How to find the inverse/unit for mod?

Question

So for this problem, its asking to find the inverses of [4], [5], and [7]. I'm trying to skim through my textbook but I can't find an example.

Like for this problem, I don't understand why the Units are 1 and 2, the textbook I have is garbage.

Answer 1

Examples:

For a) you can find the inverses like this.

For example when we want to find the inverse of $[4]$ in $\mathbb{Z}/13\mathbb{Z}$. Since $13$ is a prime we know, that $\mathbb{Z}/13\mathbb{Z}$ is a field.

To find the inverse $[4^{-1}]$ we have to solve $[4][x]=[4x]=[1]$. We can do this by brute force and just go all the way. There are just 13 possibilities for $x$.

One sees pretty fast, that $[4\cdot 10]=[40]=[1+13\cdot 3]=[1]$. Hence we have $[4^{-1}]=[10]$

This is a basic approach. An other is to use the euclidean algorithm.

Now go find the other inverses!

For b) we can use the results in a). The equations can be solved as always:

$[4]X=[7]$.

We want to isolate $X$. For that we multiply with $[4^{-1}]$, and get:

$[10][4]X=[10][7]$

$X=[70]=[5]$

Answer 2

The inverse of [7] (mod 13) is the number, x, such that 7x= 1 (mod 13) which, in turn, means that 7x= 1+ 13n for some integer n. That is the same as the "Diophantine equation", 7x- 13n= 1. As Cornman said, we can use the "Euclidean algorithm" to solve it:

7 divides into 13 once with remainder 6: 13- 7= 6 6 divides into 7 once with remainder 1: 7- 6= 1

Replace "6" in the second equation with "13- 7": 7- (13- 7)= 7(2)- 13(1)= 1.

So one solution to 7x- 13n= 1 is x= 2, n= 1.

The multiplicative inverse of 7 (mod 13) is 2.