Prove that $(−1_R)a = −a$ (edited with new attempt) Let R be a ring with unity $1_{R} ∈ R$
Prove that $(−1_R)a = −a$.
This is done like this: 
$(−1_R)a + a = (−1_R)a + 1_Ra = ((−1_R) + 1_R)a = 0_R*a = 0_R$. In short, we have,
$(−1_R)a + a = 0_R$, which implies $(−1_R)a = −a$ as required.
Maybe this is a very trivial question, but what is the purpose of proving it in this way? I mean what are we proving here, and why did we introduce this part $(−1_R)a + a = (−1_R)a + 1_Ra$ here? Please let me know if my question makes any sense to you. I just need to understand the basics to do this types of proofs. Thanks.
New attempt: is this correct: 
$(-1_R)a = 0 + (-1_R)a = ((-a)+a)+(-1_R)a = ((-a)+1_R*a)+(-1_R)a = (-a)+(1_R*a +(-1_R)a) = -a$
 A: Your question makes sense. The reason is the following: we want to prove something using just ring definition. So, we can't introduce anything. We are using distributive law(which is a ring's definition) instead.
A: First, what are we proving here? The claim is that these two operations on an element always have the same result:


*

*Take the additive inverse of the element.

*Multiply the element by a particular constant.


Since these are quite different operations (on the face of it, one involves only the ring's addition operation and the other involves only its multiplication operation), it is not obvious that they have to agree in every ring under the sun, just because it satisfies the ring axioms.
For the particular question about why to rewrite $a$ as $1_R\cdot a$, this rewriting is necessary to allow you to use the distributive law in the next step.
In all subsequent practical work with rings, you're supposed to be able to imagine such simple reinterpretations without having them written down explicitly, but at this very elementary stage in the development, part of the point is to make you aware that they're happening -- e.g., the axiom that $1_R\cdot a = a$ is actually necessary for the conclusion, and this is where you're using it.
A: The first argument is shorter:
$$
(-1_R)a+a=(-1_R)a+1_Ra=(-1_R+1_R)a=0a=0,
$$
provided you have already shown $0a=0$, which is done by
$$
0a=(0a+0a)=0a+0a
$$
and, adding $-(0a)$ to both sides gives $0=0a$.
The second argument is correct too, but much longer
$$
(-1_R)a=0+(-1_R)a=(-a+a)+(-1_R)a=-a+(a+(-1_R)a)=\\
-a+(1_Ra+(-1_R)a)=-a+(1_R+(-1_R))a=-a+0a=-a+0=-a
$$
and basically uses the same argument as before.

What's the purpose of considering $(-1_R)a+a$? You prove this is zero, so the first term is the negative of the second term. Easier than doing a lengthy computation.
