Dividing fraction trick, why does it work?

[Euler]

I am going to assume everyone was taught the method where if you want to divide to fractions then you simply flip the second and then multiply them. I've been trying to see if I could find a reason as to why it works but I can't seem to. Does anyone know why or want to give me some hints?

Thanks for any help you can offer!

 

[adjacent]

I am going to assume everyone was taught the method where if you want to divide to fractions then you simply flip the second and then multiply them. I've been trying to see if I could find a reason as to why it works but I can't seem to. Does anyone know why or want to give me some hints?

Thanks for any help you can offer!

Multiplication is the reverse of division.
If you turn the fraction upside down, then you will have to change the operation.(Multiplication to division and vise versa)

 

[D H]

$\frac a b$ is defined as $a \cdot \frac 1 b$. The latter term is the multiplicative inverse. The multiplicative inverse of a rational $\frac p q$ is $\frac q p$.

 

[Euler]

$\frac a b$ is defined as $a \cdot \frac 1 b$. The latter term is the multiplicative inverse. The multiplicative inverse of a rational $\frac p q$ is $\frac q p$.

Thank-you, that is very helpful.

 

[gopher_p]

A reasonable mathematical definition of division for real numbers is $$a\div b=c\text{ if and only if }a=c\times b.$$
The multiplicative inverse of a real number $b$ is the unique real number, denoted $b^{-1}$ or $\frac{1}{b}$, such that $b\times b^{-1}=1$. We see that $$a\times b^{-1}=(c\times b)\times b^{-1}=c\times( b\times b^{-1})=c\times 1=c$$, and so $$a\times b^{-1}=c.$$

Since both $$a\div b=c\text{ and } a\times b^{-1}=c,$$ we get that $$a\div b=a\times b^{-1},$$ and division by $b$ is the same as multiplication by the multiplicative inverse of $b$. Because of the way that multiplication of rational expressions is defined, the multiplicative inverse of a rational expression is just that expression "flipped".