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. 
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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? 
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.  
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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. 
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Dan 
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. 
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 2place 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). 
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. 
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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