Well, at least pi is real... Is that little hypocrite of i even rational? Hm?
it's definitely 5-adicly rational
Haha, well done!
Out of curiosity, I guess pi cannot be rational in any p-adic set, right?
yeah, the remainder of pi after p^2 for p>2 will always be the same as the remainder after division by p (p^2 if p=2), so it truncates to the left of the radix, and p-adic rationals can't go off infinitely to the right of the radix. also, more trivially, pi isn't algebraic so it's not the solution to any polynomial with rational coefficients, so it's also not the solution to any polynomial mod p^n.
Not sure I understand all of it, but the last part about pi not being algebraic made sense to me, at least! Thanks!
what's the remainder of pi after division by 3 and 3^2? notice how the remainder is the same -- i.e. there's no 3s digit needed. and the same will be true for all higher powers of 3. this is because pi is <3^n for all n > 1. but we failed to express the fractional part of pi. if we extend to the p-adic rationals, we can express it as an infinite expansion but that's not allowed for p-adics -- they can only have a finite number of digits in the part to the right of the radix.
it's an algebraic integer in C, and even in Q(i). So kinda.
e: I propose a solution
Naturally based
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