3 ms·

Because there is no bijection between the rationals and the reals, wouldn't that imply that there are some irrational reals with no rational between them, allow

Because there is no bijection between the rationals and the reals, wouldn't that imply that there are some irrational reals with no rational between them, allowing this function to be continuous, at least in places we can't actually compute?

btilly 2y ago
No, those are unrelated concepts.
roserevealed 2y ago
That doesn't actually follow- even though the reals are uncountable, they're what's called "separable", which means that there is a countable set which is "arbitrarily good at approximating them", basically- the rationals! So even though there are uncountable reals, for any real number, you can find rational numbers that get arbitrarily close. You can actually see this pretty easily if you think of reals in decimal notation- sqrt(2) is irrational, but 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414123, ... etc are all rational numbers that get closer and closer to sqrt(2). So you should think of real numbers as uncountable, but nonetheless surrounded by rationals no matter how close you zoom in.
- pfdietz 2y ago
  Not even the rationals are needed here. Just the rationals whose denominators are powers of two, for example, will do it (these have binary representations with a finite number of digits).
  - x3n0ph3n3 2y ago
    What about numbers with infinite digits?
    - dhosek 2y ago
      All numbers have infinite digits, it‘s just that the “terminating” numbers have an infinite string of zeroes at the end.
    - pfdietz 2y ago
      What about them?
pjbeam 2y ago
The issue there I think is the idea of two reals having a "between."
- seanhunter 2y ago
  Any two real numbers x and y are either equal or have an infinite number of reals between them don’t they? If your real numbers are x and y, then (x+y)/2 for example. The density of the rationals in the reals means that one of the numbers between them is provably a rational number.
  - pjbeam 2y ago
    For sure, I meant something more like a slot with a name.
n4r9 2y ago
I'm a bit late to see this comment, but it was disappointing that none of the replies so far show why this is not true. Take any irrational numbers x and y, and suppose for convenience that x < y. Let d = y-x i.e. the length of the interval between x and y. Let b be an integer so large that 1/b is less than d. Consider fractions of the form a/b. As a increases, a/b will eventually (strictly) exceed the value of x. Suppose we fix a to be the smallest such value, i.e. (a-1)/b < x < a/b. Adding d to both sides we get: a/b - 1/b + d < x + d Using the fact that 1/b < d and x + d = y, we get a/b < y. Therefore x < a/b < y.

btilly 2y ago

No, those are unrelated concepts.

roserevealed 2y ago

That doesn't actually follow- even though the reals are uncountable, they're what's called "separable", which means that there is a countable set which is "arbitrarily good at approximating them", basically- the rationals! So even though there are uncountable reals, for any real number, you can find rational numbers that get arbitrarily close. You can actually see this pretty easily if you think of reals in decimal notation- sqrt(2) is irrational, but 1.4, 1.41, 1.414, 1.4142, 1.41421, 1.414123, ... etc are all rational numbers that get closer and closer to sqrt(2). So you should think of real numbers as uncountable, but nonetheless surrounded by rationals no matter how close you zoom in.

pfdietz 2y ago
Not even the rationals are needed here. Just the rationals whose denominators are powers of two, for example, will do it (these have binary representations with a finite number of digits).
- x3n0ph3n3 2y ago
  What about numbers with infinite digits?
  - dhosek 2y ago
    All numbers have infinite digits, it‘s just that the “terminating” numbers have an infinite string of zeroes at the end.
  - pfdietz 2y ago
    What about them?

pjbeam 2y ago

The issue there I think is the idea of two reals having a "between."

seanhunter 2y ago
Any two real numbers x and y are either equal or have an infinite number of reals between them don’t they? If your real numbers are x and y, then (x+y)/2 for example. The density of the rationals in the reals means that one of the numbers between them is provably a rational number.
- pjbeam 2y ago
  For sure, I meant something more like a slot with a name.

n4r9 2y ago

I'm a bit late to see this comment, but it was disappointing that none of the replies so far show why this is not true. Take any irrational numbers x and y, and suppose for convenience that x < y. Let d = y-x i.e. the length of the interval between x and y. Let b be an integer so large that 1/b is less than d. Consider fractions of the form a/b. As a increases, a/b will eventually (strictly) exceed the value of x. Suppose we fix a to be the smallest such value, i.e. (a-1)/b < x < a/b. Adding d to both sides we get: a/b - 1/b + d < x + d Using the fact that 1/b < d and x + d = y, we get a/b < y. Therefore x < a/b < y.