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How to calculate with residue classes in $\mathbb{Z}{/5\mathbb{Z}}$?

  • $- \overline x \neq \overline x$ but $- \overline x = \overline{5-x}$
  • $\overline x + \overline y = \overline{x+y}$
  • $\overline x \cdot \overline y = \overline{x\cdot y}$

So the following calculation should be right?

$ \overline 2 \cdot \overline 1 + \overline 2 \cdot \overline 1 - \overline 3 \cdot \overline 4 = \overline{2\cdot1} + \overline{2\cdot 1} - \overline{3\cdot4} = \overline{2\cdot1 + 2\cdot 1 - 3 \cdot 4} = \overline{-8} = \overline{-3} = \overline{5-3} = \overline{2} $


How to use this, when solving a LES like the following with coefficients in $\mathbb{Z}_{5}$?

$ \left( \begin{array}{cccc|c} 3 & -1 & 0 & 2 & -4\\ 1 & 0 & -3 & 2 & 2\\ 2 & 2 & -3 & 0 & 1\\ \end{array} \right) \rightsquigarrow \left( \begin{array}{cccc|c} 0 & -1 & 9 & -4 & -10\\ 1 & 0 & -3 & 2 & 2\\ 0 & 0 & 3 & -4 & -1\\ \end{array} \right) $

Can I run the Gaussian algorithm without taking to much care of the residue classes and convert the integers after I've finished the algorithm?

Don't multiply a row by 5 during the Gaussian algorithm

  • 0
    For a walkthrough example see [my answer here](http://math.stackexchange.com/a/86367/11619). Unfortunately that is modulo 29 as opposed to 5, so the numbers are a bit trickier. It might help you to understand, how the logic of the elementary operations is the same over **any** field.2012-05-29

1 Answers 1

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Yeah, so if you want to construct addition and multiplication tables for $\mathbb{Z}_5$, one way is to do the following:

Write out the general forms of an element of each class. For instance, $ 5k,5k+1,5k+2,5k+3,5k+4$. Now, do additions and multiplications to construct the table that you want by looking at the resultant remainder mod 5. Now, after you do this, to do a row reduction, you reduce your numbers by writing them in that form. I.e take $\frac{n}{5}$, and look at its floor. This tells you k. Then, to get the equilivelence class, take $n-5k$ and you're done. Now, you can do this for positives and negitives. Next, when you clear a row by making the leading term a one, you look up in your table the inverse of whatever is the leading term, and multiply the row by it. Thankfully, $Z_5$ is a field, so you can always find an inverswe--except for zero, of course.

Note that to get general expressions for addition and multiplication mod n, you do the same thing but with n and k.