Stuck, finding inverse in element in ring Z

In summary, the conversation discusses finding the inverse element to 7 in the ring Z_{13}. The suggested method is to use the extended Euclidean algorithm, but it is also possible to guess the answer. The value of k represents an integer that needs to be found in order to solve the equation 7x = 1 + k*13. It is also mentioned that 13 is small enough to simply guess the answer. The inverse element to 7 in the ring Z_{13} is 2.
  • #1
jdnhldn
9
0

Homework Statement



I need to find the inverse to element in 7 in ring [tex]Z_{13}[/tex]

Homework Equations



7^-1 in [tex]Z_{13}[/tex]

The Attempt at a Solution



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

And then my notes was messed up :(
 
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  • #2
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.
 
  • #3
Dick said:
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?
 
  • #4
jdnhldn said:
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.
 
  • #5
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?
 
  • #6
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.
 

1. What is an inverse element in a ring Z?

In a ring Z (the set of integers), an inverse element is an integer that when multiplied by another integer yields the identity element, which is 1. In other words, the inverse element undoes the multiplication operation.

2. How do you find the inverse element in a ring Z?

To find the inverse element in a ring Z, you can use the Extended Euclidean Algorithm. This algorithm involves finding the greatest common divisor (GCD) of the two integers and then using the GCD to calculate the inverse element. Alternatively, you can use Fermat's Little Theorem or Euler's Theorem to find the inverse element.

3. What is the identity element in a ring Z?

The identity element in a ring Z is the integer 1. This is because when 1 is multiplied by any other integer, the result is the same integer. For example, 1 x 5 = 5.

4. Can all elements in a ring Z have an inverse?

No, not all elements in a ring Z have an inverse. Only non-zero integers that are coprime (have no common factors) with the modulus (the number being used for the ring Z) have an inverse element. For example, in ring Z with modulus 10, the inverse of 3 would be 7 because 3 and 7 are coprime.

5. Why is finding the inverse element important in ring Z?

Finding the inverse element in a ring Z is important because it allows us to solve equations involving multiplication. It also allows us to perform division operations, which are not defined in a ring Z. In addition, finding the inverse element is useful in cryptography and coding theory.

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