If you think about a set with an *infinite* number of elements, you might conclude that there is not much to say about the number of elements anymore. Infinite is infinite, right? Well, it turns out that not every infinity has the same “value”, and that there’s actually quite a lot to say about it.

Take the *natural numbers*, \(\mathbb{N}=\{0,1,2,\ldots\}\). There are an infinite number of those. The *integers* (\(\mathbb{Z}\)) are the natural numbers, together with the negatives of the natural numbers. There are, of course, also an infinite number of those. You might expect that there are twice as many integers as there are natural numbers, but you would be mistaken. We can prove this by associating each integer with a natural number, like this:

\(\mathbb{Z}\) | 0 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | … |

\(\mathbb{N}\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | … |

This shows that the integers correspond one-to-one to the natural numbers. The *cardinality* (the number of elements) of \(\mathbb{N}\) and \(\mathbb{Z}\) is the “smallest” infinity. It is written as \(\aleph_0\) (aleph-null, where aleph is a letter of the Hebrew alphabet). Infinite sets with cardinality \(\aleph_0\) are called *countable* (sometimes more specifically *countable infinite*), since the elements of such a set can be *enumerated* (as in the table above with the elements of \(\mathbb{Z}\)).

## Sorry, No Vacancies

Once things get infinite, the normal intuition about numbers no longer works. Take, for example, *Hilbert’s paradox of the Grand Hotel*: consider a hotel with a countable infinite number of rooms that are all occupied. If a new guest arrives, he or she *can* still be accommodated. We simply move the guest that is in room 1 to room 2, the one from room 2 to room 3, and so on. Room 1 is then free for the new guest. So we now have one happy guest in room 1, and a countable infinite number of slightly miffed ones in the others.

## Beyond Countable

The *rational numbers* (\(\mathbb{Q}\)) still have cardinality \(\aleph_0\). It is tempting to conclude from this that all infinite sets have the same cardinality, but that is not true. In the step to the *real numbers* (\(\mathbb{R}\)), something fundamentally changes. The reals *cannot be enumerated*, an aspect of the reals that is beautifully highlighted when they are described as the *continuum*. Not unexpectedly, the reals are called *uncountable infinite*.

But it doesn’t stop there. The cardinality of the reals is only the “smallest” uncountable infinity. The different cardinalities can actually be numbered: \(\aleph_0\), \(\aleph_1\), \(\aleph_2\),… In this list, each cardinality is the one that is just bigger than the previous one. If the *continuum hypothesis* is true (way out of the scope of this article), then the cardinality of the reals (noted as \(\mathfrak{c}\), and proven to be equal to \(2^{\aleph_0}\)) is equal to \(\aleph_1\). So, the next time you think about infinite sets, be aware that it’s more complicated than you’d expect…

Hi! I'm new here, but I have a simpler proof of the irrationality of the square root of two. First, we'll assume that p/q is in lowest terms. If this is true, then p^2/q^2 is in lowest terms. (This is true because the prime factorization of squares is the original's repeated twice.) Now, if p^2/q^2=2 and is in lowest terms, then q^2 is 1. Thus, q is 1 (or -1). Going back to the fraction p/q, we see that if q equals 1, p equals the square root of two. However, if p isn't an integer, then p/q isn't a fraction. I hope you liked my ideas! Tell me what you think.

I’m afraid there are several things wrong with this… I would suggest that you take a look at the relevant Wikipedia article, which contains several proofs (look at “Proof by unique factorization”, which seems relatively close to what you’re trying to do). You could also take a look at Proof by contradiction, where the irrationality of the square root of 2 is also used as an example. Keep up the interest for math!

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