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inverse of mod function? like y= x mod p, 1=5 mod 2 how do i get 5 from {1,2}

Mathematics
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doesnot exist as inverse only possible for one-one function
y=|x| ? inverse of this will not be a function
inverse exist x=|y| but its not a function

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Other answers:

mod is not 1-1 function
modulus is a many-one function hence it's inverse not possible
c= a mod b = -> corresponds to
NO PROBLEMAS SENIOUR @hartnn ! Let B be the set of ordered pairs of pairs-of-numbers, in other words pairs of points in the plane: \[ b \in B \,\,\iff b=(P_1, P_2) \] then the inverse function of \[Abs^{-1}(y) = ((y,y),(y,-y)) \] and a medal , naturally. forgive my humbleness...
whats and how ? \(Abs^{-1}(y) = ((y,y),(y,-y))\)
if i have y=x mod 7, how would i find its inverse ?
The Abs is of course |x| not the algebraic coincidence of names.
i eas talking about remainder
like 5 mod 2 is 1
if modulus is with only number then it is one-one and inverse will exist
x = sgn(x) .|x|
Anyway also for remainder possible to invert \[ mod:\,r_{Ideal} \rightarrow r \\ mod^{-1}: r \rightarrow r_{Ideal}\]
and a , the action that should not be asked but anyway received :)
but i didn't understand....
thx oh exalted one !
Learn about Ideals even if you dont have them :) http://en.wikipedia.org/wiki/Ideal_(ring_theory)
MEdals, Medals gentleman for the wondering scholar ....! Keep them falling !
hmmm...didn't study ideals, i'll go through them, thanks
Everybody that did - pls gratify
i think the question is maybe something different, are you asking how do you solve \[x\equiv y (mod n)\] for \(y\) if you know \(x\) ?
yes, i wrote somewhere in between = -> correcponds to
If that was the question I would not solve it - but it was not!
so for example solve \( 5\equiv x\text{ mod } 3\)
No uniqueness - only an IDEAL OF ANSWERS
there are many (or infinite ?) values of x in 5=x mod 3 ??
Yes - they form an IDEAL
ideals live in arbitrary rings, and are not necessary for understanding elementary number theory
Anyway the set pf answers here IS an Ideal.
By the way the only proper way to make equivalence necessary for "REMAINDER CALCULUS" is bu using ideals. Elementary - well yes.
ok, thank you @Mikael i'll go through IDEALS and ask u if i have any doubts. thanks to @satellite73 also :)
Firstly is the "cyclical group". It will give you the applications. Only then that. This takes you further
any good reference for such things(other than wikipedia) where these things are explained in lucid manner ?
http://en.wikipedia.org/wiki/Modular_arithmetic
1 mathdl.maa.org/images/upload_library/22/Polya/Brenton.pdf en.wikipedia.org/wiki/Elementary_group_theory math.uc.edu/~hodgestj/Abstract%20Algebra/GroupTheory512.pdf www.rowan.edu/.../Some%20Elementary%20Group%20Theory.pdf
because of those ... i cannot get the reference
only last one i couldn't get
Just ggle "elementary remainder group theory"

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