# Are real numbers Countable?

• I
I think that real number is countable. Because there is one to one correspondence from natural numbers to (0,1) real numbers.

0.1 - 1
0.2 - 2
0.3 - 3
...
0.21 - 12
...
0.123 - 321
...
0.1245 - 5421
...

I think that is a one-to-one corresepondence. Any errors here?

weirdoguy, Vanadium 50, PeroK and 1 other person

fresh_42
Mentor
Yes. They are uncountable. Assume that your list contains all real numbers. Now add 1 to the first digit of the first number, add 1 to the second digit of the second number, 1 to the third digit of the third number, and so on. If you find a 9, then turn it into a zero.

The number you get, if you concatenate all changed digits is a number that was not in your list, i.e. whichever list you choose, there is a number that is not listed. Hence the real numbers are uncountable.

dextercioby and Delta2
Yes. They are uncountable. Assume that your list contains all real numbers. Now add 1 to the first digit of the first number, add 1 to the second digit of the second number, 1 to the third digit of the third number, and so on. If you find a 9, then turn it into a zero.

The number you get, if you concatenate all changed digits is a number that was not in your list, i.e. whichever list you choose, there is a number that is not listed. Hence the real numbers are uncountable.

1 - 1
2 - 2
3 - 3
...
10 - 10
11 - 11
...
123-123
...

Assume that this list contains all natural numbers. Now add 1 to the ones place of the first number.(1+1=2), add 1 to the tens place of the second number(0+1=1), add 1 to the hundreds place of the third number(0+1=1), and so on. (If you do it third times, you get 112.)

The number you get, if you concatenate all changed digits is a number that was not in your list, i.e. whichever list you choose, there is a number that is not listed. Hence the natural numbers are uncountable.

fresh_42
Mentor

1 - 1
2 - 2
3 - 3
...
10 - 10
11 - 11
...
123-123
...

Assume that this list contains all natural numbers. Now add 1 to the ones place of the first number.(1+1=2), add 1 to the tens place of the second number(0+1=1), add 1 to the hundreds place of the third number(0+1=1), and so on. (If you do it third times, you get 112.)

The number you get, if you concatenate all changed digits is a number that was not in your list, i.e. whichever list you choose, there is a number that is not listed. Hence the natural numbers are uncountable.
No. Natural numbers are not infinitely long. My argument goes like this:
0.168435 ...
0.899958 ...
0.915476 ...
0.153798 ...
...
turns into

0.268435 ...
0.809958 ...
0.916476 ...
0.153898 ...
...
If we have e.g. 0.5, then we write it as 0.5000000 ...

You cannot do this with natural numbers.

sysprog
No. Natural numbers are not infinitely long. My argument goes like this:
0.168435 ...
0.899958 ...
0.915476 ...
0.153798 ...
...
turns into

0.268435 ...
0.809958 ...
0.916476 ...
0.153898 ...
...
If we have e.g. 0.5, then we write it as 0.5000000 ...

You cannot do this with natural numbers.
If I
1
2
3
4
...

turns into

2
12
103
1004

I can go on this process.

Natural numbers have the potential to be infinitely long.

jim mcnamara
Mentor
Yes, the reals are infinite and the natural number are infinite. Up to this point you got it.
But naturals are countably infinite. The reals are uncountably infinite. Not the same.

I do not see where things got confused, but I think that @fresh_42 did not want to explicitly inflict Cantor's proof on you. I think he is using the proof in "example mode". He can correct me.

Further reading which I hope helps you:
https://en.wikipedia.org/wiki/Countable_set

Klystron
FactChecker
Gold Member
I think that real number is countable. Because there is one to one correspondence from natural numbers to (0,1) real numbers.

0.1 - 1
0.2 - 2
0.3 - 3
...
0.21 - 12
...
0.123 - 321
...
0.1245 - 5421
...

I think that is a one-to-one corresepondence. Any errors here?
All the numbers you listed have finite length and end in infinite 0's. At what point would you count 0.1212121212...?
And there are a lot more where that came from. I think that you can see that you have only counted a tiny, tiny subset of the real numbers.
EDIT: In fact, this method misses a lot of rational numbers, which are countable. 0.12121212.. = 12/99.

Last edited:
sysprog
Dale
Mentor
2020 Award
I think that is a one-to-one corresepondence. Any errors here?
Besides the actual proofs that the reals are uncountable, it seems clear that your construction fails for real numbers greater than 1. I also don’t think it works for irrational numbers and maybe not even some rational numbers.

TeethWhitener
Gold Member
@fresh_42 left out the last step in the traditional diagonal argument:
No. Natural numbers are not infinitely long. My argument goes like this:
0.168435 ...
0.899958 ...
0.915476 ...
0.153798 ...
...
turns into

0.268435 ...
0.809958 ...
0.916476 ...
0.153898 ...
...
If we have e.g. 0.5, then we write it as 0.5000000 ...

You cannot do this with natural numbers.
After we have the second list:
0.268435 ...
0.809958 ...
0.916476 ...
0.153898 ...
we take the changed (bolded) numbers and make a new number (0.2068…) that wasn’t in the previous list. We know it wasn’t because if it were the nth number, its nth digit would be 1 less than it is.

sysprog and Delta2
It's very simple. (mirror refection with zero in center)

Every sigle-digit decimals correspond to every single-digit natural numbers(9 pieces).
(1-0.1, 2-0.2, 3-0.3, ...., 9-0.9)

Every two-digits decimals correspond to every two-digit natural numbers(90 pieces).
(10 - 0.01, 11 - 0.11, 12-0.21, ..., 97 - 0.79, 98 - 0.89, 99 - 0.99)

Every three-digits decimals correspond to every three-digit natural numbers(900 pieces).
(100 - 0.001, 101 - 0.101, 102 - 0.201, ..., 997 - 0.799, 998 - 0.899, 999 - 0.999)

And so on....

Every dicimals have countable digit decimals, aren't they?

If you let go of the mindset that Real numbers are uncountable, it's very simple and easy.
Aren't they?

TeethWhitener
Gold Member
Every dicimals have countable digit decimals, aren't they?
Prove it.

Dale
Prove it.
If natural numbers have countable digits, and every dicimal also has countable digits.
(Because it's mirror refection with zero in center).
Natural numbers are countable and natural numbers are greater than digits of natural numbers, so digits of natural numbers are countable, and also every dicimal has countable digits.

FactChecker
Gold Member
If natural numbers have countable digits, and every dicimal also has countable digits.
(Because it's mirror refection with zero in center).
Natural numbers are countable and natural numbers are greater than digits of natural numbers, so digits of natural numbers are countable, and also every dicimal has countable digits.
What would you call ##12/99 = 0.1212121212...= 0.121212\overline{12}##? Are you going to ignore all the rational numbers like that?

TeethWhitener
Gold Member
and every dicimal also has countable digits.
You can’t just assert the part I asked you to prove and expect anyone to buy it. If you want to assert that every infinite string can be mapped to the reversed infinite string, then yes, you’ve just effectively shown that you can map the real numbers to the real numbers. But you have to put “every dicimal” in 1-1 correspondence with every natural number first. You’ve been given a number of counterexamples at this point. I suggest you study them with a little bit of seriousness.

sysprog
What would you call ##12/99 = 0.1212121212...= 0.121212\overline{12}##? Are you going to ignore all the rational numbers like that?
12/99 is correspond to .....1212121212, but there is not such a natural number.

But I think if your logic is true, rational numbers are also uncountable.

In the case of rational numbers, countable means like that, it shows a possibility of one-to-one correspondence. But in the case of real numbers, the perspective changes.

mathwonk
Homework Helper
2020 Award
pl;ace an interval of length 1/4 over the natural number 1. Then place an interval of length 1/8 over the natural number 2. then place an interval of length 1/16 over the natural number 3. .... continuing, you have covered the natural numbers by a sequence of intervals of length 1/2. but the unit interval [0,1] has length 1. contradiction.

kith, dextercioby, sysprog and 2 others
FactChecker
Gold Member
12/99 is correspond to .....1212121212, but there is not such a natural number.

But I think if your logic is true, rational numbers are also uncountable.
No. It just shows that the natural numbers are not countable by your method. They are countable but your method is seriously flawed.

Last edited:
No. It just shows that the natural numbers are not countable by your method. They are countable but your method is grossly flawed.

Yes. I understand countable meaning and subtle difference between rational numbers and real numbers.

I pick a random rational number as 12/99 and if order can be given it's countable.

But if I pick π, I cannot give order to that number. So it's uncountable.

Thanks!!

FactChecker
FactChecker
Gold Member
I think it is worthwhile to realize how easy it is to create irrational numbers. Just make sequences that never start repeating, like 0.101001000100001000000100000001000000001...
And for any such number, you can add it to a list of rational numbers to get a list of irrational numbers.
IMHO, that makes it easier to accept how the irrational numbers are such a huge set compared to the rationals (although it is not a proof by any means).
Cantor's proof is genius and is worth studying. Mathematicians of his time hated and resisted his conclusion that there are higher, uncountable, orders of infinity, but it was undeniable.

Last edited by a moderator:
emptyboat
mathman
I think that real number is countable. Because there is one to one correspondence from natural numbers to (0,1) real numbers.

0.1 - 1
0.2 - 2
0.3 - 3
...
0.21 - 12
...
0.123 - 321
...
0.1245 - 5421
...

I think that is a one-to-one corresepondence. Any errors here?
You list has only numbers with finite decimal expansions - not enough.

Not only are the reals uncountably infinite; also, any non-empty interval (open or closed) subset (no matter how small as long as it has non-zero extension) of ##\mathbb{R}## (the reals) that includes all of the reals that are within itself (inclusive or exclusive of boundaries) is also uncountably infinite.

Klystron and FactChecker
Svein
And the set of all subsets of the real numbers are even more uncountable and so on...

And the set of all subsets of the real numbers are even more uncountable and so on...
I trust that you understand that 'uncountable' is a 'yes-or-no' condition, and that therefore nothing can possibly be "even more uncountable" than something else that is 'uncountable' is; however, I think that it's worth pointing out that even the infinitesimal is exactly as 'uncountably infinite' in its fractional expansion as the entirety of ##\mathbb{R}## (the reals) in its fractional expansions is (the concept of divisibility may be regarded as inconsistent with the concept of infinitesimality; i.e., it could be not unrightly said that the infinitesimal is not divisible, wherefore, there could be no 'fractional expansion' of the infinitesimal).

Last edited:
And the set of all subsets of the real numbers are even more uncountable and so on...
That reminds me of Russel's Paradox

The reals on [0,1] are easily countable, just paste them in Excel and click 'sort ascending'

FactChecker and sysprog