Real Numbers of Type Rational a plus b root 2 form Field

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Theorem

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

$\Q \sqbrk {\sqrt 2} := \set {x \in \R: x = a + b \sqrt 2: a, b \in \Q}$

that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.


Then the algebraic structure:

$\struct {\Q \sqbrk {\sqrt 2}, +, \times}$

where $+$ and $\times$ are conventional addition and multiplication on real numbers, is a field.


Corollary

The field $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ is a subfield of $\struct {\R, +, \times}$.


Proof


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By definition, $\Q \sqbrk {\sqrt 2} \subseteq \R$ where $\R$ is the set of real numbers.

Thus $+$ and $\times$ on $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ are well-defined.


Closure

Let $a_1 + b_1 \sqrt 2, a_2 + b_2 \sqrt 2 \in \Q \sqbrk {\sqrt 2}$.

Then:

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


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

So both $+$ and $\times$ are closed on $\Q \sqbrk {\sqrt 2}$.

$\Box$


Associativity

As addition and multiplication are associative on $\R$ it follows from Restriction of Associative Operation is Associative that they are also associative on $\Q \sqbrk {\sqrt 2}$.

$\Box$


Commutativity

As addition and multiplication are commutative on $\R$ it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Q \sqbrk {\sqrt 2}$.

$\Box$


Identity

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 $\Q \sqbrk {\sqrt 2}$.


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 $\paren {1 + 0 \sqrt 2} \times \paren {a + b \sqrt 2}$.

So $\paren {1 + 0 \sqrt 2}$ is the identity for $\times$ on $\Q \sqbrk {\sqrt 2}$.

$\Box$


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 $\Q \sqbrk {\sqrt 2}$.


Calculating the product inverse of $\paren {a + b \sqrt 2}$ is less trivial.


From Difference of Two Squares we have:

$\paren {a + b \sqrt 2} \paren {a - b \sqrt 2} = a^2 - 2 b^2$

which leads to:

$\paren {a + b \sqrt 2} \paren {\dfrac {a - b \sqrt 2} {a^2 - 2 b^2} } = 1 = 1 + 0 \sqrt 2$

so demonstrating that the product inverse of $\paren {a + b \sqrt 2}$ is $\dfrac a {a^2 - 2 b^2} - \dfrac {b \sqrt 2} {a^2 - 2 b^2}$.

As $a$ and $b$ are rational, it follows that so are $\dfrac a {a^2 - 2 b^2}$ and $\dfrac b {a^2 - 2 b^2}$.

So the product inverse of $\paren {a + b \sqrt 2}$ is an element of $\Q \sqbrk {\sqrt 2}$.

$\Box$


Distributivity

We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Q \sqbrk {\sqrt 2}$.

$\Box$


The result follows by putting all the pieces together.

$\blacksquare$


Sources

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