Rational Addition Identity is Zero
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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 $k \in \Z^*$.
$\blacksquare$