Multiplication by \(-1\) gives the additive inverse
An Introduction to Real Analysis
2025-10-24
Show that \((-1) \cdot m = -m\).
Using distributivity and Proposition 1.2.1 from Katznelson and Katznelson (2024, 6), since \[\begin{align*} m + (-1) \cdot m &= 1 \cdot m + (-1) \cdot m\\[0.75em] &= \underbrace{(1 + (-1))}_{= 0} \cdot m\\[0.75em] &= 0 \cdot m\\[0.75em] &= 0, \end{align*}\] and additive inverses in \(\symbb{Z}\) are unique thanks to Exercise 1.2.3 of Katznelson and Katznelson (2024), it follows that \((-1) \cdot m = -m\).
Remark 1. Although Katznelson and Katznelson (2024) only ask for a proof that multiplication by \(-1\) gives the additive inverse in \(\symbb{Z}\), the proof works just as well in any ring \(R\). The proof in \(\symbb{Z}\) uses distributivity of multiplication over addition, that anything times zero is zero, and the uniqueness of additive inverses. All of these properties are available in any ring \(R\).
Distributivity is true by definition in any ring. Anything times zero equals zero in any ring, because if \(r, 0 \in R\) then \(r \cdot 0 = r \cdot (0 + 0) = r \cdot 0 + r \cdot 0\), so then \[\begin{align*} 0 &= r \cdot 0 + (-(r \cdot 0))\\[0.75em] &= (r \cdot 0 + r \cdot 0) + (-(r \cdot 0))\\[0.75em] &= r \cdot 0 + \underbrace{(r \cdot 0 + (-(r \cdot 0)))}_{= 0}\\[0.75em] &= r \cdot 0 + 0\\[0.75em] &= r \cdot 0. \end{align*}\] This is the same as the proof of Proposition 1.2.1 by Katznelson and Katznelson (2024, 6). Lastly, additive inverses are unique in any ring because the additive structure of a ring is a group, and it was proven in Exercise 1.2.3 of Katznelson and Katznelson (2024) that inverse elements in any group are unique. \(\blacksquare\)