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Math Help - Basic Question regarding Simplification of congruences

  1. #1
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    Basic Question regarding Simplification of congruences

    Hi, I have just started learning the fundamentals of Number Theory and I'm having some trouble wrapping my head around a few steps that were presented in my textbook.

    the first question contains a simplification of:

    7x = 1(mod 8)

    to

    -x = 1(mod 8)

    a side note states that this is possible because 7=-1(mod 8)

    Now, if 7 were congruent to -1(mod 8), and since -1(mod 8) is congruent to -1 (this is what I gathered from their next step), would this imply that 7 is congruent to -1, and hence the next step simply replaces 7 with -1?



    The second question contains the following step of simplification:

    5(25^n)(mod 14)

    to :

    5(11^n)(mod 14)

    I can see that the base has reduced by 14, but I am having trouble understanding why exactly 11^n would give the same remainder as 25^n when divided by 14.

    I know that that a^n=b^n (mod 14), as well as the other "rules", but I just can't figure it out.

    Thanks in advance.
    Last edited by Wandering; April 7th 2011 at 11:55 AM.
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Wandering View Post
    Hi, I have just started learning the fundamentals of Number Theory and I'm having some trouble wrapping my head around a few steps that were presented in my textbook.

    the first question contains a simplification of:

    7x = 1(mod 8)

    to

    -x = 1(mod)

    a side note states that this is possible because 7=-1(mod 8)

    Now, if 7 were congruent to -1(mod 8), and since -1(mod 8) is congruent to -1 (this is what I gathered from their next step), would this imply that 7 is congruent to -1, and hence the next step simply replaces 7 with -1?
    you are right but one little mistake here is that you have said "...since -1(mod 8) is congruent to -1....". we never say that.
    rather you say '-1 is congruent to -1 (mod 8)'.
    you can replace 7 with -1 because of the elementary theorems of modular arithmetic. In full it will read:
     7 \equiv -1 (mod \, 8)
    and, x \equiv x(mod \, 8)
    so, 7x \equiv -x (mod \, 8)
    do you see which theorem is used here?


    Quote Originally Posted by Wandering View Post
    The second question contains the following step of simplification:

    5(25^n)(mod 14)

    to :

    5(11^n)(mod 14)

    I can see that the base has reduced by 14, but I am having trouble understanding why exactly 11^n would give the same remainder as 25^n when divided by 14.

    I know that that a^n=b^n (mod 14), as well as the other "rules", but I just can't figure it out.

    Thanks in advance.
    here:
    25 \equiv 11(mod \, 14)
    so, 25^n \equiv 11^n(mod \, 14)(why?)
    so, 5(25^n) \equiv 5(11^n) (mod \, 14)(why?)
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