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 January 29th, 2016, 11:09 AM #1 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Counting the Irrational Numbers It is only necessary to count the irrational numbers between 0 and 1. 1) Express the irrational number as a decimal. 2) Remove the decimal point. 3) The result is a natural number which gives the count from 0. The natural numbers are countable. Proof: There are 5 numbers from zero to 5: 1,2,3,4,5 Thanks to skipjack for motivating this proof, though he may not agree with it, with a penetrating question.
January 29th, 2016, 11:19 AM   #2
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 Originally Posted by zylo The result is a natural number
No it isn't. ALL THE NATURAL NUMBERS ARE FINITE which means that they have a finite number of digits. Your "thing" isn't a number. Besides, the operation "remove the decimal point" is not well-defined mathematically (except for finite decimals, where there is a very natural definition).

Since you clearly don't know what a natural number is, you don't know what a real number is, you don't know what "infinite" means and you don't understand Cantor's diagonal argument; why do you persist in writing nonsense masquerading as proofs about the subjects?

Last edited by v8archie; January 29th, 2016 at 11:28 AM.

 January 29th, 2016, 12:17 PM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 I agree that any natural number is finite. The number of natural numbers isn't finite (Archimedes, Peano) but you can count them by definition of count: 1/1 with the sequence of natural numbers. Any string of natural numbers is countable and so it is a natural number (has a count). 3/7/4/2/8/........... 1/2/3/4/5............ Zeros right after the decimal are easily dealt with, count between .1 and .2 and it's the same count as between 0 and .1. If you don't like this, there is always: ----------------------------------------- Any real number can be expressed as a decimal. The number of numbers that can be expressed by n decimal places is 10^n. 10^n is countable for all n. Proof: 10^2 is countable. 10^(n+1) = 10x10^n is countable. The reals are countable. ------------------------------------------- Last edited by skipjack; February 1st, 2016 at 07:58 AM.
January 29th, 2016, 01:17 PM   #4
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 Originally Posted by zylo I agree that any natural number is finite... Any string of natural numbers is countable and so it is a natural number (has a count). 3/7/4/2/8/........... 1/2/3/4/5............
OK. The bold bit doesn't make sense. Numbers are not countable or countable, sets are. And how can you have a finite natural number with an infinite number of digits? It's like trying to add up $\sum \limits_{n=0}^\infty a_n10^n$. It's more or less a geometric series with common ratio 10. How is that ever going to be finite?

 January 29th, 2016, 01:28 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 In a number system, any natural number expressed as a string of natural numbers is a natural number: n=13546 is a natural number system expressed as a string of natural numbers, it's the decimal system.
January 29th, 2016, 01:37 PM   #6
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Quote:
 Originally Posted by zylo Any string of natural numbers is countable and so it is a natural number (has a count).
I believe v8archie was referring to the concept of calling a string of natural numbers a natural number. A string is a set, a natural number is not.

-Dan

 January 29th, 2016, 01:42 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,683 Thanks: 2664 Math Focus: Mainly analysis and algebra Neither of those sentences make sense. I'll try to comment on what I think you meant to write. "Any string of numerals (0,1,2,3,...,9) represents a natural number". Only if it is a finite string. "n=13546 is a natural number expressed as a string of numerals in the decimal system". This is true. The string of numerals is finite in length. It's also a natural number in many other number systems, notably base 7, base 8 and base 9.
 February 1st, 2016, 07:20 AM #8 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 The Reals can be placed in countable order. The Reals can be placed in countable order Proof by induction: With 2 decimal places I can represent all 2-place decimals between 0 and 1 in countable order: 0 .01 .02 . .99 1 If I have n+1 decimal places, I can arrange n of them in countable order. Then I can add an additional decimal place to divide each interval into ten countable intervals: 0 .001 .002 . .009 .010 .011 . .019 .020 .021 . . .999 By induction, all the real decimals can be placed in countable order. Including $\displaystyle \pi$, skipjack's question. Any real number can be expressed as an integer (countable) and a decimal (countable). Last edited by skipjack; March 5th, 2016 at 11:22 AM.
 February 1st, 2016, 08:55 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 All three of your purported proofs fail for the same reason: almost all irrationals, when expressed as a decimal, have infinitely many decimal places, which means that you aren't counting them. For example, without its decimal point, $\pi$ would become 314159265... (with infinitely many digits), which you've agreed is not a natural number as it's not finite. Your inductive process instead confirms that the reals that have a decimal expansion that terminates are countable. Such reals are necessarily rational, and the rational numbers are countable. Thanks from topsquark and v8archie
 February 2nd, 2016, 08:26 AM #10 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 If you can count n decimal places you can count n+1 decimal places, therefore by induction you can count ALL (countably infinite) decimal places and therefore all real numbers. If you don't accept induction, that is another matter.

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