Undergrad Base Pi Integers: Isomorphism with Rationals?

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The discussion centers on the concept of expressing numbers in a base defined by the irrational number π, similar to how integers are expressed in base 10. It highlights that while Z_π can be constructed with elements of the form a_nπ^n + a_{n-1}π^{n-1} + ... + a_1π + a_0, the operations defined do not yield rational numbers, leading to confusion about whether this set can be considered a group. Participants debate the implications of addition in this system, noting that traditional arithmetic rules do not apply, as evidenced by the claim that 5 + 5 equals π in Z_π, which contradicts standard integer addition. The conversation reveals a fundamental disagreement on the nature of the operations and the validity of the group structure proposed. Ultimately, the discussion raises questions about the meaningfulness of using π as a base in this context.
  • #31
I think maybe you want the 1's place the be the ##\frac{\pi}{10}##'s place?
 
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  • #32
Jarvis323 said:
I think maybe you want the 1's place the be the ##\frac{\pi}{10}##'s place?
If that's true then the 100s place should be ##10\pi##, not ##\pi^2##. But then the set is just the integers scaled by ##\pi/10## which again is not interesting.
 
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  • #33
The OP seems to have bailed from this thread. It's probably not useful to try and figure out what he really meant. (Is π = 10?)
 
  • #34
Atran said:
A proper number is expressed in \pi in a similar way as a decimal integer is expressed in base 2. For example, 4375_{\pi} = 4{\pi^3}+3{\pi^2}+7{\pi}+5.
@Atran, I don't believe you understand how numbers can be represented in different bases.
In base 2, the digits are 0 and 1. Counting would proceed from 0, 1, 10, 11, 100, 101, 110, and so on, with all numbers in this list being in binary, and corresponding to the decimal number 0, 1, 2, 3, 4, 5, 6, and so on.
In base 3, the digits are 0, 1, and 2. Any integer could be represented as a sum of multiples (0, 1, or 2) powers of 3. For example, the decimal number 34 (= 27 + 6 + 1) would be written in base-3 (ternary or trinary) as 121 (## 1 \cdot 3^3 + 2 \cdot 3^1 + 1 \cdot 3^0##).
Conversion to any other positive integer base of 2 or larger would be similar.
Atran said:
The only exception I make is that the 10 digits are included when expressing a number with \pi. To clarify, the first positive such numbers are: 0,1,2,3,4,5,6,7,8,9,{\pi},{\pi}+1,{\pi}+2,...,2{\pi},2{\pi}+1,...5{\pi},5{\pi}+1,5{\pi}+2,...,9{\pi}+8,9{\pi}+9,{\pi}^2,....
This makes no sense whatsoever. In any usual number base, such as base-2, base-3, base-8, base-10, base-16, base-64, the digits used run from 0 up to, but not including, the base. In your list, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ##\pi##, your "base" lies between 3 and 4.
A major flaw in your list is that for ##10_\pi##, or ##1 \cdot \pi^1 + 0 \cdot \pi^0##, you list this as being larger than 9. In fact, as already mentioned, it lies between 3 and 4.

Thread closed.
 

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