modular arithmetic question about functionsby elegysix Tags: arithmetic, mod base functions, modular 

#1
Oct3011, 09:49 AM

P: 269

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 



#2
Oct3011, 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 realvalued 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)) 



#3
Oct3011, 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 Z_{p}[x], the set of all polynomials with coefficients in Z_{p} (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 x^{2} + 1". that is, you declare p(x) to be equivalent to q(x) if and only if: p(x) = q(x) + (x^{2}+1)k(x) we can write [p(x)] for the equivalence class of p(x). now look what happens with [x]: [x]^{2} + [1] = [x^{2}+ 1] = [0] (since x^{2}+1 = 0 + (x^{2}+1)(1), here "q(x)" is 0, and "k(x)" is 1). this means that [x]^{2} = [1] = [1]. this means that ("mod x^{2} + 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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