Using only the field axioms of real numbers, prove that $-x = (-1)x$

Question

Using only the field axioms of real numbers, prove that $-x = (-1)x$
Ths is how I attempted to solve this problem: $$1+(-1)=0 \iff x(1+(-1))=0\cdot x \iff x+(-1)x=0\iff(-1)x=-x$$
However, I am not entirely sure whether the operations I've performed are in accordance with the axioms.

Answer 1

You don't need equivalences. You were on the right track (although shorter methods are possible). I have typed out every step that's missing in your proof, but a lot can be made shorter.

Let $x \in \mathbb{R}$

Then: $1+(-1) = 0$

$\implies x.(1+(-1)) = x.0$ [multiplication is a well defined function $\mathbb{R} \times \mathbb{R} \to \mathbb{R}]$

$\implies x.(1+(-1)) = 0$ [$0.x = 0$]

$\implies x.1 + x.(-1) = 0$ [distributivity]

$\implies x + x.(-1) = 0$ [$x.1 = x$, 1 is the neutral element for multiplication]

$\implies (-x)+(x + x.(-1)) = 0 + (-x)$ [addition is a well defined function, the inverse for the addition exists]

$\implies (-x)+(x + x.(-1)) = -x$ [$0$ neutral element for addition]

$\implies ((-x)+x) + x.(-1) = -x$ [associativity]

$\implies 0 + x.(-1) = -x$ [$(-x) + x = 0]$

$\implies x.(-1) = -x$ [$0$ neutral element for addition]

$\implies (-1).x = -x$ [multiplication is commutative]

The result follows, as $x$ was chosen arbitrarily.