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Proof again: Show that a mod m = b mod m if a = b (mod m)

  1. Apr 5, 2007 #1
    1. The problem statement, all variables and given/known data

    Let m be a positive integer. Show that a mod m = b mod m if [itex]a\,\equiv\,b\left(mod\,m\right)[/itex].



    2. Relevant equations

    Congruency - [tex]a\,\equiv\,b\left(mod\,m\right)[/tex]

    a is congruent to b modulo m.



    3. The attempt at a solution

    I really, really, really suck at proofs. But here is what I tried.

    [tex]a\,\equiv\,b\left(mod\,m\right)\,\longrightarrow\,a\,mod\,m\,\equiv\,\left[b\left(mod\,m\right)\right]\,\left(mod\,m\right)[/tex]

    I have no idea how to prove anything about this stuff. Please help!!!
     
  2. jcsd
  3. Apr 5, 2007 #2

    matt grime

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    Why are you trying to prove it using symbols?

    Q. What does it mean in words for a to be congruent to b mod m?



    Q. What is a mod m, and what is b mod m?




    The definitions of these things mean there is *nothing* here that needs to be proved.
     
  4. Apr 5, 2007 #3
    a is congruent to b mod m if m divides (a - b).

    a mod m is the remainder of the division operation of [itex]\frac{a}{m}[/itex].

    b mod m is the remainder of the division operation of [itex]\frac{b}{m}[/itex].
     
  5. Apr 5, 2007 #4
    Does this count as "showing" a mod m [itex]\equiv[/itex] b mod m?

    [tex]a\,\equiv\,b\left(mod\,m\right)\,\longrightarrow\,a\,mod\,m\,\equiv\,\left[b\left(mod\,m\right)\right]\,\left(mod\,m\right)[/tex]

    By definition of mod m and congruence mod m,

    [tex]b(mod\,m)(mod\,m)\,\equiv\,b[/tex]

    So,

    [tex]a(mod\,m)\,\equiv\,b[/tex]

    Then,

    [tex]a\,\equiv\,\left[a(mod\,m)\right](mod\,m)\,\equiv\,a[/tex]

    Q.E.D.
     
  6. Apr 5, 2007 #5

    matt grime

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    Put it *simple* words. Phrases that involve the words: remainder on division by m. a=b mod m if and only if a and b have the same remainder on division by m. That's what modulo arithmetic. So, as you see, there is nothing to prove. a mod m is the remainder on division by m.

    So all it is asking is: show that if a and b have the same remainder on division by m, then the remainder of a on division by m is the same as the remainder of b on division by m. So there isn nothing to prove beyond writing out the definitions.
     
  7. Apr 5, 2007 #6
    a mod m is the remainder after a is divided by m.

    a mod m and b mod m are equal if and only if their remainders are the same.

    m divides (a - b) <---- What do I do with this?

    Is that really it?
     
    Last edited: Apr 5, 2007
  8. Apr 5, 2007 #7

    matt grime

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    nothing

    yes.
     
  9. Apr 5, 2007 #8
    Review the definitions.

    What is the mathematical definition of [itex]a \equiv b \mbox{ (mod m)}[/itex]?
     
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