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Is there an equation, that will return either 1 if a variable is positive, or 0 if the variable is negative.

For example, to see if an integer is odd or even you can use:

$$r=\frac{(-1)^n+1}{2}$$

where $r$ will be $1$ if $n$ is even, and $r$ will be $0$ if $n$ is odd.

So is there a similar equation, to see if $n$ is positive or negative?

(sorry I have no idea what to tag this)

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3 Answers 3

What you want is known as the Heaviside step function. There are more or less contrived ways to represent it as a closed formula, but usually no good reason to bother. It is much easier for everybody simply to say "1 if positive, 0 if negative" than to attempt to get that result out of a particular arithmetic expression.

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+1. One notation I like is Iverson bracket. –  Did Oct 18 '11 at 19:19
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But isn't a square bracket also used for rounding? –  Jonathan. Oct 18 '11 at 19:23
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@Jonathan, do you mean this? First, these are not the same symbol, second the arguments are different (real numbers vs properties). –  Did Oct 18 '11 at 19:37
    
@Jonathan, not universally. In general, you get to define your use of notation freely as long as you make it clear to your reader what you're doing. –  Henning Makholm Oct 18 '11 at 19:38
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I’d not understand $[x]$ as an abbreviation for $\left\lfloor x+\dfrac12\right\rfloor$ unless that meaning was explicitly stated; I would understand $[x\ge 0]$ (or square brackets around any other relational expression) as an Iverson bracket, though I’d expect a sensible author to explain the notation. –  Brian M. Scott Oct 18 '11 at 19:43

I don't know if this is what you are looking for, but maybe:

EDIT: $\frac {x+|x|}{2}$ where $|x|$is the standard absolute value, returns a $x$ if $x>0$ and a 0 if $x\leq 0$

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Ummm, that returns $x$ when $x>0$. You need to divide by $x$ to get 1, but then you are undefined if $x=0.$ –  Thomas Andrews Oct 18 '11 at 19:06
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Am I being retarded or am I misinterpreting the standard absolute value? $\frac{3+|3|}{2}$ = 3? –  user12205 Oct 18 '11 at 19:06
    
Yes, I did something dumb here; let me check. –  gary Oct 18 '11 at 19:10
    
@Gary, if $x > 0$ I get the answer as x, (e.g. $\frac{5+|5|}{2}=5$) –  Jonathan. Oct 18 '11 at 19:10
    
Jonathan: I edited , and now you get x back if x>0 and 0 otherwise. Does that work? –  gary Oct 18 '11 at 19:12

If you absolutely need an elementary formula, you can consider

$f(x)=\frac{1}{2}(\frac{\sqrt{x^2}}{x}+1)$.

Do note however that $f(0)$ is undefined.

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