# Is the set of dyadic rationals a field?

I recently learned that the dyadic rationals is the set of rational numbers of the form $$\frac{p}{2^q}$$ where $p$ is an integer and $q$ is greater than or equal to zero.

I think the set of dyadic rationals is not a field. Here's why:

One of the requirements for a set to be a field is this:

Every element $a$ in the set has exactly one reciprocal such that $a$ multiplied by the reciprocal equals $1$.

I think that this dyadic rational does not have a reciprocal: $$\frac34$$ The reciprocal is $4/3$, but that is not a dyadic rational because there is no integer $q$ greater than or equal to $0$ such that $2^q=3$.

Therefore the dyadic rationals fails one of the requirements for being a field.

Therefore the dyadic rationals are not a field.

Ha! How about that logic. Am I thinking correctly? Am I correct?

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Yes, that's correct. – Zhen Lin Aug 1 '12 at 9:48

Correct.

The ring of dyadic rationals is obtained from the integers $\Bbb Z$ inverting $2$. But that is not enough to invert all non-zero integers.

Besides, there's no smaller field of the rationals $\Bbb Q$ containing $\Bbb Z$.

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It is true that the ring $\Bbb D$ of dyadic rationals is not a field. But your argument is missing too many details to judge its correctness. To show that $\Bbb D$ is not a field it suffices to show that $\rm\:3\:$ has no inverse in $\Bbb D.\:$ Suppose $\rm\:1/3 = n/2^k.\:$ We may assume that $\rm\:n\:$ is odd by cancelling $2$'s. Therefore $\rm\:2^k = 3n\:$ is odd, so $\rm\:k=0,\:$ so $\rm\:1/3 = n/2^0 = n\in \Bbb Z,\:$ contradiction.

$\Bbb D = \Bbb Z[1/2]\:$ is the ring generated by adjoining $\,1/2\,$ to $\Bbb Z.\:$ To generate a field we need that every nonzero integer is invertible, or, equivalently, every prime is invertible. This will be true iff one adjoins to $\Bbb Z$ a set of rationals such that every prime occurs in some (lowest-terms) denominator. These matters are clarified when one studies generalizations of fraction rings (localizations) in commutative algebra.

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