I'm working on proving the commutative laws of addition and multiplication in the rationals. I've tried to make this as simple and as clear as possible.
My main questions are
In the proof of the commutative law of multiplication in $\mathbb{Q}$, am I making leaps with how I handle the multiplicative inverses? Specifically, I feel like I might be skipping something when I say $nm^{-1}pq^{-1}=(pq^{-1})(nm^{-1})$.
Does my use of the distributive law in the proof of the commutative law of addition make sense? I use it "backwards" in the fourth line, but I wanted to know if it makes sense to use it that way.
Can I even refer to multiplicative inverses or the fact that $\frac{x}{x}=1$? I'm afraid this might only work if the theorem assumed a field $R$ rather than the commutative ring of integers.
Any and all feedback is much appreciated! Thanks in advance.
Theorem If $\mathbb{Z}$ satisfies the axioms for a commutative ring, then addition and multiplication are commutative in $\mathbb{Q}$.
Definitions—If $\mathbb{Z}$ satisfies the axioms for a commutative ring, then multiplication and addition of the elements of $\mathbb{Z}$ are commutative. That is,
$$x+y=y+x\text{ for all }x, y \in \mathbb{Z}$$ $$xy=yx \text{ for all }x, y \in \mathbb{Z}$$
We need to show that these axioms hold for $\mathbb{Q}$. In $\mathbb{Q}$, addition and multiplication are defined in the following way, with $m, n, p, q \in \mathbb{Z}$:
$$\frac{n}{m}+\frac{p}{q}=\frac{nq+mp}{mq} \text{ and } \frac{n}{m} \cdot \frac{p}{q}=\frac{np}{mq}$$
Multiplication—To show that multiplication is commutative in $\mathbb{Q}$, we'll use only the axioms that define a commutative ring. We need to show that $$\frac{n}{m} \cdot \frac{p}{q}=\frac{p}{q} \cdot \frac{n}{m}.$$
With the definition of multiplication in $\mathbb{Z}$ above, we can see that
$$\frac{n}{m} \cdot \frac{p}{q}=(nm^{-1})(pq^{-1})=nm^{-1}pq^{-1}$$
Since $m, n, p, q \in \mathbb{Z}$, they satisfy the axioms for a commutative ring, so by the Commutative Law of Multiplication,
$$nm^{-1}pq^{-1}=(pq^{-1})(nm^{-1})=\frac{p}{q} \cdot \frac{n}{m}.$$
We can see now that $\mathbb{Q}$ satisfies this axiom of $\mathbb{Z}$ itself. Thus, multiplication in $\mathbb{Q}$ is commutative.
Addition—To show that addition is commutative in $\mathbb{Q}$, we need to show that $$\frac{n}{m}+\frac{p}{q}=\frac{p}{q}+\frac{n}{m}.$$ We begin by using the multiplicative identity. Then, we use the axioms for a commutative ring to show that addition is commutative in $\mathbb{Q}$.
\begin{align*} &\frac{n}{m}+\frac{p}{q} = \frac{n}{m} \Bigg(\frac{q}{q}\Bigg) + \frac{p}{q}\Bigg(\frac{m}{m}\Bigg) \tag*{Multiplicative Identity} \\ \\ & =\frac{nq}{mq} + \frac{pm}{qm} \tag*{Definition of Multiplication} \\ \\ & =\frac{nq}{mq} + \frac{pm}{mq} \tag*{Commutative Law of Multiplication} \\ \\ & =\frac{nq+pm}{mq} \tag*{Distributive Law} \\ \\ & =\frac{pm+nq}{mq} \tag*{Commutative Law of Addition} \\ \\ & =\frac{pm}{mq} + \frac{nq}{mq} \tag*{Distributive Law} \\ \\ & =\frac{pm}{qm} + \frac{nq}{mq} \tag*{Commutative Law of Multiplication} \\ \\ & =\frac{p}{q}\Bigg(\frac{m}{m}\Bigg) + \frac{n}{m}\Bigg(\frac{q}{q}\Bigg) \tag*{Definition of Multiplication}\\ \\ & =\frac{p}{q} + \frac{n}{m} \tag*{Multiplicative Identity} \end{align*}
We can now clearly see that addition in $\mathbb{Q}$ is commutative and the proof is complete.