How can we prove that $\frac{a}{b }\times\frac{c}{d} =\frac{ac}{bd}$

Question

I am slowly reading calculus by michael spivak and it is one of the problems in first chapter. however I cant prove it please help me with it...

Answer 1

To see that equality holds, we will write the elements $\frac1b$ and $\frac1d$ as $b^{-1}$ and $d^{-1}$, respectively. Thus

\begin{align*} \frac ab\cdot\frac cd&=(a\cdot b^{-1})\cdot(c\cdot d^{-1})\\ &=(a\cdot c)\cdot(b^{-1}\cdot d^{-1})\\ &=(a\cdot c)\cdot(b\cdot d)^{-1}\\ &=\frac{ac}{bd}. \end{align*}

The second line uses the associative and commutative properties of multiplication.

Answer 2

Proving something like this, something that many people are often taught as the definition of multiplying fractions, is a difficult task because you need to first take the step of specifying exactly what you know and are assuming, separating the definitions from the things you don't want to assume.

Let us take as a given that $a/b=a \cdot \frac{1}{b}$, so that we can convert the problem to something only involving recipricals and multiplication, namely,

$$ a \cdot c \cdot \frac{1}{b}\cdot \frac{1}{d} = ac \frac{1}{bd}. $$

Since, given any equation, we can always multiply both sides by ac, it suffices to show that

$$ \frac{1}{b} \cdot \frac{1}{d}=\frac{1}{bd}.$$

We can do this by multiplying both sides by $bd$ (and using the fact that, by definition, $x \frac{1}{x}=1$ for all nonzero $x$) and rearranging, using the fact that multiplication is commutative.

Explicitly,

$$1=\left(b\cdot \frac{1}{b}\right)\left(d\cdot \frac{1}{d}\right)=bd\frac{1}{b}\frac{1}{d}$$

and so dividing both sides by $bd$ (or multiplying both sides by $1/bd$), we get

$$ \left(\frac{1}{b}\cdot \frac{1}{d}\right)=\frac{1}{bd}$$.

But for the above to be considered a proof, you have to accept that $a/b=a \frac{1}{b}$ and $b\frac{1}{b}=b/b=1$. If you do not accept that as a starting point, what I have done is not enough.