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>Are irrational numbers even on a number line? Yes, e is between 2 and 3 and Pi is between 3 and 4. There are geometrical lengths corresponding to each number.

>Are irrational numbers even on a number line? Yes, e is between 2 and 3 and Pi is between 3 and 4. There are geometrical lengths corresponding to each number. >Isn’t it definitionally impossible to pick it as a “point along the line”? No, it's mathematically possible to have a random process which picks a random real between 0 and n, with equal probability. Imagine it akin to throwing a dart at a line and picking the point it lands on as the number. Since there are only countably many rationals and uncountably many irrationals (i.e. not just infinitely more, but so many that you could never pair off the rationals with the irrationals, there are just too many) on any such length of the real line, chances are the number you end up with is overwhelmingly likely to be irrational.

wat10000 2y ago
And it’s not “overwhelmingly likely” as in there’s a 99% chance or whatever. If you choose a random point on the line, the probability of choosing a rational is zero.
- IanKerr 2y ago
  Yep, exactly. I glossed over that detail a bit because explaining how a meagre set has a truly zero probability of being picked, while technically still being a possible result of a random process, is a bit messy to wrap your head around colloquially.
- alexashka 2y ago
  > If you choose a random point on the line, the probability of choosing a rational is zero. Wat? If a thing is in my pocket, there's an above zero probability of me picking it when I randomly take a thing out of my pocket. What are math people doing that's different?
  - josephcsible 2y ago
    > Do math people not feel the need to explain themselves when they state things that defy common sense everyone except math people agree upon? Is that part of thinking you're 'smart'? It's a pretty basic thing covered in undergrad prob/stats classes. We don't re-explain it every time we use it for the same reason computer scientists don't re-explain the halting problem every time it comes up.
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  - BalinKing 2y ago
    > If a thing is in my pocket, there's an above zero probability of me picking it when I randomly take a thing out of my pocket. This is only true if there are only a finite number of things in your pocket, though… I think an analogy is how we always have 1/n>0 for any finite (positive) number n—and yet, 1/infinity=0. For something more precise, I think the corresponding Wikipedia page (FWIW) is https://en.wikipedia.org/wiki/Almost_never https://en.wikipedia.org/wiki/Almost_never.
    - deleted 2y ago
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  - wat10000 2y ago
    They’re working with infinity. Your pocket is not infinite. Numbers are.
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      [deleted]
jncfhnb 2y ago
I’m struggling to understand how a thrown dart could land on an irrational number. It seems definitionally that any physically realized outcome must pertain to a rational number because it is impossible to physically measure one at any level of precision. It is possible to write a random process that returns 5 or pi with 50/50 odds so this isn’t a very compelling argument that it’s possible. I don’t feel the semantics of picking a random point along a number line is gg solved just by appealing to the existence of uncountably infinite irrationals. By most people’s definitions of random points along the number line, including the dart throw, it seems to me the probability of getting an irrational is 0. Invoking the number of possible outcomes has bad feeling implications. For example if your set is 1 2 3 pi 4, then the probability of getting an outcome in [3,4) is higher than [2,3) and that seems like it’s breaking the intuition of what the line represents. Like as a stupid example say we only include the irrational numbers between 9 and 10 and pick a random point between 1 and 10. If the random method uniformly sampled a point along the line by distance we would suggest a 90% chance of getting a rational number <= 9 and a 10% chance of getting an irrational number above 9. But if we sample by naive odds you’d probably claim there’s a near 100% chance of getting an irrational number above 9 because there’s an uncountable infinity up there. That seems dumb.
- thaumasiotes 2y ago
  > It seems definitionally that any physically realized outcome must pertain to a rational number because it is impossible to physically measure one at any level of precision. Sure, that's correct, but it isn't what people are talking about here. > By most people’s definitions of random points along the number line, including the dart throw, it seems to me the probability of getting an irrational is 0. That depends on the number line you're using. You can say that irrationals don't exist and you won't lose anything. But if your number line includes the reals, then the rationals form 0% of it. > Invoking the number of possible outcomes has bad feeling implications. That isn't how this is measured. You don't want to compare a count to an area. For probability, you need to compare like with like. A number line is one-dimensional, so we consider one-dimensional areas, or "lengths". The interval from 0 to 50 has length 50. How much of that length is occupied by rationals, and how much by irrationals? Each value is a point with no length. So, to measure the rationals, we assign to each rational point an interval that contains it. We will estimate the total length occupied by the rational numbers within the interval as being no greater than the total length of the intervals we put around each one. Since there are only countably many rationals, we can use an infinite series with a finite sum to restrict our total-length-of-intervals to a finite amount. (Rational number one gets an interval 3 units wide. Rational number two gets one 0.3 units wide. Number three gets one 0.03 units wide. What do all these intervals add up to? Four thirds.) We can scale those intervals however we like. We will scale them down. If our first set of intervals had total length 20, we can multiply them all by 1/400 and now they'll have total length 1/20. The limit of this process is a total length of zero, which is our upper bound on how much of the length of our interval is occupied by rational numbers. Since zero is also a lower bound on any length, we know that the total length of the interval occupied by rational numbers is exactly equal to 0. It is then easy to calculate the probability that a randomly chosen value from this interval will be rational: it is 0 (the amount of length occupied by rationals) over 50 (the total amount of length). > Like as a stupid example say we only include the irrational numbers between 9 and 10 and pick a random point between 1 and 10. If the random method uniformly sampled a point along the line by distance we would suggest a 90% chance of getting a rational number <= 9 and a 10% chance of getting an irrational number above 9. This seems to be just you being confused over the concept of a uniform distribution.

wat10000 2y ago

And it’s not “overwhelmingly likely” as in there’s a 99% chance or whatever. If you choose a random point on the line, the probability of choosing a rational is zero.

IanKerr 2y ago
Yep, exactly. I glossed over that detail a bit because explaining how a meagre set has a truly zero probability of being picked, while technically still being a possible result of a random process, is a bit messy to wrap your head around colloquially.
alexashka 2y ago
> If you choose a random point on the line, the probability of choosing a rational is zero. Wat? If a thing is in my pocket, there's an above zero probability of me picking it when I randomly take a thing out of my pocket. What are math people doing that's different?
- josephcsible 2y ago
  > Do math people not feel the need to explain themselves when they state things that defy common sense everyone except math people agree upon? Is that part of thinking you're 'smart'? It's a pretty basic thing covered in undergrad prob/stats classes. We don't re-explain it every time we use it for the same reason computer scientists don't re-explain the halting problem every time it comes up.
  - deleted 2y ago
    [deleted]
- BalinKing 2y ago
  > If a thing is in my pocket, there's an above zero probability of me picking it when I randomly take a thing out of my pocket. This is only true if there are only a finite number of things in your pocket, though… I think an analogy is how we always have 1/n>0 for any finite (positive) number n—and yet, 1/infinity=0. For something more precise, I think the corresponding Wikipedia page (FWIW) is https://en.wikipedia.org/wiki/Almost_never https://en.wikipedia.org/wiki/Almost_never.
  - deleted 2y ago
    [deleted]
- wat10000 2y ago
  They’re working with infinity. Your pocket is not infinite. Numbers are.
  - deleted 2y ago
    [deleted]

jncfhnb 2y ago

I’m struggling to understand how a thrown dart could land on an irrational number. It seems definitionally that any physically realized outcome must pertain to a rational number because it is impossible to physically measure one at any level of precision. It is possible to write a random process that returns 5 or pi with 50/50 odds so this isn’t a very compelling argument that it’s possible. I don’t feel the semantics of picking a random point along a number line is gg solved just by appealing to the existence of uncountably infinite irrationals. By most people’s definitions of random points along the number line, including the dart throw, it seems to me the probability of getting an irrational is 0. Invoking the number of possible outcomes has bad feeling implications. For example if your set is 1 2 3 pi 4, then the probability of getting an outcome in [3,4) is higher than [2,3) and that seems like it’s breaking the intuition of what the line represents. Like as a stupid example say we only include the irrational numbers between 9 and 10 and pick a random point between 1 and 10. If the random method uniformly sampled a point along the line by distance we would suggest a 90% chance of getting a rational number <= 9 and a 10% chance of getting an irrational number above 9. But if we sample by naive odds you’d probably claim there’s a near 100% chance of getting an irrational number above 9 because there’s an uncountable infinity up there. That seems dumb.

thaumasiotes 2y ago
> It seems definitionally that any physically realized outcome must pertain to a rational number because it is impossible to physically measure one at any level of precision. Sure, that's correct, but it isn't what people are talking about here. > By most people’s definitions of random points along the number line, including the dart throw, it seems to me the probability of getting an irrational is 0. That depends on the number line you're using. You can say that irrationals don't exist and you won't lose anything. But if your number line includes the reals, then the rationals form 0% of it. > Invoking the number of possible outcomes has bad feeling implications. That isn't how this is measured. You don't want to compare a count to an area. For probability, you need to compare like with like. A number line is one-dimensional, so we consider one-dimensional areas, or "lengths". The interval from 0 to 50 has length 50. How much of that length is occupied by rationals, and how much by irrationals? Each value is a point with no length. So, to measure the rationals, we assign to each rational point an interval that contains it. We will estimate the total length occupied by the rational numbers within the interval as being no greater than the total length of the intervals we put around each one. Since there are only countably many rationals, we can use an infinite series with a finite sum to restrict our total-length-of-intervals to a finite amount. (Rational number one gets an interval 3 units wide. Rational number two gets one 0.3 units wide. Number three gets one 0.03 units wide. What do all these intervals add up to? Four thirds.) We can scale those intervals however we like. We will scale them down. If our first set of intervals had total length 20, we can multiply them all by 1/400 and now they'll have total length 1/20. The limit of this process is a total length of zero, which is our upper bound on how much of the length of our interval is occupied by rational numbers. Since zero is also a lower bound on any length, we know that the total length of the interval occupied by rational numbers is exactly equal to 0. It is then easy to calculate the probability that a randomly chosen value from this interval will be rational: it is 0 (the amount of length occupied by rationals) over 50 (the total amount of length). > Like as a stupid example say we only include the irrational numbers between 9 and 10 and pick a random point between 1 and 10. If the random method uniformly sampled a point along the line by distance we would suggest a 90% chance of getting a rational number <= 9 and a 10% chance of getting an irrational number above 9. This seems to be just you being confused over the concept of a uniform distribution.