Stuck, finding inverse in element in ring Z

[jdnhldn]

Homework Statement

I need to find the inverse to element in 7 in ring $$Z_{13}$$

Homework Equations

7^-1 in $$Z_{13}$$

The Attempt at a Solution

Needs to find X so that, 7x=1 in $$Z_{13}$$ => 7x=1+k*13

And then my notes was messed up :(

 

[Dick]

The general method is the extended Euclidean algorithm. But Z_13 is small enough it's probably easier to just guess the answer. Try that first.

 

[jdnhldn]

The general method is the extended Euclidean algorithm. But Z_13 is small enough it's probably easier to just guess the answer. Try that first.

This was a note from the class, I've forgotten what k means in this one. May I please ask you what you think it means?

 

[Dick]

This was a note from the class, I've forgotten what k means in this one. May I please ask you what you think it means?

It's some integer that you want to find, just like x. If you can find integers x and k such that 7x=1+k*13 then if you reduce both side mod 13, you'll see x is 7^(-1). Like I said, see if you can find values by guessing.

 

[HallsofIvy]

You want to find x and k such that 7x= 13k+ 1. That is the same as 7x- 13k= 1.
7 divides into 13 once with remainder 6. That says 13= (1)(7)+ 6 or (1)(13)+ (-1)(7)= 6

6 divides into 7 once with remainder 1. That says 7= (1)(6)+ 1 or (1)(7)+ (-1)(6)= 1.
Replacing "6" in that from the previous equation, (1)(7)+ (-1)((1)(13)+ (-1)(7))= (2)(7)- 13(1)= 1.

But, as Dick said, 13 is small enough that it's probably simpler to just look at 7(1), 7(2), 7(3), etc. You do know what 7 times 2 is, don't you?

 

[jdnhldn]

So z13= 0 to 13 in the ring. First I didn't get what a ring was, then I saw an example of the clock that z12 is 12=0 and 11+1=12=0 and 12+1=13-1=0. Something in that style. Now I got it.

inverse elements to 7 in ring z13 = 2

Thanks guy.