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Modular arithmetic question about functions

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Oct30-11, 09:49 AM
P: 344
Hello, I'm new to modular arithmetic, but I was wondering -

Given that
b = R mod(n) where b,R, and n are all integers

Is it plausible to consider
G(x) = R(x) mod(f(x)) with G,R,f all functions of x?

Has this been done or does this not even make sense?
If its been done, what is it called?
If it doesn't work, could you explain a little?

thanks :D
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Oct30-11, 11:25 AM
P: 366
Yes, it does make sense. Have you studied abstract algebra yet? Modular arithmetic is a special case of quotient objects, in this case a quotient ring. On Wikipedia they have some good examples involving polynomials and smooth functions.

For another example, let [itex] R := \{f \in C[0,1] : f \quad \text{is continuous} \} [/itex] be the ring of continuous real-valued functions defined on [0,1]. Given [itex] f \in R [/itex] we can define an equivalence relation by g~h iff [itex]g(x) - h(x) =a(x)f(x) [/itex] for some function [itex] a(x) \in R [/itex]. Another way to write this would be g(x) = h(x) mod (f(x))
Oct30-11, 01:08 PM
Sci Advisor
P: 906
not only is it possible, but in fact, it's one of the ways in which finite fields are created.

normally, what one does, is first create the integers mod p, where p is a prime.

then one considers Zp[x], the set of all polynomials with coefficients in Zp (integers mod p).

and THEN, one picks an "irreducible" polynomial (for our purposes, this just means "no factoring possible"), say g(x), and considers the polynomials "mod g(x)".

doing that, gives a finite field, which is surprising, because it's not immediately intuitive that identifying a certain subset of the infinite number of polynomials, gives you such a small (finite) set of "remainders".

one surprisingly familar example of a set of functions (mod f(x)) is the complex numbers. first, you take the set of all real polynomials, R[x].

then, you "mod x2 + 1". that is, you declare p(x) to be equivalent to q(x) if and only if:

p(x) = q(x) + (x2+1)k(x)

we can write [p(x)] for the equivalence class of p(x). now look what happens with [x]:

[x]2 + [1] = [x2+ 1] = [0]

(since x2+1 = 0 + (x2+1)(1), here "q(x)" is 0, and "k(x)" is 1).

this means that [x]2 = -[1] = [-1].

this means that ("mod x2 + 1") [x] is a square root of [-1].

so we have (there's some technical details i'm glossing over) the bijection:

a+bi <---> [a+bx] = [a][1] + [b][x], that is [x] acts "just like" i.

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