0
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
  • 1
    $\begingroup$ Have you tried to prove that the multiplicative inverse is unique? $\endgroup$ Feb 22 '14 at 7:04
  • 1
    $\begingroup$ 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
1
$\begingroup$

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,:

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.