3 ms·

This actually helps a lot with the intuition, thanks! So in a sense every irrational number is an infinite number because there is no way to actually represen

• colanderman 13y ago
  Almost! See pesenti's link about definable numbers: most irrationals have no finite representation, although almost all the ones we are familiar with (e.g. sqrt(2), π, e) do: they can be represented as mathematical formulæ. (And indeed, note that such "definable" reals are countable!) So in my example above, π would not work if, say, your tree was a parse tree of mathematical formulæ: one could locate the formula for π in finite time. Unfortunately I cannot give an example of a single number which does not have such a finite parse tree, as, by its very nature, I cannot write down a finite description of it!
• jfarmer 13y ago
  Not quite. I can represent pi in a finite way, like so: π. That's about as finite as it gets! However, if you fix a finite alphabet beforehand, you will never be able to describe all real numbers using strings constructed from this alphabet. You can always add a new symbol to your alphabet to represent some new irrational number, but no matter how many symbols you add you'll never be able to reach "every" irrational number. Put another way, for a given alphabet there's no way to represent every irrational number in a finite way. However, you could have an irrational number that can be finitely represented in one alphabet, but unable to be finitely represented in another.