You are watching: Can decimals be even or odd

I am not saying that this is a way to consider "even" and "odd" numbers, but I think it"s quite cool that it looks somehow as a generalization of this notion.

It doesn"t make sense to look at the digits of an irrational number to determine its "parity", hence let"s just stick with rational numbers.

When we were looking at integers, we said that an integer was *even* when it was divisible by $2$ and *odd* when it is not. Instead of using words, we could give much better information, which is to define a map $$\begin{align}\nu_2 : \bsci-ch.orgbb N & \to \bsci-ch.orgbb N \cup \{ 0 \} \\n=2^k m & \mapsto k\end{align}$$where in the latter, $m$ is odd. The map $\nu$ then measures how *even* the integer $n$ is, in the sense that when there are more powers of 2 that divide $n$, it is considered "more divisible by two".

This concept can be generalized to the rational numbers : write $x = a/b$ with $a$ and $b \neq 0$ integers. Write $a = 2^{k_1} a"$ and $b=2^{k_2} b"$, so that $x = 2^{k_1 - k_2} (a"/b")$ and now both $a"$ and $b"$ are odd. Now define $\nu$ again in a similar manner :$$\begin{align}\nu_2 : \bsci-ch.orgbb Q & \to \bsci-ch.orgbb Z \\x = \frac{2^{k_1}a"}{2^{k_2}b"} & \mapsto k_1 - k_2.\end{align}$$When $x$ is an integer, we recover the map we had before, so this can be considered as an *extension* of this map to the rational numbers.

I don"t know if you have done some abstract algebra, but I am going to work things out a little more so that readers that might be interested find good information here.

If we replace $\nu_2$ by $\nu_p$, where $p$ is a prime number, all that we did works just fine and we get a map from $\bsci-ch.orgbb Q$ to $\bsci-ch.orgbb Z$ which indicates how much can a rational be divisible by $p$ (by allowing "negative divisibility" in some sense, because it also measures how many times $p$ divides the denominator). We are getting close to the concept of a *discrete valuation*, which is defined as the following : given a field $K$, a discrete valuation $\nu$ is a function from $K^{\times}$, the group of units, to $\bsci-ch.orgbb Z$, which satisfies the following properties :$$\begin{align}\nu(ab) & = \nu(a) + \nu(b) \\\nu(a+b) & \ge \min \{ \nu(a),\nu(b) \}\end{align}$$and must also be surjective. This function has many interesting algebraic properties. Recalling informations we had before such as "the set of all even integers form a ring", and just by considering this map and its properties, we can show that $$\{ x \in K^{\times} \, | \, \nu(x) \ge 0 \} \cup \{ 0 \}$$ is a subring of $K$ which contains the identity of $K$.

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For some examples of how this is relevant, consider this : if we take the map $\nu_2$ we had before (over $\bsci-ch.orgbb Q$), proving this fact means that the set of all rational numbers with odd denominator form a subring of $\bsci-ch.orgbb Q$ (and so does the set of all rational numbers with denominators coprime to $p$). Valuations are extensively studied, and you can probably look it up on Wikipedia or ask for a reference if you are more interested in this.