Die Seite wurde neu angelegt: „{{quiz-Header|Buchseite=Kanalcodierung/Einige Grundlagen der Algebra }} [[File:|right|]] ===Fragebogen=== <quiz display=simple> {Multiple-Choice Frage…“
In the adjacent graph, the (partially incomplete) addition and multiplication tables for $q = 5$ and $q = 6$ are given, where both addition ("$+$") and multiplication ("$\hspace{0.05cm}\cdot\hspace{0.05cm}$") modulo $q$ are to be understood.
To be checked is whether the number sets $Z_5$ and $Z_6$ satisfy all the conditions of a Galois field $\rm GF(5)$ and $\rm GF(6)$, respectively.
In the [[Channel_Coding/Some_Basics_of_Algebra#Definition_of_a_Galois_field|"theory section"]] a total of eight conditions are mentioned, all of which must be met. You are to check only two of these conditions:
$\rm(D)$ For all elements there is an <b>additive inverse</b> (Inverse for "$+$"):
* validity of commutative law, associative law and distributive law
are satisfied by both, $Z_5$ and $Z_6$.
Hints: The exercise refers to the chapter [[Channel_Coding/Some_Basics_of_Algebra| "Some Basics of Algebra"]].
===Fragebogen===
<quiz display=simple>
{Multiple-Choice Frage
|type="[]"}
- Falsch
+ Richtig
{Input-Box Frage
===Questions===
<quiz display=simple>
{Complete the addition table for $q = 5$. Enter the following values:
|type="{}"}
|type="{}"}
$\alpha$ = { 0.3 }
$A_{04} \ = \ ${ 4 }
$A_{14} \ = \ ${ 0. }
$A_{44} \ = \ ${ 3 }
{Complete the multiplication table for $q = 5$. Enter the following values:
|type="{}"}
$M_{04} \ = \ ${ 0. }
$M_{14} \ = \ ${ 4. }
$M_{44} \ = \ ${ 1. }
{Does the $Z_5$ set satisfy the conditions of a Galois field?
|type="[]"}
+ Yes.
- No, there is not an additive inverse for all elements $(0, \hspace{0.05cm}\text{...} \hspace{0.1cm}, 4)$ .
- No, the elements $1, \hspace{0.05cm}\text{...} \hspace{0.1cm}, 4$ do not all have a multiplicative inverse.
{Does the $Z_6$ set satisfy the conditions of a Galois field?
|type="[]"}
- Yes.
- No, there is not an additive inverse for all elements $(0, \hspace{0.05cm}\text{...} \hspace{0.1cm}, 5)$ .
+ No, the elements $1, \hspace{0.05cm}\text{...} \hspace{0.1cm}, 5$ do not all have a multiplicative inverse.
{The number sets $Z_2, \ Z_3, \ Z_5$ and $Z_7$ yield a Galois field, but the sets $Z_4, \ Z_6, \ Z_8, \ Z_9$ do not. What do you conclude from this?
*Since both the required additive and multiplicative inverses exist ⇒ $Z_5$ describes a Galois field $\rm GF(5)$
*Correct is the <u>proposed solution 1</u>.
'''(4)''' From the blue addition table on the statement page, we see that all numbers $(0, \, 1, \, 2, \, 3, \, 4, \, 5)$ of the set $Z_6$ have an additive inverse
⇒ in each row (and column) there is exactly one zero.
*On the other hand, a multiplicative inverse ${\rm Inv}_{\rm M}(z_i)$ exists only for $z_i = 1$ and $z_i = 5$, viz.
*For $z_i = 2, \ z_i = 3$ and $z_i = 4$, we find no element $z_j$, so that $(z_i \cdot z_j) \, {\rm mod} \, 6 = 1$.
*Correct is the <u>proposed solution 3</u> ⇒ the blue tables for $q = 6$ do not yield a Galois field $\rm GF(6)$.
[[Category:Aufgaben zu Kanalcodierung|^2.1 Einige Grundlagen der Algebra
'''(5)''' Correct is the <u>proposed solution 2</u>:
*A finite number set $Z_q = \{0, \, 1, \hspace{0.05cm} \text{...} \hspace{0.1cm} , \, q-1\}$ of natural numbers leads to a Galois field only if $q$ is a prime number.
*Of the number sets mentioned above, this is true only for $Z_{11}$.
{{ML-Fuß}}
^]]
[[Category:Channel Coding: Exercises|^2.1 Some Basics of Algebra^]]
[[de:Aufgaben:Aufgabe 2.2: Eigenschaften von Galoisfeldern]]
In the adjacent graph, the (partially incomplete) addition and multiplication tables for $q = 5$ and $q = 6$ are given, where both addition ("$+$") and multiplication ("$\hspace{0.05cm}\cdot\hspace{0.05cm}$") modulo $q$ are to be understood.
To be checked is whether the number sets $Z_5$ and $Z_6$ satisfy all the conditions of a Galois field $\rm GF(5)$ and $\rm GF(6)$, respectively.
In the "theory section" a total of eight conditions are mentioned, all of which must be met. You are to check only two of these conditions:
$\rm(D)$ For all elements there is an additive inverse (Inverse for "$+$"):
Since both the required additive and multiplicative inverses exist ⇒ $Z_5$ describes a Galois field $\rm GF(5)$
Correct is the proposed solution 1.
(4) From the blue addition table on the statement page, we see that all numbers $(0, \, 1, \, 2, \, 3, \, 4, \, 5)$ of the set $Z_6$ have an additive inverse
⇒ in each row (and column) there is exactly one zero.
On the other hand, a multiplicative inverse ${\rm Inv}_{\rm M}(z_i)$ exists only for $z_i = 1$ and $z_i = 5$, viz.
For $z_i = 2, \ z_i = 3$ and $z_i = 4$, we find no element $z_j$, so that $(z_i \cdot z_j) \, {\rm mod} \, 6 = 1$.
Correct is the proposed solution 3 ⇒ the blue tables for $q = 6$ do not yield a Galois field $\rm GF(6)$.
(5) Correct is the proposed solution 2:
A finite number set $Z_q = \{0, \, 1, \hspace{0.05cm} \text{...} \hspace{0.1cm} , \, q-1\}$ of natural numbers leads to a Galois field only if $q$ is a prime number.
Of the number sets mentioned above, this is true only for $Z_{11}$.
