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In this document, there are two outlines of ways to prove that
multiplication of our defined integers is independent of choice of
representatives. The problem was definitely too hard without an
extensive hint!
First proof:
Define $(m,n) \,R\, (p,q)$ as $m+q=n+p$ for all natural numbers $m,n,p,q$.
Assume $(m,n)=(m',n')$ and $(p,q)=(p',q')$. Show that
$(mp+nq,mq+np)\,R\,(m'p'+n'q',m'q'+n'p')$.
$m+n'=n'+m$ and $p+q'=p'+q$ by our hypotheses.
$mp+nq+m'q'+n'p' = mq+np+m'p'+n'q'$ is what we want to show.
$mp+nq+m'q'+n'p'+
n'p+m'q+m'p+n'q=mq+np+m'p'+n'q'+n'p+m'q+m'p+n'q$
is equivalent (cancellation property of addition: I added
$n'p+m'q+m'p+n'q$ to both sides).
$$mp+nq+m'q'+n'p'+n'p+m'q+m'p+n'q =$$
by algebra
$$(m+n')p+(n+m')q+m'(q'+p)+n'(p'+q) =$$ by hypotheses: if you look at
this step you will see why I added the weird thing I added to both
sides: it makes it possible to apply the assumptions.
$$(n+m')p+(n'+m)q+(p'+q)m'+(q'+p)n' =$$
by algebra
$$mq +np +m'p'+n'q'+m'p+n'q+m'q+n'p$$
which is what we want.
Second proof:
This is more natural (there is less sense of having pulled a rabbit out of
a hat) but it requires more work. It also has the unfortunate feature of
breaking into cases.
Lemma: If $(m,n) = (m',n')$, then either $m=m'$ and $n=n'$ or there is
a $k$ such that either $m'=m+k$ and $n'=n+k$ or $m=m'+k$ and $n=n'+k$.
Proof: If $m=m'$, then $m+n'=n+m'$, so $n'+m=n+m$, so $n'=n$.
If $m'>m$, then $m'=m+k$, and $m+n'=m'+k+n$, so $(n+k)+m = n'+m$, so $n+k=n'$.
The proof that if $m>m'$ that $m=m'+k$ and $n=n'+k$ goes in exactly
the same way.
We can assume that either $(m,n)=(m',n')$ or $(m+k,n+k)=(m',n')$, because we
can stipulate that the primed letters are larger without loss of generality,
and similarly either $(p,q)=(p',q')$ or $(p',q') = (p+r,q+r)$.
Really, there are just three cases to consider:
1. both of $m=m'$ and $p=p'$: there is nothing to show, as in this
case $(m,n) = (m',n')$ and $(p,q)=(p',q')$. It is easy to see that if
we drop all primes the two products are the same.
2. one of $m=m'$ and $p=p'$ is true; without loss of generality, we can
assume $m=m',n=n'$ and $p'=p+k,q'=q+k$, and show
$mp+nq+m'q'+n'p' = mq+np+m'p'+n'q'$
by showing $mp+nq+m'q'+n'p' = mp+nq+m(q+k)+n(p+k) = mp+nq+mq+mk+np+nk =
mq+np+m(p+k)+n(q+k) = mq+np+m'p'+n'q'$
3. both of $m=m'$ and $p=p'$ are false: so we can assume
$m'=m+k,n'=n+k,p'=p+r,q'=q+r$, and
show $mp+nq+m'q'+n'p' = mq+np+m'p'+n'q'$
$$mp+nq+m'q'+n'p' =$$
by facts about primed letters
$$mp+nq+(m+k)(q+r)+(n+k)(p+r)=$$
by algebra
$$mp+nq+mq+mr+kq+kr+np+nr+kp+kr =$$
by algebra (regroup to swap terms with $p$ and $q$)
$$mq+np+mp+mr+kp+kr+nq+nr+kq+kr =$$
by algebra (reversing the distribution step above)
$$mq+np+(m+k)(p+r)+(n+k)(q+r)=$$
by facts about primed letters
$$mq+np+m'p'+n'q'$$
This proof is somewhat more natural.
This was not a good homework problem without some kind of major hint
about how to proceed!
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