Mathematical Paradoxes of Infinity

There are a large number of mathematical puzzles associated with infinity, and because I don’t accept the notion of a completed infinity, I feel that most of them are not actually paradoxical.  Many people (even mathematicians who do not study mathematical philosophy) are not clear about the difference between potential and actual (completed) infinity: here is the first paragraph on that difference from wikipedia (

In the philosophy of mathematics, the abstraction of actual infinity involves the acceptance (if the axiom of infinity is included) of infinite entities, such as the set of all natural numbers or an infinite sequence of rational numbers, as given, actual, completed objects. This is contrasted with potential infinity, in which a non-terminating process (such as “add 1 to the previous number”) produces a sequence with no last element, and each individual result is finite and is achieved in a finite number of steps.

It is important to note that both systems are happy to use the term infinity: if they did not believe in infinity, then they would hold a different mathematical position called finitism.  They use the term in significantly different ways, however. Mathematicians who believe in completed infinities (like Aleph Null) treat infinity like a number.  Mathematicians who do not accept completed infinity, and believe only in potential infinity, treat infinity like a limit process (which has a more complicated representation than a finite number).

The subject of this article is a paradoxical statement which is widely held to be true by mathematicians who believe in a completed infinity:

1) $latex 1.0 = 0.99\overline{9} &s=1$

Most people are not intuitively comfortable with this statement, because it implies that two different numerical (decimal) representations correspond to a single number.  The “proof” looks as follows:

2) $latex x = 0.99\overline{9} &s=1$

3) Multiply both side by 10 to get : $latex 10x = 9.9\overline{9} &s=1$

4) Subtract (3) from (2) to get: $latex 10x-x = 9x = 9.9\overline{9} – 0.99\overline{9} = 9 &s=1$

5) Therefore, $latex x = 1 &s=1$

There are at least two ways out of this paradox.  Most mathematician these days believe in completed infinity, so they are forced to accept (2) & (5) and therefore (1).  If you hold an intuitionistic view of mathematics, however, you can get rid of the paradox by handling infinite representations as limit series.

Before we go farther, I should mention that whether or not my reasoning is sound, it is mathematically novel (at least according to some of my mathematician friends).  If you happen to have conversations about intuitionism with narrow-minded mathematicians, they will not be sympathetic to your view (despite the fact that a number of genius mathematicians have advocated this stance, such as Arend HeytingL. E. J. BrouwerStephen Kleene, and Henri Poincare).

To return to how an intuitionist might represent a number which has “infinite length”.  Obviously, a representation like $latex 0.99\overline{9} &s=1$ where the nines run “to infinity” will not be acceptable.  However, we can turn it into a limit series as follows:

5) $latex \sum\limits_{n=1}^{\infty} \frac{9}{10^n} &s=1$

In other words, repeating decimal representations are written as a limit where the number of terms approaches infinity.  The significant difference between the limit representation and the repeating decimal representation is that the former is more clearly seen to be a process, not a “number whose length has a cardinality of aleph null”, which again is disallowed (in intuitionism). Given this representation, let’s re-write (4):

6) Subtract (3) from (2) to get: $latex 10x-x = 9x = 10 * \sum\limits_{n=1}^{\infty} \frac{9}{10^n} – \sum\limits_{n=1}^{\infty} \frac{9}{10^n} = \sum\limits_{n=1}^{\infty} {\frac{9*10}{10^n} – \frac{9}{10^n}} = \sum\limits_{n=1}^{\infty} \frac{81}{10^n} &s=1$

The result from (6) is not 9, it is (commonly written as) $latex 8.9\overline{9} &s=1$, and for an intuitionist, it is 8.1 after one term, approaching a limit of 9 only “at” infinity (which, on that account, we never reach, and therefore we do not find that (1)). Now, why did we arrive at a different result when we used a limit formula, instead of the subtraction formula? My “intuitive” answer is that we paid attention to the number of terms in the series expansion. In other words, when in (4) we perform the subtraction $latex 9.9\overline{9} – 0.99\overline{9} &s=1$, we are subtracting term i in the second series expansion from term i+1 in the first series expansion (i.e. we are ignoring the first term). While this might be OK if you have a completed infinity of number of terms, an intuitionist account must yield a correct series at every (finite) step, and combine terms under the appropriate loop variable, i. That means that after three terms, we have:

7) $latex 9.99 – 0.999 = 9 – 0.009 = 8.991 &s=1$

In other words, $latex 9x \neq 9 &s=1$, and as a result, intuitionists who do not believe in a completed infinity do not have to accept that $latex 1.0 = 0.\overline{9} &s=1$.

OK, mathematicians who can either logically tear me apart or pat me on the back, I welcome your feedback.


4 responses to “Mathematical Paradoxes of Infinity”

  1. I received two replies to this in a separate thread. More easily dismissed was the second of the two: “If we give up infinity, then we can’t resolve Zeno’s paradox”. Intuitionists do not give up infinity, they give up the notion of a completed infinity. We are happy to construct limit series that describe the spatial path of the arrow, as well as the temporal intervals that correspond to that spatial path (such that a constant velocity is maintained).

    The second response is a bit more challenging for me. Because I am defending the uniqness of $latex 0.9\overline{9} &s=1$ and $latex 1.0 &s=1$, I was asked for the decimal representation of $latex 1 – 0.9\overline{9} &s=1$. If I cannot give a decimal representation of that number, then I must concede that not all numbers have decimal representations (which I probably have to concede for irrational numbers anyway, since that seems to require decimal strings which have a length corresponding to a completed infinity). That said, it is a good question: is there a decimal representation of $latex 1 – 0.9\overline{9} &s=1$, and if so, what is it?

    1. Let’s begin the challenge by writing the two numbers in decimal form:

      1) $latex x_1 = 0.9\overline{9} = \sum\limits_{n=1}^{\infty} \frac{9}{10^n} &s=1$

      2) $latex x_2 = 1.0 = 1+\sum\limits_{n=1}^{\infty} \frac{0}{10^n} &s=1$

      3) $latex x_2 – x_1 = 1.0 – 0.9\overline{9} = 1-\sum\limits_{n=1}^{\infty} \frac{9}{10^n} &s=1$

      If we write out the continuing approximation of this series for n=1,2,3,…, we have:

      3) $latex 1, 0.1, 0.01, 0.001, 0.0001, … &s=1$

      To express this as a series, we have:

      4) $latex 1 + \sum\limits_{n=1}^{\infty} \frac{1}{10^n} – \frac{1}{10^(n-1)} &s=1$

      I think this is a good and explicit representation: it is explicit in the sense that it does not express a completed infinity. I feel that the overbar notation, understood as an infinite decimal string, leads to mathematical errors. However, limit series are inconvenient to write, so the following shorthand for exactly the equations given above seems reasonable:

      4) $latex x_2 – x_1 = 1.00 – 0.9\overline{9} = 1.00 – 0.\overline{9}9 = 0.\overline{0}1 &s=1$

      I have not seen the overbar used like this before, and in fact, I am not sure if the overbar notation has a formal definition (it was explained to me as “repeat whatever decimals are under the bar infinitely“). However, it suggests that a decent definition (for an intuitionist) would be something like:

      5) $latex \overline{x} := \sum\limits_{n=1}^{\infty} {x^n} = x, xx, xxx, … &s=1$

      Of course, this needs to be understood as a string manipulation, since it will have different effects whether it is used before or after the decimal point. Its use here under a summation is important because it indicates the importance of the loop variable, $latex n &s=1$. To reiterate the issue that began this thread, the steps toward convergence of the series are essential for intuitionist philosophers, because we never reach a completed infinity. For that reason, please reflect on the necessity of the following inequality for mathematicians who do not believe in a completed infinity:

      6) $latex 0.9\overline{9} \neq 0.\overline{9} &s=1$

      Although in some sense they can be said to be equal at infinity, they are not equal for any finite-length expansion of the index n (which is implicit in the overbar notation).

  2. Jim Conant Avatar
    Jim Conant

    It occurs to me that we should review how real numbers are standardly encoded by infinite decimal representations. In fact, let me switch to binary as it’s easier to state. Let’s look at the interval [0,1] and show how to assign an infinite string of binary digits to any $$x\in [0,1]$$. Step 1 is to cut the interval in two equal pieces. If $$x$$ lies in the left interval, assign the first digit to be 0. If it lies in the right, assign the first digit to be 1. Now look at whichever subinterval you chose, and cut it in half and repeat the process. So if $$x$$ lies in the left subinterval, assign the second digit to be 0 and if it lies in the right assign the second digit to be 1. Repeat ad infinitum. Now there is only one ambiguity in the above definition. $x$ may lie on the boundary between two intervals. In this case, you could assign the nth digit to be either 0 or 1. However, once you make that choice, all subsequent choices are fixed. If you pick $$0$$ then in every subsequent interval $$x$$ will be at the right, so all subsequent digits are $$1$$. Similarly, if you pick $$1$$ then all subsequent digits will be $$0$$. So there is an ambiguity for numbers that have terminating binary representations. Any 1 at then end of a representation can be replaced by 0111… So, from this perspective, the ambiguity of .1000 and .0111… is an artifact of the encoding of reals. If $$.1000\neq.0111…$$, then you need to explain how you are encoding reals as decimals or in binary in a way that is different from this method.

    1. There is an assumption in this argument that would not be accepted by intuitionists: “x may lie on the boundary between two intervals”.

      For the intuitionist argument to be plausible, the interval between 0 and 1 cannot consist of an infinite number of points. Mereotopology seems to be the standard answer, which takes ‘part’ as primitive instead of ‘point’. Intuitively, a space of a given dimensionality cannot consist of parts of a lower dimensionality, as is the case with infinitely many points creating a line.

      The real numbers can represent points of division of a continuous interval, but since they have zero width, they are nominal (merely imputed) on the interval. I continue to think that all decimal representations which are different from one another will divide that interval at different places.

      This point of view might be able to be summed up topologically by saying that all intervals are open intervals, but in this case open intervals are said to be touching if they share a boundary (which in the linear case is a point). That feels desirable for intuitionists, because it avoids the topological paradoxes of objects with open/closed boundaries being unable to be in contact with one another. I am not sure if doing this creates other difficulties, but there are no associated paradoxes of which I am aware.

Leave a Reply

Your email address will not be published. Required fields are marked *