1. ## 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

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

Thanks for any help understanding what is going on here.

2. Originally Posted by imind
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

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

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
$[x]=[11]$