Intuitive way to explain |Q| = aleph zero?

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The discussion focuses on intuitive explanations for the cardinality of rational numbers and perfect squares, demonstrating that both sets can be mapped to natural numbers (N). For perfect squares, a one-to-one correspondence is established through square roots, while for rationals, a systematic listing and a diagonal counting method illustrate their countability. The conversation emphasizes that both proofs, although differing in complexity, reveal the same underlying principle of countability. Additionally, a method using prime decomposition is mentioned as a bijection from rational numbers to a subset of natural numbers. Overall, the thread highlights various approaches to understanding the concept of cardinality in a calculus-grade context.
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What is an intuitive, calculus-grade way to explain that rationals have the same cardinality as N? (Same question for perfect squares.)
 
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I'm not trying to be difficult, but aren't the standard proofs pretty inuitive? Especially for the perfect squares. To every element of the set of perfect squares you can correspond exactly on element of the set of naturals: its square root. The same goes for the other way around. So you have a one to one correspondance between the squares and naturals, so they have the same cardinality. The proof is nearly identical for the rationals, except that the bijection is somewhat more complicated
 
Look at http://home.att.net/~numericana/answer/sets.htm in the section called "The countability of rational numbers".
 
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Heres a pictorial way to show it for the rationals. Start with the positive rationals and start listing them like so
1/1 1/2 1/3 1/4 1/5...
2/1 2/2 2/3 2/4 2/5 ...
3/1 3/2 3/3 3/4 3/5 ...

Now, to count, follow this sequence: 1/2, 1/2, 2/1, 3/1, 2/2, 1/3, 1/4, 2/3,... (I hope you can see what I'm doing). Now, it should be clear that this list is countable. Now, insert the negative rationals just before their positive counterparts, i.e. -1/1, 1/1, -1/2,..., and just put zero at the top left. Now, using the same arguement, the full set of Q is countable.

I'm not sure whether this was something like what you were after, but it seems quite intuitive!
 
Ah, diagonal method - possibly because a proof that Q is countable is called diagonal, and that R isn't yet has the same adjective, people have wrong impressions.One way is a standard method using prime decomposition.

Send a/b to 2^a*3^b (for a and b >0) or 2^a*3^b*5 for negative rationals. This is a bijection from Q onto a subset of N. This proof shows that any finite product of countable sets is countable, and shows an important difference between union and product (in the categorical sense, perhaps).
 
I thought both the "perfect squares" and the "rationals" demonstration can use the cross-tabulation square where row and column headings are the natural numbers. For the perfect squares, each cell holds the product "row # * column #"; all perfect squares are on the diagonal, which you can count linearly.

For the rationals, each cell is row #/column # (or the inverse), but I wasn't sure how to count them. Now I know. Thanks.
 
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