $b$ is positive. +If the element $0$ exists and $a+b = 0$, then $a + b + a = a$ and so $b+a = 0$. +Therefore, $a$ and $b$ commute. + +If $a + b$ is positive, then +\begin{align} +(a+b) + (a+b) &> a + b \\ +(b + a) + b &> b \\ +b + a &> 0 +\end{align} + +We already showed that positive elements commute and so +\[ (b+a) + b = b + (b + a)\] + +Now we look at the case where $a+b < b+a$. +Then we have that +\begin{align} +2(a + b) &= a + ((b+a) + b) \\ +&= a + (b + (b + a)) \\ +&= (a + b) + (b + a) \\ +&> (a + b) + (a + b) \\ +&= 2(a + b) +\end{align} +Which is a contradiction. +We get a similar contradiction in the case that $b+a < a+b$. +Therefore, $a+b = b+a$. + +The case where $a+b$ is negative follows similarly. +The case where $b$ is negative and $a$ is positive is symmetric. +\end{proof} \ No newline at end of file diff --git a/blueprint/web/dep_graph_document.html b/blueprint/web/dep_graph_document.html index 6789cdf..6b977c9 100644 --- a/blueprint/web/dep_graph_document.html +++ b/blueprint/web/dep_graph_document.html @@ -547,7 +547,7 @@

Dependencies

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0.2 Content

# + + + @@ -500,6 +503,40 @@

0.2 Content

+
+
+ + Proof + + ▶ +
+
+

Let \(a\) and \(b\) be elements of an ordered semigroup \(S\). If either \(a\) or \(b\) is zero, then they commute.

+

We begin with the case that \(a\) and \(b\) are positive. If \(a + b {\lt} b + a\), then for all \(n\in \mathbb {N}^+\), we have that

+
+ \begin{align} (n+1)(a+b) & = a + n(b+a) + b \\ & {\gt} n(b+a) + b \\ & {\gt} n(b+a) \\ & {\gt} n(a+b) \end{align} +
+

And so \(a+b\) and \(b+a\) form an anomalous pair.

+

The same for the case that \(b + a {\lt} a + b\). Therefore, we must have that \(a+b = b+a\).

+

The case where \(a\) and \(b\) are negative follows similarly.

+

We now look at the case where \(a\) is negative and \(b\) is positive. If the element \(0\) exists and \(a+b = 0\), then \(a + b + a = a\) and so \(b+a = 0\). Therefore, \(a\) and \(b\) commute.

+

If \(a + b\) is positive, then

+
+ \begin{align} (a+b) + (a+b) & {\gt} a + b \\ (b + a) + b & {\gt} b \\ b + a & {\gt} 0 \end{align} +
+

We already showed that positive elements commute and so

+
+ \[ (b+a) + b = b + (b + a) \] +
+

Now we look at the case where \(a+b {\lt} b+a\). Then we have that

+
+ \begin{align} 2(a + b) & = a + ((b+a) + b) \\ & = a + (b + (b + a)) \\ & = (a + b) + (b + a) \\ & {\gt} (a + b) + (a + b) \\ & = 2(a + b) \end{align} +
+

Which is a contradiction. We get a similar contradiction in the case that \(b+a {\lt} a+b\). Therefore, \(a+b = b+a\).

+

The case where \(a+b\) is negative follows similarly. The case where \(b\) is negative and \(a\) is positive is symmetric.

+ +
+