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\title{Additional Math 187 Notes for Jan 26}
\author{Dr. Holmes}
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\maketitle
These additional notes include some discussion of definitions and
proofs discussed in the 4.2 lecture, especially those which do not
actually occur in the book.
\begin{description}
\item[Definition:]
For natural numbers $m,n$, we define $mn$ as $nn \vee m=n$.
\item[Axiom 10:]
Let $m$ and $n$ be natural numbers. Then exactly one of the following
statements is true: $mn$).
\end{description}
We use Axiom 10 to prove the cancellation property of addition:
\begin{description}
\item[Theorem:] For any natural numbers $m,n,r$, if $m+r=n+r$ then $m=n$.
\item[Proof:]
We prove the contrapositive: ``if $m\neq n$, then $m+r \neq n+r$.
Assume $m\neq n$.
By Axiom 10, either $mn$. If $m>n$,
there is an $x$ such that $x+n=m$ (definitions of $<$ and $>$). To
show that $x=m-n$, we need to show that there is only one such $x$.
Suppose $x+n = m$ and $y+n=m$. Then we have $x+n=y+n$ (things equal
to the same thing are equal to each other) and so $x=y$ by the
cancellation property of addition (just proved above). So we have the
following: if $x+n = m$, then $x=m-n$, so $(m-n)+n = n$.
\end{description}
As an example of reasoning with subtraction, we prove the distributive
property of multiplication over subtraction.
\begin{description}
\item[Theorem:] Let $a,b,c$ be natural numbers. If $b-c$ is defined,
then $ac-bc$ is defined and $a(b-c) = ab-ac$.
\item[Proof:] Assume that $b-c$ is defined.
$$(b-c)+c = b$$ by the definition of subtraction,
$$a((b-c)+c) = ab$$ by properties of equality,
$$a(b-c) + ac = ab$$ by distributivity of multiplication over addition,
and finally
$$a(b-c) = ab-ac$$ by the definition of subtraction (really by the
observation that follows it).
\end{description}
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