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How many digits has $\sqrt{2}$? Countable or uncountable many?

nonuser
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  • Can you elaborate? what do you mean by how many decimals? –  Dec 18 '19 at 20:48
  • Countably infinite many. –  Dec 18 '19 at 20:49
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    Well, so $\sqrt 2 = 1+\sum_{n=1}^{\infty} \frac {a_n}{10^n}$ for $a_n\in {0,1,2,3,4,5,6,7,8,9}$. The ${a_n}$ are countable, clearly. Is this what you meant? – lulu Dec 18 '19 at 20:50
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    Every real number has countably many digits in its decimal expansion. Did you really mean to ask whether or not it is repeating? Note that a real number has an eventually repeating decimal expansion if and only if it is rational. This characterizes rational numbers. – MPW Dec 18 '19 at 20:55
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    I am curious how you would write down an uncountable number of digits? – copper.hat Dec 18 '19 at 20:57
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    @copper.hat: Simply define a function $f:[0,1]\to{0,1,2,3,4,5,6,7,8,9}$? – Blue Dec 18 '19 at 21:00
  • @copper.hat And how would you write down a countable number of digits? – nonuser Dec 18 '19 at 21:01
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    @Aqua: Write down one digit every minute :-). – copper.hat Dec 18 '19 at 21:06
  • @copper.hat OK, then please do write down it decimal expansion. – nonuser Dec 18 '19 at 21:20
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    Still not sure the question is clear. With the standard definition of "digit", it's clear that the answer is "countable". Or were you asking about a non-standard use of the term? – lulu Dec 18 '19 at 21:25
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    @Aqua: It is straightforward to write down an algorithm to print the digits of any number $x$ whose appropriately weighted sum converges to $x$. Are you asking if all digits are zero after a certain point? Presumably you are aware that any real has a decimal (or binary, or whatever) expansion. – copper.hat Dec 18 '19 at 21:26

2 Answers2

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$\sqrt{2}$ can be viewed as the limit of the sequence $1, 1.4, 1.41,...$ so countably many.

Note that any real number has thus at most countably many digits since it is the limit of the sequence that consists of its digits i.e. if $a\in \mathbb{R}$ then $a = \lim_{n\to +\infty}{\frac{\lfloor{10^{n}a}\rfloor}{10^{n}}}$.

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There is a first digit, a second digit, a third, etc., and there are no other digits. So, countably infinite.

More formally: we can put the digits in $\sqrt{2}$ and their position within it in a bijection with the natural numbers. So, countably infinite.

\begin{array}{cccc} 1 & 4 & 1 & 4 & 2 &...\\ \updownarrow & \updownarrow & \updownarrow & \updownarrow & \updownarrow &...\\ 1&2&3&4&5&... \end{array}

Bram28
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