We mark and sort the summands according to their square root factor:
\[\begin{gather*} \color{green}{aa'} + \color{gold}{ab'\sqrt{2}} + \color{blue}{ac'\sqrt{3}} + \color{pink}{ad'\sqrt{6}} + \color{gold}{b\sqrt{2}a'} + \\ \color{green}{b\sqrt{2}b'\sqrt{2}} + \color{pink}{b\sqrt{2}c'\sqrt{3}} + \color{blue}{b\sqrt{2}d'\sqrt{6}} + \color{blue}{c\sqrt{3}a'} + \color{pink}{c\sqrt{3}b'\sqrt{2}} + \\ \color{green}{c \sqrt{3} c'\sqrt{3}} + \color{gold}{c\sqrt{3}d'\sqrt{6}} + \color{pink}{d\sqrt{6}a'} + \color{blue}{d\sqrt{6} b'\sqrt{2}} + \color{gold}{d\sqrt{6} c' \sqrt{3}} + \\ \color{green}{d\sqrt{6} d' \sqrt{6}} \end{gather*}\]
Thus the quadrupel is the following:
\[\begin{gather*} (\color{green}{aa'} + \color{green}{b\sqrt{2}b'\sqrt{2}} + \color{green}{c \sqrt{3} c'\sqrt{3}} + \color{green}{d\sqrt{6} d' \sqrt{6}},\\ \color{gold}{ab'\sqrt{2}} + \color{gold}{b\sqrt{2}a'} + \color{gold}{c\sqrt{3}d'\sqrt{6}} + \color{gold}{d\sqrt{6} c'}, \\ \color{blue}{ac'\sqrt{3}} + \color{blue}{b\sqrt{2}d'\sqrt{6}} + \color{blue}{d\sqrt{6} b'\sqrt{2}} + \color{blue}{c\sqrt{3}a'}, \\ \color{pink}{ad'\sqrt{6}} + \color{pink}{b\sqrt{2}c'\sqrt{3}} + \color{pink}{c\sqrt{3}b'\sqrt{2}} + \color{pink}{d\sqrt{6}a'}) \end{gather*}\]
Let \(S = \{a+b\sqrt{2}|a,b \in \mathbb{Q}\}\),
\(S\) is trivially closed under addition by \(\mathbb{Q}\) being a field.
As can be shown analogously to 101), \(S\) is closed under multiplication.
Each element in \(S\) has a multiplicative inverse, i.e.: For any \(a+ \sqrt{2}b \neq 0\) there exists \(c + \sqrt{2}d\) such that \((a+\sqrt{2}b)(c + \sqrt{2}d = 1\) meaning \(ac + 2bd + \sqrt{2}ad + \sqrt{2}bc = 1\) thus we get the set of equations:
\[\begin{gather*} ad + bc = 0\\ ac + 2bd = 1 \end{gather*}\]
We can use matrices to solve this system of equations:
\[\begin{gather*} \begin{bmatrix} a & 2b\\ b & a \end{bmatrix} \cdot \begin{bmatrix} c \\ d \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \iff \begin{bmatrix} c\\ d \end{bmatrix}= \begin{bmatrix} a & 2b\\ b & a \end{bmatrix}^{-1} \cdot \begin{bmatrix} 1 \\ 0 \end{bmatrix} \end{gather*}\]
and
\[\begin{gather*} \begin{bmatrix} a & 2b\\ b & a \end{bmatrix}^{-1} = \frac{1}{a^2-2b^2}\begin{bmatrix} a & -2b\\ -b & a \end{bmatrix}^{-1} \end{gather*}\]
thus
\[\begin{gather*} \begin{bmatrix} c \\ d \end{bmatrix} = \frac{1}{a^2-2b^2} \begin{bmatrix} a \\ -b \end{bmatrix} \end{gather*}\]
meaning \(c = \frac{a}{a^2-2b^2}, d = \frac{-b}{a^2-2b^2}\) which are both in \(\mathbb{Q}\) since \(a = b = 0\) is not possible by construction.
Compute \((3-5\sqrt{2})^{-1}\).
\((3-5\sqrt{2})^{-1}\) is thus \(c = \frac{3}{-41}, d = \frac{-5}{41}\).
Hint: Use the binomial theorem and consider the equation \(\binom{p}{k} = p \cdot \frac{(p-1)!}{k!(p-k)!}\) for \(0 < k < p\). Show that \(\binom{p}{k} \in \mathbb{N}\) implies that the fraction on the right-hand side must be an integer, too, since the factors in the denominator do not divide \(p\).
We apply the binomial theorem:
\[\begin{gather*} (a+b)^p = \sum_{k=0}^p \binom{p}{k} a^{p-k} b^k \end{gather*}\]
we split the sum in order to be able to apply the next step of the hint:
\[\begin{gather*} \binom{p}{0} a^pb^0 + \sum_{k=1}^{p-1} p\cdot \frac{(p-1)!}{k!(p-k)!} a^{p-k}b^k + \binom{p}{p}a^0 b^p = \\ = a^p + \sum_{k=1}^{p-1} p \cdot \frac{(p-1)!}{k!(p-k)!} \cdot a^{p-k} \cdot b^k + b^p \end{gather*}\]
Now, we know \(\binom{p}{k} = p \cdot \frac{(p-1)!}{k!(p-k)!} \in \mathbb{N}\) and we know \(k!(p-k)!\) does not divide \(p\) as \(k \leq p-1 < p\) and thus each \(k (k-1) \cdots 1\) does not divide \(p\), analogously this holds for \((p-k)!\)’s factors. Since \(p\) is prime, no number smaller than \(p\) divides \(p\), therefore \(\frac{(p-1)!}{k!(p-k)!} \in \mathbb{N}\) as \(p \in \mathbb{N}\) and \(\binom{p}{k} \in \mathbb{N}\). Thus \(p \cdot \frac{(p-1)!}{k!(p-k)!}\) can be written as \(p \cdot x\) where \(x \in \mathbb{N}\) and thus \(x \in K\) as well.
This means \(px = (x + x + \dots + x) = x(1+ \dots + 1) = 0\) by definition of \(char(K) = p\).
\[\begin{gather*} \bar{b(x)} = \bar{b_7 x^7 + b_6 x^6 + \dots + b_1x + b_0} \end{gather*}\]
and can be identified with a byte \(b_7b_6 \dots b_1b_0\). Compute the sum of the two bytes \(10010101\) and \(11001100\) in the field.
\[\begin{align} &10010101\\ +&11001100\\ \hline &01011001 \end{align}\]
no carry by addition over polynoms and xor since \(\mathbb{Z}_2\). Thus \(x^6 + x^4 + x^3 +1\) mod \(m(x)\).
We show \(C\) is a linear code by defining a generating matrix of dimensions \(k \times r \cdot k\) for it:
\[\begin{gather*} \begin{bmatrix} 1^r & 0^r & 0^r & \dots \\ 0^r & 1^r & 0^r & \dots \\ 0^r & 0^r & 1^r & \dots \\ \vdots & \vdots & \vdots & \ddots\\ \end{bmatrix} \end{gather*}\]
and the check matrix of dimensions \((r-1)k \times (r\cdot k)\) by:
\[\begin{gather*} \begin{bmatrix} m & m_0 & m_0 & \dots \\ m_0 & m & m_0 & \dots \\ m_0 & m_0 & m & \dots \\ \vdots & \vdots & \vdots & \ddots\\ \end{bmatrix} \end{gather*}\]
where each \(m,m_0\) represents a block of size \((r-1)\cdot r\) and \(m_0\) is defined by a block of only zeros while \(m\) is defined by:
\[\begin{gather*} \begin{matrix} 1 & 1 & 1 & \dots \\ -1 & 0 & 0 & \dots \\ 0 & -1 & 0 & \dots \\ 0 & 0 & -1 & \dots \\ \vdots & \vdots & \vdots & \ddots\\ \end{matrix} \end{gather*}\]
each of these blocks represents a system of equations which check all \(r\) bits are the same, i.e. bit \(1\) is the same as bit \(2\) etc.
We want \(d=3\) in order to be able to correct \(1\) error.
\[\begin{gather*} G = \pmatrix{ 0 & 1 & 1 & 0 & 1\\ 1 & 0 & 0 & 1 & 1 } \end{gather*}\]
\[\begin{gather*} H = \pmatrix{ 1 & 1 & 0 & 1 & 0\\ 1 & 0 & 1 & 1 & 0\\ -1 & 0 & -1 & 0 & 1\\ -1 & -1 & 0 & 0 & 1\\ 0 & 0 & 0 & -1 & -1 } \end{gather*}\]
from the system of equations:
\[\begin{gather*} b_1 + b_2 = b_3 + b_4\\ b_1 = b_4\\ b_2 = b_3\\ b_1 + b_2 = b_5\\ b_3 + b_4 = b_5 \end{gather*}\]