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Question about mod function and congruences

  1. Apr 23, 2013 #1
    Im having a bit of hard time understanding how is that two intergers (a and b) divided by a common divisor (m) have the same remainder imply that the difference of (a and b) will aslo be divisible by m?

    Essentially what im asking is:

    a [itex]\equiv[/itex] b (mod m) [itex]\Rightarrow[/itex] m|(a-b)

    the "|" means divides just incase you aren't fimiliar with that symbol.

    a [itex]\equiv[/itex] b (mod m) says a/m and b/m will have the same remainder. Since, they have the same remainder (a - b) will also be be divisible by m.

    example 1) 29 [itex]\equiv[/itex] 15 (mod 7) [itex]\Rightarrow[/itex] 7|(29 -15)

    Why is the difference of 29 -15 also divisible by 7?

    Is it because when since 29 and 15 have the same reminder means that we are simply taking out factors of 7 and the common reminder from the 29 and 15?

    29 - 15
    [7(4) + 1] - [ 7(2) + 1]

    = 7(2) = 14 which is divisible by 7


    example 2) 11 [itex]\equiv[/itex] 4 (mod 7) [itex]\Rightarrow[/itex] 7|(11 - 4)
    11 - 4
    [7(1) + 4] - [7(0) + 4]

    = 7 which is divisible by 7


    Even if my reasoning is correct, please try to explain in your own way. I can do it mathmetically but that problem im having is understanding it.
     
    Last edited: Apr 23, 2013
  2. jcsd
  3. Apr 23, 2013 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi John112! :smile:

    Yes, your reasoning is correct.

    If a and b are both = 4 (mod 7),

    then there exist integers p and q such a = 7p + 4, b = 7q + 4,

    so a - b = 7(p - q). :wink:
     
  4. Apr 23, 2013 #3
    Thanks for that clear explanation tiny-tim!
     
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