2.1#

We consider \((\mathbb{R} \setminus \{-1\}, \star)\), where

\[ a \star b := ab + a + b, \qquad a,b \in \mathbb{R} \setminus \{-1\} \]

2.1a#

Show that \((\mathbb{R} \setminus \{-1\}, \star)\) is an Abelian group.

In order to show that \((\mathbb{R} \setminus \{-1\}, \star)\) in an Abelian group, we need to first show that it fulfills the conditions to be a group.

Proving that \(\star\) is closed under \(\mathbb{R} \setminus \{-1\}\)#

Note that \(\star\) contains the operations of addition and multiplication. Because \(\mathbb{R}\) is a ring, by definition \(\mathbb{R}\) is closed under addition and multiplication and thus closed under \(\star\).

However, we need to further prove that \(\star\) cannot produce \(-1\). We will prove this by contradiction.

Suppose \(a \star b = ab + a + b = -1\) for \(a,b \in \mathbb{R} \setminus \{-1\}\).

\[\begin{aligned} ab + a + b &= -1 \\ ab + a + b + 1 &= 0 \\ (a+1)(b+1) &= 0 \\ \end{aligned}\]

Note that for \(ab + a + b = -1\), both \(a\) and \(b\) must equal \(-1\). However, \(-1 \not \in \mathbb{R} \setminus \{-1\}\). So we proved that \(a \star b = ab + a + b \neq -1\) for \(a,b \in \mathbb{R} \setminus \{-1\}\) via contradiction, and thus \(\star\) is closed under \(\mathbb{R} \setminus \{-1\}\).

Proving that \(\star\) is associative under \(\mathbb{R} \setminus \{-1\}\)#

We want to prove that \((a \star b) \star c = a \star (b \star c)\) for \(a,b,c \in \mathbb{R} \setminus \{-1\}\). We will prove this by changing the form of \((a \star b) \star c\) and \(a \star (b \star c)\) to be a sum of products and showing that they are equal.

\((a \star b) \star c\) (left side)#
\[\begin{aligned} (a \star b) \star c &= (ab + a + b) \star c \end{aligned}\]

Let \(d = ab + a + b\) so that \((a \star b) \star c = d \star c\).

\[\begin{aligned} (a \star b) \star c = d \star c &= dc + d + c \\ &= (ab + a + b)c + (ab + a + b) + c \\ &= abc + ac + bc + ab + a + b + c \\ &= \underline{abc + ab + ac + bc + a + b + c} \end{aligned}\]
\(a \star (b \star c)\) (right side)#
\[\begin{aligned} a \star (b \star c) = a \star (bc + b + c) \end{aligned}\]

Let \(e = bc + b + c\) so that \(a \star (b \star c) = a \star e\).

\[\begin{aligned} a \star (b \star c) = a \star e &= ae + a + e \\ &= a(bc + b + c) + a + (bc + b + c) \\ &= abc + ab + ac + a + bc + b + c \\ &= \underline{abc + ab + ac + bc + a + b + c} \end{aligned}\]

Thus, \((a \star b) \star c = a \star (b \star c)\).

Proving that \((\mathbb{R} \setminus \{-1\}, \star)\) has a neutral element#

We want to show that \(\exists e \in \mathbb{R} \setminus \{-1\}, \: \forall x \in \mathbb{R} \setminus \{-1\}: x \star e = e \star x = x\).

Suppose \(e = 0\).

\[\begin{aligned} x \star 0 &= (x \cdot 0) + x + 0 = x \\ 0 \star x &= (0 \cdot x) + 0 + x = x \end{aligned}\]

Therefore, \(\mathbb{R} \setminus \{-1\}\) has a neutral element.

Proving that \((\mathbb{R} \setminus \{-1\}, \star)\) has an inverse element#

We want to show that \(\forall x \in \mathbb{R} \setminus \{-1\}, \: \exists y \in \mathbb{R} \setminus \{-1\}\) such that \(x \star y = y \star x = e\) where \(e = 0\) (as proven above).

We’re first going to solve for the inverse element \(y\) and then substitute it back into the expression \(x \star y\) to prove that \(x \star y = 0\) is true.

Solving for the inverse element \(y\)#
\[\begin{aligned} x \star y = xy + x + y &= 0 \\ xy + y &= -x \\ y(x + 1) &= -x \\ y &= \frac{-x}{x+1} \end{aligned}\]
Proving \(x \star y = 0\)#
\[\begin{aligned} x \star y = x \star \frac{-x}{x+1} &= 0 \\ x(\frac{-x}{x+1}) + x - \frac{x}{x+1} &= 0 \\ \frac{-x^2}{x+1} + \frac{x(x+1)}{x+1} - \frac{x}{x+1} &= 0 \\ -x^2 + x^2 + x - x &= 0 \\ 0 = 0 \end{aligned}\]

Note: for brevity, I am skipping the proof for \(y \star x = 0\).

Therefore, \(\mathbb{R} \setminus \{-1\}\) has an inverse element for every \(x \in \mathbb{R} \setminus \{-1\}\).

Proving \((\mathbb{R} \setminus \{-1\}, \star)\) is an Abelian group#

Now that we have proven that \((\mathbb{R} \setminus \{-1\}, \star)\) is indeed a group, we can now determine whether it is Abelian. To do so, we need to show that \(a \star b = b \star a\) for all \(a,b \in \mathbb{R} \setminus \{-1\}\).

\[\begin{aligned} a \star b &= ab + a + b \\ b \star a &= ba + b + a \end{aligned}\]

Because \(\mathbb{R}\) is a commutative ring, addition and multiplication are commutative, so \(\mathbb{R} \setminus \{-1\}\) is an Abelian group. \(\square\)

2.1b#

Solve \(3 \star x \star x = 15\) in the Abelian group \((\mathbb{R} \setminus \{-1\}, \star)\).

\[\begin{aligned} 3 \star x \star x &= 15 \\ (3x + 3 + x) \star x &= 15 \\ (3x + 3 + x)x + (3x + 3 + x) + x &= 15 \\ 3x^2 + 3x + x^2 + 3x + 3 + x + x &= 15 \\ 4x^2 + 8x + 3 &= 15 \\ 4x^2 + 8x &= 12 \\ x^2 + 2x &= 3 \\ x^2 + 2x - 3 &= 0 \\ (x+3)(x-1) &= 0 \\ \fbox{x = \{-3, 1\}} \end{aligned}\]

2.2 (in-progress)#

Let \(n\) be in \(\mathbb{N} \setminus \{0\}\). Let \(k,x \in \mathbb{Z}\). We define the congruence class \(\bar{k}\) of the integer \(k\) as the set

\[\begin{aligned} \bar{k} &= \{x \in \mathbb{Z} \mid x - k = 0 \text{ (mod } \textit{n} \text{)}\} \\ &= \{x \in \mathbb{Z} \mid (\exists a \in \mathbb{Z}): \: (x -k = n \cdot a)\} \end{aligned}\]

We now define \(\mathbb{Z}/n\mathbb{Z}\) (sometimes written as \(\mathbb{Z}_n\)) as the set of all congruence classes modulo \(n\). Euclidean division implies that this set is a finite set containing \(n\) elements:

\[ \mathbb{Z}_n = \{\bar{0}, \bar{1}, ..., \overline{n-1}\} \]

For all \(\bar{a}, \bar{b} \in \mathbb{Z}_n\), we define

\[ \bar{a} \oplus \bar{b} := \overline{a+b} \]

2.2a#

Show that \((\mathbb{Z}_n, \oplus)\) is a group. Is it Abelian?

In order to show that \((\mathbb{Z}_n, \oplus)\) in an Abelian group, we need to first show that it is a group.

Intuiting congruence classes#

Congruence classes were a bit difficult for me to wrap my head around so I needed to write this section for myself. This is not required for proving that \((\mathbb{Z}_n, \oplus)\) is a group.

Let’s try to look at congruence classes in a different way to get a better intuition of what they actually look like in practice. Let’s consider the example:

\[ \mathbb{Z}_4 = \{\bar{0}, \bar{1}, \bar{2}, \bar{3}\} \]

In this example \(n = 4\). Let’s look at the congruence class \(\bar{2}\). From the definition of congruence classes, this looks like:

\[ \bar{2} = \{x \in \mathbb{Z} \mid (\exists a \in \mathbb{Z}): \: (x - 2 = 4 \cdot a)\} \]

Let’s actually put some numbers in this set. As you can see, \(\bar{2}\) is an infinite set, so let’s just start with \(a=\{1,2,3\}\).

\[\begin{aligned} \bar{2} &= \{..., \: 4 \cdot 1 + 2, \: 4 \cdot 2 + 2, \: 4 \cdot 3 + 2, \: ...\} \\ &= \{..., \: 6, \: 10, \: 14, \: ...\} \end{aligned}\]

Working from this example, we can see the general form.

\[\begin{aligned} \bar{k} &= \{..., \: n(-2)+k, \: n(-1)+k, \: n(0)+k, \: n(1)+k, \: n(2)+k, \: ...\} \\ \bar{k} &= \{..., \: -2n+k, \: -n+k, \: k, \: n+k, \: 2n+k, \: ...\} \end{aligned}\]

Proving \(\oplus\) is closed under \(\mathbb{Z}_n\)#

To show that \(\oplus\) is closed under \(\mathbb{Z}_n\), we want to show that \((\bar{a} \oplus \bar{b}) \in \mathbb{Z}_n\).

From the general form of the congruence class, we can see that:

\[ \bar{a} \oplus \bar{b} = \overline{a+b} = \{..., \: -2n+(a+b), \: -n+(a+b), \: a+b, \: n+(a+b), \: 2n+(a+b), \: ...\} \]

Note, that \(\bar{a}, \bar{b} \in \mathbb{Z}_n \implies a,b \leq n-1\).

\[ a,b \in \mathbb{Z}_n, \: a+b \geq n \mid (\exists c \in \mathbb{Z}): \: c \lt n, \: \bar{c} = \overline{a+b} \]

That is, for any set \(a+b \geq n \mid \overline{a+b}\), there is equivalent set \(\bar{c}\) where \(c \lt n\).

Notice that the elements in the set \(\overline{a+b}\) differ by \(n\). So \(\exists x \in \mathbb{Z} \mid \overline{a+b} = \overline{a+b-xn}\) and \(a+b-xn \lt n\). Because we have shown that we can find an equivalent congruence class for any \(\overline{a+b}\) that is in \(\mathbb{Z}_n\), we have shown that \(\oplus\) is closed under \(\mathbb{Z}_n\).

Proving \(\oplus\) is associative under \(\mathbb{Z}_n\)#

Addition is associative for \(\mathbb{Z}\) because \(\mathbb{Z}\) is a commutative ring, therefore \(\oplus\) (e.g. \(a \oplus b := \overline{a+b}\)) is associative under \(\mathbb{Z}_n\).

Proving that \((\mathbb{Z}_n, \oplus)\) has a neutral element#

We want to show that \(\exists \bar{e} \in \mathbb{Z}_n, \forall \bar{k} \in \mathbb{Z}_n : \: \bar{k} \oplus \bar{e} = \bar{e} \oplus \bar{k} = \bar{k}\).

Consider \(\bar{e} = \bar{0}\).

\[ \bar{k} \oplus \bar{0} = \bar{0} \oplus \bar{k} = \overline{k+0} = \bar{k} \]

Therefore, the neutral element in \((\mathbb{Z}_n, \oplus)\) is \(\bar{0}\).

Proving that \(\mathbb{Z}_n, \oplus\) has an inverse element#

To-Do

2.3 (todo)#

To-Do

2.4#

Compute the following matrix products, if possible:

2.4a#

\[ \begin{bmatrix} 1 & 2 \\ 3 & 4 \\ 7 & 8 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \\ \end{bmatrix} \]

Dimensions of matrices are incompatible because the number of columns in the left matrix does not equal the number of rows in the right matrix.

2.4b#

\[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \]
\[\begin{aligned} (0,0) &= 1 \cdot 1 + 2 \cdot 0 + 3 \cdot 1 = 1 + 0 + 3 = 4 \\ (0,1) &= 1 \cdot 1 + 2 \cdot 1 + 3 \cdot 0 = 1 + 2 + 0 = 3 \\ (0,2) &= 1 \cdot 0 + 2 \cdot 1 + 3 \cdot 1 = 0 + 2 + 3 = 5 \\ \newline (1,0) &= 4 \cdot 1 + 5 \cdot 0 + 6 \cdot 1 = 4 + 0 + 6 = 10 \\ (1,1) &= 4 \cdot 1 + 5 \cdot 1 + 6 \cdot 0 = 4 + 5 + 0 = 9 \\ (1,2) &= 4 \cdot 0 + 5 \cdot 1 + 6 \cdot 1 = 0 + 5 + 6 = 11 \\ \newline (2,0) &= 7 \cdot 1 + 8 \cdot 0 + 9 \cdot 1 = 7 + 0 + 9 = 16 \\ (2,1) &= 7 \cdot 1 + 8 \cdot 1 + 9 \cdot 0 = 7 + 8 + 0 = 15 \\ (2,2) &= 7 \cdot 0 + 8 \cdot 1 + 9 \cdot 1 = 0 + 8 + 9 = 17 \end{aligned}\]\[\begin{bmatrix} 4 & 3 & 5 \\ 10 & 9 & 11 \\ 16 & 15 & 17 \end{bmatrix}\]

2.4c#

\[ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \]
\[\begin{aligned} (0,0) &= 1 \cdot 1 + 1 \cdot 4 + 0 \cdot 7 = 1 + 4 + 0 = 5 \\ (0,1) &= 1 \cdot 2 + 1 \cdot 5 + 0 \cdot 8 = 2 + 5 + 0 = 7 \\ (0,2) &= 1 \cdot 3 + 1 \cdot 6 + 0 \cdot 9 = 3 + 6 + 0 = 9 \\ \newline (1,0) &= 0 \cdot 1 + 1 \cdot 4 + 1 \cdot 7 = 0 + 4 + 7 = 11 \\ (1,1) &= 0 \cdot 2 + 1 \cdot 5 + 1 \cdot 8 = 0 + 5 + 8 = 13 \\ (1,2) &= 0 \cdot 3 + 1 \cdot 6 + 1 \cdot 9 = 0 + 6 + 9 = 15 \\ \newline (2,0) &= 1 \cdot 1 + 0 \cdot 4 + 1 \cdot 7 = 1 + 0 + 7 = 8 \\ (2,1) &= 1 \cdot 2 + 0 \cdot 5 + 1 \cdot 8 = 2 + 0 + 8 = 10 \\ (2,2) &= 1 \cdot 3 + 0 \cdot 6 + 1 \cdot 9 = 3 + 0 + 9 = 12 \end{aligned}\]\[\begin{bmatrix} 5 & 7 & 9 \\ 11 & 13 & 15 \\ 8 & 10 & 12 \end{bmatrix}\]

2.4c#

\[ \begin{bmatrix} 1 & 2 & 1 & 2 \\ 4 & 1 & -1 & -4 \end{bmatrix} \begin{bmatrix} 0 & 3 \\ 1 & -1 \\ 2 & 1 \\ 5 & 2 \end{bmatrix} \]
\[\begin{aligned} (0,0) &= 1 \cdot 0 + 2 \cdot 1 + 1 \cdot 2 + 2 \cdot 5 = 0 + 2 + 2 + 10 = 14 \\ (0,1) &= 1 \cdot 3 + 2 \cdot -1 + 1 \cdot 1 + 2 \cdot 2 = 3 - 2 + 1 + 4 = 6 \\ (1,0) &= 4 \cdot 0 + 1 \cdot 1 + -1 \cdot 2 + -4 \cdot 5 = 0 + 1 - 2 - 20 = -21 \\ (1,1) &= 4 \cdot 3 + 1 \cdot -1 + -1 \cdot 1 + -4 \cdot 2 = 12 -1 - 1 - 8 = 2 \end{aligned}\]\[\begin{bmatrix} 14 & 6 \\ -21 & 2 \end{bmatrix}\]

2.4e#

\[ \begin{bmatrix} 0 & 3 \\ 1 & -1 \\ 2 & 1 \\ 5 & 2 \end{bmatrix} \begin{bmatrix} 1 & 2 & 1 & 2 \\ 4 & 1 & -1 & -4 \end{bmatrix} \]
\[\begin{aligned} (0,0) &= 0 \cdot 1 + 3 \cdot 4 = 12 \\ (0,1) &= 0 \cdot 2 + 3 \cdot 1 = 3 \\ (0,2) &= 0 \cdot 1 + 3 \cdot -1 = -3 \\ (0,3) &= 0 \cdot 2 + 3 \cdot -4 = -12 \\ \newline (1,0) &= 1 \cdot 1 - 1 \cdot 4 = -3 \\ (1,1) &= 1 \cdot 2 - 1 \cdot 1 = 1 \\ (1,2) &= 1 \cdot 1 - 1 \cdot -1 = 2 \\ (1,3) &= 1 \cdot 2 - 1 \cdot -4 = 6 \\ \newline (2,0) &= 2 \cdot 1 + 1 \cdot 4 = 6 \\ (2,1) &= 2 \cdot 2 + 1 \cdot 1 = 5 \\ (2,2) &= 2 \cdot 1 + 1 \cdot -1 = 1 \\ (2,3) &= 2 \cdot 2 + 1 \cdot -4 = 0 \\ \newline (3,0) &= 5 \cdot 1 + 2 \cdot 4 = 13 \\ (3,1) &= 5 \cdot 2 + 2 \cdot 1 = 12 \\ (3,2) &= 5 \cdot 1 + 2 \cdot -1 = 3 \\ (3,3) &= 5 \cdot 2 + 2 \cdot -4 = 2 \end{aligned}\]\[\begin{bmatrix} 12 & 3 & -3 & -12 \\ -3 & 1 & 2 & 6 \\ 6 & 5 & 1 & 0 \\ 13 & 12 & 3 & 2 \end{bmatrix}\]