Talk:Frivolous theorem of arithmetic

Countability of rational numbers

????????Since between each consecutive integer there is the same uncountably infinite number of rational and irrational numbers????????

The algebraic numbers (which include the rational numbers) are countable, while the irrational numbers have the cardinality of the continuum (uncountable infinity).

—Daniel Forgues 20:42, 24 August 2010 (UTC)

In that case, we should amend that line thus: "Since between each consecutive integer there is the same uncountably infinite number of rational and irrational numbers ..."

But I'm no Georg Cantor, and it'll take me a while to understand why this is so. In the meantime, you can go ahead with the amendment if you're sure of it. Alonso del Arte 21:12, 25 August 2010 (UTC)

You probably have been led to believe, wrongly, that there are as many rational numbers as irrational numbers because when we consider the real line (which is a totally ordered set), we have

Between any two rational numbers, one may find at least one (actually an infinity) irrational number; between any two irrational numbers, one may find at least one (actually an infinity) rational number.

but one may not assume that the above mentioned infinities are the same! Also, the cardinality of a set (where the order of elements does not matter) allows for any ordering of its elements.— Daniel Forgues 07:17, 14 November 2012 (UTC)

---

The proof of the countability of the rationals is pretty simple, here it is:

The set $\scriptstyle \mathbb {Q} \,$ of rational numbers is countable.

—Daniel Forgues 22:23, 25 August 2010 (UTC)

That's a broken link now. But maybe your addition of "As we let m^2 go to infinity" could be construed to turn my thought experiment into an actual proof. Alonso del Arte 23:45, 13 November 2012 (UTC)

Look for orderings of rational numbers, where we have a 1-to-1 correspondence with the set of positive integers, thus implying equicardinality between the two sets, which means that the rational numbers are denumerable (countable). — Daniel Forgues 07:17, 14 November 2012 (UTC)

---

The proof of the countability of the algebraic numbers is very similar. The real numbers are uncountable only because the transcendental numbers are uncountable. Virtually all the real numbers are transcendental.

—Daniel Forgues 22:57, 25 August 2010 (UTC)

Look for orderings of algebraic numbers, where we have a 1-to-1 correspondence with the set of positive integers, thus implying equicardinality between the two sets, which means that the algebraic numbers are denumerable (countable). — Daniel Forgues 07:17, 14 November 2012 (UTC)

---

I've seen those proofs in books and in journals, and I've thought to myself, "That makes sense." But that doesn't relieve me of the fear that in talking about infinity I will make some small mistake that renders my remarks monumentally (and perhaps infinitely) wrong. It may have already happened in actuality.

But the point here is that there are no more integers between one integer and the next than there are between any other integer and the integer following that one. If I'm right about that, that's the point I wanted to make before going into the quasi-proof of the theorem. Alonso del Arte 22:15, 26 August 2010 (UTC)