Why any rational number can be written as $P/Q$? Why any rational number can be written as $P/Q$?
and how to prove that where $p,q$ are integers and at least one of $p,q$ is odd?
I can see that at least one of $p,q$ is odd, but I don't how to write down the proof properly.
I did is:
Let $x$ be a rational number.
Case 1: $x$ is integer
then $x = x/1$ where $1$ is an odd number.
Case 2: $x$ is not integer
write $x = m/n$
if $m,n$ are both even, then the $2$ cancels out, and repeat the case $2$ process until at least one of the two numbers is odd.
 A: By the fundamental theorem of arithmetic (uniqueness and existence of decomposition into prime factors), any integer $N\ne0$ can be written as
$$
N=2^an
$$
where $n$ is odd (the exponent $a$ is a non negative integer and it can be $0$, precisely when $N$ is already odd). The case $P/Q$ where $P=0$ is easy: $P/Q=0/1$. So we can assume also $P\ne0$.
Now we have $P=2^ap$ and $Q=2^bq$, with $p$ and $q$ odd, three cases are possible:


*

*$a>b$ and $\dfrac{P}{Q}=\dfrac{2^ap}{2^bq}=\dfrac{2^{a-b}p}{q}$ where the denominator is odd;

*$a=b$ and $\dfrac{P}{Q}=\dfrac{2^ap}{2^bq}=\dfrac{p}{q}$ where both the numerator and the denominator are odd;

*$a<b$ and $\dfrac{P}{Q}=\dfrac{2^ap}{2^bq}=\dfrac{p}{2^{b-a}q}$ where the numerator is odd.
This is just a better formalized version of your argument: the key is that you can't keep dividing by $2$ ad infinitum, so the process eventually stops.
A: Because a rational number can be expressed as a ratio, that is to say, any rational number can be represented $\frac{p}{q}$.
One of $p,q$ must be odd, because if they were both even, they wouldn't be coprime, and therefor, not simplified. (Assuming $q\ne 0$, but if $q=0$, it's already not a rational number.)
