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Math Help - Multiplicative Inverse

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    Multiplicative Inverse

    While working to learn modern algebra (abstract algebra) in preparation for next year I stumbled across the section on multiplicative inverses. The textbook is pretty solid and I can see how the concepts from the previous sections are relating to this one. However I got stuck on Excercise 10 because I skipped Excercise 9 (I didn't understand it) ... Now I know the answer for Excercise 10 is right but I'm unable to relate "why" with my answer and I was wondering if I could get a little clarification. Below I've typed the excercises for reference... Thanks for the help

    Excercise 9 (I skipped this one):

    Let [a] be an element of Z_{n} that has a multiplicative inverse [a]^{-1} in Z_{n}. Prove that [x] = [a]^{-1}[b] is the unique solution in Z_{n} to the equation [a][x]=[b]

    Excercise 10

    Solve each of the following equations by finding [a]^{-1} and using the result in Excercise 9.

    (a) [4][x] = [5] in Z_{13}

    This format looked VERY similar to a previous excercise in which I used the Extended Eclidean Algorithm to solve so I first tried that

    1 = 4m + 13n

    13 = (4)(3) + (1)(1)
    4 = (1)(4) + 0

    1 = (13)(1) - (4)(3)
    1 = (13)(1) + (4)(-3)

    5 = (13)(5) + (4)(-15)

    5 \equiv (4)(-15) mod 13

    -15 = (13)(-2) + 11

    Thus x = 11 is a solution. Now from here I checked my answer in the book and it is indeed right however they list it a separate way and I'm wondering what exactly their method says and if it's shorter. The method I've used worked for all 8 of the parts in this excercise, so the methodology is right, but I'm kind of lost as to what it means haha

    Book answer:

    [x] = [4]^{-1}[5] = [10][5] = [11]


    Thanks for any help understanding what is going on here.
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    Quote Originally Posted by imind View Post
    While working to learn modern algebra (abstract algebra) in preparation for next year I stumbled across the section on multiplicative inverses. The textbook is pretty solid and I can see how the concepts from the previous sections are relating to this one. However I got stuck on Excercise 10 because I skipped Excercise 9 (I didn't understand it) ... Now I know the answer for Excercise 10 is right but I'm unable to relate "why" with my answer and I was wondering if I could get a little clarification. Below I've typed the excercises for reference... Thanks for the help

    Excercise 9 (I skipped this one):

    Let [a] be an element of Z_{n} that has a multiplicative inverse [a]^{-1} in Z_{n}. Prove that [x] = [a]^{-1}[b] is the unique solution in Z_{n} to the equation [a][x]=[b]

    Excercise 10

    Solve each of the following equations by finding [a]^{-1} and using the result in Excercise 9.

    (a) [4][x] = [5] in Z_{13}

    This format looked VERY similar to a previous excercise in which I used the Extended Eclidean Algorithm to solve so I first tried that

    1 = 4m + 13n

    13 = (4)(3) + (1)(1)
    4 = (1)(4) + 0

    1 = (13)(1) - (4)(3)
    1 = (13)(1) + (4)(-3)

    5 = (13)(5) + (4)(-15)

    5 \equiv (4)(-15) mod 13

    -15 = (13)(-2) + 11

    Thus x = 11 is a solution. Now from here I checked my answer in the book and it is indeed right however they list it a separate way and I'm wondering what exactly their method says and if it's shorter. The method I've used worked for all 8 of the parts in this excercise, so the methodology is right, but I'm kind of lost as to what it means haha

    Book answer:

    [x] = [4]^{-1}[5] = [10][5] = [11]


    Thanks for any help understanding what is going on here.
    Since 4 and 13 are coprime when you have this step

    1 = (13)(1) - (4)(3) \iff 1=1\cdot 13+(-3)\cdot 4

    This gives that -3 is the multiplicative inverse of 4, but -3=10\text{mod}13

    So 10 is the inverse

    So if you multiply both sides of the equation above you get

    [10][4][x]=[10][5]

    [40][x]=[50] now reducing modulo 13 gives
    [x]=[11]
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