Proving that $-a=(-1)\cdot a$

abstract-algebra field-theory

$$a+(-1)a=(1)a+(-1)a=(1+(-1))a=0\cdot a=0$$Hence by definition we have that $-a=(-1)a$

Added: We also want to show that the additive inverse of $a$ has no ambiguities, i.e., if $x$ and $y$ are the additive inverses of $a$ then $x=y$ and hence we can just denote this element by $-a$. Proof: $$x=x+0=x+(a+y)=(x+a)+y=0+y=y$$


Hint $\ $ Both are inverses of $\rm\:a\:$ so they are equal by uniqueness of inverses, viz.

$$\rm \begin{array}{} a'+a\ =\: 0\\ \rm \: a+a'' =\:\! 0\end{array}\quad\Rightarrow\quad a'\: =\ a' + (a + a'')\ =\ (a' + a) + a'' =\: a''$$ Using that, it suffices to prove that $\rm\:\ (-1)\:a + a = 0,\ $ i.e. $\rm\:(-1)\:a\:$ is an inverse of $\rm\:a\:.$ This follows easily by the ring axioms (hint: scale $\rm\: -1\: +\: 1\: =\: 0\:$ by $\rm\:a\:$).

It is useful to abstract out the lemma on uniqueness of inverses since it is ubiquitous in algebra.

Also, as I often emphasize, uniqueness theorems provide powerful tools for proving equalities.

Related

challenging alternating infinite series involving $\ln$
Odd order moments of a symmetrical distribution
Importance of Least Upper Bound Property of $\mathbb{R}$
How do you prove the $p$-norm is not a norm in $\mathbb R^n$ when $0<p<1$?
Does $abab=baba$ imply commutativity in a Group of uneven order?
Intuition for cross product of vector with itself and vector with zero vector
What is the main use of Lie brackets in the Lie algebra of a Lie group?
Can we get un-obvious results by defining sophisticated topologies?
Cyrillic alphabet in math
Showing that $\sum_{k=0}^{\infty} a^{k} \cos(kx) = \frac{1- a \cos x}{1-2a \cos x + a^{2}}$ without using complex variables
Can asymptotes be curved?
Simplification of an expression containing $\operatorname{Li}_3(x)$ terms