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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.

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    It is by definition written $p\over q$. – 2017-01-10
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    If $\dfrac p q$ with both even than $\dfrac p q = \dfrac {2 p_1}{2q_1}=\dfrac {p_1}{q_1}$... – 2017-01-10
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    If $q$ is the *least* denominator then one of $\,p,q\,$ is odd, else you can cancel $2$ from both, contra leastness. That's one way to *formalize* the argument using induction (in well-order form, or least number principal). Similarly for any common divisor, so least terms implies the numerator and denominator are coprime. – 2017-01-10
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    By *definition* a rational number is one that can be written as a ratio. And your proof is just fine, though you said 'even' at the end that should be 'odd', which I corrected. – 2017-01-10

2 Answers 2

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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:

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

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

  3. $a

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.

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    The first sentence only holds for $N \ne 0$. Dividing by $2$ could occur ad infinitum if it were only up to the numerator, so it's a good thing the denominator is never $0$ :). – 2017-01-10
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    @ErickWong Yes, I'll add the assumption that $N\ne0$. – 2017-01-10
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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.)

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    That proof is already in the question. – 2017-01-10
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    @BillDubuque reworeded it, hopefully different enough to be considered "different" proof – 2017-01-10