Basic Field Properties: multiplication

I am struggling with the proofs:

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

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

I have done the rest of the theorem but it is just these two that are difficult. To prove them you can only use the axioms of multiplication: Associative, Commutative, One is a real and Multiplicative inverse.

If anyone can help me out that would be much appreciated.

• What definition of exponentiation are you using? Feb 22 '14 at 7:01
• we are only given that: for each a within R with a not equal to 0 there is a^-1 such that a.a^-1 =1 Feb 22 '14 at 7:02
• Have you tried to prove that the multiplicative inverse is unique? Feb 22 '14 at 7:04
• Let $e\in \mathbb{F}$ and $f,g$ be their multiplicative inverses. Then $f=f1=f(eg)=(fe)g=g$, then $f=g$ as desired. Now since $a^{-1}a=1$ and $(a^{-1})^{-1}(a^{-1})=1$, then $(a^{-1})^{-1}$ is the multiplicative inverse of $a^{-1}$ and by uniqueness it follows that $a=(a^{-1})^{-1}$ Feb 22 '14 at 7:11

a) In order to prove it using the axioms of multiplication first we use the axiom that states that there exists $1\in \mathbb R$ such that for every $x\in \mathbb R$ it holds that $x*1=x$ ,so we have:

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

then we use the axiom that states that for ever $x\in \mathbb R$ there exists $(x^{-1})$ such that $x*(x^{-1})=1$,:

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

Now we use the associative law,:

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

a) $a^{-1} \cdot a = 1 \Rightarrow \left(a^{-1}\right)^{-1} = a$. This follows from the definition of the multiplicative inverse.

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

Then $ab = 1$, so $(-a)(-b) = ab = 1$.

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

• Note that $(-a)(-b)=ab$ is not something that can be assumed without proof (it is provable from the axioms and definitions). Feb 22 '14 at 7:09
• Right. I skip it, but can provide details. Feb 22 '14 at 7:10
• Your proof of (a) is incomplete if you don't use uniqueness ...and in fact also that of (b). Feb 22 '14 at 10:27