The ring of dyaidc rationals is $R=\{\frac{a}{b}:a,b\in\mathbb{Z}, b=2^n, n\in\mathbb{N}\}$.

I want to be able to say that if we have a rational $\frac{a}{b}$ in which $a=2^{j}e$ where $e$ is odd and $b=2^i$ then $\frac{a}{b}$ is dyadic if and only if $j<i$, because otherwise we could reduce to $\frac{a}{b}=2^{j-i}e$ and it would not be in R.

However, if we insist upon writing things in this "simplest form" then I don't see how we could have that $1 \in R$, i.e. that $R$ is a ring with identity.

I've been asked to discuss the units (among other things) in this ring, but I can't make sense of that task for the reason I laid out above.

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First, the definition as given does not make it clear whether $\mathbb{N}$ includes $0$ or not; if it does, then this immediately gives you that denominators equal to $1$ are allowable. But even if it doesn't:
Second, nowhere in the definition does it require $\gcd(a,b)=1$. Note that a fraction can be written with a denominator a power of $2$, whether or not it is in least terms, if and only if, when written in least terms, the denominator is a nonnegative power of $2$.
Third: what you "want to say" is incorrect. Even setting aside the issue of $1$ not being in the ring, such a set is not even closed under sums: $\frac{3}{2}$ and $\frac{1}{2}$ would both be in the set, but their sum would not be.
So: a rational number is "dyadic" if and only if it can be written as $\frac{a}{b}$, with $a$ and $b$ integers, and $b$ a power of $2$. We do not require the fraction to be in least terms, only that it can be written as a fraction whose denominator is a power of $2$. In particular, this includes numbers like $\frac{2}{2}$, $\frac{7}{2}$, $\frac{1024}{8}$, $\frac{3072}{8}$, etc. It also includes $\frac{9}{6}$, since $\frac{9}{6}$ can be written as $\frac{3}{2}$, which is of the desired form.