In this exercise we prove some implications of the axioms of multiplication over a field.
This is pretty much an exercise in practicing notating formal proofs. There is nothing enlightening here and we can use proposition (1.14) in the book as an outline.
In all of the parts below, \(x, y, z \in F\), where \(F\) is a field with additive identity \(0\) and multiplicative identity \(1\).
(a) If \(x \ne 0\) and \(xy = xz\) then \(y=z\)
\[
\begin{align}
y &= 1y & \text{(identity element)}\\
&= \left( \frac{1}{x} x \right) y & \text{(existence and definition of multiplicative inverse)}\\
&= \frac{1}{x} \left(x y\right) & \text{(associativity of multiplication)}\\
&= \frac{1}{x} \left(x z\right) & \text{(by assumption)}\\
&= \left(\frac{1}{x} x\right)z & \text{(associativity of multiplication)}\\
&= 1z & \text{(definition of multiplicative inverse)}\\
&= z & \text{(definition of multiplicative identity)}
\end{align}
\]
(b) If \(x \ne 0\) and \(xy=x\) then \(y=1\)
\[
\begin{align}
xy &= x & \text{(by assumption)}\\
xy &= x1 & \text{(by definition of the multiplicative identity)} \\
y &= 1 & \text{(cancellation by part (a) above with }z = 1\text{)}
\end{align}
\]
(c) If \(x \ne 0\) and \(xy=1\) then \(y=1/x\)
\[
\begin{align}
xy &= 1 & \text{(by assumption)} \\
\left(\frac{1}{x}\right)\left(xy\right) &= \left(\frac{1}{x}\right) 1 & \text{(existence of }1/x\text{)} \\
\left( x \left(\frac{1}{x}\right)\right)y &= \left(\frac{1}{x}\right) 1 & \text{(associativity)} \\
1 y &= \left(\frac{1}{x}\right) 1 & \text{(definition of }1/x\text{)} \\
y &= \frac{1}{x} & \text{(definition of the multiplicative identity)}
\end{align}
\]
(d) If \(x \ne 0\) then \(1/(1/x) = x\)
In order to avoid confusion we relabel (c) as:
If \(\alpha \ne 0\) and \(\alpha\beta=1\) then \(\beta=1/\alpha\)
Now let \(\alpha = 1/x\) and \(\beta = x\) then \(\alpha\beta = \left(\frac{1}{x}\right)x = 1\), therefore by (c) \(\beta = 1/\alpha\) or substituting \( x = 1/(1/x)\).