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\title{Math 171 Class Notes}
\author{Lucas Garron}
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\section*{Mar 29, 2011}

\subsection*{Reals}
We will state several axioms. Any set with these properties is called the \emph{set of real numbers}. (HW: Show that the set is unique).

\subsubsection*{Axiom Structure}
\li
\item Field Axoms (algebra), $+, \*$
\item Order Axioms, $>$
\item Completeness Axioms, related to limits
\lid

\subsubsection*{Algebraic structures:}
{\definition Semigroup.} A semigroup $(G, *)$ is a set $G$ with a map $*:G \times G \-> G$ ($*(g, g') = g*g'$ - $g, g' \in G$)

$*$ should be associative. $(g_1*g_2)*g_3 = g_1*(g_2*g_3)$\\
There is an identity $e \in G$ such that $g*e = g = e*g$ for all $g \in G$.

{\lemma The identity in a semigroup is unique.} Suppose both $e$ and $e'$ are identity elements. Then $e = e*e' = e'$.

{\definition Commutative (or Abelian) Semigroup} A commutative semigroup is a semigroup $(G, *)$ such that $g*g' = g'*g$ for all $g, g' \in G$.

{\definition Left inverse}. If $(G, *)$ is a semigroup, $g \in G$ is a left inverse for $g$ is an element $g' \in G$ such that $g'*g = e$.

{\definition Right inverse}. Same with $g*g' = e$.

{\definition Invertible} $g \in G$ if it has a left inverse and a right inverse.

{\lemma If $(G, *)$ is a semigroup and $g \in G$ is invertible, then any left inverse equals any right inverse.} Suppose $g \in G$, $a$ a left inverse, $r$ a right inver for $g$. Then $a = a*e = a*(g*b)=(a*g)*b=e*b=b$

{\definition Group} A group $(G, *)$ is a semigroup all of whose elements are invertible.\\
{\definition Commutative Group} A commutative group is a group if $g*g' = g'*g$ for all $g, g' \in G$.
{\definition Field.} A field $(\F, +, \*)$ is a set $F$ and two maps
\li
\item $+: \F\times \F \-> \F$
\item $\*: \F\times \F \-> \F$
\lid
...such that  $(\F, +)$ is a commutative group, $(F, \*)$ is a commutative semigroup.\\
$0$ in the identity of $(\F, +)$, $1$ in the identity of $(\F, +)$, every $x \in \F$ ($x \neq 0$) is invertible in $(F, \*)$, $1 \neq 0$, and the distributive law holds:
$(x+y)\*z = (x\*z)+(y\*z)$ for all $x, y, z \in =\F$.


{\lemma If $\F$ is a field, then for all $x \in \F, x\*0 = 0$.}\\

{\notation The inverse of $x$ in $(\F, +)$ is written as $-x$. For $(\F, \*)$, $x\i = {1\/x}$}\\

Proof of Lemma: $0 = x\*0 + (-(x\*0)) = x\*(0+0)+(-(x\*0)) = (x\*0)+(x\*0)+(-(x\*0)) = (x\*0)+((x\*0)+(-(x\*0))) = (x\*0) + 0 = x\*0$

{\example Other algebraic statements:}\\
$-x = (-1)\*x$ ($x \in \F$)\\
$(-x)\*(-y)=x\*y$ ($x, y \in \F$)

{\example $\Q, \R, \C, \Z/2\Z$}

\subsection*{Order Axioms}
{\definition Ordered Field} An ordered field $\F, +, \*, P$ is a field $(\F, +, \*)$ and a subset $P$ of $\F$ such that:
\li
\item $x \in \F \imp$ exactly one of the following holds: $x \in P, -x \in P, x = 0$
\item $x, y \in P \imp x+y \in Y, x\*y \in P$
\lid
$P$ is called the positive elements of $\F$.

{\example $\R, \Q, \{a+b\sqrt{2}: a, b \in Q\}\subset\R $}

{\lemma $1 \in P$.}
Proof: By $1.$ of the definition, exactly one of the following holds: $1=0, 1 \in P, -1 \in P$.\\
Since we have a field, $1 \neq 0$.\\
Assume $-1 \in P$. Then $(-1)\*(-1) = 1\in P$ by $2.$ Contradiction.

{\definition We write $x > y$ if $x-y \in P$}
{\definition We write $x < y$ if $y-x \in P$}
This has the usual properties. (Note: $x \in P$ means $x>0$, $x \not \in P$ means $x<0$)
(e.g. $x>y \imp x+z>y+z$, $x, y, z \in \F$)

\subsection*{Completeness}
Problem: Fields like $\Q$ have sequences that don't converge inside the field.

{\definition Upper bound, lower bound} Suppose $\F$ is an ordered field and $A \subset \F$. We say that $x \in \F$ is an upper bound for $A$ if $a \in A \imp a \leq x$ (lower bound: $a \geq x$)

{\definition Least upper bound} Suppose $A \subset \F, A \neq \0$. We say that $x \in A$ is a least upper bound for $A$ if:
\li
\item $x$ is an upper bound of $A$
\item if $y$ is an upper bound of $A$, then $y \geq x$
\lid

{\definition Greatest upper bound} Complementary definition.

{\lemma If $A \subset \F, A \neq \0$ has a least upper bound, it is unique.} Suppose not, $x, y \in \F$ are both least upper bounds. Then $x \leq y, y \leq x \imp x=y$

{\notation If it exists, then the unique least upper bound is denoted $sup A$ (supremum)}
{\notation Greatest lower bound: $inf A$ (infimum)}

{\definition Reals} The reals are the ordered field $(\R, +, \*, P)$ such that if $A \subset \R, A \neq \0$ and $A$ has an upper bound, then $A$ has a least upper bound.

{\example In $\Q$, truncated decimal expansions of $\sqrt 2$ have no least upper bound. In $\R$, they do.}

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