Why do we invert then multiply when dividing fractions? I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?
 A: $\frac{x}{\frac{1}{2}}=\frac{x}{\frac{1}{2}}\cdot1=\frac{x}{\frac{1}{2}}\cdot\frac{2}{2}=\frac{x\cdot 2}{\frac{1}{2}\cdot2}=\frac{2x}{1}$
Generally, 
$\frac{a}{\frac{b}{c}}=\frac{a}{\frac{b}{c}}\cdot\frac{c}{c}=\frac{ac}{b}=a\cdot\frac{c}{b}$
A: Several answers focused on the fact that division is really just multiplication with the inverse, and that is perfectly fine.
Here's another aproach:
When you calculate what $\frac{a}{b}$ is, what are you really doing? You are asking yourself 

"How many pieces do I get if I divide $a$ into $b$ equal pieces?"

so you are asking yourself 

"by how much must I multiply $b$ to get $a$?"

For example, if you are calculating $\frac{10}2$, you want to know how many groups of $2$ you can make out of $10$ people, or, equivalently, what you need to multiply by $2$ to get $10$, and the answer is $5$ because $2\cdot 5=10$.

So, the  bottom line here is that $x=\frac{a}{b}$ is defined as the unique number that satisfies the equation $bx = a$.
Using this definition, you can now see what happens if $b\frac12$. Well, in that case, by how much must I multiply $\frac 12$ to get $a$? In other words, what is the solution to the equation $$\frac12 \cdot x = a?$$
You can easily see that the answer is obvious when you multiply the equation by $2$ to get $x=2a$.
A: Unfortunately, nobody tells you the definition of $x/y$ nowadays. (Did they ever?) The simplest definition is surely that if $x$ is a real number and $y$ is a non-zero real number, then $x/y$ is--by definition-- a mere shorthand for the expression $x y^{-1}.$ So whenever $y$ is non-zero, we have:
$$\frac{x}{y} = x/y = xy^{-1}$$
This viewpoint is very powerful; for example, your question now becomes very easy to answer. Observe that if $a$ is a real number and $c$ and $d$ are non-zero real numbers, then we may argue as follows:
$$\frac{a}{c/d} = \frac{a}{c d^{-1}} = a(cd^{-1})^{-1} = a(c^{-1}d) = adc^{-1} = \frac{ad}{c}$$
The above chain of reasoning makes use of the following rules about raising to the power of $-1$, which hold for all real numbers distinct from $0$:


*

*$(xy)^{-1} = x^{-1} y^{-1}$

*$(x^{-1})^{-1} = x$


Other important laws (that I didn't use) include:


*$x x^{-1} = 1$

*$1x = x$ (this last one holds even if $x$ equals $0$.)


In summary:

Theorem. For all real numbers $a$ and all non-zero real numbers $c$ and $d$,
  we have: $$\frac{a}{c/d} = \frac{ad}{c}$$

Applying this to your problem:
$$\frac{x}{1/2} = \frac{x \cdot 2}{1} = 2x$$
The take home message is that before trying to understand division, you should try to understand raising to the power of $-1$. The expressions $x/y$ and $$\frac{x}{y}$$ are best viewed as nifty notation for $x y^{-1}$, nothing more.
A: Suppose you have $2$ pies and $x$ people at a party each want a piece, so each  pie is cut into pieces, each a fraction $2/x$ of a pie. Half of the party leaves and takes their pieces with them, thus taking away a whole pie.  The remaining people are now sad and lonely, but they all still get the same size piece of pie, $1$ pie $/$  $(x/2)$ people.
A: Division can be defined as multiplication by the inverse. Explicitly, for any associative division algebra (including real numbers, rational numbers, complex numbers, quaternions, etc.) we can define division by
$$a \div b = \frac{a}{b} := ab^{-1}$$
for any $b \neq 0$. If the algebra is commutative (that is $ab = ba$ for any $a, b$), which the real numbers, rationals, and complex numbers are, then we must have
$$\frac{a}{b} \div \frac{c}{d}= ab^{-1}\div (cd^{-1})= ab^{-1}(cd^{-1})^{-1}=ab^{-1}(dc^{-1})=adb^{-1}c^{-1}=(ad)(bc)^{-1}= (ad) \div (bc) =\frac{ad}{bc}= \frac{a}{b} \cdot \frac{d}{c}.$$
Thus, when you divide fractions you "flip and multiply".
A: Recall that if $ABC$ and $A'B'C'$ are similar triangles, then the ratio of $BC$ to $AB$ is the same as $B'C'$ to $A'B'$.  If the length of $AB$ is $\frac{1}{2}$, and the length of $A'B'$ is $1$, then the length of $B'C'$ is twice the length of $BC$.
