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 [itex]\frac{a/b}{c/d} = \frac{ad}{bc}[/itex]

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:

[itex]\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}}[/itex]. Now, what I want to do is [itex]\frac{\frac{a}{b}}{c d^{-1}} = \frac{\frac{a}{b}}{\frac{c}{d}}[/itex]. But I am having trouble justifying the step [itex]\frac{1}{d^{-1}} =
\frac{1}{\frac{1}{d}}[/itex]

 

[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 [itex]\frac{a/b}{c/d} = \frac{ad}{bc}[/itex]

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:

[itex]\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}}[/itex]. Now, what I want to do is [itex]\frac{\frac{a}{b}}{c d^{-1}} = \frac{\frac{a}{b}}{\frac{c}{d}}[/itex]. But I am having trouble justifying the step [itex]\frac{1}{d^{-1}} =
\frac{1}{\frac{1}{d}}[/itex]

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##.