Is \(\frac{a/b}{c/d} = \frac{ad}{bc}\) Always True in Mathematics?

[embphysics]

Homework Statement

The statement that is purported to be true is $\frac{a/b}{c/d} = \frac{ad}{bc}$

Homework Equations

The Attempt at a Solution

So, I am going along with my proof, and I believe it to be going nicely. However, there is one step that I am unsure of:

$\frac{\frac{a}{b} d}{c} = \frac{\frac{a}{b} d}{c} \cdot d \cdot d^{-1} \Rightarrow \frac{\frac{a}{b} d \cdot d^{-1}}{c d^{-1}} = \frac{\frac{a}{b}}{c d^{-1}}$. Now, what I want to do is $\frac{\frac{a}{b}}{c d^{-1}} = \frac{\frac{a}{b}}{\frac{c}{d}}$. But I am having trouble justifying the step $\frac{1}{d^{-1}} = <br /> \frac{1}{\frac{1}{d}}$

 

[tiny-tim]

hi embphysics!

i'm not sure what formulas you're allowed to use

anyway, why not just multiply the RHS, ad/bc, by the bottom of the LHS, c/d ?

 

[embphysics]

Well, the formulas and properties I am permitted to use are given in the first chapter of Spivak's Calculus. So, tiny-tim, I am not certain that that manipulation is defined.

 

[Ray Vickson]

Homework Statement

The statement that is purported to be true is $\frac{a/b}{c/d} = \frac{ad}{bc}$

Homework Equations

The Attempt at a Solution

So, I am going along with my proof, and I believe it to be going nicely. However, there is one step that I am unsure of:

$\frac{\frac{a}{b} d}{c} = \frac{\frac{a}{b} d}{c} \cdot d \cdot d^{-1} \Rightarrow \frac{\frac{a}{b} d \cdot d^{-1}}{c d^{-1}} = \frac{\frac{a}{b}}{c d^{-1}}$. Now, what I want to do is $\frac{\frac{a}{b}}{c d^{-1}} = \frac{\frac{a}{b}}{\frac{c}{d}}$. But I am having trouble justifying the step $\frac{1}{d^{-1}} = <br /> \frac{1}{\frac{1}{d}}$

Why not just use the definition? $A/B$ is that number $X$ which, when multiplied by $B$, gives you $A$; that is, it is the solution of the equation $BX = A$.