Solving Modulo Arithmetic Equations in Z/5Z^2

  • Thread starter Thread starter mtayab1994
  • Start date Start date
  • Tags Tags
    Arithmetic Classes
Click For Summary

Homework Help Overview

The discussion revolves around solving a system of equations in the context of modulo arithmetic within the set K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z}. The equations involve variables x and y, with specific notation indicating numbers with a bar over them.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss solving the system of equations by substituting expressions for x and y. There is a mention of obtaining a solution set and questioning its correctness. Some participants also explore the equivalence of certain values in modulo 5.

Discussion Status

There are multiple attempts to solve the equations, with some participants affirming the correctness of the solutions presented. Additional information about notation and set definitions has been shared, contributing to the discussion.

Contextual Notes

Participants note the challenges with notation in LaTeX and express concerns about the lack of guidance from their instructor regarding the mathematical concepts involved.

mtayab1994
Messages
584
Reaction score
0

Homework Statement



Solve in K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z} the following system of equations:

\begin{cases}<br /> 2^{-}x-3^{-}y=1^{-} &amp; 1^{-}x+2^{-}y=2^{-}\end{cases}
Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

The Attempt at a Solution



I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

Is that solution correct??
 
Last edited:
Physics news on Phys.org
mtayab1994 said:

Homework Statement



Solve in K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z} the following system of equations:

\begin{cases}<br /> 2^{-}x-3^{-}y=1^{-} &amp; 1^{-}x+2^{-}y=2^{-}\end{cases}



Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

The Attempt at a Solution



I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

Is that solution correct??

I get (4, 4) as the only solution. Since -6 ##\equiv -1## (mod 5) ## \equiv ## 4 (mod 5), I think we arrived at the same thing.
 
mtayab1994 said:

Homework Statement



Solve in K=\mathbb{Z}/_{5}\mathbb{Z}\cdot\mathbb{Z}/_{5}\mathbb{Z} the following system of equations:

\begin{cases}<br /> 2^{-}x-3^{-}y=1^{-} &amp; 1^{-}x+2^{-}y=2^{-}\end{cases}
Not that the ^- means a number with a bar over it. ( I don't know how to input it in the latex software that I use.

The Attempt at a Solution



I solve this system of equations by using the second equation to get x=2-2y and i substituted it for x in the first equation and i got a solution of S={(-6^-,4^-)}

Is that solution correct??

As Mark already said, your solution is correct. :smile:For your information, \bar or \overline will put a bar on top of a symbol.
Like this: ##\bar{12}## or ##\overline{12}##.Btw, your set should be ##\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}##.

##5\mathbb{Z}## is the set with the number 5 multiplied with any whole number (the 5-folds).

##\mathbb{Z}/5\mathbb{Z}## is the so called quotient set with all 5-folds "divided away".

##A \times B## is the so called cartesian product of sets A and B that consists of ordered pairs.
 
I like Serena said:
As Mark already said, your solution is correct. :smile:


For your information, \bar or \overline will put a bar on top of a symbol.
Like this: ##\bar{12}## or ##\overline{12}##.


Btw, your set should be ##\mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z}##.

##5\mathbb{Z}## is the set with the number 5 multiplied with any whole number (the 5-folds).

##\mathbb{Z}/5\mathbb{Z}## is the so called quotient set with all 5-folds "divided away".

##A \times B## is the so called cartesian product of sets A and B that consists of ordered pairs.

Thanks for the extra information. Too bad my teacher doesn't tell us any of this stuff.
 
mtayab1994 said:
Thanks for the extra information. Too bad my teacher doesn't tell us any of this stuff.

Ah well, at least we can give some added value then, here on PF. :wink:
 
I like Serena said:
Ah well, at least we can give some added value then, here on PF. :wink:

YEP:redface:
 

Similar threads

Replies
2
Views
1K
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
8
Views
2K
Replies
4
Views
2K
Replies
7
Views
2K
Replies
6
Views
2K