# Function that normalizes a result between 0 and 1 [closed]

is there some function or procedure that I can apply to normalize the result of a division (or any other operation, really) to values between 0 and 1? I.e., given two values X and Y, is there some "func(X/Y)" whose result is between 0 and 1?

The reason I need this is quite nerdy: I'm preparing a RPG setting where interdimensional travel is possible, "dimension" meaning another universe where physics are different from our own. To measure how different the "target" universe is, a product of fundamental constants (i.e., gravitational force, speed of light, etc.) is used, and compared between the two universes. For example, the product of the target universe contants divided by the product of ours.

Ideally, this comparison would yield a result between "1" (totally compatible universe, laws equal to our own) and "0" (totally different universe). But I don't know how to achieve this.

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## closed as not a real question by Jasper Loy, Noah Snyder, no identity, Arkamis, rschwiebOct 16 '12 at 13:15

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I'm not sure what you want but $x \mapsto \frac{x}{1 + x}$ yields a map $\mathbf R^+ \to [0, 1]$. –  user25784 Oct 15 '12 at 11:44
You may want to consider entropy. However, you should be more specific, because the real question is: What is different? In mathematics this definition is crucial, since many of the object's properties depend on some form of measurement of a distance and the like. –  user13655 Oct 15 '12 at 11:44
@user13655 I don't really care that much about what's "different". I would only use that to add flavour, so I can qualify universes in some way. –  rsuarez Oct 15 '12 at 11:56
What the heck is "qualifying universes"? –  user13655 Oct 15 '12 at 12:37
@user13655 sorry, sometimes I'm quite cryptic because I believe everyone can read my mind :-) I mean that I only need a way to differentiate universes, and that'd be this number. For example: "a 0.923 K-compatible universe" (being K our universe's constant), or a "74% compatible universe". –  rsuarez Oct 15 '12 at 14:33

$$f(x,y) = 1 - e^{-|x-y|}$$
$\arctan(x/y)$ is between $-\pi/2$ and $\pi/2$, so to get something between 0 and 1 you could use $(1/2)+(1/\pi)\arctan(x/y)$.