(My intuition tells me that if you choose a subset of the primes dividing the base, you can somehow obtain from that a value of √1 in a way that maps “none” to 1, and “all” to −1, and the remaining combinations to the irrational integers. No more specific ideas.)
That is correct! In base pe11⋅...⋅penn (for distinct primes p1,...,pn), an integer is determined by an integer in each of the bases p1,...,pn, essentially by the Chinese remainder theorem. In other words, Qpe11⋅...⋅penn=Qp1×...×Qpn. In prime base, 1 and −1 are the only two square roots of 1. In arbitrary base, a number squares to 1 iff the number it reduces to in each prime factor of the base also squares to 1, and there’s 2 options for each of these factors.
Sounds convincing (although maybe that’s because you agree with me), but I have no idea how specifically one could “reduce a non-periodic infinite number in a prime factor”.
If you have a 10-adic integer, and you want to reduce it to a 5-adic integer, then to know its last n digits in base 5, you just need to know what it is modulo 5n. If you know what it is modulo 10n, then you can reduce it module 5n, so you only need to look at the last n digits in base 10 to find its last n digits in base 5. So a base-10 integer ending in …93 becomes a base-5 integer ending in …33, because 93 mod 25 is 18, which, expressed in base 5, is 33.
The Chinese remainder theorem tells us that we can go backwards: given a 5-adic integer and a 2-adic integer, there’s exactly one 10-adic integer that reduces to each of them. Let’s say we want the 10-adic integer that’s 1 in base 5 and −1 in base 2. The last digit is the digit that’s 1 mod 5 and 1 mod 2 (i.e. 1). The last 2 digits are the number from 0 to 99 that’s 1 mod 25 and 3 mod 4 (i.e. 51). The last 3 digits are the number from 0 to 999 that’s 1 mod 125 and 7 mod 8 (i.e. 751). And so on.
That is correct! In base pe11⋅...⋅penn (for distinct primes p1,...,pn), an integer is determined by an integer in each of the bases p1,...,pn, essentially by the Chinese remainder theorem. In other words, Qpe11⋅...⋅penn=Qp1×...×Qpn. In prime base, 1 and −1 are the only two square roots of 1. In arbitrary base, a number squares to 1 iff the number it reduces to in each prime factor of the base also squares to 1, and there’s 2 options for each of these factors.
Sounds convincing (although maybe that’s because you agree with me), but I have no idea how specifically one could “reduce a non-periodic infinite number in a prime factor”.
If you have a 10-adic integer, and you want to reduce it to a 5-adic integer, then to know its last n digits in base 5, you just need to know what it is modulo 5n. If you know what it is modulo 10n, then you can reduce it module 5n, so you only need to look at the last n digits in base 10 to find its last n digits in base 5. So a base-10 integer ending in …93 becomes a base-5 integer ending in …33, because 93 mod 25 is 18, which, expressed in base 5, is 33.
The Chinese remainder theorem tells us that we can go backwards: given a 5-adic integer and a 2-adic integer, there’s exactly one 10-adic integer that reduces to each of them. Let’s say we want the 10-adic integer that’s 1 in base 5 and −1 in base 2. The last digit is the digit that’s 1 mod 5 and 1 mod 2 (i.e. 1). The last 2 digits are the number from 0 to 99 that’s 1 mod 25 and 3 mod 4 (i.e. 51). The last 3 digits are the number from 0 to 999 that’s 1 mod 125 and 7 mod 8 (i.e. 751). And so on.