For any real number $x$, rational or irrational, $\frac{x}2=\frac12x$ is a real number, and $2\cdot\frac{x}2=x$. If $x$ is rational, $\frac{x}2$ is also rational; if $x$ is irrational, $\frac{x}2$ is also irrational; but in both cases it exists.
Your difficulty seems to arise from trying to apply the notion evenly divisible to real numbers in general, when in fact it applies only to integers. It is perfectly true that if $n$ is an integer, then there is an integer $m$ such that $n=2m$ if and only if $n$ is even, so that some integers — the odd ones — are not two times some integer; and it’s often useful to distinguish the even integers from the odd integers. The odd integers are, however, still two times some rational number, albeit not an integer. E.g., $5=2\cdot\frac52$.
When we move from the integers to the rational numbers, matters change: if $p$ is rational, so is $\frac{p}2$, so every rational number is twice some rational numbers. And when we move on to the real numbers, we retain this property: if $x$ is a real number, so is $\frac{x}2$, so every real number is twice a real number.
$x$ and $y$ are commensurable if there exists a real number, $r$ and positive integers $m$ and $n$ such that $x = mr$ and $y=nr$. If such an $r$ exists, it is called a common measure. If $x$ and $y$ are commensurable, we can aviod mention of a common measure by writing $x : y :: m : n$, or $\dfrac xy = \dfrac mn$. The value of $r$ depends on the values of $x$ and $y$. Saying, for example, that $\sqrt 2$ is irrational is equivalent to saying that $\sqrt 2$ and $1$ are incommensurable.
Best Answer
Let $n$ be an integer and $p:=\frac{n}{n+1},q:=\frac{1}{n+1}$ then $p/q=n$. Note that both $p$ and $q$ are non-integers (as $\gcd (n,n+1)=1$).
Also if $r:=\frac{n^2}{n+1},s:=\frac{n+1}{n}$ then $rs=n$. Clearly both $r$ and $s$ are non-integers, because $\gcd (n^2,n+1)=\gcd (n,n+1)=1.$
So for every integer $n$ you can find non-integral rationals whose quotient is $n$ and non-integral rationals whose product is $n$.