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Can we draw this function? $f\colon\mathbb{R}\to\mathbb{R}$, given by $$f(x) = \left\{\begin{array}{ll} 1 &\mbox{if $x\in\mathbb{Q}$;}\\ 0 &\mbox{if $x\notin\mathbb{Q}$.} \end{array}\right.$$

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    Depends what you mean by "draw"; Plank's constant would get in the way if you try to accurately picture it; the rationals and the irrationals are both dense on the line, so any drawing in which a point in the graph has any area (no matter how small) would simply "look" like two horizontal lines, one at height $0$ and one at height $1$.2011-04-15
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    @Arturo @Ross Millikan: Actually, I find the idea of defining what it means for a function to be "drawable" to be kind of interesting. I'd submit that a good definition would be that you could low-pass filter the function and write the result as a piecewise smooth function.2011-04-15
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    If you want to draw dotted where and how? so your observation doesnt give any further information, just discontuity of function. and an interesting drawing-curve must give more information than that.2011-04-15
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    I know that it's discontinuous2011-04-15
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    @Mopzer: Is your first comment addressed at Ross's answer? If so you should comment on his answer, not your question. If it's addressed at me, I don't understand the comment.2011-04-16
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    @Mopzer: There *isn't* that much information asbout this function: it takes exactly two values, both of them on dense sets; it is discontinuous everywhere, but that is not something draw a "drawing" will let you see. There are **physical limits** to what you can represent pictorially, and an accurate representation of this function is well beyond those limits.2011-04-16

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