Solving Modulo Arithmetic Equations in Z/5Z^2

[mtayab1994]

Homework Statement

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

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

 

[Mark44]

Homework Statement

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

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



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.

 

[I like Serena]

Homework Statement

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

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

 

[mtayab1994]

As Mark already said, your solution is correct.


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.

 

[I like Serena]

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.

 

[mtayab1994]

Ah well, at least we can give some added value then, here on PF.

YEP