Prove that $(-a).b=-(a.b)$ , $(-a)^{-1}=-(a^{-1})$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.

Question

Prove that

1)$\forall a, b \in \mathbb{R}$, $(-a).b=-(a.b)$.

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

$=(-a).b+[a.b+(-(a.b))]$

$=[(-a).b+a.b]+(-(a.b))$

$=[((-a)+a)b]+(-(a.b))$

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

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

$=-(a.b)$

2) $\forall a\in \mathbb{R}\setminus \lbrace 0\rbrace, (-a)^{-1}=-(a^{-1})$.

$(-a)^{-1}=(-a)^{-1}.1$

$=(-a)^{-1}.[(-a).(-a^{-1})]$

$=[(-a)^{-1}.(-a)].(-a^{-1})$

$=1.(-a^{-1})$

$=-a^{-1}$

3) If $a\neq 0$ and $b \neq 0$, then $ab \neq 0$ and $(a.b)^{-1}=a^{-1}.b^{-1}$.

$(a.b)^{-1}=(a.b)^{-1}.1$

$=(a.b)^{-1}.[a.b.b^{-1}.a^{-1}]$

$=(a.b)^{-1}.[(a.b).(b^{-1}.a^{-1})]$

$=[(a.b)^{-1}.(a.b)].(b^{-1}.a^{-1})$

$=1.(b^{-1}.a^{-1})$

$=b^{-1}.a^{-1}$

$=a^{-1}.b^{-1}$

Is that true please? Thanks.

I used these axioms:

Answer 1

Your proofs are mostly good.

But before you can prove 1) you must prove:

$0\cdot b = 0$ for all $b$.

Pf: $0\cdot b = (0 + 0)\cdot b$ (as $0 = 0 + 0$)$

$= 0\cdot b + 0\cdot b$ (via distribution)

So $0 = 0\cdot b + (-0\cdot b) $ (as every number has an additive inverse)

$= (0\cdot b + 0\cdot b) + (-0\cdot b) $ (by substitution)

$= 0\cdot b + (0\cdot b + (-0\cdot b))$ (associativity)

$= 0\cdot b + 0$ (definition of inverse)

$= 0\cdot b$ (definition of 0)$

In your prove of 2) you substituted $1$ with $(-a)(-a^{-1})$ which I don't think is justified without proof.

But you have proven in 1) that $(-a)(-a^{-1}) = -(a(-a^{-1}))$ and by commutativity $-(a(-a^{-1})) = -((-a^{-1})a)$ and by 1 again $= -(-(a^{-1}a)) = -(-1)$.

But we still need to prove that $-(-a) = a$ which can be done as

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

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

$-(-a) + 0 = a$

$-(-a) = a$.

Your 3) is just fine

Answer 2

Why not: $(-a)b + ab = (-a + a)b= 0\cdot b = 0$ (assuming you know the latter, which si not an axiom but a consequence of them, hopefully shown before), so that $ab$ is the additive inverse of $(-a)b$ and as inverses are unique in any group we have $(-a)b = -(ab)$.

For the second $a\cdot a^{-1}=1$ by definition. Assuming you have shown $(-x)\cdot(-y)=xy$ we write $(-a) \cdot -(a^{-1})=1$ and so $-(a^{-1})$ is the multiplicative inverse of $-a$ so by unicity again $(-a)^{-1} = -(a^{-1})$.

For 3, $(ab)(a^{-1}b^{-1}) =1 $ by applying commutativity and the definition of inverse twice. So again by unicity of multiplicative inverses: $(ab)^{-1} = a^{-1} \cdot b^{-1}$.

So as a basic algebra lemma for all of this: if $(G, \ast,^{-1}, e)$ is a group, inverses are unique: if $a \ast b=e$ then $b=a^{-1}$, where $a^{-1}$ is promised by the group axioms. In a field $F$, $(F, + ,-,0)$ forms a group and so does $(F\setminus \{0\},\cdot,^{-1}, 1)$, so we can apply this.