Using only the axioms for a field, give a formal proof for $\frac{1}{z_{1}z_{2}}=\frac{1}{z_{1}}\cdot \frac{1}{z_{2}}$.

Question

Using only the axioms for a field, give a formal proof for $\frac{1}{z_{1}z_{2}}=\frac{1}{z_{1}}\cdot \frac{1}{z_{2}}$, where $z_{1},z_{2}\in \mathbb{C}$.

This is my attempt:

$\frac{1}{z_{1}z_{2}}=\frac{1}{z_{1}z_{2}}\cdot 1=\frac{1}{z_{1}z_{2}}\cdot z_{1}z_{1}^{-1}=\frac{1}{z_{2}}z_{1}^{-1}\cdot 1=\frac{1}{z_{2}}z_{1}^{-1}\cdot z_{2}z_{2}^{-1}=z_{1}^{-1}z_{2}^{-1}=\frac{1}{z_{1}}\cdot \frac{1}{z_{2}}$

Is this correct though? Any critique is appreciated!

Answer 1

Your proof assumes that$$\frac1{z_1z_2}\cdot z_1=\frac1{z_2}.$$How do you know that?

I would use the fact that\begin{align}\left(\frac1{z_1}\cdot\frac1{z_2}\right)\cdot(z_1\cdot z_2)&=\left(\frac1{z_1}\cdot\frac1{z_2}\right)\cdot(z_2\cdot z_1)\\&=\left(\left(\frac1{z_1}\cdot\frac1{z_2}\right)\cdot z_2\right)\cdot z_1\\&=\left(\frac1{z_1}\cdot\left(\frac1{z_2}\cdot z_2\right)\right)\cdot z_1\\&=\left(\frac1{z_1}\cdot1\right)\cdot z_1\\&=\frac1{z_1}\cdot z_1\\&=1.\end{align}

Answer 2

The inverse of an element is unique, therefore it suffices to show that $$\frac{1}{z_1}\cdot\frac{1}{z_2} (z_1 z_2) = 1$$ which is not too hard.

Proof that the inverse is unique:

Suppose $xy=1=xz$, then $y=x^{-1}xy = x^{-1}xz = z$.