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I am struggling to prove that if $a$ is a real number, then

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

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 I can only use the axioms of multiplication: multiplication is associative and commutative, "one" is a real number, and every non-zero real number has a multiplicative inverse.

Xander Henderson
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Bryce Ramgovind
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  • What definition of exponentiation are you using? – Zubin Mukerjee Feb 22 '14 at 07:01
  • 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 – Bryce Ramgovind Feb 22 '14 at 07:02
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    Have you tried to prove that the multiplicative inverse is unique? – Jose Antonio Feb 22 '14 at 07:04
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    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 $a=(a^{-1})^{-1}$ – Jose Antonio Feb 22 '14 at 07:11

2 Answers2

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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,:

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

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

Zubin Mukerjee
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DeepSea
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