You are watching: Can decimals be even or odd
I am no saying the this is a method to consider "even" and also "odd" numbers, yet I think it"s rather cool the it watch somehow as a generalization the this notion.
It doesn"t make feeling to look at the digits of an irrational number to determine its "parity", therefore let"s just stick through rational numbers.
When us were looking in ~ integers, we said that one integer to be even once it to be divisible by $2$ and also odd when it is not. Rather of using words, we can give much far better information, i beg your pardon is to specify a map $$\beginalign\nu_2 : \bsci-ch.orgbb N & \to \bsci-ch.orgbb N \cup \ 0 \ \\n=2^k m & \mapsto k\endalign$$where in the latter, $m$ is odd. The map $\nu$ then measures how even the essence $n$ is, in the feeling that as soon as there are more powers the 2 that divide $n$, the is taken into consideration "more divisible by two".
This ide can be generalized to the rational numbers : compose $x = a/b$ v $a$ and also $b \neq 0$ integers. Create $a = 2^k_1 a"$ and $b=2^k_2 b"$, so that $x = 2^k_1 - k_2 (a"/b")$ and now both $a"$ and $b"$ space odd. Now define $\nu$ again in a comparable manner :$$\beginalign\nu_2 : \bsci-ch.orgbb Q & \to \bsci-ch.orgbb Z \\x = \frac2^k_1a"2^k_2b" & \mapsto k_1 - k_2.\endalign$$When $x$ is one integer, we recoup the map we had before, so this have the right to be taken into consideration as one extension of this map come the reasonable numbers.
I don"t know if you have actually done some abstract algebra, however I am going to job-related things the end a little more so the readers that can be interested find an excellent information here.
If we change $\nu_2$ 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 just how much deserve to a rational be divisible through $p$ (by permitting "negative divisibility" in some sense, due to the fact that it likewise measures how plenty of times $p$ divides the denominator). We are getting close come the principle of a discrete valuation, i beg your pardon is identified as the adhering to : provided a field $K$, a discrete valuation $\nu$ is a duty from $K^\times$, the team of units, to $\bsci-ch.orgbb Z$, i beg your pardon satisfies the adhering to properties :$$\beginalign\nu(ab) & = \nu(a) + \nu(b) \\\nu(a+b) & \ge \min \ \nu(a),\nu(b) \\endalign$$and must likewise be surjective. This function has countless interesting algebraic properties. Recalling informations us had before such as "the set of all also integers form a ring", and also just by considering this map and its properties, we can present that $$\ \, \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 exactly how this is relevant, take into consideration this : if we take the map $\nu_2$ us had prior to (over $\bsci-ch.orgbb Q$), prove this fact method that the set of all rational numbers v odd denominator kind a subring that $\bsci-ch.orgbb Q$ (and for this reason does the set of every rational numbers v denominators coprime come $p$). Price are broadly studied, and also you deserve to probably watch it increase on Wikipedia or ask for a referral if girlfriend are much more interested in this.