May 27th, 2016, 05:28 AM  #11 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90  
May 27th, 2016, 05:40 AM  #12 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,965 Thanks: 2282 Math Focus: Mainly analysis and algebra 
There isn't one, obviously. A natural number $n$ is finite by the definition of finite given many times before. It has at most $1+\log n \le n$ digits with equality only for $n=1$. Thus every natural number has a finite number of digits. This means that an infinite string of digits cannot be a natural number. The corollary to this is that given a string of digits of length $k$, the minimum natural number it can be is $10^{k1}$. So for you to have a natural number with an infinite quantity of digits, you must produce a finite number $n \ge 10^{k1}$ for which $k$ is infibite. Since $n \gt 10^{k1} \gt k$, this means that you need to find a finite natural number $n$ that is greater than an infinite one $k$. But our definition of "finite" states that if there is a natural number greater than $ k$, then $k$ is finite. So $k$ is both finite and infinite. This is clearly impossible (by our definition of infinite), so we have a contradiction. Thus, the assumption that $k$ is infinite must be flawed to avoid the logical progression that $k$ finite implies $k$ infinite. It is actually your argument that requires the existence of a largest finite natural number, not mine. Just supposing that your argument were correct and each real number were two natural numbers put together. Can you really imagine a world in which nobody who has studied mathematics over the last 150 years didn't notice? Not one mathematician has spotted two strings of numbers in a real number and thought: I wonder if both of these are natural numbers? Last edited by v8archie; May 27th, 2016 at 05:56 AM. 
May 27th, 2016, 05:51 AM  #13 
Global Moderator Joined: Dec 2006 Posts: 18,034 Thanks: 1393  That example uses a finite number of digits. For a number such as 3.14159265... (i.e. $\pi$), you would need 14159265... > .14159265..., but the lefthand side isn't a natural number. For every natural number, n, there is a larger natural number: n+1, so there is no largest natural number. Every natural number has a finite number of significant digits. You've already posted words to that effect here, zylo, when you defined a binary number, remember? 
May 27th, 2016, 05:52 AM  #14 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 
Why isn't 3333333333..... with a countably infinite number of 3's not a natural number?

May 27th, 2016, 05:59 AM  #15 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,965 Thanks: 2282 Math Focus: Mainly analysis and algebra 
See my edit of post 12. It contains a proof.

May 27th, 2016, 06:37 AM  #16 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90  I agree with you and skipjack. My terminology was incorrect. Thanks. Let me rephrase it: Given all decimal fractions, every sequence of digits appears in the sequence of the natural numbers, ie, there is a 1:1 correspondence of the reals with the natural numbers. By definition, the reals are countable. Last edited by skipjack; May 31st, 2016 at 09:19 AM. 
May 27th, 2016, 07:01 AM  #17 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,965 Thanks: 2282 Math Focus: Mainly analysis and algebra 
No. Only finite sequences appear in the rationals. The reals contain infinite sequences such as $(\pi  3)$.

May 27th, 2016, 07:10 AM  #18  
Global Moderator Joined: Dec 2006 Posts: 18,034 Thanks: 1393  Quote:
I gave as an example the sequence 14159265... (corresponding to $\pi$), which doesn't appear in the "sequence of the natural numbers". The reals corresponding to sequences of finite length are countable (by your method), but these don't include any irrational real number.  
May 27th, 2016, 07:45 AM  #19 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,155 Thanks: 90 
The rationals, infinite repeating sequences of natural numbers (minus the decimal point), are countable. That leaves all the infinite nonrepeating sequences of natural numbers. 
May 27th, 2016, 08:06 AM  #20 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,965 Thanks: 2282 Math Focus: Mainly analysis and algebra 
Yes. The irrationals are uncountable.


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