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Math Help - Algebra Help. How to write answer in modulo 4 manner?

  1. #1
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    Algebra Help. How to write answer in modulo 4 manner?

    If x ≡ 3 (mod 4) and y ≡ 2 (mod 4), write in the modulo 4 mannner for
    (a) x + y
    (b) y - x
    (c) xy
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  2. #2
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Hey mous99.

    There are things called congruence laws that deal with all of these results. Are you aware of these results?
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  3. #3
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Quote Originally Posted by mous99 View Post
    If x ≡ 3 (mod 4) and y ≡ 2 (mod 4), write in the modulo 4 mannner for
    (a) x + y
    (b) y - x
    (c) xy
    (a) x≡3 (mod 4)
    x = 4p + 3

    y ≡ 2 (mod 4)
    y = 4q + 2

    x + y = ( 4p + 3) + ( 4q + 2)
    = 4 ( p+q) + (3 + 2)
    = 4 ( p+q) + 5
    = 5 (mod 4)



    Am I right???
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  4. #4
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Just one more step: What is 5 (mod 4) in terms of a (mod 4) where a is the principal modulus (between 0 and 3 inclusive)?
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  5. #5
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Quote Originally Posted by chiro View Post
    Just one more step: What is 5 (mod 4) in terms of a (mod 4) where a is the principal modulus (between 0 and 3 inclusive)?
    ''5 (mod 4) in terms of a (mod 4) where a is the principal modulus (between 0 and 3 inclusive)?''

    what does this mean??
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  6. #6
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    5 (mod 4) = 4 + 1 (mod 4) = 1 (mod 4). Principle modulus means the one between 0 and n-1 inclusive (mod n).
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  7. #7
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Quote Originally Posted by chiro View Post
    5 (mod 4) = 4 + 1 (mod 4) = 1 (mod 4). Principle modulus means the one between 0 and n-1 inclusive (mod n).
    (a) x≡3 (mod 4)
    x = 4p + 3

    y ≡ 2 (mod 4)
    y = 4q + 2

    x + y = ( 4p + 3) + ( 4q + 2)
    = 4 ( p+q) + (3 + 2)
    = 4 ( p+q) + 5
    = 5 (mod 4)
    = 4 + 1 (mod 4)
    = 1 (mod 4)


    Right?
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  8. #8
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Yeah pretty much.

    What you wrote wasn't wrong by the way: it's just that usually answers are written in the principal branch just for future reference.
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  9. #9
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    y-x
    = (4q+2) - (4p+3)
    = 4(q-p) + (2-3)
    = 4(q-p) -1

    How to write answer in modulo 4 manner?
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  10. #10
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    Recall that 4x - 1 = 4x + 4 - 4 - 1 = 4(x-1) + 3.
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  11. #11
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    Re: Algebra Help. How to write answer in modulo 4 manner?

    in general:

    if x is congruent to a mod n, and y is congruent to b mod n, then:

    x+y is congruent to a+b mod n
    xy is congruent to ab nod n.

    therefore, mod 4:

    if x ≡ 3 and y ≡ 2:

    x+y ≡ 3+2 ≡ 5 ≡ 1 (mod 4)
    x-y ≡ x+(-y) ≡ 3+(-2) ≡ 3+3 ≡ 6 ≡ 2 (mod 4)
    xy ≡ (3)(2) ≡ 6 ≡ 1 (mod 4)

    one doesn't need the "extra variables p and q" because ultimately we don't care what they are.

    these facts about congruences are easy to prove:

    x ≡ a (mod n) means: x - a = tn, for some integer t.
    y ≡ b (mod n) meaus: x - a = un, for some integer u.

    therefore:

    x + y = a + tn + b + un = (a + b) + (t + u)n

    that is:

    (x + y) - (a + b) = (t + u)n, and since t+u is an integer if t and u are:

    x + y ≡ a + b (mod n).

    similarly:

    xy = (a + tn)(b + un) = ab + (au + bt + tun)n, and au+bt+tun is also an integer (since all of a,b,t,u and n are) so:

    xy - ab = (au + bt + tun)n and xy ≡ ab (mod n).

    often, when one wants to distinguish between an INTEGER x and it's equivalence class modulo n, one writes: [x] instead of x. this lets us define two NEW operations on just the equivalence classes (often denoted by + and *, but these are "addition mod n" and "multiplication mod n" not the usual ones) by:

    [x] + [y] = [x+y]
    [x]*[y] = [xy]

    in a standard abuse of notation, it is frequent to see "a (mod n)" used in place of [a] (or sometimes just "a" itself) but until one is used to it, it can be unsettling to see things like:

    2 + 3 = 1 (mod 4), since this is certainly NOT true for the integers 2,3 and 1.
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