Numbers of Type Integer a plus b root 2 form Subdomain of Reals/Proof 1

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\Z \sqbrk {\sqrt 2}$ denote the set:

$\Z \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Z}$

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:

\(\displaystyle \paren {a_1 + b_1 \sqrt 2} + \paren {a_2 + b_2 \sqrt 2}\) \(=\) \(\displaystyle \paren {a_1 + a_2} + \paren {b_1 + b_2} \sqrt 2\)
\(\displaystyle \) \(\in\) \(\displaystyle \Z \sqbrk {\sqrt 2}\)


\(\displaystyle \paren {a_1 + b_1 \sqrt 2} \times \paren {a_2 + b_2 \sqrt 2}\) \(=\) \(\displaystyle \paren {a_1 \times a_2 + 2 \times b_1 \times b_2} + \paren {a_1 \times b_2 + b_1 \times a_2} \sqrt 2\)
\(\displaystyle \) \(\in\) \(\displaystyle \Z \sqbrk {\sqrt 2}\)

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:

\(\displaystyle \paren {a + b \sqrt 2} + \paren {0 + 0 \sqrt 2}\) \(=\) \(\displaystyle \paren {a + 0} + \paren {b + 0} \sqrt 2\)
\(\displaystyle \) \(=\) \(\displaystyle a + b \sqrt 2\)

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:

\(\displaystyle \paren {a + b \sqrt 2} \times \paren {1 + 0 \sqrt 2}\) \(=\) \(\displaystyle \paren {a \times 1 + 2 \times b \times 0} + \paren {b \times 1 + a \times 0} \sqrt 2\)
\(\displaystyle \) \(=\) \(\displaystyle a + b \sqrt 2\)

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:

\(\displaystyle \paren {a + b \sqrt 2} + \paren {-a + \paren {-b} \sqrt 2}\) \(=\) \(\displaystyle \paren {a - a} + \paren {b - b} \sqrt 2\)
\(\displaystyle \) \(=\) \(\displaystyle 0 + 0 \sqrt 2\)

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