Ok, so this problem popped up in my mind in Real Analysis lecture. Here goes:

Lets say I take a sheet of white paper(Any size say 2X1 units). Let this sheet of paper can be divided into infinitely many strips horizontally. Let each strip denote a real number between 0 to 2 (both included). Now, each strip is also a real number line segment of interval [0,1] such that each strip has all the real numbers between [0,1] (including 0 and 1).

So, If i remove all the parts of the paper with irrational points(I just put a hole at every point where irrational number lies), How would that paper look like? Of course I will be able to see the paper, but would I see the whole sheet or a semi transparent sheet? What If I put it in a dark room and shine a torch through it, what will be the intensity of light passing through? How would the mass of the paper be affected.

The confusion arises because a set of irrationals is uncountable while rationals is countable! I wonder if the paper will be mass less, yet visible.

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    $\begingroup$ Possible duplicate of Rational vs Irrational distribution but I do not think there is a reasonable answer to this question $\endgroup$ Jan 19, 2018 at 5:55
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    $\begingroup$ Not all math ideas can be literally transfer to the real world. That's why mathematicians aren't all physicists. $\endgroup$ Jan 19, 2018 at 5:55
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    $\begingroup$ This seems more of a physics question than a math one. The simplest answer is that your desired construction is impossible to create due to the discrete nature of how atoms in our universe act. In the real numbers, there is no "smallest nonzero real number" but in our physical universe there absolutely is a smallest size object (or width, etc...). $\endgroup$
    – JMoravitz
    Jan 19, 2018 at 5:56
  • $\begingroup$ If you were to ignore this, then you'd have to define what "weight" and "visibility" mean. Under most convenient definitions, they would have to correspond to some sort of measure or probability measure. Given that the measures you would usually use, such as the lebesgue measure, have the rationals as being of measure zero, it would seem to me at least that the paper would be completely invisible and massless (not to mention unable to maintain form or be held, etc..) $\endgroup$
    – JMoravitz
    Jan 19, 2018 at 6:07
  • $\begingroup$ I'd imagine mass is a lebesgue measure so weightless, but imagine if visibility is assumed (but it must be stated) to be able detect points that you'd see the rational points and the background behind the irrationals and they'd appear superimpossed. But really the mechanism of "seeing" points must be explained. $\endgroup$
    – fleablood
    Jan 19, 2018 at 7:36

1 Answer 1


The real world is finite in many ways. As a result, the tools that we use to approximate the world as infinitely divisible—real numbers, continuous surfaces, integrals, and so on—are useful, but sometimes not applicable.

Of course there are problems with puncturing a piece of paper infinitely many times at mathematically perfectly correct locations. With a real piece of paper and real tools it cannot be done. However, we can make a mathematically ideal model of your paper, in which case our infinite-world approximation tools apply.

  • In such a model, we can say that your paper is essentially like the unit square $S = [0,1]\times [0,1]$ in $\mathbb{R}^2$.

  • $S$ represents an infinitely thin, infinitely divisible two dimensional surface.

  • We can cut out the irrational points, resulting in the perforated piece of paper $P\equiv S \cap (\mathbb{Q}\times \mathbb{Q})$. (Here, $\mathbb{Q}$ denotes the set of rational numbers, which we want to retain, getting rid of the remaining irrational points.)

  • Excitingly, the tools of real analysis are exactly the ones we use to model the properties you care about:

    • The area of $P$, as determined by an integral, tells you how "much" of the paper is there. You can use this as a measure of the mass of the paper remaining.
    • You can take the ratio of the area of $P$ and the original area of $S$ to compute the "density" of the remaining paper in the space it originally occupied. This can tell you something about the "opacity" of $P$— an idealized measure of what fraction of uniformly-bright light falling on the original area $S$ will pass through $P$, assuming for simplicity that the paper is made of a material that transmits no light itself (the paper itself is perfectly opaque, so light only gets through empty spaces).
    • This density constitutes a kind of transparency, if you like— not strictly because the medium becomes more translucent to light the way greased paper does, but rather because it is full of holes.
  • And hence we can answer all of these questions in a similar way. One of our most tools for measuring area in this context is the Lebesgue measure: it extends the Riemann integral studied in elementary calculus in such a way that you can compute the area of functions like:

    $f(x)=\begin{cases}1&\text{if }x\text{ is irrational}\\0&\text{if }x\text{ is rational}\end{cases}$

  • Now for some notation. If $A$ is any set in $\mathbb{R}^n$, let ${1}_A$ denote the characteristic function $$1_A(x) = \begin{cases}1 & x \in A\\ 0 & x \notin A\end{cases}$$

  • So if $U=[a,b]$ is an interval on the real line, then $\int_{-\infty}^\infty 1_U(x)\,dx = \int_a^b 1 dx = b-a.$

    The general rule is the integral of a set's characteristic function represents the size of the set.

  • As another example, the original "size" (area) of your paper $S$ is: $$\iint_{\mathbb{R}^2} 1_{S} = \int_{0}^1\int_0^1 1 \,dx = 1$$ as you'd expect

  • Now we can figure out the "size" of your paper $P$. We want: $$\iint_{\mathbb{R}^2} 1_{P} = \iint_{[0,1]^2 \cap (\mathbb{Q}\times\mathbb{Q})} 1$$

  • Without explaining how to compute this Lebesgue integral (which takes some developed machinery to show), it turns out that this integral is 0. As a result, a reasonable mathematically-ideal answer to your questions is as follows:

    • The perforated paper has exactly zero total area.
    • The fraction of paper you've removed is essentially all of it.
    • There is technically some paper remaining ($P$ is not an empty set), but—in a way that cannot happen in real life—so little paper remains that it has no mass. In real analysis, this is referred to as a set of measure zero.
    • As such, exactly zero percent of incoming light will reflect off the paper and into your eyes. In any reasonable sense, you will not be able to see anything because there is essentially no paper left to see.
    • Similarly, light shining from behind the paper will essentially not run into anything, so the light will not be obstructed or diminished in any way. The paper will be perfectly "transparent".
    • And the paper will be perfectly massless.

These should be taken to be surprising results about our models involving real numbers, rational numbers, irrational numbers, and integration, rather than claims about what happens with genuine perforated pieces of paper. Taken as mathematical results, they offer neat insights into the way we've formalized our ideas of number, divisibility, and size, and the ways in which those formalisms sometimes defy our geometric intuitions. (For another similar surprising result, see for example the Banach-Tarski paradox.)

Good question, and good luck on your studies of real analysis!

  • $\begingroup$ I think this explanation perfectly acceptable. I realize it now that mathematical world is sometimes impossible to be modeled in real world. $\endgroup$ Jan 19, 2018 at 17:06
  • $\begingroup$ Its kind of funny, you subtract an infinity from another infinity and you know something is there, that is also an infinity, but at the same time its nothing! $\endgroup$ Jan 19, 2018 at 17:08

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