Proof of $ \forall a \in \Bbb{R}: -a = (-1) a $.

Question

Problem. Prove that $ \forall a \in \Bbb{R}: -a = (-1) a $.

I already have a proof but I would like to see another. :)

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Proof by contradiction

Suppose that $ -a \neq (-1) a $. Then \begin{align} -a & < (-1) a, \\ -a + a & < (-1) a + a, \\ -a + a & < (-1) a + (1) a, \\ 0 & < ((-1) + 1) a, \\ 0 & < (0) a, \\ 0 & < 0. \quad (\text{Contradiction!!!}) \end{align} Therefore, $ \forall a \in \Bbb{R}: -a = (-1) a $. $ \quad \blacksquare $

Is my proof correct?

Answer 1

Well, you haven't considered the case $-a > (-1)a$...

But why not run your proof in reverse:

$0 = 0\\0 = a0\\0 = a(-1 + 1)\\-a + a = a(-1) + a(1)$

$-a + a = a(-1) + a\\(-a + a) + (-a) = (a(-1) + a) + (-a)\\-a +(a + (-a)) = a(-1) + (a + (-a))$

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

...?

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As you can see (and is remarked in the comments above), this depends on the fact that $a0 = 0$.

Let's prove this:

$a0 = a(0+0) = a0 + a0\\a0 + (-(a0)) = (a0 + a0) + (-(a0))\\0 = a0 + (a0 + (-(a0)))$

$0 = a0 + 0 = a0$.