3
$\begingroup$

For any positive rational numbers $p/q$ and $m/n$, I can decide which is bigger by comparing $pqn$ and $mqn$.

For irrational numbers, we can't use this test. Furthermore, almost all irrational numbers can't be specified by a finite formula in a given symbolic system (i.e. computed to arbitrary precision with an algorithm whose definition is finite). Some can: $\sqrt[5]{2}$ and $\ln{\pi}$ are two examples. There are algorithms available to compute either of these to arbitrary precision, and at some point, their decimal expansions diverge and we can conclude which one is bigger. This is equivalent to finding a rational number which is bigger than one but less than the other.

However, is it possible to prove a fact like $\sqrt[5]{2} > \ln{\pi}$ without obtaining a rational number between them?

More generally, say we can describe two expressions for real numbers, $r$ and $s$, in terms of $+,-,\times,\div$, some transcendental functions, infinite series, etc., such that we could approximate each of their decimal expansions. Is it possible in general to compute/prove which of the expressions $r$ and $s$ is bigger without finding a decimal expansion or otherwise finding a rational number in between them?

$\endgroup$
  • 1
    $\begingroup$ This is an interesting class of problems but I think it's too open ended to have a single answer aside from "look at the decimal expansions" or some variation. Possibly you could come up with a solution on a reasonable subclass (say, rational powers of rationals, maybe with some pi in there, and try to work in log, and so on). But I think it would have to be iteratively expanding the class of problems you can solve, not doing it all at once. $\endgroup$ – Richard Rast Oct 18 '18 at 23:01
  • $\begingroup$ As a side note I once spent three pages proving an inequality like that, and it involved some nontrivial integration and so on. There may have been an easier way but I never found it $\endgroup$ – Richard Rast Oct 18 '18 at 23:02
3
$\begingroup$

The set of numbers for which there is a formula to approximate their decimal expansions is called the computable numbers. The order relation on the computable numbers is non-computable. If the two numbers are known to be unequal, then it is computable, but the method is successive approximation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.