[lrc]

It is better stated that: since there are "so many more" irrational numbers than rational ones, if you were to pick a real number "at random," the probability that it would be rational is zero. The "many more" and "random" ideas are made precise in measure theory (and elsewhere).

[apetersonBFI]

If you actually look at how real numbers are constructed. They are quite bizarre. The simple concept of the number line becomes a quite complicated set of sets that follow certain conditions.

(Sqrt(2) as a real number, is actually encoded as the set of all rationals less than sqrt(2) on the number line).

[thaumasiotes]

> The "many more" and "random" ideas are made precise in measure theory (and elsewhere).

Well... one of the consequences of that precision is the theorem that there is no such concept as choosing a real number "at random".

[jaza]

There is an infinite quantity of both rational and irrational numbers, so isn't it therefore impossible for there to be more of one than of the other? Or is the reasoning that, because there is an infinite quantity of irrational numbers between any two given rational numbers, there are therefore many more irrational numbers than rational numbers? I would have thought that there being an infinite quantity of both, makes it impossible to compare the quantities.

[defrost]

There is a mapping from counting numbers (1, 2, 3, ...) to rationals and back again that shows these quantities are the same; for every element in set A there's an element in set B and vice versa.

This is not the case for irrationals... therefore it is concluded that the infinity of irrationals is a larger infinity than the infinity of rationals.

See:

https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument

https://mathworld.wolfram.com/CantorDiagonalMethod.html

[SantalBlush]

There are also infinitely many rationals between any two distinct irrationals.

My favorite way of visualizing the difference uses the fact that every rational has a repeating decimal after some nth decimal place, and no irrational has a repeating decimal. Say you want to construct a number x, where 0 < x < 1, by drawing integers 0 through 9 randomly from a hat. Each integer drawn from the hat is placed at the end of the decimal; for example, if you draw 1,3,7,4 then the decimal becomes 0.1374. You then draw, say, 1, and it becomes 0.13741, and so on. If you could draw infinitely many times from the hat, what is the probability that you'll construct a number with a repeating sequence? That would give a rational number.

[eru]

> There is an infinite quantity of both rational and irrational numbers, so isn't it therefore impossible for there to be more of one than of the other?

Mathematicians can even meaningfully compare infinities.

See eg https://www.cantorsparadise.com/this-may-seem-more-irrationa... or https://math.stackexchange.com/questions/474415/intuitive-ex...

You can also look at eg a uniform random variable on the interval between 0 to 1. The probability of hitting a rational number is 0%. The probability of hitting an irrational number is 100%.

> Or is the reasoning that, because there is an infinite quantity of irrational numbers between any two given rational numbers, there are therefore many more irrational numbers than rational numbers?

No, that's not enough. There are also an infinitely many rational numbers between any two given irrational numbers.

[Tao3300]

> because there is an infinite quantity of irrational numbers between any two given rational numbers

Indeed, you've grasped the core of it. There's no rule you can write for irrational numbers such that "b is the next number after a", because there are infinitely many numbers between a and b that you'd be missing. You can't count them, i.e. you can't map them to integers.

Uncountable Infinities > Countable Infinities

[wheels]

Counter-intuitive as it may seem, math does have the notion of larger infinities.