division by fraction proof I was trying to figure out the property of dividing by fraction : $\frac{x}{\frac{a}{b}} = \frac{x\cdot b}{a}$
An other representation for this problem is to show $X\div (A\div B) = (X\cdot A)\div B$.
A simple proof, is by multiplying and dividing by $\frac{b}{a}$ which leads to the wished result of $x\cdot \frac{b}{a}$. 
My questions are, Is there a reason why this method of proving works? for me it seems like a "trick". A second question, do you have an other way to prove it? My searching so far led mostly to intuitive explanations about the concept of dividing. 
 A: The simplest approach is to multiply the whole fraction by $\frac b b$, which is the same as multiplying by $1$.
A: You can only use this "trick", because of a number of old assumptions that you're making. Namely, your elements a and b are real numbers, and the real numbers form a Field. In a field, I am guaranteed an inverse for every element, so 1/a and 1/b are what we call these inverses. If you wanted a more thorough proof, you would go back and show why you're guaranteed inverses in a Field. 
A: ${1\over {b\over a}} \cdot x={x\over{b\over a}}$ and ${a\over b}\cdot{b\over a} = 1$ the rest follows. 
A: After the integers one usually looks at ratios of integers, the fractions that form the rational numbers. The notation $\frac{m}{n}$ is too cool for words! You think about how may times the integer on the bottom (denominator) goes into the top (numerator), and you now know what division is.
The best part is even if you see nested fractions like
$\frac{x}{\frac{a}{b}}$
you can use all those same rules. There is only one way it could work, and it does work! So you just have to respect the 'main/first dividing line' (nesting) and instead of integers you can have real or complex numbers. Darn, you can even put in polynomials!
All you have to remember is $\frac{m}{n} = m n^{-1}$ and the laws of numbers gives you everything you need to derive any 'trick'. So if you want to add two fractional expressions, find a common denominator.
