Nope, that doesn''t mean the density is the same, because this theorem says "there is at least one" of each between two irrational numbers, it does not specify how many of each, or their densities.
Similarly, for each irrational number x , there exists an integer n so that n > x, and for each integer n there exists an irrational number x so that x > n. But IR \ Q is far more infinite than IN.
Where does the "more infinite" result come from? I don''t know the english version of the world, but it could be something like "countability". What the? Well, it means that you can count all elements in a set (that is, tie at least one number to each element) without forgetting any. A countable set is therefore less infinite than an uncountable set. We have two important results :
Q is countable ( 0 for 0, 1 for 1, 2 for -1, 3 for 1/2,4 for -1/2, 5 for 1/3, 6 for -1/3, 7 for 2/3...) : using that method you can tie one integer to each rational number.
IR is uncountable : suppose it was, and write each real number xn in [0,1] as the series of digits 0,xn1xn2xn3...
We therefore have a series of real numbers xn where n is an integer tied to xn. We can build the real y = 0,y1y2y3y4y5... where yk = 0 if xkk > 0 and yk = 1 if xkk = 0. This number has at least one digit of difference with ANY of the reals in the series xn. But since we counted all the elements in [0,1], and y is not in the series, y is not in that range. But IT IS! Therefore, IR is uncountable.
Since Q is countable and IR is uncountable, then IR\Q must be uncountable too (or we''d have a way to count IR), which means IR\Q is "more infinite" than Q.
ToohrVyk
