1. Properties of Elements in Fields $\mathbb{F}$

1.1 $ \forall a, b \in \mathbb{F} \quad a + b = b+a$

1.2 $\forall a, b, c \in \mathbb{F} \quad a + (b + c) = (a + b) + c$

1.3 $\exists 0 \in \mathbb{F} \: \forall a \in \mathbb{F} \quad a + 0 = a$

1.3.1 0 is unique. Let there be another element 0' such that $a + 0' = a$. Then,
\begin{align*}
0 &= 0 + 0' && \text{(by the assumed property of 0')}\\
&= 0' + 0 &&\text{(by 1.1)}\\
&= 0' &&\text{(by 1.3)}
\end{align*}

1.4 $\forall a \in \mathbb{F} \: \exists b \in \mathbb{F} \quad a + b = 0$

1.4.1 Such b is unique. Let b' be another one. Then
\begin{align*}
b & = b + 0 &&\text{(by 1.3)}\\
& = b+(a+b') &&\text{(by the assumed property of b')}\\
& = (b+a)+b' &&\text{(by 1.2)}\\
& = (a+b)+b' &&\text{(by 1.1)}\\
& =0 + b' &&\text{(by 1.4)}\\
& = b'+0 &&\text{(by 1.1)}\\
& = b' &&\text{(by 1.3)}
\end{align*}

1.5 $ \forall a, b \in \mathbb{F} \quad ab=ba$

1.6 $ \forall a, b, c \in \mathbb{F} \quad a(bc)=(ab)c$

1.7 $ \exists 1 \in \mathbb{F} \: \forall a \in \mathbb{F} \quad a1 = a$

1.7.1 1 is unique. Proof: Suppose 1' has the same property as 1. Then,
\begin{align*}
1 &= 11' &&\text{(by the assumed property of 1')}\\
&=1'1 &&\text{(by 1.5)}\\
&=1' &&\text{(by 1.7)}\\
\end{align*}

1.8 $ \forall a \in \mathbb{F} \: \exists b \in \mathbb{F} \quad ab = 1$

1.8.1 Such b is unique. Let b' be another one. Then $b = b1 = b(ab') = (ba)b' = (ab)b' = 1b' = b'1 = b'$

1.9 $ \forall a, b, c \in \mathbb{F} \quad a(b+c) = ab + ac$