Problems with inverses in arithmetic in ring z

In summary, To find the multiplicative inverse of 11, mod 20, solve the diophantine equation 11m- 20n= 1. In this case, the solution is m= -9, n= -5, which means the multiplicative inverse of 11, mod 20, is 11 itself. This can be verified by checking 11(11)= 1 (mod 20).
  • #1
morrowcosom
54
0

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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  • #2
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?
 
  • #3
morrowcosom said:
bump

You can 'bump' a post if no one has given any response. I don't think that's the case here.
 
  • #4
The first thing you absolutely need to do is to figure out what 11^-1 mod 20 is.
 
  • #5
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?
 
  • #6
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).
 

FAQ: Problems with inverses in arithmetic in ring z

What is the ring Z?

The ring Z, also known as the ring of integers, is the set of all whole numbers including positive, negative, and zero. It is denoted by the symbol Z and is an important concept in number theory and abstract algebra.

What are the problems with inverses in arithmetic in ring Z?

The main problem with inverses in arithmetic in ring Z is that not all elements have an inverse. In other words, not all numbers have a multiplicative inverse in the ring Z. This means that division is not always possible in this ring.

Can you give an example of an element in ring Z that does not have an inverse?

Yes, zero (0) is an element in ring Z that does not have an inverse. This is because any number multiplied by zero will result in zero, but there is no number that can be multiplied by zero to get a non-zero result.

How do we determine if an element in ring Z has an inverse?

In order for an element in ring Z to have an inverse, it must be a unit. A unit is an element that has a multiplicative inverse. In ring Z, the only units are 1 and -1. Therefore, any other element does not have an inverse.

Are there any other types of numbers that have problems with inverses in arithmetic?

Yes, the concept of inverses in arithmetic is not limited to ring Z. Other examples include matrices, where not all matrices have an inverse, and complex numbers, where division by zero is undefined. These types of numbers are studied in abstract algebra and have their own set of rules and properties.

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