Is it true, that $(a:b) \cdot 2=a:(b:2)$, when b is even? I was doing my maths homework and I found that $(a:b) \cdot 2=a:(b:2)$ when b is even!I tried to improve it with examples, but I am not sure if it is true.Can you put it to the test and tell me if I have found a formula!
 A: It is in fact true for all $a,b$ but if $b$ is odd, then you will get a non-integer in the denominator. In fact a more general statement is true: 

$$a :(b:c)=(a:b)\times c$$

To see this, note that I can multiply with $c : c$, which is just 1.
$$a :(b:c) \times c:c =(a:b)\times c$$
But as you probably know, we do denominator times denominator and numerator times numerator when multiplying fractions. So we get
$$(a\times c) :(b:c \times c) =(a:b)\times c$$
$$(a\times c) :b =(a:b)\times c$$
$$(a:b)\times c =(a:b)\times c$$
which is clearly true. When you are formally writing a proof, you should write this form the end to the beginning, but I think this is easier for you to understand. 
A: That is true for any number and can be proven using the associativity of the product (https://en.wikipedia.org/wiki/Associative_property). You just have to consider that dividing is the same as multiplying by the inverse of any number.
In your particular situation:
$$(a/b) \cdot 2=(a \cdot 1/b) \cdot 2 = a \cdot (2 \cdot 1/b) = a \cdot (2/b) = \frac{a}{b/2}$$
The only thing is that $b/2$ is an integer only when $b$ is even.
