30

Is it possible to obtain $\pi$ from finite amount of operations $\{+,-,\cdot,\div,\wedge\}$ on $\mathbb{N}$ (or $\mathbb{Q}$, the answer will still be the same)? Note that the set of all real numbers obtainable this way contains numbers that are not algebraic (for example $2^{2^{1/2}}$ is transcendental).

Bonus: If it happens that the answer is no, is it a solution to some equation generated that way (those $5$ operations performed finitely many times on elements on $\mathbb{N}$) ?

Adam Zalcman
  • 3,446
Mathemagician
  • 443
  • Related: https://math.stackexchange.com/questions/2611084/is-it-possible-to-express-pi-as-ab-for-a-and-b-non-transcendental-numb – user202729 Jul 28 '18 at 13:34
  • 4
    (this one is more general. If the answer to this one is "no" then the answer to the other one is "no" as well) – user202729 Jul 28 '18 at 13:35
  • 6
    Of course not, but I can't prove it. There are only countably many finite expressions of this sort, so you can only express countably many real numbers this way. Unless a real has some reason to be expressible it almost certainly can't be. – Ross Millikan Jul 28 '18 at 14:36
  • 3
    Yes, there are only countably many of them, so most real numbers do not fall into this category. Still, not much can be deduced about any specific number from that fact. π is arguably somewhat special. – Mathemagician Jul 28 '18 at 14:57
  • Could Machin-like formulas be of any use? – sam wolfe Sep 05 '19 at 17:54
  • It can be obtained using calculus, check Wikipedia. – Anas Khaled Oct 31 '19 at 17:55
  • If you only use +,-,x,/, you can't get $\pi$ because it will only yield a rational number, so you have to use the exponential operator sometime. – Some Guy Jan 05 '21 at 04:27
  • The bonus question is the same question except allowing the use of $\log$, since all the operations are invertible, and their inverse is included in the allowed operations - except for $^$, which needs $\log$. – SomeCallMeTim Dec 07 '23 at 13:29
  • As mentioned above, it is probably false, but a disproof is difficult. A hope could be to describe a subfield of $\Bbb{R}$ containing all elements generated by the method you describe, but not containing $\pi$. How such a field would look is difficult to imagine though. – SomeCallMeTim Dec 07 '23 at 13:44

0 Answers0