Problems with inverses in arithmetic in ring z

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Homework Help Overview

The discussion revolves around calculating an expression involving inverses in arithmetic within the group Z20, specifically the expression 7*11 + 9*11^-1. Participants are exploring the concept of multiplicative inverses in modular arithmetic.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the expression but expresses confusion regarding their result compared to a computer program's output. They question the difference between ring congruence arithmetic and regular congruence arithmetic.
  • Some participants question the clarity of the original poster's notation and suggest focusing on finding the inverse of 11 mod 20 first.
  • Others provide detailed reasoning on how to find the multiplicative inverse of 11 mod 20, including steps involving the Euclidean algorithm and Diophantine equations.

Discussion Status

Participants are actively engaging with the problem, with some providing guidance on how to approach finding the multiplicative inverse. There is a mix of interpretations regarding the original poster's calculations and notation, but no consensus has been reached on the final answer.

Contextual Notes

There is a mention of a computer program providing a different answer, which raises questions about the correctness of the original poster's calculations. The discussion also touches on the definitions and implications of ring congruence versus regular congruence arithmetic.

morrowcosom
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Homework Statement


Calculate 7*11 + 9*11^-1 in the group Z20


Homework Equations





The Attempt at a Solution


77+ (9*1/11) in group Z20
77 + 9/11
17 +11x= 20mod+9
My solution was 12, this makes 149 on both sides when you multiply the mod times 7.
I am doing independent study and the computer program I am using says the answer is 16.
What am I doing wrong? Btw, what is the difference between ring congruence arithmetic and regular congruence arithmetic? They seem pretty similar.

 
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I'm not really sure what you are doing. What is 20mod+9 supposed to mean? At some point you need to figure out what 11^(-1) mod 20 is. Why not do it first?
 
morrowcosom said:
bump

You can 'bump' a post if no one has given any response. I don't think that's the case here.
 
The first thing you absolutely need to do is to figure out what 11^-1 mod 20 is.
 
The first thing you absolutely need to do is to figure out what 11^-1 mod 20 is

original problem: calculate 7*11+9*11^-1 with mod 20

So in order to figure out the inverse of 11^-1 mod 20, I would do the typical
11x=mod 20+1, and from here would I go 17+99x=mod 20+1(I get 16 on this one, the supposedly correct solution), or some other route?
 
To find the multiplicative inverse of 11 mod 20, you want a number, m, such that 11m= 1 (mod 20) or 11m= 1+ 20n for some integer n.

That is the same as solving the diophantine equation 11m- 20n= 1.

11 divides into 20 once with remainder 9: 20- 11= 9.

9 divides into 11 once with remainder 2: 11- 9= 2.

2 divides into 9 four times with remainder 1: 9- 4(2)= 1.

Replace the "2" in that equation with 11- 9: 9- 4(11- 9)= 5(9)- 4(11)= 1.

Replace the "9" in that equation with 20- 11: 5(20- 11)- 4(11)= 5(20)- 9(11)= 1.
(Of course: 100- 99= 1.)

A solution to 11m- 20n= 1 is m= -9, n= -5. Thus, the multiplicative inverse of 11, mod 20, is -9= 20- 9= 11 (mod 20). That is, the multiplicative inverse of 11, mod 20, is 11 itself.

Check: 11(11)= 121= 6(20)+ 1= 1 (mod 20).
 

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