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I have to prove that

$$-a = (-1)a$$

how I went about it was I added $0a$ to either side, and then used the distributive property to factor out the $(-1 + 0)$

1. $-a + 0a = (-1)a + 0a$

2. $(-1 + 0)a = (-1 + 0)a$

3. $a=a$

However I'm not sure if this is valid because in the step 2, in order to pull the $-$ sign out of $-a$, I had to basically assume what the proof is trying to prove, which is that $-a=(-1)a$.

Is is fine, or should I be approaching it a different way?

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    Consider that $a + (-1)a = 1a + (-1)a = (1 + -1)a = 0a = 0 $. – 2017-02-14
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    @gj255 So I can assume that $(a + -a) = 0$ even before confirming that $-a = (-1)a$? I guess my real question is "What does $-a$ even mean if it's not defined to be $(-1)a$? – 2017-02-14
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    $(-1)a$ is defined as the product of $(-1)$ and $a$, where $(-1)$ is the inverse of $1$ in the additive group of the ring. $-a$ is the additive inverse of $a$. It is not immediately obvious that those are the same, apart from some suggestive algebraic notation. Therefore it needs to be proven. – 2017-02-14
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    @Arthur ahhh I've never thought about it that way...I'm very new to this, but that makes a lot of sense now. – 2017-02-14
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    I think you should be approaching it a different way - in what way multiplication is "positive" and increasing at the same time? – 2017-02-14

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Indeed you have assumed the result you were trying to prove. Let's take a step back and ask what we mean by $-a$. This should be the number which, when added to $a$, yields zero. That is to say, we define the object $-a$ to be the additive inverse of $a$.

With this in mind, what we ought to consider is the sum $$ (-1)a + a \,,$$ since if this is zero, then we've demonstrated that indeed, $(-1)a$ acts as an additive inverse of $a$ (and by uniqueness it is the additive inverse of $a$). We have

$$a + (-1)a = 1a + (-1)a = (1 + -1)a = 0a = 0 \,, $$ as required.