I think I heard that all numbers in the real world are irrational. So that means most of math is not real, except of course the irrationals like pi :)
There are incomputable models.
Pi shows up in many physics equations, but that’s entirely due to our choice of units.
There is nothing that says that the distribution between rational and irrational numbers that show up in nature is the same as the distribution in our construction of the real numbers.
It's a bit more complicated than that.
Similar, an electron in a single isolated atom has specific quantised energy levels. But if you look at the electrons in a hunk of copper, they are essentially free to absorb and emit energy in almost arbitrary amounts.
An even stronger example is time: as far as I am aware, time is not quantised in any of our accepted theories.
There's some reasonable speculation that ultimately everything is quantised at the Planck scale, including time. But that's just a very reasonable hunch, not something that 'physics has shown'. (And you already point out that trying to marry quantum mechanics with general relativity is a hot mess.)
I agree that 'quantisation all the way down' is the way to bet. But that's just speculation.
(But I strongly disagree with your claim that physics is build on integers. Yes, it might be discretised, but there are plenty of discrete structures that are not integers. Look at a Rubik's cube for a simple example. On top of that: almost any real world measurement is better described by a probability distribution than by single number, be that an integer or otherwise.)
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:
That is true, but it's never taught. I don't even know what that mapping is, though I've seen it mentioned once in a popular treatment.
What's taught is always the mapping from naturals to rationals that overcounts the rationals, hitting them all an infinite number of times. (Because it's very easy to show a bijection between the naturals and the ordered pairs, but while (2,3) and (4,6) are distinct ordered pairs, they do not represent distinct rationals.)
But then all you've shown is that the naturals are at least as numerous as the rationals. To show that the naturals and the rationals have the same cardinality, you either rely on the idea that the naturals are the smallest infinite set, or you appeal to the fact that the naturals are a subset of the rationals.
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.
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
The argument as to why the irrational numbers are uncountable and the rationals are countable is more involved than what you've made out. But very simply you can think of it as you need an infinite string of digits to describe each irrational number, but each rational number can be written as two finite strings of digits (in the form A/B, where A and B are integers). So to write our the irrationals you have an infinite number of strings, where each string is also infinitely long, while with the rationals you have an infinite number of strings, but each string is finite.
That's literally the same thing. What is counting if it isn't being able to say what the next thing is? Do you have a mapping to integers or not? If so, then every n has n+1.
I know it was more complicated, but jaza had the essence of it. Without what they observed the whole thing falls apart. Yeah, it still needs proof, but I'm pretty sure five other comments went there.
> So to write our the irrationals you have an infinite number of strings, where each string is also infinitely long, while with the rationals you have an infinite number of strings, but each string is finite.
You've set the table but forgotten the feast! You're missing the step where you demonstrate that there's a number that isn't in this list. (Hint: think diagonally.)
The point I was trying to make is that there is no concept of 'next' inherent to the rationals, nor is there any natural or canonical ordering. The ordering and what comes 'next' is entirely a property of which arbitrary mapping you choose (I'm partial to Gödel numbering). The resultant order that your mapping imposes on the rationals is rarely useful or meaningful.
That ordering doesn't have the property that all sets of rationals contain a least element, or that any rational has a successor rational. (That would make them "well ordered".) But it's a natural ordering.
>> The argument as to why the irrational numbers are uncountable and the rationals are countable is more involved than what you've made out. But very simply you can think of it as you need an infinite string of digits to describe each irrational number, but each rational number can be written as two finite strings of digits (in the form A/B, where A and B are integers). So to write our the irrationals you have an infinite number of strings, where each string is also infinitely long, while with the rationals you have an infinite number of strings, but each string is finite.
This argument doesn't actually work. If there were only a countable number of irrational numbers, you could specify them all fully by doing no more than a countable amount of work, even stipulating that describing a single irrational number requires listing a countably infinite number of digits.
> Indeed, you've grasped the core of it.
What? That's not the core of anything. It tells you that the irrationals are dense in the real number line. You know what other set is dense in the real line? The rationals.
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.
Well... one of the consequences of that precision is the theorem that there is no such concept as choosing a real number "at random".
(Sqrt(2) as a real number, is actually encoded as the set of all rationals less than sqrt(2) on the number line).
The "real" in "real numbers" has ultimately not much to do with our everydays notion of real. I'd rather treat it as an arbitrary name. You could as well call them "asdfasdf numbers" and they'd remain the same.