## Definition

$$\dfrac{p}{q} + \dfrac{r}{s} = \dfrac{ps + rq}{qs}$$
THEOREM

Proof.     Ron Freiwald
LEMMA

Zero is a left identity element of rational number addition.

$$\forall x \in \mathbb{Q} : 0 + x = x$$
Proof 1.     ProofWiki
Proof 2.     \begin{align} 0 + x &= \dfrac{0}{k} + \dfrac{p}{q} \\[1em] &= \dfrac{0q + pk}{kq} \\[1em] &= \dfrac{pk}{qk} \\[1em] &= \dfrac{p}{q} \\[1em] &= x \end{align}
LEMMA

Zero is a right identity element of rational number addition.

$$\forall x \in \mathbb{Q} : x + 0 = x$$
Proof 1.     ProofWiki
Proof 2.     \begin{align} x + 0 &= \dfrac{p}{q} + \dfrac{0}{k} \\[1em] &= \dfrac{pk + 0q}{qk} \\[1em] &= \dfrac{pk}{qk} \\[1em] &= \dfrac{p}{q} \\[1em] &= x \end{align}
THEOREM

Zero is an identity element of rational number addition.

$$\forall x \in \mathbb{R} : 0 + x = x + 0 = x$$
Proof.

Zero is both a left and right identity element of rational number addition. Therefore, zero is an identity element of rational number addition.

THEOREM

$$\forall x,y \in \mathbb{Q} : x + y = y + x$$
Proof 2.     \begin{align} x + y &= \dfrac{p}{q} + \dfrac{r}{s} \\[1em] &= \dfrac{ps + rq}{qs} \\[1em] &= \dfrac{rq + ps}{sq} \\[1em] &= \dfrac{r}{s} + \dfrac{p}{q} \\[1em] &= y + x \end{align}
$$\forall x,y,z \in \mathbb{Q} : x + y + z = x + (y + z)$$
Proof 2.     \begin{align} x + y + z &= \dfrac{p}{q} + \dfrac{r}{s} + \dfrac{t}{u} \\[1em] &= \dfrac{ps + rq}{qs} + \dfrac{t}{u} \\[1em] &= \dfrac{(ps + rq)u + t(qs)}{qsu} \\[1em] &= \dfrac{p(su) + (ru + ts)q}{q(su)} \\[1em] &= \dfrac{p}{q} + \dfrac{ru + ts}{su} \\[1em] &= \dfrac{p}{q} + \left(\dfrac{r}{s} + \dfrac{t}{u}\right) \\[1em] &= x + (y + z) \end{align}