A math problem that kills brain cells

This proof is not difficult, but the cardinality of infinite set is really an interesting thing.

For example, even numbers (2,4,6 ... 2n) are positive integers (1, 2,3, ... n).

However, through: 2←→ 1, 4←→2, 6←→3, ..., 2n←→n, ... a one-to-one mapping is established between them, so that even numbers and positive integers have the same cardinality, which is what we know: "The whole is greater than it. Ha ha laugh

Cut the crap and start to prove it now:

The set of rational numbers in (0, 1) =B;

Set of natural numbers A = {0, 1, 2, 3, ... n, n+ 1, ...}

Then I only need to arrange B into a sequence containing all rational numbers in (0, 1), and I can prove that A and B have the same cardinality.

That is, b = {r 1, R2, R3, ... rn, r (n+ 1)...} where r1≠ R2 ≠ ... ≠ rn ≠ r (n+650 ..

Establish sequence x:

1/2, 1/3,2/3, 1/4,2/4,3/4, 1/5,2/5,3/5,4/5, 1/6,2/6...5/6,.... 1/n，2/n....(n- 1)/n ....

The numerator and denominator in X can be reduced (for example, 2/4, 2/6, 3/6 ...).

You get the sequence rn,

Rn includes all rational numbers in (0, 1) and r 1 ≠ R2 ≠ ... ≠RN≠R(n+ 1)≦. ..

r 1= 1/2←→0

r2= 1/3←→ 1