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let $0\neq a\in \mathbb{F}$ Prove: $$(a^{-1})^{-1}=a$$

I have tried to start with the definition of an inverse, For all $0\neq a\in \mathbb{F}$ we have:

$$a\cdot a^{-1}=a^{-1}\cdot a=1$$ But could not find a way to continue (tried multiplying by $(a^{-1})^{-1}$

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  • $\begingroup$ What is $\mathbb{F}$, a ring, a field, or still something else? You'll start understanding mathematics when you always start with a definition. $\endgroup$
    – user436658
    Commented Dec 23, 2017 at 18:24
  • $\begingroup$ @ProfessorVector it is a field $\endgroup$
    – newhere
    Commented Dec 23, 2017 at 18:25
  • $\begingroup$ Definitions belong into the question, not into some comment. $\endgroup$
    – user436658
    Commented Dec 23, 2017 at 18:26
  • $\begingroup$ I know that the assumption of being in a field is common, but this can be weakened to a division ring, where multiplication is not necessarily commutative. $\endgroup$ Commented Dec 23, 2017 at 18:29
  • $\begingroup$ @newhere Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Commented Mar 27, 2018 at 16:20

5 Answers 5

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

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  • $\begingroup$ $$(a^{-1})^{-1}(a^{-1})=1\implies (a^{-1})^{-1}(a^{-1})a=(a^{-1})^{-1}a=a$$ $\endgroup$ Commented Dec 23, 2017 at 18:26
  • $\begingroup$ @GNUSupporter yes! I fixed the typo meanwhile! thanks $\endgroup$
    – user
    Commented Dec 23, 2017 at 18:27
  • $\begingroup$ But how can I know that $(a^{-1})^{-1}$ is the inverse of $a^{-1}$? $\endgroup$
    – newhere
    Commented Dec 23, 2017 at 18:27
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    $\begingroup$ Is that not what the inverse notation means? They always exist, are unique, so that's how we denote it. $\endgroup$ Commented Dec 23, 2017 at 18:27
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By definition of the inverse the identity $a \cdot a^{-1}=a^{-1} \cdot a=1$ implies that $a$ is the inverse of $a^{-1}$.

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\begin{align} a&=1\cdot a\tag{definition of 1}\\ &=((a^{-1})^{-1}\cdot (a^{-1}))\cdot a\tag{definition of inverse}\\ &=(a^{-1})^{-1}\cdot ((a^{-1})\cdot a)\tag{associativity}\\ &=(a^{-1})^{-1}\cdot 1\tag{definition of inverse}\\ &=(a^{-1})^{-1}\tag{definition of 1} \end{align}

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There are several correct answers here. I will try another, using more words, in hopes of making things clearer.

By definition, the inverse of an element $a$ is the thing you multiply $a$ by to get $1$. Suppose the inverse of $a$ is $b$. That means $ab = 1$. So $a$ is what you multiply $b$ by to get $1$ so $a$ is the inverse of $b$. Inverses come in pairs, e.g. $2$ and $1/2$ in the rational numbers.

Now if you choose to name the inverse of $x$ with the notation $x^{-1}$ then it follows from the previous statement that $(a^{-1})^{-1}= b^{-1} = a$.

Since you are working in a field, you do not need to worry about whether an inverse exists or is unique or works with multiplication in either order. Those complications come up in other algebraic structures.

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Since a has an inverse then we have $$a•a^{-1}=a^{-1}•a=1$$ Then we just replace a by b to get $$ b•a^{-1}=a^{-1}•b=1$$ Which gives us that b is actually the inverse of $a^{-1}$

Thus: $$ b={a^{-1}}^{-1}$$ Hence done. Hope it helps

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