Let \(t = \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}+
2\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}\) be a \(3\)-tuple with
entries from \(\mathbb{R}\).
What is \(t_2^2\)?

Recall that \(t_2\) is the second entry of \(t\). Let us perform
tuple arithmetic to obtain \(t\) in the simplest possible form:
\[t = \begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}+
2\begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}
=
\begin{bmatrix} 0 \\ 1 \\ 2\end{bmatrix}+
\begin{bmatrix} 2 \\ -2 \\ 2 \end{bmatrix}
= \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix}.
\]
Hence, \(t_2^2 = (-1)^2 = 1\).

Example 2

Write out all the elements of the set
\(\{n \in \mathbb{N} : (n-2)^2 \leq 10\}\).

Note that if \(n\in\mathbb{N}\), then \(n-2\) is an integer.
So the only possibilities for \((n-2)^2\) to be less than or equal to \(10\)
are \(0,1,4,9\). Thus, the possibilities for \(n-2\) are
\(0, \pm 1, \pm 2, \pm 3\).
As \(n\) must be a natural number, the elements of the set are just
\(1, 2, 3, 4, 5\).

Example 3

Write out all the elements of the set
\(\left\{\begin{bmatrix} a \\ b \end{bmatrix} \in \mathbb{Z}^2 : |a+b|=1,~a\geq 0,~b \geq -1\right\}\).

We consider two cases based on the value of \(b\).

Case 1. \(b \geq 0\).
Since \(a \geq 0\), \(|a+b|=1\) can be simplified as \(a+b=1\).
Because \(a\) and \(b\) are required to be nonnegative integers, we must have
\[a = 0, b = 1,\]
or \[a = 1, b = 0.\]

Case 2. \(b = -1\).
Hence, we need \(a\) to satisfy \(|a-1|=1\).
The only possibilities are \(a = 0\) and \(a = 2\).

Combining the two cases, we see that all the elements of the given set are
\(\begin{bmatrix} 0\\ 1\end{bmatrix}\),
\(\begin{bmatrix} 1\\ 0\end{bmatrix}\),
\(\begin{bmatrix} 0\\ -1\end{bmatrix}\), and
\(\begin{bmatrix} 2\\ -1\end{bmatrix}\).

Example 4

Evaluate the expression \(1 + (1 + (1 + 1)\cdot 1)\) over \(GF(2)\).

Give the multiplicative inverse of \(4-\sqrt{5}\) in the form
\(a+b\sqrt{5}\) where \(a,b \in \mathbb{Q}\).

Recall that the multiplicative inverse of \(4-\sqrt{5}\) is
a number \(\alpha\) such that \(\alpha (4- \sqrt{5}) = 1\).
In this question, we need to give \(\alpha\) in the
form \(a + b\sqrt{5}\) with \(a,b\in \mathbb{Q}\).

Hence, we want to find \(a,b\in \mathbb{Q}\) such that
\((a+b\sqrt{5})(4-\sqrt{5}) = 1\).

Expanding the left-hand side gives:
\begin{eqnarray*}
\\
(a+b\sqrt{5})(4-\sqrt{5})
& = & a(4-\sqrt{5}) +b\sqrt{5}(4-\sqrt{5}) \\
& = & 4a-a\sqrt{5} +4b\sqrt{5}-5b \\
& = & (4a-5b) + (-a+4b)\sqrt{5}.
\end{eqnarray*}
So, we need
\((4a-5b) + (-a+4b)\sqrt{5}\) to be equal to \(1\).
Because \(\sqrt{5}\) is irrational and \(-a+4b\) is rational,
we must have \(-a+4b = 0\). This means that
we need \(4a-5b=1\).

Now, \(-a + 4b = 0\) means \(a = 4b\). Substituting this into \(4a-5b=1\),
we get \(4(4b)-5b=1\), implying that \(11b = 1\). So, \(b = \frac{1}{11}\).
Hence, \(a = \frac{4}{11}\). Therefore, the answer is
\(\frac{4}{11}+\frac{1}{11}\sqrt{5}\).

(One can in fact arrive at the answer much more quickly as follows:
In effect, we want to clear the radical
in the denominator of \(\frac{1}{4-\sqrt{5}}\). To do so, we simply
compute \(\frac{1}{4-\sqrt{5}}\cdot \frac{4+\sqrt{5}}{4+\sqrt{5}}
= \frac{4+\sqrt{5}}{4^2 - (\sqrt{5})^2}
= \frac{4+\sqrt{5}}{16 - 5} = \frac{4+\sqrt{5}}{11}.\)
Here, \(4 + \sqrt{5}\) is called the conjugate of \(4-\sqrt{5}\).)

Example 6

Let \(a \in \mathbb{F}\) where \(\mathbb{F}\) denotes a field.
Suppose that \(a \neq 0\).
Show that \(a^{-k} = \left(a^{-1}\right)^k\) for every natural number \(k\).
(You may use the fact that \(a^k b^k = (ab)^k\) for all \(a,b \in \mathbb{F})\).

Since \(a \neq 0\), we know that \(a^{-1}\) exists and that \(a^{-1} a=1\).
Let \(b = \left(a^{-1}\right)^k\).
Then \(ba^k = \left(a^{-1}\right)^k a^k =
\left(a^{-1}a\right)^k = 1^k = 1\).
Hence, by definition, \(b\) is the multiplicative inverse of \(a\).

Example 7

Let \(\mathbb{Q}(\sqrt{2})\) denote the set of
numbers of the form \(a + b\sqrt{2}\) where
\(a\) and \(b\) are rational numbers.
Show that \(\mathbb{Q}(\sqrt{2})\) under the usual addition and
multiplication for real numbers form a field.

First, we need to check that \(\mathbb{Q}(\sqrt{2})\) is closed under
addition and multiplication; that is,
adding two numbers in \(\mathbb{Q}(\sqrt{2})\)
results in another number in \(\mathbb{Q}(\sqrt{2})\)
and multiplying two numbers in \(\mathbb{Q}(\sqrt{2})\)
also results in another number in \(\mathbb{Q}(\sqrt{2})\).

For addition, note that
\( (a+b\sqrt{2}) + (c + d\sqrt{2}) = (a+c) + (b+d)\sqrt{2}\).
If \(a,b,c,d\) are rational, so are \(a+c\) and \(b+d\).

For multiplication, note that
\( (a+b\sqrt{2}) \cdot (c + d\sqrt{2}) = (ac+2bd) + (ad+bc)\sqrt{2}\).
If \(a,b,c,d\) are rational, so are \(ac+2bd\) and \(ad+bc\).

Hence, adding or multiplying numbers in \(\mathbb{Q}(\sqrt{2})\) gives
numbers that belong to the set.

Next, we need to check that properties of a field are satisfied:

For all \(a, b,\) and \(c \in \mathbb{F}\),
the following hold:
\(a + (b+c) = (a+b)+ c\) and \(a\cdot (b\cdot c) = (a\cdot b)\cdot c\).
(Associativity of addition and multiplication.)

For all \(a\) and \(b \in \mathbb{F}\),
the following hold:
\(a + b = b + a\) and \(a\cdot b = b\cdot a\).
(Commutativity of addition and multiplication.)

For all \(a, b,\) and \(c \in \mathbb{F}\),
the following holds:
\(a \cdot (b+c) = a \cdot b+ a\cdot c\).
(Left-distributivity.)

There exists an element of \(\mathbb{F}\), denoted by 0,
such that for all \(a \in \mathbb{F},
a + 0 = a\).
(Existence of additive identity.)

There exists an element of \(\mathbb{F}\) not equal to 0, denoted by 1,
such that for all \(a \in \mathbb{F}, a \cdot 1 = a\).
(Existence of multiplicative identity.)

For each \(a \in \mathbb{F}\), there exists an element
\(-a \in \mathbb{F}\), such that \(a+(-a) = 0\).
(Existence of additive inverse.)

For each \(a \in \mathbb{F}\) not equal to 0, there exists an element
\(a^{-1} \in \mathbb{F}\), such that \(a\cdot a^{-1} = 1\).
(Existence of multiplicative inverse.)

It seems like a monumental task to check all the listed properties.
However, \(\mathbb{Q}(\sqrt{2})\)
is a subset of the set of real numbers.
If we use the fact that the real numbers form a field, then
properties 1 through 5 come for free.

So we need to check properties 6 and 7.
The additive inverse of \(a + b\sqrt{2}\) is simply
\( (-a) + (-b)\sqrt{2} \). As \(-a\) and \(-b\) are rational numbers,
\( (-a) + (-b)\sqrt{2} \) is an element of \(\mathbb{Q}(\sqrt{2}\).
Hence, additive inverses exist for all elements of
\(\mathbb{Q}(\sqrt{2}\).

We now find the multiplicative inverse
of \(a + b\sqrt{2}\) when \(a + b \sqrt{2} \neq 0\).

Note that not both \(a\) and \(b\) can be zero.
Using the irrationality of \(\sqrt{2}\), we see that
\(a - b\sqrt{2} \neq 0\).
Then
\begin{eqnarray*}
\frac{1}{a + b \sqrt{2}} & = &
\frac{1}{a + b \sqrt{2}} \cdot \frac{a-b\sqrt{2}}{a - b \sqrt{2}} \\
& = & \frac{a-b\sqrt{2}}{a^2 - 2b^2} \\
& = & \frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}
\end{eqnarray*}
As \(\frac{a}{a^2-2b^2}\) and \(\frac{b}{a^2-2b^2}\)
are rational numbers,
\(\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}\)
is the multiplicative inverse of
\(a + b\sqrt{2}\) from \(\mathbb{Q}(\sqrt{2})\).

Example 8

Consider the function
\(T:\mathbb{R}^2 \rightarrow \mathbb{R}^2\) given by
\[
T\left( \begin{bmatrix} x \\ y \end{bmatrix} \right)
= \begin{bmatrix} 2x - y \\ x + y \end{bmatrix}.\]
Show that \(T\) is surjective.

We need to show that for every
\(\begin{bmatrix} a\\ b\end{bmatrix} \in \mathbb{R}^2\),
there exist real numbers \(x\) and \(y\) such that
\(T\left( \begin{bmatrix} x \\ y \end{bmatrix} \right)=\begin{bmatrix}
a \\ b \end{bmatrix}\).

It suffices to show that the following system always has a solution:
\begin{eqnarray*}
2x - y & = & a \\
x + y & = & b
\end{eqnarray*}
Adding the two equations gives \(3x = a+b\), implying that
\(x = \frac{a+b}{3}\).
From the second equation, we get that \(y = b - x = \frac{-a + 2b}{3}\).

Thus,
\(T\left( \begin{bmatrix}
\frac{a+b}{3}
\\
\frac{-a+2b}{3}
\end{bmatrix} \right)=\begin{bmatrix}
a \\ b \end{bmatrix}\) for all \(a,b\in\mathbb{R}\).