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

Question

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

The teacher gave us a proof but I would like to see another :)

Proof by contradiction

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

Is my proof correct?

I already have the proof of $0a=0$

$(1)a + (-1)a = a(1-1)=0$ by associativity. Is this ok? – Maffred Jan 29 '16 at 05:07 — Maffred, Jan 29 '16 at 05:07
it is not by associativity, it is by Distributivity – Jose Vega Jan 30 '16 at 22:58 — Jose Vega, Jan 30 '16 at 22:58

score 1 · Accepted Answer · answered Jan 29 '16 at 04:51

1

This proof looks okay, but the main thing to be concerned about at this level is circularity, and we can't really check that. If you're sure you don't need this result to prove the used results, then you're good.

answered Jan 29 '16 at 04:51

Stella Biderman

31,155

I was actually just thinking the same thing. – M10687 Jan 29 '16 at 04:53

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

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