# Wrong in counting real number

I know that real number is uncountable,the same as the closed interval [0,1].However,I don't know what is wrong with my method of counting numbers in [0,1]:

0.1,0.2…0.9

0.01,0.02…0.99

0.01,0.02,…0.999

....

Can anyone tell me what'wrong with it?Thanks.

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## 2 Answers

You are merely counting rational numbers, not all real numbers between $0$ and $1$. The problem with the real numbers is that from any list of real numbers, using Cantor's diagonal argument, one can show that there exists a real number that is not in that list. This is why the real numbers are uncountable.

Have fun reading this : http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

Hope that helps,

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and even not all rationals, just with denominator $2^a5^b$. – Berci Oct 20 '12 at 1:41
I know what you mean thanks. – 89085731 Oct 20 '12 at 1:45
Now I know what is called countable set.That means that you devise a method to count with which every element can be placed in a particular position n. – 89085731 Oct 20 '12 at 1:52
@89085731 : Yes, exactly. The "method of counting" would be the bijection between your countable set $S$ and the set of positive integers $\mathbb N$, and if you count the elements by $S = \{x_1, \dots, x_n, \dots \}$, then the bijection is $i \mapsto x_i$. By the way, if the answer I gave you answers your question you should click on the check button under the downvote button. – Patrick Da Silva Oct 20 '12 at 2:59

When does your enumeration reach pi? Or Sqrt(2)? Or even 1/13? It doesn't.

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$\pi \notin [0,1]$, nor does $\sqrt 2$. But you're right about $\frac 1{13}$. – Patrick Da Silva Oct 20 '12 at 2:54
@PatrickDaSilva: true, but he could ask about $\pi -3$ and $\sqrt 2 -1$ – Ross Millikan Oct 20 '12 at 3:10
Sure, I was just mentioning. – Patrick Da Silva Oct 20 '12 at 17:34