nyuuzyou/wikis huggingface dataset:
Real Numbers are Uncountably Infinite/Cantor's First Proof

== Theorem ==
== Proof ==
We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N}$ omits at least one $x \in \R$.
Let $\sequence {x_k}_{k \mathop \in \N}$ be a sequence of distinct real numbers.
Let a sequence of closed real intervals $\sequence {I_n}$ be defined as follows:
Let:
Since the terms of $\sequence {x_k}_{k \mathop \in \N}$ are distinct, $a_k \ne b_k$.
Thus $I_k$ is not a singleton.
Let:
:$I_{n - 1} = \closedint {a_{n - 1} } {b_{n - 1} }$
It can be assumed that infinitely many of the $x_k$ lie inside $I_{n - 1}$.
Otherwise the proof is complete.
Let $y$ and $z$ be the first two such terms of $\sequence {x_k}_{k \mathop \in \N}$.
Let:
Thus we have sequences:
:$\sequence {a_k}_{k \mathop \in \N}$
:$\sequence {b_k}_{k \mathop \in \N}$
with:
:$ a_1 < a_2 < \cdots < b_2 < b_1$
So $\sequence {a_k}_{k \mathop \in \N}$ and $\sequence {b_k}_{k \mathop \in \N}$ are monotone, and bounded above and bounded below respectively.
Therefore by the Monotone Convergence Theorem (Real Analysis) both are convergent.
Let:
Clearly we have $A \le B$.
So:
:$\closedint A B \ne \O$
Let $h \in \closedint A B$.
Then:
:$\forall k: h \ne a_k, b_k$
We claim that $h \ne x_k$ for all $k$. 
Suppose that $h = x_k$ for some $k$. 
Then there are only finitely many terms in the sequence before $h$ occurs.
Therefore only finitely many of the $a_k$ precede $h$.
Let $a_d$ be the last term of the sequence $\sequence {a_k}_{k \mathop \in \N}$ preceding $h$.
We defined $a_{d + 1}$, $b_{d + 1}$ to be interior points of $I_d$, and also $h \in I_{d + 1}$ by the definition of $h$.
Therefore $a_{d + 1}$ must precede $h$ in the sequence, for the sequence is increasing.
This is a contradiction.
== Historical Note ==
Category:Real Numbers are Uncountably Infinite
Real Numbers are Uncountably Infinite/Cantor's Second Proof

== Theorem ==
== Proof ==
By definition, a perfect set is a set $X$ such that every element $x \in X$ is the limit of a sequence of elements of $X$ distinct from $x$.
From Real Numbers form Perfect Set, $\R$ is perfect.
Therefore it is sufficient to show that a perfect subset of $X \subseteq \R^k$ is uncountable.
We prove the equivalent result that every sequence $\sequence {x_k}_{k \mathop \in \N} \subseteq X$ omits at least one element of $X$.
Let $y_1 \in X$.
Let $B_1:= \map {\BB_r} {y_1}$ be a closed ball whose center is at $y_1$.
Consider a closed ball:
:$B_{n - 1} \supseteq B_n := \map {\BB_{\map \delta n} } {y_n}$
such that:
:$\map \delta n \le \dfrac {\map \delta {n - 1} } 2$
:$y_n \in X$
:$x_n \notin B_n$
Note that $\map \delta 1 = r$.
We can satisfy the condition $x_n \notin B_n$ because $X$ is perfect, so every closed ball whose center is at an element of $X$ contains infinitely many elements of $X$.
Since $\map \delta n \le \dfrac r {2^{n - 1} }$, $y_n$ is Cauchy.
Therefore from Perfect Set is Closed, we may let:
:$\ds Y = \lim_{n \mathop \to \infty} y_n \in X$
For $n \in \N$ we have:
:$\set {y_m: m > n} \subseteq B_n$
so $Y \in B_n$.
But by construction:
:$\forall n \in \N: x_n \notin B_n$
Therefore:
:$\forall n \in \N: Y \ne x_n$
and the proof is complete.
== Historical Note ==
Category:Real Numbers are Uncountably Infinite
Real Numbers are Uncountably Infinite/Historical Note

== Historical Note on Real Numbers are Uncountably Infinite ==
The fact that the Real Numbers are Uncountably Infinite was first demonstrated by in $1874$.
Cantor's first and second proofs given above are less well known than the diagonal argument, and were in fact downplayed by Cantor himself: the first proof was given as an aside in his paper proving the countability of the algebraic numbers.
 conjectures that this was because Cantor feared opposition from among other contemporaries who aggressively dismissed Cantor's ideas.
In particular Cantor's first proof is worth reading; several texts reject the first proof as being more complicated and less instructive, but this seems to have arisen because the Diagonal argument has proven to be a more versatile tool and thus the others forgotten and dismissed. 
In this instance the first and second proofs of Cantor are of equal merit to the diagonal argument, and the three together present clearly the flavor of the theorem.
== Sources ==
* : $\S 1.1$ Set Theory and the Foundations of Mathematics
Category:Historical Notes
Real Numbers are Uncountably Infinite/Proof 1 using Ternary Notation

== Theorem ==
== Proof ==
It is sufficient to show that the real interval $I = \set {x \in \R: 0 < x \le 1}$ is uncountable.
Let $x \in I$.
From Existence of Base-N Representation, $x$ has a unique representation of the form:
:$x = \dfrac {\epsilon_1} 3 + \dfrac {\epsilon_2} {3^2} + \dfrac {\epsilon_3} {3^3} + \cdots$ 
where $\epsilon_k = 0, 1$ or $2$ and an infinite number of $\epsilon_k$ are different from $0$.
Let $S \subseteq I$ be countably infinite.
Let $S = \set {x_1, x_2, \ldots}$.
Let $\epsilon_{k 1}, \epsilon_{k 2}, \epsilon_{k 3}, \ldots$ be the ternary digits of $x_k$.
Let $\epsilon_k = 1 + 2 \epsilon_{k k} - \epsilon_{k k}^2$ so that:
:$\epsilon_k = 1$ if $\epsilon_{k k} = 0$ or $\epsilon_{k k} = 2$
:$\epsilon_k = 2$ if $\epsilon_{k k} = 1$
Then:
:$(1): \quad \forall k: \epsilon_k \ne 0$
:$(2): \quad \forall k: \epsilon_k \ne \epsilon_{k k}$
We have that $0 < x \le 1$ so $x \in I$.
But the real number $\displaystyle x = \sum \epsilon_k 3^{-k}$ is different from every $x_k \in I$.
Thus we have found an element of $I$ which is not an element of $S$.
Therefore $S$ is a proper subset of $I$.
It follows by definition that $I$ is uncountable.
== Sources ==
* : Introduction to Set Theory: $2$. Set Theoretical Equivalence and Denumerability
Category:Real Numbers are Uncountably Infinite
Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation

== Theorem ==
=== Lemma ===
== Proof ==
Define a mapping $f: \powerset {\N_{>0} } \to \R$ thus:
:$\map f S = 0.d_1 d_2 \ldots$, interpreted as a ternary expansion where $\sequence {d_n}$ is the characteristic function of $S$.
That is:
:$\ds \map f S = \sum_{i \mathop \in S} 3^{-i}$
By the lemma, $f$ is an injection.
 that $\R$ is countable.
Then there is an injection $g: \R \to \N$.
By Composite of Injections is Injection, $g \circ f: \powerset \N \to \N$ is an injection.
But this contradicts No Injection from Power Set to Set.
Hence, by Proof by Contradiction, $\R$ is not countable.
Category:Real Numbers are Uncountably Infinite
Real Numbers are Uncountably Infinite/Proof 2 using Ternary Notation/Lemma

== Lemma to Real Numbers are Uncountably Infinite: Proof 2 using Ternary Notation ==
Let $\sequence {d_n}$ and $\sequence {e_n}$ be infinite sequences in $\set {0, 1}$ such that:
:$\exists m \in \N: d_m \ne e_m$
That is, the sequences $\sequence {d_n}$ and $\sequence {e_n}$ are different in at least one term.
Then the ternary representations $D = 0.d_1 d_2 \ldots$ and $E = 0.e_1 e_2 \dots$ represent distinct real numbers.
== Proof ==
Let $\sequence {d_n} \ne \sequence {e_n}$.
By the Well-Ordering Principle, there is a smallest $n \in \N_{>0}$ such that $d_n \ne e_n$.
, suppose that $d_n = 0$ and $e_n = 1$.
Let:
:$\ds K = 0.d_1 d_2 \ldots d_{n - 1} = \sum_{i \mathop = 1}^{n - 1} d_i 3^{-i}$
:$\ds D := K + \sum_{i \mathop = n + 1}^\infty d_i 3^{-i}$
:$\ds E := K + 3^{-n} + \sum_{i \mathop = n + 1}^\infty e_i 3^{-i} \ge K + 3^{-n}$
But then:
Thus $D < E$, so $D \ne E$.
Category:Real Numbers are Uncountably Infinite
Real Numbers are Uncountably Infinite/Set-Theoretical Approach: Proof 1

== Theorem ==
== Proof ==
By Surjection from Natural Numbers iff Countable, a set $A$ is countable there exists a surjection $f: \N \to A$.
Suppose there exists a surjection $f: \N \to \R$.
Then $\forall x \in \R: \exists n \in \N: \map f n = x$ as $f$ is surjective.
Let $d_{n, 0}$ be the integer before the decimal point of $\map f n$.
Similarly, for all $m > 0$, let $d_{n, m}$ be the $m$th digit in the decimal expansion of $\map f n$.
Let $e_0$ be an integer different from $d_{0, 0}$.
Similarly, for all $m > 0$, let $e_m$ be an integer different from $d_{m, m}$.
Specifically, we can define $e_0$ to be $d_{0, 0} + 1$, and:
:$e_m = \begin {cases} 1 & : d_{m, m} \ne 1 \\ 2 & : d_{m, m} = 1 \end {cases}$
Now consider the real number $\ds x = e_0 + \sum_{n \mathop = 1}^\infty \frac {e_n} {10^n}$.
Its decimal expansion is:
:$x = \sqbrk {e_0 . e_1 e_2 e_3 \ldots}_{10}$
Since $e_0 \ne d_{0, 0}$, $x \ne \map f 0$.
Similarly, for each $n \in \N$ such that $n \ge 1$, we have that $e_n \ne d_{n, n}$ and so $x \ne \map f n$.
Thus $x$ is a real number which is not in the set $\set {\map f n: n \in \N}$.
Hence $f$ can not be surjective.
== Sources ==
* : $\S 2.4$: Theorem $5$
Category:Real Numbers are Uncountably Infinite
Real Numbers are Uncountably Infinite/Set-Theoretical Approach: Proof 2

== Theorem ==
== Proof ==
By Cantor's Theorem there is no surjection:
:$\N \twoheadrightarrow \powerset \N$
Additionally, we have Power Set of Natural Numbers is not Countable.
Therefore, if we can show that $\powerset \N$ injects into $\R$ then there is no injection $\R \hookrightarrow \N$ and $\R$ is uncountable.
To prove the theorem we construct an injection $f: \powerset \N \to \R$.
For a subset $S \subseteq \N$, let $\chi_S$ be the characteristic function of $S$, and let $d_i = \map {\chi_S} i$ for all $i \in \N$.
By the definition of characteristic function, $\sequence {d_i}_{i \mathop \in \N}$ is an infinite sequence of $1$s and $0$s.
There are two cases: $\sequence {d_i}_{i \mathop \in \N}$ terminates in an infinite sequence of $1$s, or it does not.
Suppose $\sequence {d_i}_{i \mathop \in \N}$ does not terminate in an infinite sequence of $1$s.
Then $\map f S$ is the binary expansion of the following number in $\hointr 0 1$:
:$0.d_1 d_2 d_3 d_4 \ldots$
Otherwise $\sequence {d_i}_{i \mathop \in \N}$ does terminate in an infinite sequence of $1$s.
Then $\map f S$ is the integer expressed in binary as:
:$1 d_1 d_2 d_3 \ldots d_k$
where $d_k$ is the last member of the sequence not equal to $1$.
In either case, every subset of $\N$, that is, element of $\powerset \N$, is mapped to an element of $\R$.
That $f$ is an injection follows from the uniqueness statement of Existence of Base-N Representation.
Category:Real Numbers are Uncountably Infinite
Real Numbers are not Well-Ordered under Conventional Ordering

== Theorem ==
Let $\struct {\R, \leqslant}$ be the ordered structure consisting of the real numbers under the usual ordering.
Then $\struct {\R, \leqslant}$ is not a well-ordered set.
== Proof ==
;Proof by Counterexample
Consider the set:
:$A := \set {x \in \R: x > 1}$
Suppose $a$ were the smallest element of $A$.
Then $a > 1$.
But then:
:$1 < \dfrac {a + 1} 2 < a$
and so $a$ has been shown not to be the smallest element of $A$ after all.
Hence $A$ has no smallest element.
Thus there exists a subset of $\R$ which has no smallest element.
Thus $\R$ is not a well-ordered under $\leqslant$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Real Numbers
Category:Well-Orderings
Real Numbers form Field

== Theorem ==
The set of real numbers $\R$ forms a field under addition and multiplication: $\struct {\R, +, \times}$.
== Proof ==
From Real Numbers under Addition form Infinite Abelian Group, we have that $\struct {\R, +}$ forms an abelian group.
From Non-Zero Real Numbers under Multiplication form Abelian Group, we have that $\struct {\R_{\ne 0}, \times}$ forms an abelian group.
Next we have that Real Multiplication Distributes over Addition.
Thus all the criteria are fulfilled, and $\struct {\R, +, \times}$ is a field.
== Also see ==
* Definition:Field of Real Numbers
* Real Numbers form Totally Ordered Field
== Sources ==
* : $\text I$. Linear Algebra: $\text A$. Fields
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $2$
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $15$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields
* : $\S 19$: Properties of $\Z_m$ as an algebraic system
* : $\S 55$. Special types of ring and ring elements: $(4)$
* : $\S 2.4$: The rational numbers and some finite fields
* 
* 
Category:Examples of Fields
Category:Real Numbers
Real Numbers form Integral Domain

== Theorem ==
The set of real numbers $\R$ forms an integral domain under addition and multiplication: $\struct {\R, +, \times}$.
== Proof ==
Recall that Real Numbers form Field.
The result then follows directly from Field is Integral Domain.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain
Category:Examples of Integral Domains
Category:Real Numbers
Real Numbers form Ordered Integral Domain

== Theorem ==
The set of real numbers $\R$ forms an ordered integral domain under addition and multiplication: $\struct {\R, +, \times, \le}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $9$
Category:Examples of Ordered Integral Domains
Category:Real Numbers
Category:Real Numbers form Ordered Integral Domain
Real Numbers form Ordered Integral Domain/Proof 2

== Theorem ==
== Proof ==
We have that the real numbers $\struct {\R, +, \times}$ form an integral domain.
It remains to specify a property $P$ on $\R$ such that:
:$(1): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a + b}$
:$(2): \quad \forall a, b \in \R: \map P a \land \map P b \implies \map P {a \times b}$
:$(3): \quad \forall a \in \R: \map P a \lor \map P {-a} \lor a = 0$
We have that the integers $\struct {\Q, +, \times}$ form an ordered integral domain, where $\Q$ denotes the set of rational numbers.
Let $P'$ be the (strict) positivity property on $\struct {\Q, +, \times}$.
Let us define the property $P$ on $\R$ as:
:$\forall a \in \R: \map P a \iff a = \eqclass {\sequence {a_n} } {}: \forall n \in \N: \map {P'} {a_n}$
That is, an element $a = \eqclass {\sequence {a_n} } {}$ has $P$ $a_n$ has (strict) positivity property in $\Q$ for all $n \in \N$.
Now let $a = \eqclass {\sequence {a_n} } {}$ and $b = \eqclass {\sequence {b_n} } {}$ such that $\map P a$ and $\map P b$.
Then by definition of real addition:
:$\eqclass {\sequence {a_n} } {} + \eqclass {\sequence {b_n} } {} = \eqclass {\sequence {a_n + b_n} } {}$
and real multiplication:
:$\eqclass {\sequence {a_n} } {} \times \eqclass {\sequence {b_n} } {} = \eqclass {\sequence {a_n \times b_n} } {}$
It can be seen from the definition of (strict) positivity $P'$ on $\Q$ that $\map P {a + b}$ and $\map P {a \times b}$.
It can be seen that if $\map P a$ then $\neg \map P {-a}$ and vice versa.
Also we note that $\neg \map P 0$ and of course $\neg \map P {-0}$.
So the property $P$ we defined fulfils the criteria for the (strict) positivity property.
Hence the result.
Category:Real Numbers form Ordered Integral Domain
Real Numbers form Perfect Set

== Theorem ==
Consider the set of real numbers $\R$ as a (complete) metric space with the usual (Euclidean) metric.
Then $\R$ forms a perfect set.
== Proof ==
By definition, a perfect set is a set which equals its set of limit points.
Let $x \in \R$.
Consider the sequence:
:$\sequence {y_k} = x + \dfrac 1 k$
Then as $\sequence {z_k} = \dfrac 1 k$ is a basic null sequence it follows that:
:$\ds \lim_{n \mathop \to \infty} \sequence {y_k} = x$
Thus we see that $x$ is a limit point of $S$.
Category:Real Numbers
Category:Perfect Sets
Real Numbers form Ring

== Theorem ==
The set of real numbers $\R$ forms a ring under addition and multiplication: $\struct {\R, +, \times}$.
== Proof ==
From Real Numbers under Addition form Infinite Abelian Group, $\struct {\R, +}$ is an abelian group.
We also have that:
:Real Multiplication is Closed:
::$\forall x, y \in \R: x \times y \in \R$
:Real Multiplication is Associative:
::$\forall x, y, z \in \R: x \times \paren {y \times z} = \paren {x \times y} \times z$
Thus $\struct {\R, +}$ is a semigroup.
Finally we have that Real Multiplication Distributes over Addition:
Hence the result, by definition of ring.
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $4$
* : $\S 54$. The definition of a ring and its elementary consequences
Category:Examples of Rings
Category:Real Numbers
Real Numbers form Subfield of Complex Numbers

== Theorem ==
The field of real numbers $\struct {\R, +, \times}$ forms a subfield of the field of complex numbers $\struct {\C, +, \times}$.
== Proof ==
From Additive Group of Reals is Normal Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
From Multiplicative Group of Reals is Normal Subgroup of Complex, $\struct {\R, \times}$ is a subgroup of $\struct {\C, \times}$.
The result follows from the Subfield Test via the One-Step Subgroup Test and Two-Step Subgroup Test.
== Sources ==
* : Chapter $4$: Fields: $\S 16$. Subfields: Example $21$
* : $\S 2.1$: Subrings: Examples $1$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
* : $\S 56$. Subrings and Subfields
* : $\S 2.4$: The rational numbers and some finite fields
* 
* 
Category:Examples of Subfields
Category:Real Numbers
Category:Complex Numbers
Real Numbers form Totally Ordered Field

== Theorem ==
The set of real numbers $\R$ forms a totally ordered field under addition and multiplication: $\struct {\R, +, \times, \le}$.
== Proof ==
From Real Numbers form Field, we have that $\struct {\R, +, \times}$ forms a field.
From Ordering Properties of Real Numbers we have that $\struct {\R, +, \times, \le}$ is a totally ordered field.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers
* 
* 
Category:Examples of Fields
Category:Ordered Fields
Category:Real Numbers
Real Numbers form Valued Field

== Theorem ==
The set of real numbers $\R$ forms a valued field under addition, multiplication and absolute value: $\struct {\R, +, \times, \size{\,\cdot\,}}$.
== Proof ==
From Real Numbers form Field, we have that $\struct {\R, +, \times}$ forms a field.
From Absolute Value is Norm, we have that $\size{\size{\,\cdot\,}}$ is a norm on $\struct {\R, +, \times}$.
Hence $\struct {\R, +, \times, \size{\,\cdot\,}}$ is a valued field by definition.
== Also see ==
* Definition:Field of Real Numbers
* Real Numbers form Totally Ordered Field
Category:Examples of Fields
Category:Real Numbers
Real Numbers form Vector Space

== Theorem ==
The set of real numbers $\R$, with the operations of addition and multiplication, forms a vector space.
== Proof ==
Let the field of real numbers be denoted $\struct {\R, +, \times}$.
From Real Vector Space is Vector Space, we have that $\struct {\R^n, +, \cdot}$ is a vector space, where:
:$\mathbf a + \mathbf b = \tuple {a_1 + b_1, a_2 + b_2, \ldots, a_n + b_n}$
:$\lambda \cdot \mathbf a = \tuple {\lambda \times a_1, \lambda \times a_2, \ldots, \lambda \times a_n}$
where:
:$\mathbf a, \mathbf b \in \R^n$
:$\lambda \in \R$
:$\mathbf a = \tuple {a_1, a_2, \ldots, a_n}$
:$\mathbf b = \tuple {b_1, b_2, \ldots, b_n}$
When $n = 1$, the vector space degenerates to:
:$\mathbf a + \mathbf b = \tuple {a + b}$
:$\lambda \cdot \mathbf a = \tuple {\lambda \times a}$
where:
:$\mathbf a, \mathbf b \in \R$
:$\lambda \in \R$
:$\mathbf a = \tuple a$
:$\mathbf b = \tuple b$
Thus it can be seen that the vector space $\struct {\R^1, +, \cdot}$ is identical with the field of real numbers denoted by $\struct {\R, +, \times}$.
== Also see ==
* Properties of Real Numbers
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.2$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $4$. LINEAR VECTOR SPACE: Example $2$
* : Chapter $1.1$: Normed and Banach spaces. Vector Spaces
Category:Linear Algebra
Category:Real Numbers
Category:Examples of Vector Spaces
Real Numbers form only Ordered Field which is Complete

== Theorem ==
The set of real numbers $\R$ is the only ordered field which also satisfies the Continuum Property.
== Proof ==
From Real Numbers form Totally Ordered Field we have that $\R$ forms an totally ordered field.
From the Continuum Property we have that $\R$ is complete.
It remains to be shown that any ordered field which also satisfies the Continuum Property is isomorphic to $\R$.
== Sources ==
* 
* 
Category:Real Numbers
Category:Ordered Fields
Category:Continuum Property
Real Numbers of Type Rational a plus b root 2 form Field

== 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 ===
== Proof ==
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:
So both $+$ and $\times$ are closed on $\Q \sqbrk {\sqrt 2}$.
=== Associativity ===
As addition and multiplication are associative on $\R$ it follows from Restriction of Associative Operation is Associative that they are also associative on $\Q \sqbrk {\sqrt 2}$.
=== Commutativity ===
As addition and multiplication are commutative on $\R$ it follows from Restriction of Commutative Operation is Commutative that they are also commutative on $\Q \sqbrk {\sqrt 2}$.
=== Identity ===
We have:
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:
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 $\Q \sqbrk {\sqrt 2}$.
=== Inverses ===
We have:
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 $\Q \sqbrk {\sqrt 2}$.
=== Distributivity ===
We have that Real Multiplication Distributes over Addition, so by Restriction of Operation Distributivity, $\times$ is distributive over $+$ on $\Q \sqbrk {\sqrt 2}$.
The result follows by putting all the pieces together.
== Sources ==
* : Chapter $4$: Fields: $\S 16$. Subfields: Example $22$
Category:Examples of Fields
Real Numbers of Type Rational a plus b root 2 form Field/Corollary

== Corollary to Real Numbers of Type Rational a plus b root 2 form Field ==
Let $\Q \sqbrk {\sqrt 2}$ denote the set:
:$\Q \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.
The field $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ is a subfield of $\struct {\R, +, \times}$.
== Proof ==
So $\Q \sqbrk {\sqrt 2} \subseteq \R$.
From Real Numbers of Type Rational a plus b root 2 form Field, $\struct {\Q \sqbrk {\sqrt 2}, +, \times}$ is a field.
As stated in the proof of the Real Numbers of Type Rational a plus b root 2 form Field, numbers of the form $a + b \sqrt 2$ are real.
Hence the result by definition of subfield.
== Sources ==
* : Chapter $4$: Fields: $\S 16$. Subfields: Example $22$
Category:Examples of Subfields