Quadratic Integers over 2 form Subdomain of Reals/Proof 1
Theorem
Let $\Z \sqbrk {\sqrt 2}$ denote the set of quadratic integers over $2$:
- $\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$
That is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are integers.
Then the algebraic structure:
- $\struct {\Z \sqbrk {\sqrt 2}, +, \times}$
where $+$ and $\times$ are conventional addition and multiplication on real numbers, form an integral subdomain of the real numbers $\R$.
Proof
$\Z \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of real numbers, so we immediately have that addition and multiplication are well-defined.
Closure
Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Z \sqbrk {\sqrt 2}$.
Then:
\(\ds \paren {a_1 + b_1 \sqrt 2} + \paren {a_2 + b_2 \sqrt 2}\) | \(=\) | \(\ds \paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 2\) | ||||||||||||
\(\ds \) | \(\in\) | \(\ds \Z \sqbrk {\sqrt 2}\) |
\(\ds \paren {a_1 + b_1 \sqrt 2} \times \paren {a_2 + b_2 \sqrt 2}\) | \(=\) | \(\ds \paren {a_1 \times a_2 + 2 \times b_1 \times b_2} + \paren {a_1 \times b_2 + b_1 \times a_2} \sqrt 2\) | ||||||||||||
\(\ds \) | \(\in\) | \(\ds \Z \sqbrk {\sqrt 2}\) |
So both the operations $+$ and $\times$ are closed on $\Z \sqbrk {\sqrt 2}$.
$\Box$
Associativity
We have that addition and multiplication are associative on $\R$.
Therefore it follows from Restriction of Associative Operation is Associative that they are also associative on $\Z \sqbrk {\sqrt 2}$.
$\Box$
Commutativity
We have that addition and multiplication are commutative on $\R$.
Therefore it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Z \sqbrk {\sqrt 2}$.
$\Box$
Identities
We have:
\(\ds \paren {a + b \sqrt 2} + \paren {0 + 0 \sqrt 2}\) | \(=\) | \(\ds \paren {a + 0} + \paren {b + 0} \sqrt 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a + b \sqrt 2\) |
and similarly for $\paren {0 + 0 \sqrt 2} + \paren {a + b \sqrt 2}$.
So $\paren {0 + 0 \sqrt 2}$ is the identity for $+$ on $\Z \sqbrk {\sqrt 2}$.
Then:
\(\ds \paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}\) | \(=\) | \(\ds \paren {a \times 1 + 2 \times b \times 0} + \paren {b \times 1 + a \times 0} \sqrt 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a + b \sqrt 2\) |
and similarly for $\paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}$.
So $\paren {1 + 0 \sqrt 2}$ is the identity for $\times$ on $\Z \sqbrk {\sqrt 2}$.
$\Box$
Inverses
We have:
\(\ds \paren {a + b \sqrt 2} + \paren {-a + \paren {-b} \sqrt 2}\) | \(=\) | \(\ds \paren {a - a} + \paren {b - b} \sqrt 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0 + 0 \sqrt 2\) |
and similarly for $\paren {-a + \paren {-b} \sqrt 2} + \paren {a + b \sqrt 2}$.
So $\paren {-a + \paren {-b} \sqrt 2}$ is the inverse of $\paren {a + b \sqrt 2}$ for $+$ on $\Z \sqbrk {\sqrt 2}$.
We have no need to investigate inverses for $\times$ (which is convenient as $\times$ happens not to be closed for inverses on $\Z \sqbrk {\sqrt 2}$).
$\Box$
Distributivity
We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Z \sqbrk {\sqrt 2}$.
$\Box$
Divisors of Zero
By Real Numbers form Field, $\R$ is a field and so by Field is Integral Domain is itself an integral domain.
Hence it has no proper zero divisors.
$\Box$
The result follows by putting all the pieces together.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $1$: Integral Domains: $\S 5$. Further Examples of Integral Domains: Example $4$