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hartnn

  • 2 years ago

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

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  1. suvesh253
    • 2 years ago
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    doesnot exist as inverse only possible for one-one function

  2. mukushla
    • 2 years ago
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    y=|x| ? inverse of this will not be a function

  3. mukushla
    • 2 years ago
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    inverse exist x=|y| but its not a function

  4. ganeshie8
    • 2 years ago
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    mod is not 1-1 function

  5. suvesh253
    • 2 years ago
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    modulus is a many-one function hence it's inverse not possible

  6. hartnn
    • 2 years ago
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    c= a mod b = -> corresponds to

  7. Mikael
    • 2 years ago
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    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...

  8. hartnn
    • 2 years ago
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    whats and how ? \(Abs^{-1}(y) = ((y,y),(y,-y))\)

  9. hartnn
    • 2 years ago
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    if i have y=x mod 7, how would i find its inverse ?

  10. Mikael
    • 2 years ago
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    The Abs is of course |x| not the algebraic coincidence of names.

  11. hartnn
    • 2 years ago
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    i eas talking about remainder

  12. hartnn
    • 2 years ago
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    like 5 mod 2 is 1

  13. suvesh253
    • 2 years ago
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    if modulus is with only number then it is one-one and inverse will exist

  14. estudier
    • 2 years ago
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    x = sgn(x) .|x|

  15. Mikael
    • 2 years ago
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    Anyway also for remainder possible to invert \[ mod:\,r_{Ideal} \rightarrow r \\ mod^{-1}: r \rightarrow r_{Ideal}\]

  16. Mikael
    • 2 years ago
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    and a , the action that should not be asked but anyway received :)

  17. hartnn
    • 2 years ago
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    but i didn't understand....

  18. Mikael
    • 2 years ago
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    thx oh exalted one !

  19. Mikael
    • 2 years ago
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    Learn about Ideals even if you dont have them :) http://en.wikipedia.org/wiki/Ideal_(ring_theory)

  20. Mikael
    • 2 years ago
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    MEdals, Medals gentleman for the wondering scholar ....! Keep them falling !

  21. hartnn
    • 2 years ago
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    hmmm...didn't study ideals, i'll go through them, thanks

  22. Mikael
    • 2 years ago
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    Everybody that did - pls gratify

  23. satellite73
    • 2 years ago
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    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 \(x\) ?

  24. hartnn
    • 2 years ago
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    yes, i wrote somewhere in between = -> correcponds to

  25. Mikael
    • 2 years ago
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    If that was the question I would not solve it - but it was not!

  26. satellite73
    • 2 years ago
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    so for example solve \( 5\equiv x\text{ mod } 3\)

  27. Mikael
    • 2 years ago
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    No uniqueness - only an IDEAL OF ANSWERS

  28. hartnn
    • 2 years ago
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    there are many (or infinite ?) values of x in 5=x mod 3 ??

  29. Mikael
    • 2 years ago
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    Yes - they form an IDEAL

  30. satellite73
    • 2 years ago
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    ideals live in arbitrary rings, and are not necessary for understanding elementary number theory

  31. Mikael
    • 2 years ago
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    Anyway the set pf answers here IS an Ideal.

  32. Mikael
    • 2 years ago
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    By the way the only proper way to make equivalence necessary for "REMAINDER CALCULUS" is bu using ideals. Elementary - well yes.

  33. hartnn
    • 2 years ago
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    ok, thank you @Mikael i'll go through IDEALS and ask u if i have any doubts. thanks to @satellite73 also :)

  34. Mikael
    • 2 years ago
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    Firstly is the "cyclical group". It will give you the applications. Only then that. This takes you further

  35. hartnn
    • 2 years ago
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    any good reference for such things(other than wikipedia) where these things are explained in lucid manner ?

  36. Mikael
    • 2 years ago
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    http://en.wikipedia.org/wiki/Modular_arithmetic

  37. Mikael
    • 2 years ago
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    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

  38. hartnn
    • 2 years ago
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    because of those ... i cannot get the reference

  39. hartnn
    • 2 years ago
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    only last one i couldn't get

  40. Mikael
    • 2 years ago
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    Just ggle "elementary remainder group theory"

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