How to deduce the "divide by a fraction" formula from the definition of division

Question

From the comments I got, my question amonts to : can the " inverse of inverse" law be derived from the definition of division.

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Division is defined as : $\dfrac AB = A.\dfrac1B$, that is,

" dividisng by A by B is, by definition, mutiplicating A by the inverse of B".

My question is:

How do I derive from this definition, the equality:

$\frac{a}{b/c}$ = $\frac{ac}{b}$

I tried this:

$\frac{a}{b/c}$

= $\frac{a}{b\times1/c}$ ( applying the definition of division to the denominator)

=$\frac{a\times1}{b\times1/b}$ ( using " 1 is the identity for multiplication")

= $\frac ab$$\times$$\frac{1}{1/c}$ ( using $\frac{ab}{cd}$ = $\frac{a\times b}{c\times d}$ in the reverse sense)

But could not go further.

How to recover $\frac {c}{1}$ from $\frac{1}{1/c}$ using exclusively the definition of division $\frac AB$ = A.$\frac1B$ ?

It seems to me I am moving in a circle, since apparently I would need the formula I want to prove to obtain the last equality I want.

Answer 1

\begin{align} \dfrac{1}{\left\{\dfrac CD \right\}} &= \dfrac{1}{\left\{\dfrac CD \right\}} \cdot 1 \\ &= \dfrac{1}{\left\{\dfrac CD \right\}} \cdot \dfrac DD \\ &= \dfrac{1 \cdot D}{\dfrac CD\cdot D } \\ &= \dfrac DC \end{align}

Therefore

$$\dfrac AB \div \dfrac CD = \dfrac AB \cdot \dfrac{1}{\left\{\dfrac CD \right\}} = \dfrac AB \cdot \dfrac DC = \dfrac{AD}{BC}$$

Answer 2

$1/(1/c)$

Here, according to the definition, we need to multiply 1 with the multiplicative inverse of (1/c)

Multiplicative inverse of a number is a number which when multiplied by the given number yields 1

So, here multiplicative inverse of (1/c) is c (you can check it by calculating),

Hence, $1/(1/c)$ = c

I would not say the comments to the question are less informative or different from my answer but I presented this answer in my own way to make it clear.

Answer 3

Just do it:

$\frac a {\frac bc}$ will by definition be $a\cdot \frac 1{\frac bc}$

So we need to figure out what $\frac 1{\frac bc}$ is.

And it is the value $k$ so that $\frac bc\cdot k = 1$.[1]

So we need to solve for $(\frac bc)k =(b\cdot \frac 1c)\cdot k = 1$.

We know that $(b\cdot \frac 1c)\cdot c = b\cdot(\frac 1c\cdot c)= b\cdot 1 = b$.

And so $(b\cdot \frac 1c)\cdot (c \cdot \frac 1b) = b\cdot (\frac 1c \cdot c)\cdot \frac 1b = b\cdot 1 \cdot \frac 1b = b\cdot \frac 1b = 1$.

So $\frac 1{\frac bc} = c\cdot \frac 1b = \frac cb$.

So $\frac a{\frac bc} = a\cdot \frac 1{\frac bc}=a\frac cb = a(c\cdot \frac 1b)=(ac)\frac 1b = \frac {ac}b$

......

[1]

This assumes that for any $m \ne 0$ that there exists a $k$ that $mk =1$ and that $k$ is unique.

That such a $k$ exists is the definition of $\mathbb Q$ is a field.

We can prove $k$ is unique.

If $k$ and $j$ are both inverses of $m$ then $mk =1=km $ and $mj = 1=jm$.

That means that $kmj =(km)j=1\cdot j = j$ but $kmj=k(mj) =k\cdot 1 =k$. SO $j = k$.

And inverses are unique.

We can use that to prove that $\frac 1{\frac 1c} = c$.

because $c \times \frac 1c = 1$ that means.... by definition that $c$ is the inverse of $\frac 1c$. SO $c = \frac 1{\frac 1c}$.