Can we draw the characteristic function of the rationals?

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 $1$. – Arturo Magidin Apr 15 '11 at 20:22
@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. – Colin K Apr 15 '11 at 20:38
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. – Mopzer Moreena Apr 15 '11 at 22:22
I know that it's discontinuous – Mopzer Moreena Apr 15 '11 at 22:25
@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. – Arturo Magidin Apr 16 '11 at 1:46

It depends upon what you mean by "draw". You could draw a dotted line along $y=0$ and another along $y=1$ with the dots meaning that the function is not continuous and so some of the points you (seem to) have covered are not part of the function. If by draw you mean a continuous curve in the plane, no you cannot, because the function is not continuous.