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\noindent {\bf Question}
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\noindent Prove that each of the following statements holds in a
field $F$, using only the axioms of a field.
\begin{enumerate}
\item $a\cdot (-b) =(-a)\cdot b =-(a\cdot b)$ for all $a$, $b\in F$;
\item $(-a)\cdot (-b) =a\cdot b$ for all $a$, $b\in F$;
\item $(-1)\cdot a =-a$ for all $a\in F$;
\item $(-1)\cdot (-1) =1$.
\end{enumerate}

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\noindent {\bf Answer}
\begin{enumerate}
\item Since $F$ is a commutative group under addition, $a + (-a) =
0$.  Multiplying on the right by $b$ and applying the above fact
that $0\cdot b = 0$, we get $(a + (-a))\cdot b = 0$.  Apply the
distributive law to get $a\cdot b + (-a)\cdot b = 0$.  Adding the
additive inverse $-(a\cdot b)$ of $a\cdot b$ to both sides and
using the two facts that $-(a\cdot b) + a\cdot b = 0$ and that $0$
is the additive identity, we obtain $(-a)\cdot b = -(a\cdot b)$.
(Similarly, starting with $b + (-b) = 0$ and multiplying on the
left by $a$, we get that $a\cdot (-b) = -(a\cdot b)$.)  (And as
above, since both $(-a)\cdot b$ and $a\cdot (-b)$ are equal to
$-(a\cdot b)$, they are equal to each other.)
\item Start with $a + (-a) = 0$, and multiply both sides on the right
by $b + (-b)$.  Expanding out, we get $a\cdot b + a\cdot (-b) +
(-a)\cdot b + (-a)\cdot (-b) = 0$.  Since $a\cdot (-b) = (-a)\cdot
b = -(a\cdot b)$, this becomes $a\cdot b + (-(a\cdot b)) +
(-(a\cdot b)) + (-a)\cdot (-b) = 0$.  Since $-(a\cdot b)$ is the
additive inverse for $a\cdot b$, this becomes $-(a\cdot b) +
(-a)\cdot (-b) = 0$.  Adding $a\cdot b$ to both sides and
simplifying, this becomes $(-a)\cdot (-b) = a\cdot b$, as desired.
\item Start with $1 + (-1) = 0$, and multiply on the right by $a$.
Since $0\cdot a = 0$, this becomes $(1 + (-1))\cdot a = 0$.
Expanding out, this becomes $1\cdot a + (-1)\cdot a = 0$.  Since
$1$ is the multiplicative identity, this becomes $a + (-1)\cdot a
= 0$.  Adding $-a$ to both sides and simplifying, this becomes
$(-1)\cdot a = -a$, as desired.
\item Since we know already that $(-a)\cdot (-b) = a\cdot b$, we can
take $a =1$ and $b =1$ to get $(-1)\cdot (-1) = 1\cdot 1 = 1$,
with this last equality following from the fact that $1$ is the
multiplicative identity.
\end{enumerate}

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