Zero is unique
An Introduction to Real Analysis
2025-10-24
Show that if \(n + m = m\), then \(n = 0\). This proves that the \(0\) element in \(\symbb{Z}\) is unique.
Since \[\begin{align*} n + m &= m\\[0.75em] \implies (n + m) + (-m) &= m + (-m)\\[0.75em] \implies n + \underbrace{(m + -m)}_{= 0} &= \underbrace{m + (-m)}_{= 0}\\[0.75em] \implies \underbrace{n + 0}_{= n} &= 0\\[0.75em] \implies n &= 0, \end{align*}\] \(0 \in \symbb{Z}\) is unique.
Remark 1. Although Katznelson and Katznelson (2024) only ask for a proof that the additive identity in the integers is unique, the proof works just as well for the identity element in any group \(G\). The proof only uses the existence of the inverse element \(-m\), associativity of the binary operation \(+\), and the definition of additive identity when concluding that \(n + 0 = n\). All of these properties are available in any group. \(\blacksquare\)