# Distribution of types of numbers in the real line

I am learning about number systems in mathematics.

I am told that prime numbers are distributed in a specific pattern amongst the natural numbers, and that there is a notion of density amongst the primes, and that the primes are less dense, the farther you are along the natural numbers.

So if this is the case:

Let [x,y] be notation for an interval of natural numbers: the set whose elements are x, y, and every natural number between x and y, if the expression makes sense.

The length of an interval [x,y] is y - x.

If we have 2 intervals of natural numbers of the same length, say [a,b] and [c,d]. Then they may each contain a different number of prime numbers. If c is much larger than a, then its likely there are more primes in [a,b] than in [c,d].

Is this correct?

If we now let [a,b] and [c,d] be intervals of real numbers of the same length, then analogously, does the position of [a,b] and [c,d] in the number line affect the "density" of certain types of irrational numbers in the intervals?

For example are Quadratic Surds, or transcendental, or constructible numbers more "prevalent" earlier in the number-line (closer to 0) than farther out (as in the case of primes)?

I'm trying to see if intervals of real numbers are the "same" with regards to diversity of types of number. Or equivalently, if the density of the various types of special rational and irrational numbers is constant throughout the number-line (unlike the case of primes in the natural numbers).

• all dense countable subsets of $– Jorge Fernández Hidalgo Jan 11 '18 at 3:33 • Any open real interval of non-zero length contains a countably infinite set of rationals and a countably infinite set of irrational algebraics . The set of all algebraics is countable. Any open real interval of non-zero length contains a set of transcendentals whose cardinal is the cardinal of the reals. – DanielWainfleet Jan 11 '18 at 15:43 • To give a reasoned mathematical argument requires a definition of "evenly distributed". Counting integers in a bounded region of the real numbers at least gives a finite number, but when we try to argue about the rational numbers we have the difficulty that already any interval with positive length contains infinitely many of them. So comparing "how many" requires a choice of measuring stick. – hardmath Jan 13 '18 at 1:31 ## 1 Answer It's correct that primes get less frequent and that two "intervals" of natural numbers of the same length will contain different numbers of primes and that$[a,a+nm]$in many cases will contain more primes that$[b,b+m]$if$a<b$. Using the word "likely" or generalising to intervals of real numbers requires a more formalised way of measuring. I'm not sure what "Quadratic Surds" (english is not my native language, and by "surd" I mostly understand the symbol) mean, but$[2,3]$only contain$6$numbers that are square roots og natural numbers, while$[102,103]$contain$206$numbers with that property, so in a way that is a class of numbers that get more frequent. But both intervals contain infinitely many numbers, so you have be careful with any attempt to formalise it, because you quickly end up dividing by infinity and comparing the results. Almost all (that can be formalised, but let's skip that for now) real numbers are transcendental, and if$x$is transcendental so is$x+G$for any natural (or rational or ...) number$G\$, so any two intervals of real numbers contain the same infinte number (another concept that needs formalising that I'm going to skip here) of transcendental numbers.

The preceding paragraph (except the "..."-part) holds equally well if we substitute non-constructible for transcendental.

• So "most" real numbers are transcendental, and "most" real numbers also happen to be non-constructible? – trynalearn Jan 27 '18 at 21:09