Legal proof that set of Rational numbers is countable?

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SUMMARY

The set of rational numbers (Q) is countable, as established through a bijective function mapping positive integers (Z) to rational numbers. The proof utilizes a matrix listing method, where each rational number is represented as p/q, with p and q being positive integers. The Cantor diagonal argument demonstrates the countability by systematically listing the rationals while skipping non-lowest terms. This method confirms that a one-to-one correspondence exists between the set of positive integers and the set of rational numbers.

PREREQUISITES
  • Understanding of rational numbers and their representation as fractions (p/q).
  • Familiarity with positive integers and their properties.
  • Knowledge of bijective functions and one-to-one correspondences.
  • Basic concepts of Cantor's diagonal argument and countability.
NEXT STEPS
  • Study Cantor's proof of the countability of rational numbers in detail.
  • Explore the concept of bijective functions and their applications in set theory.
  • Learn about different methods of listing rational numbers, including the matrix approach.
  • Investigate the implications of countability in other mathematical contexts, such as real numbers.
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Mathematicians, educators, students in advanced mathematics, and anyone interested in set theory and the properties of rational numbers.

Liquid7800
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Hello,

Just wondering if this is a correct way to say the set of RATIONAL numbers is countable:

Rationals (Q) is countable , because for every Q = p/q , such that p & q are positive INTS (Z)

and since the set of positive INTs (Z) is countable ( a 1:1 correspondence) Q is countable because it is a SUBSET of a countable set...

it can be listed by listing those Q's with
denominator q = 1 in the first row of a listing matrix
denominator q =2 in the second row and so on...
with p1 = 1 , p2 =2, etc... for each row...

will eventually cover ALL rationalsREMARK:
Ive seen the little N x N matrix and path listing for each rational...but I would like to know...
How would you show this as a mapped 1:1 function?? such that A ---> B showing f(a) = some b??Thanks for any input...
 
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Liquid7800 said:
How would you show this as a mapped 1:1 function?? such that A ---> B showing f(a) = some b??
Thanks for any input...

You might be asking for an onto function. 1:1 means if f(a)=f(b), then a = b.
 
Thanks for the reply...
However, when I said
How would you show this as a mapped 1:1 function?? such that A ---> B showing f(a) = some b??
I shouldve said:
A 1:1 correspondence function or a bijective function...
 
The standard proof (Cantor) is as follows: Let a(i,j)=i/j. Then the order to show countable is a(1,1), a(2,1), a(1,2), a(3,1), [a(2,2)], a(1,3), a(4,1), a(3,2), a(2,3), a(1,4),... skipping terms in [] since these are not in lowest terms.
 

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