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May 27th, 2016, 05:28 AM   #11
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Quote:
Originally Posted by v8archie View Post
...an infinite string of digits is not a natural number because a natural number is finite.
So what is the largest natural number?
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May 27th, 2016, 05:40 AM   #12
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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 $10^{k-1}$. So for you to have a natural number with an infinite quantity of digits, you must produce a finite number $n \ge 10^{k-1}$ for which $k$ is infibite. Since $n \gt 10^{k-1} \gt k$, 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.
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May 27th, 2016, 05:51 AM   #13
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Quote:
Originally Posted by zylo View Post
1845->.1845
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 left-hand 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?
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May 27th, 2016, 05:52 AM   #14
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Why isn't 3333333333..... with a countably infinite number of 3's not a natural number?
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May 27th, 2016, 05:59 AM   #15
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See my edit of post 12. It contains a proof.
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May 27th, 2016, 06:37 AM   #16
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Originally Posted by v8archie View Post
See my edit of post 12. It contains a proof.
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.
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May 27th, 2016, 07:01 AM   #17
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No. Only finite sequences appear in the rationals. The reals contain infinite sequences such as $(\pi - 3)$.
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May 27th, 2016, 07:10 AM   #18
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Quote:
Originally Posted by zylo View Post
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.
You've just stated that you agree with me.

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.
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May 27th, 2016, 07:45 AM   #19
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The rationals, infinite repeating sequences of natural numbers (minus the decimal point), are countable.

That leaves all the infinite non-repeating sequences of natural numbers.
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May 27th, 2016, 08:06 AM   #20
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Yes. The irrationals are uncountable.
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