**Proof plan.** Mimic the construction of $\mathbb{Q}$ from $\mathbb{Z}$: form equivalence classes of pairs $(a, b)$ with $b \neq 0$ under the relation $(a,b) \sim (c,d) \iff ad = bc$, define ring operations, verify field axioms, and embed $R$ injectively. **Step 1: Define the [equivalence relation](/page/Equivalence%20Relation).** Let $S = \{(a, b) \in R \times R : b \neq 0\}$. Define $(a, b) \sim (c, d)$ iff $ad = bc$. [claim: Equivalence Relation] $\sim$ is an equivalence relation on $S$. [/claim] [proof] Reflexivity: $ab = ba$ (commutativity). Symmetry: obvious. Transitivity: if $(a,b) \sim (c,d)$ and $(c,d) \sim (e,f)$, then $ad = bc$ and $cf = de$. Multiplying: $adf = bcf = bde$. Since $d \neq 0$ and $R$ is an [integral](/page/Integral) domain, we may cancel $d$: $af = be$, so $(a,b) \sim (e,f)$. [/proof] **Step 2: Define $F = S/\!\!\sim$ with operations.** Write $\frac{a}{b}$ for the class of $(a,b)$. Define: \begin{align*} \frac{a}{b} + \frac{c}{d} &= \frac{ad + bc}{bd}, \qquad \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}. \end{align*} Note $bd \neq 0$ since $R$ is an integral domain. [claim: Operations Are Well-Defined] The operations are independent of the choice of representative. [/claim] [proof] If $(a, b) \sim (a', b')$ then $ab' = a'b$. For multiplication: $\frac{ac}{bd}$ vs $\frac{a'c}{b'd}$; we need $acb'd = a'cbd$, i.e. $acb'd = a'cbd$. Cancelling $cd$: $ab' = a'b$. ✓ Addition is similar. [/proof] **Step 3: $F$ is a field.** [claim: F Is a Field] $(F, +, \cdot, \frac{0}{1}, \frac{1}{1})$ is a field. [/claim] [proof] The ring axioms are straightforward. For inverses: if $\frac{a}{b} \neq \frac{0}{1}$ then $a \neq 0$ (otherwise $a \cdot 1 = 0 \cdot b$, so $a = 0$, contradiction). Then $\frac{b}{a} \in F$ and $\frac{a}{b} \cdot \frac{b}{a} = \frac{ab}{ba} = \frac{1}{1}$. [/proof] **Step 4: $R$ embeds into $F$.** [claim: Injective Embedding] The map $\iota : R \to F$ defined by $\iota(r) = \frac{r}{1}$ is an injective ring homomorphism. [/claim] [proof] $\iota$ preserves addition, multiplication, and sends $1_R \mapsto \frac{1}{1} = 1_F$. If $\iota(r) = \frac{0}{1}$, then $r \cdot 1 = 0 \cdot 1$, so $r = 0$. Hence $\iota$ is injective. [/proof] **Step 5: Every element of $F$ is a fraction of elements of $R$.** $\frac{a}{b} = \frac{a}{1} \cdot \left(\frac{b}{1}\right)^{-1} = \iota(a) \cdot \iota(b)^{-1}$. $\square$