A gap in Cantor's diagonal argument ?
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From: Reinhard Fischer on 6 Aug 2010 05:01 Yes, I can see. If a natural number had an infinite number of digits, infinity would be a natural number. And this is not possible. This makes sense even without the arguments of set theory. This in turn means, a terminating decimal at no time will convert into a nonterminating decimal, no matter how many decimal places you add. Thanks.
From: Robert H. Lewis on 6 Aug 2010 06:44 You wrote: "a terminating decimal at no time will convert into a nonterminating decimal, no matter how many decimal places you add." What does "convert" mean? What about 4.0 = 3.9999999... Robert H. Lewis Fordham University
From: jbriggs444 on 6 Aug 2010 11:47 On Aug 6, 9:01 am, Reinhard Fischer <reinhard_fisc...(a)arcor.de> wrote: > Yes, I can see. If a natural number had an infinite number of digits, infinity would be a natural number. And this is not possible. This makes sense even without the arguments of set theory. This in turn means, a terminating decimal at no time will convert into a nonterminating decimal, no matter how many decimal places you add. Thanks. Numbers are what they are. They do not change. Not even a "sequence" of numbers changes. It is what it is. Every position in the sequence has a fixed, unchanging value. Yes, there is an argument that can be made along the lines of: 1 is finite. The successor of a finite number is finite. It doesn't matter how many times you apply the successor operation, the result is always finite. Every number you can get by a terminating sequence of successor operations is finite. Or, taking it more in the dierction you're after... Any number you can express in a finite number of decimal digits is finite. 1 is finite. 10 is finite. 10^10 is finite. 10^(10^10) is finite. Every number you can get to by a terminating sequence of exponentiations is finite. It turns out that these arguments are not rigorous unless you carefully phrase the assumptions (axioms) on which they are based. Peano phrased some intuitively appealing axioms in a way that completely avoids mention of "time" or "sequence", "changing" or "adding". The fifth one is crucial to the arguments above. See also mathematical induction.
From: Arturo Magidin on 6 Aug 2010 14:12 On Aug 6, 1:24 am, Reinhard Fischer <reinhard_fisc...(a)arcor.de> wrote: > For a better understanding of "countable/uncountable": > > Which of the following statements is true/false? > > 1. > Terminating decimals are always countable, no matter how many decimal places they consist of. The statement is incoherent. You mean, perhaps, that a set (of real numbers) that consists only of real numbers that have a terminating decimal expansion is countable. If so, then "true". > 2. > Decimals are uncountable, if they consist of infinitely many decimal places. As above; "decimals are uncountable" is incoherent. If you mean, will a set that consists only of real numbers with infinite decimal expansions always be uncountable, the answer is "No". If you mean, "does the set of *all* real numbers that have infinite decimal expansion have uncountable cardinality", then the answer is "yes". > 3. > Numbers are countable, if they consist of a finite number of decimal places. Incoherent as above. Numbers are numbers. "Countability" applies to sets. You need to be clear and precise in your statements, or they are "not even wrong". > 4. > Numbers are uncountable, if they consist of an infinite number of decimal places. As above. "Not even wrong". > 5. > Natural numbers are countable, if they consist of a finite number of digits. The *set* of natural numbers is countable, and *all* natural numbers have only finitely many digits (use induction). You are dangerously close to "not even wrong" here as well. > 6. > Natural numbers are countable, even if they consist of an infinite number of digits. Each natural number is finite. All natural numbers have expressions with finitely many digits. A string of infinitely many digits does not represent any natural number. Close to "not even wrong". -- Arturo Magidin
From: Arturo Magidin on 6 Aug 2010 14:14
On Aug 6, 8:01 am, Reinhard Fischer <reinhard_fisc...(a)arcor.de> wrote: > Yes, I can see. If a natural number had an infinite number of digits, infinity would be a natural number. If a natural number had an infinite number of digits, then black is white. False implies anything. >And this is not possible. This makes sense even without the arguments of set theory. It has nothing to do with set theory, and everything to do with the definition of natural number. By Peano's fifth postulate, the set of natural numbers with expressions consisting of only finitely many digits equals the set of all natural numbers. > This in turn means, a terminating decimal at no time will convert into a nonterminating decimal, no matter how many decimal places you add. This is nonsense as far as I can tell. Note that every real number that has an expression with a terminating decimal expansion *also* has an expression with a *non-terminating* decimal expansion. -- ArturO Magidin |