0

First, I should preface this by saying that I am not incredibly familiar with the mathematics at play here; I am simply trying to obtain a conceptual understanding of the answer, since my current conceptual understanding would entail that real numbers are countable. As such I will likely not be using the proper mathematical language in all cases; please bear with me.

From the Wikipedia article 'Countable set':

Whether finite or infinite, the elements of a countable set can always be counted one at a time and, although the counting may never finish, every element of the set is associated with a natural number.

The issue here is that I think one can assign a unique natural number to every real number in a manner such that you could map the real and natural numbers bidirectionally and uniquely.

I would do so with the idea that you can represent every real number with a decimal base and a decimal exponent (base * 10^exponent). The base would be the set of all integers without leading zeros; that is, where the digit at the 1's place is not 0. The exponent would be the set of all integers, excluding 0. These are both subsets of the set of integers, which is countable, and I am given to understand that a subset of a countable set is countable. Thus, for every possible base and exponent, you could assign a unique natural number.

I am also given to understand that you can assign a unique natural number to every ordered pair of two natural numbers; so, one should be able to assign a unique natural number to every pair of base and exponent (which can both be assigned unique natural numbers).

That would leave every real number (excluding 0) tied uniquely to a natural number. You could then form a set of all real numbers by 'appending' 0, and increment all the other natural number bindings by 1. At this point, you would have uniquely assigned a natural number to every real number, making the set of real numbers countable.

So... what's wrong here?

Asaf Karagila
  • 405,794
lcmylin
  • 119
  • 2
    You can't represent $\pi = 3.14159\ldots$ in such a system. –  Feb 26 '16 at 20:17
  • 1
    This way, you only obtain the decimal numbers, since you consider only real numbers with a finite number of decimals. Decimal numbers are only a subset of rational numbers. – Bernard Feb 26 '16 at 20:19
  • 3
    Hmm, what about $\frac13$? Am I missing something? – MPW Feb 26 '16 at 20:19
  • there is nothing wrong with considering only the definable real numbers, and in general only definable sets. those sets are closed for any definable operation, so.. – reuns Feb 26 '16 at 20:30
  • @reuns what do you mean by "definable"? – Brondahl Aug 21 '22 at 15:23

2 Answers2

6

you can represent every real number with a decimal base and a decimal exponent (base * 10^exponent)

No, you can't. Only if it has a finite decimal representation. That is, if it's a rational number that, when writen as an irreducible fraction, has only $2$ and $5$ as prime factors in the denominator.

The set of such numbers, then, is a subset of the rational numbers. That's indeed countable.

leonbloy
  • 66,202
  • Why can you not represent an irrational number (no finite decimal representation) as long as you have an infinitely large base and exponent? And those are included in the set of integers. – lcmylin Feb 26 '16 at 20:23
  • 8
    That's a common error. To be able to pick an "arbitrarily large" integer is not the same as to pick an "infinitely large" integer (the later doesn't mean anything, actually) – leonbloy Feb 26 '16 at 20:26
  • 3
    @Textfield Nothing infinitely large is contained in the set of integers. Every integer is finite right? If not can you write down for us one that isn't? – Gregory Grant Feb 26 '16 at 20:27
  • So, that is to say, the set of all integers does not contain the set of all strings of integers 0-9 (decimals)? – lcmylin Feb 26 '16 at 20:28
  • 4
    @Textfield It contains the set of all finite strings of digits, for sure. – leonbloy Feb 26 '16 at 20:29
  • Though you can get as close as you want to any irrational number. How do you prove that you don't reach it "in the limit" ? –  Feb 29 '16 at 07:18
  • @YvesDaoust You do reach it in the limit. More precisely, any real number can be expressed as the limit of a sequence of numbers that have a finite decimal representation. So what? – leonbloy Feb 29 '16 at 11:46
  • So can you affirm that you can't count it ? –  Feb 29 '16 at 13:02
  • @YvesDaoust I don't get you. I can count it. Let $S$ be the set in question (numbers that have a finite decimal representation). $S$ is countable. The closure of $S$ (which is $\mathbb{R}$) is uncountable. https://en.wikipedia.org/wiki/Closure_(topology)#Closure_of_a_set – leonbloy Feb 29 '16 at 13:14
  • How do you prove that the closure isn't countable ? –  Feb 29 '16 at 13:18
  • http://math.stackexchange.com/questions/237300/why-is-the-set-of-all-real-numbers-uncountable – leonbloy Feb 29 '16 at 13:32
2

List all the real numbers in the order that you like.

Say

$$0.\underline435\cdots\\ 0.6\underline54\cdots\\ 0.55\underline4\cdots\\\cdots$$

Then consider a number $\delta$ such that

  • its first decimal differs from the first decimal of the first number,

  • its second decimal differs from the second decimal of the second number,

  • its third decimal differs from the third decimal of the third number,

and so on.

Say $$0.565\cdots$$

Oh, but what happens ? You didn't list $\delta$ ?!