Intuitive way to explain |Q| = aleph zero?

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SUMMARY

The discussion focuses on the intuitive explanations for the cardinality of rational numbers (Q) and perfect squares in relation to natural numbers (N). It establishes that both sets are countable through bijections: perfect squares correspond to their square roots, while rational numbers can be arranged in a systematic listing that demonstrates their countability. The diagonal method is referenced as a common proof technique, emphasizing the distinction between countable and uncountable sets. Additionally, prime decomposition is suggested as a method to establish a bijection from Q to a subset of N.

PREREQUISITES
  • Understanding of cardinality and countability in set theory
  • Familiarity with bijections and one-to-one correspondences
  • Basic knowledge of rational numbers and perfect squares
  • Concept of prime decomposition and its application in proofs
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  • Study the diagonal argument for proving the uncountability of real numbers
  • Explore the concept of countable versus uncountable sets in depth
  • Learn about prime factorization and its role in set theory
  • Investigate additional methods for demonstrating the countability of rational numbers
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Mathematicians, educators, and students interested in set theory, particularly those exploring concepts of cardinality and countability in mathematics.

EnumaElish
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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 argument, 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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