Rational Addition Identity is Zero

Theorem

The identity of rational number addition is $0$:

$\exists 0 \in \Q: \forall a \in \Q: a 0 = a = 0 a$

Proof

From the definition, the field $\struct {\Q, , \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, , \times}$ of integers.

From Zero of Inverse Completion of Integral Domain, for any $k \in \Z^*$, the element $\dfrac {0_D} k$ of $\Q$ serves as the zero of $\struct {\Q, , \times}$.

Hence $\dfrac 0 k$ is the identity for $\left({\Q, }\right)$:

\(\ds \frac a b \frac 0 k\)

\(=\)

\(\ds \frac {a \times k b \times 0} {b \times k}\)

\(\ds \)

\(=\)

\(\ds \frac {a \times k} {b \times k}\)

\(\ds \)

\(=\)

\(\ds \frac a b\)

Equal Elements of Field of Quotients

Similarly for $\dfrac 0 k \dfrac a b$.

Next we note that it is a zero:

\(\ds \frac a b \times \frac 0 k\)

\(=\)

\(\ds \frac {a \times 0} {b \times k}\)

\(\ds \)

\(=\)

\(\ds \frac 0 {b \times k}\)

\(\ds \)

\(=\)

\(\ds \frac {0 \times b} {k \times b}\)

\(\ds \)

\(=\)

\(\ds \frac 0 k\)

Equal Elements of Field of Quotients

Hence we define the zero of $\struct {\Q, , \times}$ as $0$ and identify it with the set of all elements of $\Q$ of the form $\dfrac 0 k$ where $k \in \Z^*$.

$\blacksquare$