How to prove " multiplicative inverse of inverse of $a$ is $a$ itself" ( with $a$, say, a real number) in basic arithmetics? ( $\frac{1}{1/a}$ = $a$.)

Question

Suppose I want to treat basic arithmetics on real numbers as a little deductive system ( without using abstract algebra).

In order to prove the " divide by a fraction " rule ( $\frac{a}{b/c}$ = $\frac{ac}{b}$), I need the " inverse of inverse rule" ( How to deduce the "divide by a fraction" formula from the definition of division), namely :

$\frac{1} {1/a}$ = $a$ ( provided a is not equal to 0).

How can this rule be proved without using the " divide by a fraction" rule?

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I have done this :

Assuming

• for all $a$, $\frac aa$$=$$1$ ( provided $a$ is not null)

• for all $a$, $b$, $\frac ab$$=$$a\times\frac 1b$ ( provided $b$ is not null)

• number $1$ is the identity element for multiplication and for division.

• for all $a,b,c,d$, $\frac {ac}{bd}$ $=$ $\frac ab\times\frac cd$ ( with $c,d$ not equal to $0$).

$\frac{1}{\frac 1a}$= $\frac{\frac aa}{\frac 1a}$= $\frac{\frac a1\times\frac 1a} {1\times\frac 1a}$= $\frac {\frac a1}{1}\times\frac{\frac 1a}{\frac 1a}$= $\frac a1\times1$= $a\times1$= $a$

provided $a$ is not null.

Answer 1

Suppose $a \ne 0$. Then there exists a real number, $\dfrac 1a$, such that $$a \cdot \dfrac 1a = \dfrac 1a \cdot a = 1$$ Since $\dfrac 1a = 0$ would lead to the contradiction $1 = a \cdot\dfrac 1a = 0$, we must have $\dfrac 1a \ne 0$. So there exists a real number $b$ such that $\dfrac 1a \cdot b = b \cdot \dfrac 1a = 1$ and, by definition, $b = \dfrac{1}{1/a}$. So $\dfrac 1a \cdot a = 1 = \dfrac 1a \cdot \dfrac{1}{1/a}$. Hence

$$\dfrac 1a \cdot a = \dfrac 1a \cdot \dfrac{1}{1/a}$$

Multiplying both sides, on the left, by $a$ and simplifying results in $a = \dfrac{1}{1/a}$.

Answer 2

Multiply both sides by the inverse of $a$. You know from the definition of inverse that $$a\cdot\frac 1a=\frac1a \cdot a=1$$ Then the right hand side is $1$ On the left hand side use that the inverse of $a$ is $b$. then you have $$\frac 1{1/a}\cdot \frac 1a=\frac 1b\cdot b=1$$ Subtract first term and last term from the two lines and you get $$\left(\frac1{1/a}-a\right)\frac 1a=0$$ We multiply both sides by $a$, then move $a$ to the right hand side.

The only thing that I used are associativity, commutativity, definition of inverse, $a\ne 0$ and $1/a\ne 0$.

Answer 3

Let $b$ denote the multiplicative inverse of $a$, so $xb=bx=1$ with $x=a$. But the first of these equations in $x$ is the definition of the multiplicative inverse of $b$.