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.

  • $\begingroup$ What definition of exponentiation are you using? $\endgroup$ Feb 22 '14 at 7:01
  • $\begingroup$ 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 $\endgroup$ Feb 22 '14 at 7:02
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    $\begingroup$ Have you tried to prove that the multiplicative inverse is unique? $\endgroup$ Feb 22 '14 at 7:04
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    $\begingroup$ 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 $a=(a^{-1})^{-1}$ $\endgroup$ 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:


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


Now we use the associative law,:



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

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

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