
#1
Mar2708, 02:04 PM

P: 50

The set N of natural numbers = {1, 2, 3, 4, ...}.
But there exists one (1) number N, such that N = 12345678910111213... (where the unit's place is at infinity). A good example of an irrational number then would be 1.234567891011121314... 



#3
Mar2708, 04:09 PM

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so, put a decimal in front of it.
oh he did that. 



#4
Mar2708, 05:13 PM

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there exists one number N 



#5
Mar2708, 06:05 PM

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#6
Mar2708, 07:36 PM

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#7
Mar2808, 06:54 AM

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If it were possible to construct any irrational number by putting a decimal into some positive integer, that would imply that the set of irrational numbers is countable.




#8
Mar2808, 01:23 PM

P: 50

good question 



#9
Mar2808, 01:24 PM

P: 50





#10
Mar2808, 01:25 PM

P: 50





#11
Mar2808, 01:27 PM

P: 50





#12
Mar2808, 01:32 PM

P: 50

1.234567891011... 



#13
Mar2808, 01:57 PM

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#14
Mar2808, 06:01 PM

P: 891





#15
Mar3008, 06:39 AM

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#16
Mar3108, 04:53 PM

P: 50

If f(x) = x, then lim (of f(x) as x approaches infinity) = infinity = N. (but the unit's place of N is at infinity.) 



#17
Mar3108, 06:12 PM

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But since "infinity" is not an integer, you know that N isn't an integer.




#18
Apr208, 09:39 AM

P: 41

1234567891011... is not a conventional way of representing real numbers, so unless you introduce your own convention, it doesn't mean anything. Whereas if you put a disimal point somewhere, it represents a real number in a conventional sense. Because, by convention, 1.234567... represents some real number to which the sequence, 1, 1.2, 1.23, 1.234, ... converges. This is what we call the completeness of R. If we agree to say that 1234567891011... represents where the sequence 1, 12, 123, 1234, ... go, then we may call it infinity, or more precisely, we introduce the concept of infinity.



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