Proof that $(-1)\cdot a=(-a)$

Question

I'm supposed to prove this. The answer key doesn't tell me anything more than to add zero:

$(a+(-a))$

to:

$(-1)\cdot a$

and use that:

$a=a\cdot 1$

I have these axioms to my disposal:

(A1) for all $x,y$ we have $+y=y+x$

(A2) for all $x,y,z$ we have $(x+y)+z=x+(x+z)$

(A3) there exists a number $0$ so that for all $x$ we have $x+0=x$

(A4) for all $x$ there exists a number $a$ so that $x+a=0$

(M1) for all $x,y$we have $x*y=y*x$

(M2) for all $x,y,z$ we have $(x*y)*z=x*(y*z)$

(M3) there exists a number $1=/=0$ so that for all $x$ we have $x(1/x)=1$

(M4) for all $x=/=0$ there exists a number $b$ so that $x(1/x)=1$

(AM) for all $x,y,z$ we have $z*(x+y)=z*x+z*y)$

$a ∈ R$

Answer 1

1. To prove: $(-1)\cdot a = (-a)$. In other words, that $-a$ (the opposite of $a$) is equal to the product of $-1$ (the opposite of the multiplicative unit) and $a$ itself.

2. Let's start with the expression $(a + (-a)) + (-1)\cdot a$

3. Because $a$ and $-a$ are opposites, by definition their sum is zero. Therefore, one way of evaluating this expression yields: $$(a + (-a)) + (-1)\cdot a = 0 + (-1)\cdot a = (-1)\cdot a$$ where we've used the fact that adding 0 to something doesn't change it.

4. But we can evaluate this expression another way, too. Because addition is commutative, we can change the order of terms that are added together. Because addition is associative, we can drop parentheses on things that are added together [as in $(a+b)+c = a+b+c$]. This allows us to write: $$(a + (-a)) + (-1)\cdot a = (-a) + a + (-1)\cdot a$$

5. We can use the fact that multiplying by 1 doesn't change anything. Therefore, $a = 1\cdot a$: $$(-a) + a + (-1)\cdot a = (-a) + (1 \cdot a) + (-1)\cdot a$$

6. Another important property that addition has is distribution over multiplication. This means that $(a+b)\cdot c = a\cdot c + b\cdot c$. We can apply that rule here to change $1\cdot a + (-1)\cdot a$ into $[1 + (-1)]\cdot a$.

   $$(-a) + (1 \cdot a) + (-1)\cdot a = (-a) + [1 + (-1)] \cdot a$$

7. What is $1 + (-1)$, and how can we prove it? Well, 1 is the multiplicative unit, and $(-1)$ by definition is its opposite. The sum of any number and its opposite is zero. Also, multiplying anything by zero yields zero itself, and adding zero to anything doesn't change it. Therefore, we can write: $$(-a) + [1 + (-1)] \cdot a = (-a) + 0\cdot a = (-a)+ 0 = -a$$

8. Therefore, another way of evaluating this expression yields $-a$.

9. Because one way of evaluating yields $(-1)\cdot a$ and another way yields $-a$, it follows that $(-1)\cdot a = a$.

Answer 2

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

you also have $(-1).a+a+(-a)=(-1).a+(a+(-a))=(-1).a+0=(-1).a$.