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 zylo January 29th, 2016 11:09 AM

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.

 v8archie January 29th, 2016 11:19 AM

Quote:
 Originally Posted by zylo (Post 520266) 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?

 zylo January 29th, 2016 12:17 PM

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.
-------------------------------------------

 v8archie January 29th, 2016 01:17 PM

Quote:
 Originally Posted by zylo (Post 520274) 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?

 zylo January 29th, 2016 01:28 PM

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.

 topsquark January 29th, 2016 01:37 PM

Quote:
 Originally Posted by zylo (Post 520274) 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

 v8archie January 29th, 2016 01:42 PM

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.

 zylo February 1st, 2016 07:20 AM

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).

 skipjack February 1st, 2016 08:55 AM

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.

 zylo February 2nd, 2016 08:26 AM

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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