Difference between revisions of "Aufgaben:Exercise 2.6: GF(P power m). Which P, which m?"
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===Solution=== | ===Solution=== | ||
{{ML-Kopf}} | {{ML-Kopf}} | ||
− | '''(1)''' Each element consists of two ternaries ⇒ $\underline{P = 3}, \ \underline{m = 2}$. There are $q = P^m = 3^8 = \underline{9 \rm \hspace{0. | + | '''(1)''' Each element consists of two ternaries ⇒ $\underline{P = 3}, \ \underline{m = 2}$. There are $q = P^m = 3^8 = \underline{9 \rm \hspace{0.2cm}elements}$. |
− | '''(2)''' Correct is the <u>proposed solution 1</u>: | + | '''(2)''' Correct is the <u>proposed solution 1</u>: |
− | *The neutral element of the addition $(N_{\rm A})$ satisfies for all $z_i ∈ {\rm GF}(P^m)$ the condition $z_i + N_{\rm A} = z_i$. | + | *The neutral element of the addition $(N_{\rm A})$ satisfies for all $z_i ∈ {\rm GF}(P^m)$ the condition $z_i + N_{\rm A} = z_i$. |
− | *From the addition table it can be read that "$0\hspace{0.03cm}0$" satisfies this condition. | + | *From the addition table it can be read that "$0\hspace{0.03cm}0$" satisfies this condition. |
− | '''(3)''' Correct is the <u>proposed solution 2</u>: | + | '''(3)''' Correct is the <u>proposed solution 2</u>: |
− | *The neutral element of the multiplication $(N_{\rm M})$ must always satisfy the condition $z_i \cdot N_{\rm M} = z_i$. | + | *The neutral element of the multiplication $(N_{\rm M})$ must always satisfy the condition $z_i \cdot N_{\rm M} = z_i$. |
− | *From the multiplication table, $N_{\rm M} = \, "0\hspace{0.03cm}1"$. | + | |
− | *In polynomial notation, this corresponds to $k_1 = 0$ and $k_0 = 1$: | + | *From the multiplication table, $N_{\rm M} = \, "\hspace{-0.15cm}0\hspace{0.03cm}1\hspace{-0.1cm}"$. |
+ | |||
+ | *In polynomial notation, this corresponds to $k_1 = 0$ and $k_0 = 1$: | ||
:$$k_1 \cdot \alpha + k_0 = 1 \hspace{0.05cm}.$$ | :$$k_1 \cdot \alpha + k_0 = 1 \hspace{0.05cm}.$$ | ||
− | '''(4)''' With the polynomial representation, the following calculations result: | + | '''(4)''' With the polynomial representation, the following calculations result: |
− | :$${\rm Inv_A}("\hspace{-0.05cm}0\hspace{0.03cm}2") \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2) = (-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.25cm}\Rightarrow \hspace{0.25cm}{\rm | + | :$${\rm Inv_A}("\hspace{-0.05cm}0\hspace{0.03cm}2") \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2) = (-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.25cm}\Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.1cm}0\hspace{0.03cm}1\hspace{-0.1cm}"\hspace{0.05cm},$$ |
:$${\rm Inv_A}("\hspace{-0.05cm}1\hspace{0.03cm}1")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(\alpha + 1) = \big[(-\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + | :$${\rm Inv_A}("\hspace{-0.05cm}1\hspace{0.03cm}1")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(\alpha + 1) = \big[(-\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + | ||
− | \big[(-1) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =2\alpha + 2 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm | + | \big[(-1) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =2\alpha + 2 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}\hspace{-0.05cm}2\hspace{0.03cm}2\hspace{-0.1cm}"\hspace{0.05cm},$$ |
:$${\rm Inv_A}("\hspace{-0.05cm}2\hspace{0.03cm}2")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2\alpha + 2) = \big[(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + | :$${\rm Inv_A}("\hspace{-0.05cm}2\hspace{0.03cm}2")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2\alpha + 2) = \big[(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + | ||
− | \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =\alpha + 1 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm | + | \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =\alpha + 1 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}\hspace{-0.05cm}1\hspace{0.03cm}1\hspace{-0.1cm}"\hspace{0.05cm}.$$ |
+ | |||
+ | Consequently, only the <u>first two proposed solutions</u> are correct. | ||
− | + | However, the exercise can also be solved without calculation using the addition table alone: | |
+ | *For example, you can find the inverse of "$2\hspace{0.03cm}2$" by looking for the column with the entry "$0\hspace{0.03cm}0$" in the last row. | ||
+ | |||
+ | *You find the column labeled "$1\hspace{0.03cm}1$" and thus ${\rm Inv_A}("\hspace{-0.15cm}2\hspace{0.03cm}2\hspace{-0.15cm}") = \, "\hspace{-0.15cm}1\hspace{0.03cm}1\hspace{-0.15cm}"$. | ||
− | |||
− | |||
− | |||
+ | '''(5)''' Multiplying $\alpha$ $($vector "$1\hspace{0.03cm}0$"$)$ by itself gives $\alpha^2$. | ||
+ | * If the first proposed solution were valid, the condition $\alpha^2 + 2 = 0$ and thus $\alpha^2 = (-2) \, {\rm mod} \, 3 = 1$, thus yielding the vector "$0\hspace{0.03cm}1$". | ||
− | + | * Assuming the second proposed solution, it follows from the condition $\alpha^2 + 2\alpha + 2 = 0$ in polynomial notation: | |
− | + | :$$\alpha^2 = \big [(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big ] = \alpha + 1 $$ | |
− | * Assuming the second proposed solution, it follows from the condition $\alpha^2 + 2\alpha + 2 = 0$ in polynomial notation | + | :and thus the coefficient vector "$1\hspace{0.03cm}1$". |
− | :$$\alpha^2 = [(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3] + [(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3] = \alpha + 1 $$ | ||
− | :and thus the coefficient vector "$1\hspace{0.03cm}1$". | ||
− | In the multiplication table, in row 4, column 4, we find exactly the entry "$1\hspace{0.03cm}1$" & | + | *In the multiplication table, in row 4, column 4, we find exactly the entry "$1\hspace{0.03cm}1$" ⇒ So, the correct answert is the <u>proposed solution 2</u>. |
− | '''(6)''' The multiplicative inverse to "$1\hspace{0.03cm}2$" can be found in row 6 of the multiplication table as the column with the entry "$0\hspace{0.03cm}1$" | + | '''(6)''' The multiplicative inverse to "$1\hspace{0.03cm}2$" can be found in row 6 of the multiplication table as the column with the entry "$0\hspace{0.03cm}1$" |
+ | *So the <u>proposed solution 2</u> is correct in contrast to proposal 3. Namely, ${\rm Inv_M}("\hspace{-0.15cm}21\hspace{-0.15cm}") = \, "\hspace{-0.15cm}2\hspace{0.03cm}0\hspace{-0.1cm}"$ holds. | ||
− | We check these results considering $\alpha^2 + 2\alpha + 2 = 0$ by multiplications: | + | *We check these results considering $\alpha^2 + 2\alpha + 2 = 0$ by multiplications: |
− | :$$"1\hspace{0.03cm}2" \hspace{0.05cm}\cdot \hspace{0.05cm}"1\hspace{0.03cm}0" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (\alpha + 2) \cdot \alpha = \alpha^2 + 2\alpha = (-2\alpha-2) + 2\alpha = -2 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} {\rm | + | :$$"\hspace{-0.15cm}1\hspace{0.03cm}2\hspace{-0.1cm}" \hspace{0.05cm}\cdot \hspace{0.05cm}"\hspace{-0.15cm}1\hspace{0.03cm}0\hspace{-0.1cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (\alpha + 2) \cdot \alpha = \alpha^2 + 2\alpha = (-2\alpha-2) + 2\alpha = -2 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} {\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}0\hspace{0.03cm}1\hspace{-0.15cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$ |
− | :$$"2\hspace{0.03cm}1" \hspace{0.05cm}\cdot \hspace{0.05cm}" | + | :$$"\hspace{-0.15cm}2\hspace{0.03cm}1\hspace{-0.1cm}" \hspace{0.05cm}\cdot \hspace{0.05cm}"\hspace{-0.15cm}1\hspace{0.03cm}2\hspace{-0.1cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (2\alpha + 1) \cdot (\alpha + 2) = 2 \alpha^2 + \alpha + 4\alpha + 2 = 2 \alpha^2 + 5\alpha + 2 = 2 \cdot (-2\alpha - 2) + 5\alpha + 2 = (\alpha - 2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = \alpha +1 $$ |
− | + | :$$\hspace{2.725cm} \Rightarrow \ {\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}1\hspace{0.03cm}1\hspace{-0.15cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm no\hspace{0.15cm}multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$ | |
− | :$$\hspace{2.725cm} \Rightarrow \ {\rm | ||
− | The solution suggestion 1 is therefore not correct, because there is no multiplicative inverse for "$00$". | + | *The solution suggestion 1 is therefore not correct, because there is no multiplicative inverse for "$00$". |
'''(7)''' The two expressions agree ⇒ <u>YES</u>, as the following calculations show: | '''(7)''' The two expressions agree ⇒ <u>YES</u>, as the following calculations show: | ||
− | :$$("20" + "12") \ \cdot "12" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "02"\cdot "12" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "21"\hspace{0.05cm},$$ | + | :$$("\hspace{-0.15cm}20\hspace{-0.1cm}" + "\hspace{-0.15cm}12\hspace{-0.1cm}") \ \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}02\hspace{-0.1cm}"\cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}21\hspace{-0.1cm}"\hspace{0.05cm},$$ |
− | :$$"20" \cdot "12" + "12" \cdot "12" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "02" + "22" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "21"\hspace{0.05cm}.$$ | + | :$$"\hspace{-0.15cm}20\hspace{-0.1cm}" \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" + "\hspace{-0.15cm}12\hspace{-0.1cm}" \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}02\hspace{-0.1cm}" + "\hspace{-0.15cm}22\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}21\hspace{-0.1cm}"\hspace{0.05cm}.$$ |
This means: The distributive law has been proved at least on a single example. | This means: The distributive law has been proved at least on a single example. |
Latest revision as of 15:29, 4 October 2022
A Galois field ${\rm GF}(q)$ with $q = P^m$ elements defined by the adjacent tables is to be analyzed
- for addition $($marked with "$+$"$)$, and
- for multiplication $($marked with "$\hspace{0.05cm}\cdot\hspace{0.05cm}$)".
This Galois field ${\rm GF}(q) = \{\hspace{0.1cm}z_0,\hspace{0.1cm} z_1,\hspace{0.05cm} \text{...} , \hspace{0.1cm}z_{q-1}\}$ satisfies all the requirements for a finite field listed in the chapter "Some Basics of Algebra" . Thus, commutative, associative and distributive laws are also satisfied.
Furthermore there is
- a neutral element with respect to addition ⇒ $N_{\rm A}$:
- $$\exists \hspace{0.15cm} z_j \in {\rm GF}(q)\text{:} \hspace{0.25cm}z_i + z_j = z_i \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_j = N_{\rm A} \hspace{0.25cm}{\rm (zero\hspace{0.15cm}element)} \hspace{0.05cm},$$
- a neutral element with respect to multiplication ⇒ $N_{\rm M}$:
- $$\exists \hspace{0.15cm} z_j \in {\rm GF}(q)\text{:} \hspace{0.25cm}z_i \cdot z_j = z_i \hspace{0.3cm} \Rightarrow \hspace{0.3cm} z_j = N_{\rm M} \hspace{0.25cm}{\rm (identity\hspace{0.15cm}element)} \hspace{0.05cm},$$
- for all elements $z_i$ an additive inverse ⇒ ${\rm Inv_A}(z_i)$:
- $$\forall \hspace{0.15cm} z_i \in {\rm GF}(q)\hspace{0.15cm} \exists \hspace{0.15cm} {\rm Inv_A}(z_i) \in {\rm GF}(q)\text{:}$$
- $$z_i + {\rm Inv_A}(z_i) = N_{\rm A} = {\rm "\hspace{-0.15cm}0\hspace{-0.1cm}"}\hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm short}\text{:}\hspace{0.15cm} {\rm Inv_A}(z_i) = - z_i \hspace{0.05cm}, $$
- for all elements $z_i$ except the zero element a multiplicative inverse ⇒ ${\rm Inv_M}(z_i)$:
- $$\forall \hspace{0.15cm} z_i \in {\rm GF}(q),\hspace{0.15cm} z_i \ne N_{\rm A} \hspace{0.15cm} \exists \hspace{0.15cm} {\rm Inv_M}(z_i) \in {\rm GF}(q)\text{:}$$
- $$z_i \cdot {\rm Inv_M}(z_i) = N_{\rm M} = {\rm "\hspace{-0.15cm}1\hspace{-0.1cm}"} \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm short}\text{:}\hspace{0.15cm} {\rm Inv_M}(z_i) = z_i^{-1} \hspace{0.05cm}. $$
Hints:
- This exercise belongs to the chapter "Extension Field".
- In the tables, the elements $z_0, \hspace{0.05cm} \text{...} \hspace{0.1cm} , \ z_8$ are called "coefficient vectors".
- For example, "$2 \hspace{0.03cm}1$" stands for "$2 \cdot \alpha + 1$".
Questions
Solution
(2) Correct is the proposed solution 1:
- The neutral element of the addition $(N_{\rm A})$ satisfies for all $z_i ∈ {\rm GF}(P^m)$ the condition $z_i + N_{\rm A} = z_i$.
- From the addition table it can be read that "$0\hspace{0.03cm}0$" satisfies this condition.
(3) Correct is the proposed solution 2:
- The neutral element of the multiplication $(N_{\rm M})$ must always satisfy the condition $z_i \cdot N_{\rm M} = z_i$.
- From the multiplication table, $N_{\rm M} = \, "\hspace{-0.15cm}0\hspace{0.03cm}1\hspace{-0.1cm}"$.
- In polynomial notation, this corresponds to $k_1 = 0$ and $k_0 = 1$:
- $$k_1 \cdot \alpha + k_0 = 1 \hspace{0.05cm}.$$
(4) With the polynomial representation, the following calculations result:
- $${\rm Inv_A}("\hspace{-0.05cm}0\hspace{0.03cm}2") \hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2) = (-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.25cm}\Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.1cm}0\hspace{0.03cm}1\hspace{-0.1cm}"\hspace{0.05cm},$$
- $${\rm Inv_A}("\hspace{-0.05cm}1\hspace{0.03cm}1")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(\alpha + 1) = \big[(-\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-1) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =2\alpha + 2 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}\hspace{-0.05cm}2\hspace{0.03cm}2\hspace{-0.1cm}"\hspace{0.05cm},$$
- $${\rm Inv_A}("\hspace{-0.05cm}2\hspace{0.03cm}2")\hspace{-0.15cm} \ = \ \hspace{-0.15cm} {\rm Inv_A}(2\alpha + 2) = \big[(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] =\alpha + 1 \hspace{0.25cm} \Rightarrow \hspace{0.25cm}{\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}\hspace{-0.05cm}1\hspace{0.03cm}1\hspace{-0.1cm}"\hspace{0.05cm}.$$
Consequently, only the first two proposed solutions are correct.
However, the exercise can also be solved without calculation using the addition table alone:
- For example, you can find the inverse of "$2\hspace{0.03cm}2$" by looking for the column with the entry "$0\hspace{0.03cm}0$" in the last row.
- You find the column labeled "$1\hspace{0.03cm}1$" and thus ${\rm Inv_A}("\hspace{-0.15cm}2\hspace{0.03cm}2\hspace{-0.15cm}") = \, "\hspace{-0.15cm}1\hspace{0.03cm}1\hspace{-0.15cm}"$.
(5) Multiplying $\alpha$ $($vector "$1\hspace{0.03cm}0$"$)$ by itself gives $\alpha^2$.
- If the first proposed solution were valid, the condition $\alpha^2 + 2 = 0$ and thus $\alpha^2 = (-2) \, {\rm mod} \, 3 = 1$, thus yielding the vector "$0\hspace{0.03cm}1$".
- Assuming the second proposed solution, it follows from the condition $\alpha^2 + 2\alpha + 2 = 0$ in polynomial notation:
- $$\alpha^2 = \big [(-2\alpha) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big] + \big[(-2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3\big ] = \alpha + 1 $$
- and thus the coefficient vector "$1\hspace{0.03cm}1$".
- In the multiplication table, in row 4, column 4, we find exactly the entry "$1\hspace{0.03cm}1$" ⇒ So, the correct answert is the proposed solution 2.
(6) The multiplicative inverse to "$1\hspace{0.03cm}2$" can be found in row 6 of the multiplication table as the column with the entry "$0\hspace{0.03cm}1$"
- So the proposed solution 2 is correct in contrast to proposal 3. Namely, ${\rm Inv_M}("\hspace{-0.15cm}21\hspace{-0.15cm}") = \, "\hspace{-0.15cm}2\hspace{0.03cm}0\hspace{-0.1cm}"$ holds.
- We check these results considering $\alpha^2 + 2\alpha + 2 = 0$ by multiplications:
- $$"\hspace{-0.15cm}1\hspace{0.03cm}2\hspace{-0.1cm}" \hspace{0.05cm}\cdot \hspace{0.05cm}"\hspace{-0.15cm}1\hspace{0.03cm}0\hspace{-0.1cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (\alpha + 2) \cdot \alpha = \alpha^2 + 2\alpha = (-2\alpha-2) + 2\alpha = -2 \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = 1 \hspace{0.15cm} \Rightarrow \hspace{0.15cm} {\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}0\hspace{0.03cm}1\hspace{-0.15cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$
- $$"\hspace{-0.15cm}2\hspace{0.03cm}1\hspace{-0.1cm}" \hspace{0.05cm}\cdot \hspace{0.05cm}"\hspace{-0.15cm}1\hspace{0.03cm}2\hspace{-0.1cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm} (2\alpha + 1) \cdot (\alpha + 2) = 2 \alpha^2 + \alpha + 4\alpha + 2 = 2 \alpha^2 + 5\alpha + 2 = 2 \cdot (-2\alpha - 2) + 5\alpha + 2 = (\alpha - 2) \hspace{0.1cm}{\rm mod} \hspace{0.1cm} 3 = \alpha +1 $$
- $$\hspace{2.725cm} \Rightarrow \ {\rm vector}\hspace{0.15cm}"\hspace{-0.15cm}1\hspace{0.03cm}1\hspace{-0.15cm}" \hspace{0.15cm} \Rightarrow \hspace{0.15cm}{\rm no\hspace{0.15cm}multiplicative \hspace{0.15cm}inverse}\hspace{0.05cm}.$$
- The solution suggestion 1 is therefore not correct, because there is no multiplicative inverse for "$00$".
(7) The two expressions agree ⇒ YES, as the following calculations show:
- $$("\hspace{-0.15cm}20\hspace{-0.1cm}" + "\hspace{-0.15cm}12\hspace{-0.1cm}") \ \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}02\hspace{-0.1cm}"\cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}21\hspace{-0.1cm}"\hspace{0.05cm},$$
- $$"\hspace{-0.15cm}20\hspace{-0.1cm}" \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" + "\hspace{-0.15cm}12\hspace{-0.1cm}" \cdot "\hspace{-0.15cm}12\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}02\hspace{-0.1cm}" + "\hspace{-0.15cm}22\hspace{-0.1cm}" \hspace{-0.15cm} \ = \ \hspace{-0.15cm} "\hspace{-0.15cm}21\hspace{-0.1cm}"\hspace{0.05cm}.$$
This means: The distributive law has been proved at least on a single example.