nyuuzyou/wikis huggingface dataset:
Smallest Arguments for given Multiplicative Persistence

== Sequence ==
Let $P \left({n}\right)$ denote the multiplicative persistence of a natural number $n$.
Let $a: \N \to \N$ be the partial mapping defined as:
:$\forall n \in \N: a \left({n}\right) = \text{the smallest $m \in \N$ such that $P \left({m}\right) = n$}$
The sequence of $a \left({n}\right)$ for successive $n$ begins as follows:
:
$n$ 	$a \left({n}\right)$
$0$ 	$0$
$1$ 	$10$
$2$ 	$25$
$3$ 	$39$
$4$ 	$77$
$5$ 	$679$
$6$ 	$6788$
$7$ 	$68 \, 889$
$8$ 	$2 \, 677 \, 889$
$9$ 	$26 \, 888 \, 999$
$10$	$3 \, 778 \, 888 \, 999$
$11$	$277 \, 777 \, 788 \, 888 \, 899$
It is not known what $a \left({12}\right)$ is, but it is known to be greater than $10^{200}$.
== Sources ==
* : $10$
* : $10$
* : $277,777,788,888,899$
Category:Multiplicative Persistence
Smallest Consecutive Even Numbers such that Added to Divisor Count are Equal

== Theorem ==
$30$ is the smallest positive even integer $n$ with the property:
where:
:$m \in \Z_{>0}$ is some positive integer
:$\map {\sigma_0} n$ is the divisor count function: the number of divisors of $n$.
In this case, where $n = 30$, we have that $m = 38$.
== Proof ==
From Divisor Count Function from Prime Decomposition, we have:
:$\ds \map {\sigma_0} n = \prod_{j \mathop = 1}^r \paren {k_j + 1}$
where the prime decomposition of $n$ is:
:$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$
== Sources ==
* : $38$
Category:Divisor Count Function
Category:Even Integers
Smallest Cube whose Sum of Divisors is Cube

== Theorem ==
The smallest cube $N$ such that $\map {\sigma_1} N$ is also a cube is:
:$27 \, 418 \, 521 \, 963 \, 671 \, 501 \, 273 \, 905 \, 190 \, 135 \, 082 \, 692 \, 041 \, 730 \, 405 \, 303 \, 870 \, 249 \, 023 \, 209$
where $\map {\sigma_1} N$ denotes the divisor sum of $N$.
== Proof ==
We have that:
Then from :
== Historical Note ==
== Sources ==
* : $3^9 7^3 11^3 13^3 17^3 41^3 43^3 47^3 443^3 499^3 3583^3$
Category:Cube Numbers
Category:Divisor Sum Function
Smallest Cube whose Sum of Divisors is Cube/Historical Note

== Historical Note on Smallest Cube whose Sum of Divisors is Cube ==
According to in his of $1997$, this result is due to , and can be found in , Volume $27$, on page $229$.
However, this has not been corroborated.
== Sources ==
* : $3^9 7^3 11^3 13^3 17^3 41^3 43^3 47^3 443^3 499^3 3583^3$
Category:Historical Notes
Smallest Cunningham Chain of the First Kind of Length 12

== Theorem ==
The smallest Cunningham chain of the first kind of length $12$ is:
:$554 \, 688 \, 278 \, 429$, $1 \, 109 \, 376 \, 556 \, 859$, $2 \, 218 \, 753 \, 113 \, 719$, $4 \, 437 \, 506 \, 227 \, 439$,
::$8 \, 875 \, 012 \, 454 \, 879$, $17 \, 750 \, 024 \, 909 \, 759$, $35 \, 500 \, 049 \, 819 \, 519$, $71 \, 000 \, 099 \, 639 \, 039$,
::$142 \, 000 \, 199 \, 278 \, 079$, $284 \, 000 \, 398 \, 556 \, 159$, $568 \, 000 \, 797 \, 112 \, 319$, $1 \, 136 \, 001 \, 594 \, 224 \, 639$
== Proof ==
Let $C$ denote the sequence in question.
We have that $554 \, 688 \, 278 \, 429$ is prime.
First note that:
:$\dfrac {554 \, 688 \, 278 \, 429 - 1} 2 = 277 \, 344 \, 139 \, 214 = 2 \times 138 \, 672 \, 069 \, 607$
and so is not prime.
Thus $554 \, 688 \, 278 \, 429$ is not a safe prime, and thus fulfils the requirement for $C$ to be a Cunningham chain of the first kind.
Then:
Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $12$ can be done by a computer search.
== Sources ==
* : $554,688,278,429$
Category:Cunningham Chains
Smallest Cunningham Chain of the First Kind of Length 6

== Theorem ==
The smallest Cunningham chain of the first kind of length $6$ is:
:$\tuple {89, 179, 359, 719, 1439, 2879}$
== Proof ==
By definition, a Cunningham chain of the first kind is a sequence of prime numbers $\tuple {p_1, p_2, \ldots, p_n}$ such that:
: $p_{k + 1} = 2 p_k + 1$
: $\dfrac {p_1 - 1} 2$ is not prime
: $2 p_n + 1$ is not prime.
Thus each term except the last is a Sophie Germain prime.
Let $P: \Z \to \Z$ be the mapping defined as:
:$\map P n = 2 n + 1$
Applying $P$ iteratively to each of the smallest Sophie Germain primes in turn:
Thus $\tuple {2, 5, 11, 23, 47}$ is a Cunningham chain of the first kind of length $5$.
It is noted that $\dfrac {89 - 1} 2 = 44$ which is not prime.
Hence the sequence of $6$:
:$\tuple {89, 179, 359, 719, 1439, 2879}$
== Sources ==
* : $89$
* : $89$
Category:Cunningham Chains
Smallest Cunningham Chain of the First Kind of Length 7

== Theorem ==
The smallest Cunningham chain of the first kind of length $7$ is:
:$\left({1 \, 122 \, 659, 2 \, 245 \, 319, 4 \, 490 \, 639, 8 \, 981 \, 279, 17 \, 962 \, 559, 35 \, 925 \, 119, 71 \, 850 \, 239}\right)$
== Proof ==
Let $C$ denote the sequence in question.
We have that:
:$\dfrac {1 \, 122 \, 659 - 1} 2 = 561 \, 329 = 83 \times 6763$
and so is not prime.
Thus $1 \, 122 \, 659$ is not a safe prime, as is required for $C$ to be a Cunningham chain of the first kind.
Then:
Establishing that this is indeed the smallest such Cunningham chain of the first kind of length $7$ can be done by a computer search.
== Sources ==
* : $1,122,659$
* 
* : $1,122,659$
Category:Cunningham Chains
Smallest Cunningham Chain of the Second Kind of Length 13

== Theorem ==
The smallest Cunningham chain of the second kind of length $13$ is:
:$758 \, 083 \, 947 \, 856 \, 951$, $1 \, 516 \, 167 \, 895 \, 713 \, 901$, $3 \, 032 \, 335 \, 791 \, 427 \, 801$, $6 \, 064 \, 671 \, 582 \, 855 \, 601$, $12 \, 129 \, 343 \, 165 \, 711 \, 201$, $24 \, 258 \, 686 \, 331 \, 422 \, 401$, $48 \, 517 \, 372 \, 662 \, 844 \, 801$, $97 \, 034 \, 745 \, 325 \, 689 \, 601$, $194 \, 069 \, 490 \, 651 \, 379 \, 201$, $388 \, 138 \, 981 \, 302 \, 758 \, 401$, $776 \, 277 \, 962 \, 605 \, 516 \, 801$, $1 \, 552 \, 555 \, 925 \, 211 \, 033 \, 601$, $3 \, 105 \, 111 \, 850 \, 422 \, 067 \, 201$
== Proof ==
Let $C$ denote the sequence in question.
We have that $758 \, 083 \, 947 \, 856 \, 951$ is prime.
First note that:
:$\dfrac {758 \, 083 \, 947 \, 856 \, 951 + 1} 2 = 379 \, 041 \, 973 \, 928 \, 476 = 2^2 \times 94 \, 760 \, 493 \, 482 \, 119$
and so is not prime.
Thus $758 \, 083 \, 947 \, 856 \, 951$ fulfils the requirement for $C$ to be a Cunningham chain of the second kind.
Then:
Establishing that this is indeed the smallest such Cunningham chain of the second kind of length $13$ can be done by a computer search.
== Sources ==
* : $758,083,947,856,951$
Category:Cunningham Chains
Smallest Differences between Fractional Parts of Square and Cube Roots

== Theorem ==
Apart from $6$th powers, the value of $n$ less than $50 \, 000$ for which the difference between the fractional parts of $\sqrt n$ and $\sqrt [3] n$ is smallest is $30 \, 739$.
The next integer to produce a smaller difference above that is $62 \, 324$.
== Proof ==
== Historical Note ==
== Sources ==
* : $30,739$
* : $30,739$
Category:Square Roots
Category:Cube Roots
Smallest Element/Examples/Finite Subsets of Natural Numbers

== Examples of Minimal Elements ==
Let $\FF$ denote the set of finite subsets of the natural numbers $\N$.
Consider the ordered set $\struct {\FF, \subseteq}$.
Then $\struct {\FF, \subseteq}$ has a smallest element, and that is the empty set $\O$.
== Proof ==
We have that $\O$ is a finite set.
By Empty Set is Subset of All Sets it follows that $\O$ is a subset of $\N$.
Hence $\O \in \FF$ by definition of $\FF$.
Let $A \in \FF$ be some finite subset of $\N$.
Then by Empty Set is Subset of All Sets:
:$\O \subseteq A$
Hence $\O$ is the smallest element of $\struct {\FF, \subseteq}$ by definition.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(a)}$
Category:Examples of Smallest Elements
Smallest Element/Examples/Finite Subsets of Natural Numbers less Empty Set

== Examples of Smallest Elements ==
Let $\FF$ denote the set of finite subsets of the natural numbers $\N$.
Let $\GG$ denote the set $\FF \setminus \set \O$, that is, $\FF$ with the empty set excluded.
Consider the ordered set $\struct {\GG, \subseteq}$.
$\struct {\GG, \subseteq}$ has no smallest element.
== Proof ==
From Minimal Element: Finite Subsets of Natural Numbers less Empty Set, $\struct {\GG, \subseteq}$ has more than one minimal element.
The result follows from Ordered Set with Multiple Minimal Elements has no Smallest Element.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $6 \ \text {(a)}$
Category:Examples of Smallest Elements
Smallest Element WRT Restricted Ordering

== Theorem ==
Let $S$ be a set or class.
Let $\preceq$ be an ordering on $S$.
Let $T$ be a subset or subclass of $S$.
Let $\preceq'$ be the restriction of $\preceq$ to $T$.
Let $m \in T$.
Then $m$ is the $\preceq$-smallest element of $T$ $m$ is the $\preceq'$-smallest element of $T$.
== Proof ==
Category:Smallest Elements
Smallest Element is Infimum

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a smallest element $m$.
Then $m$ is the infimum of $T$ in $S$.
== Proof ==
Let $M$ be the smallest element of $T$.
Then by definition:
:$\forall x \in T: m \preceq x$
By definition of infimum, it is necessary to show that:
:$(1): \quad m$ is a lower bound of $T$ in $S$
:$(2): \quad L \preceq m$ for all lower bounds $L$ of $T$ in $S$.
By Smallest Element is Lower Bound, $m$ is a lower bound of $T$ in $S$.
It remains to be shown that:
:$L \preceq m$ for all lower bounds $L$ of $T$ in $S$.
Let $L \in S$ be a lower bound of $T$ in $S$.
By definition of lower bound:
:$\forall t \in T: L \preceq t$
We have that $m \in T$.
Therefore:
:$L \preceq m$
Hence the result.
== Examples ==
== Also see ==
* Greatest Element is Supremum
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum
Category:Infima
Category:Smallest Elements
Smallest Element is Infimum/Examples/Arbitrary Example 1

== Example of Use of Smallest Element is Infimum ==
Let $S$ be the subset of the real numbers $\R$ defined as:
:$S = \set {1, 2, 3}$
Then the smallest element of $S$ is $1$.
From Smallest Element is Infimum it follows that:
:$\inf S = 1$
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.8$: Example $\text{(ii)}$
Category:Examples of Infima
Smallest Element is Infimum/Examples/Arbitrary Example 2

== Example of Use of Smallest Element is Infimum ==
Let $T$ be the subset of the set of real numbers $\R$ defined as:
:$T := \set {x \in \R: 1 \le x < 2}$
$T$ has a smallest element $1$.
Hence from Smallest Element is Infimum it follows that $1$ is also the infimum of $T$.
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.8$: Example $\text{(ii)}$
Category:Examples of Infima
Smallest Element is Initial Object

== Theorem ==
Let $\mathbf P$ be an order category.
Suppose the objects $\mathbf P_0$ of $\mathbf P$, considered as an ordered set, have a smallest element $p$.
Then $p$ is an initial object of $\mathbf P$.
== Proof ==
Since $p$ is the smallest element of $\mathbf P_0$, we have:
:$\forall q \in \mathbf P_0: p \le q$
that is, for every object $q$ of $\mathbf P$ there is a unique morphism $p \to q$.
That is, $p$ is initial.
== Sources ==
* : $\S 2.2$: Example $2.11$: $5$
Category:Order Categories
Smallest Element is Lower Bound

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $T \subseteq S$.
Let $T$ have a smallest element $m \in T$.
Then $m$ is a lower bound of $T$.
It follows by definition that $T$ is bounded below.
== Proof ==
Let $m \in T$ be a smallest element of $T$.
By definition:
:$\forall y \in T: m \preceq y$
But as $T \subseteq S$, it follows that $m \in S$.
Hence:
:$\exists m \in S: \forall y \in T: m \preceq y$
Thus $T$ is bounded below by the lower bound $m$.
== Also see ==
* Greatest Element is Upper Bound
== Sources ==
* : $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum
Category:Smallest Elements
Category:Boundedness
Smallest Element is Minimal

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set which has a smallest element.
Let $m$ be the smallest element of $\struct {S, \preceq}$.
Then $m$ is a minimal element.
== Proof ==
By definition of smallest element:
:$\forall y \in S: m \preceq y$
Suppose $y \preceq m$.
As $\preceq$ is an ordering, $\preceq$ is by definition antisymmetric.
Thus it follows by definition of antisymmetry that $m = y$.
That is:
:$y \preceq m \implies m = y$
which is precisely the definition of a minimal element.
== Also see ==
* Greatest Element is Maximal
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations
Category:Smallest Elements
Category:Minimal Elements
Smallest Element is Unique

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
If $S$ has a smallest element, then it can have only one.
That is, if $a$ and $b$ are both smallest elements of $S$, then $a = b$.
=== Class Theoretical Formulation ===
== Proof ==
Let $a$ and $b$ both be smallest elements of $S$.
Then by definition:
:$\forall y \in S: a \preceq y$
:$\forall y \in S: b \preceq y$
Thus it follows that:
:$a \preceq b$
:$b \preceq a$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $a = b$.
== Also see ==
* Greatest Element is Unique
== Sources ==
* : $\S 14$: Order
* : $\S 2.3$
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $5$
* 
* : $1$: Set Theory and Logic: $\S 3$: Relations
* : $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.11$
* 
Category:Smallest Elements
Category:Smallest Element is Unique
Smallest Element is Unique/Class Theory

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be an ordering.
Let $A$ be a subclass of the field of $\RR$.
Suppose $A$ has a smallest element $s$ $\RR$.
Then $s$ is unique.
That is, if $s$ and $t$ are both smallest elements of $A$, then $s = t$.
== Proof ==
Let $s$ and $t$ both be smallest elements of $A$.
Then by definition:
:$\forall y \in A: s \mathrel \RR y$
:$\forall y \in A: t \mathrel \RR y$
Thus it follows that:
:$s \preceq t$
:$t \preceq s$
But as $\preceq$ is an ordering, it is antisymmetric.
Hence by definition of antisymmetric, $a = b$.
== Also see ==
* Greatest Element of Class is Unique
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering
Category:Smallest Element is Unique
Smallest Element of Minimally Closed Class under Progressing Mapping/Proof

== Theorem ==
== Proof ==
 $b$ is not the smallest element of $N$.
Then there exists $m \in N$ such that $b \nsubseteq m$.
In particular:
:$m \ne b$
Let $B$ be the subclass of $A$ defined as:
:$B = \set {x \in A: \paren {x = b} \lor \paren {\exists y \in B: x = \map g y} }$
This is a subclass of $A$ containing $b$ which is closed under $g$.
Let $z \in B$.
We have that:
:$b \nsubseteq m$
Suppose that:
:$z \nsubseteq m$
As $g$ is a Progressing Mapping, we have that:
:$z \subseteq \map g z$
It follows that:
:$\map g z \nsubseteq m$
By the Principle of General Induction for Minimally Closed Class it follows that:
:$\forall z \in B: m \nsubseteq \map g z$
and in particular:
:$\forall z \in B: m \ne \map g z$
In summary:
:$m \ne b$
and:
:$\not \exists z \in B: m = \map g z$
It follows that:
:$m \notin B$
Thus we have created a proper subclass $B$ of $A$ containing $b$ which is closed under $g$.
This contradicts the assumption that $A$ is a minimally closed class under $g$ with respect to $b$
Hence, by Proof by Contradiction, our supposition that $b$ is not the smallest element of $N$ was false.
The result follows.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications: Exercise $4.1$
Category:Minimally Closed Classes under Progressing Mapping
Smallest Equilateral Triangle with Internal Point at Integer Distances from Vertices

== Theorem ==
The smallest equilateral triangle with sides of integer length which contains a point which is an integer distance from each vertex has a side length $112$:
:
There exists a point inside it which is $57$, $65$ and $73$ away from the three vertices.
== Proof ==
== Sources ==
* 
* : $112$
Category:Equilateral Triangles
Category:112
Category:57
Category:65
Category:73
Smallest Even Integer whose Euler Phi Value is not the Euler Phi Value of an Odd Integer

== Theorem ==
The smallest even integer whose Euler $\phi$ value is shared by no odd integer is $33 \, 817 \, 088$.
== Proof ==
We have:
Consider the equation:
:$(1): \quad \map \phi x = 2^{16} \times 257$
Let $p$ be an odd prime factor of $x$.
Then as Euler Phi Function is Multiplicative, we have:
:$p - 1 \divides p^{k - 1} \paren{p - 1} = \map \phi {p^k} \divides \map \phi x$
where $k$ is the highest power of $p$ dividing $x$.
Therefore, by considering divisors of $\map \phi x$, either:
:$p - 1 = 2^j$ for some $j \in \Z$ such that $1 \le j \le 16$
which corresponds to the Fermat primes, or:
:$p - 1 = 2^j \times 257$ for some $j \in \Z$ such that $1 \le j \le 16$
But $2^j \times 257 + 1$ is composite for $1 \le j \le 16$:
So the only possible odd prime factors of $x$ are the Fermat primes not exceeding $2^{16} + 1$:
:$3, 5, 17, 257, 65 \, 537$
such that:
:$257$ occurs with multiplicity $2$, since we need $257$ to be a factor of $\map \phi x$ but not $257^2$;
:all other prime factors occurs with multiplicity $1$, since we do not want these primes to be factors of $\map \phi x$.
As we have:
it follows that $257$ and $65 \, 537$ cannot appear together.
We can now write:
:$x = 2^y \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$
where each $\epsilon_i = 0$ or $1$.
 $x$ is odd, that is, $y = 0$.
We have that:
which is a contradiction.
Thus there can be no odd integer $x$ satisfying $(1)$, that is, $y \ge 1$.
Note that:
Therefore even solutions to $(1)$ are of the form:
:$x = 2^{9 - \epsilon_0 - 2 \epsilon_1 - 4 \epsilon_2} \times 3^{\epsilon_0} \times 5^{\epsilon_1} \times 17^{\epsilon_2} \times 257^2$
Since each of:
:$2^{-1} \times 3^1$
:$2^{-2} \times 5^1$
:$2^{-4} \times 17^1$
are greater than $1$, taking $\epsilon_i = 0$ for $i = 0, 1, 2$ would result in the smallest even integer satisfying $(1)$, that is:
:$x = 2^9 \times 257^2 = 33 \, 817 \, 088$
 
It can be established by computer that $33 \, 817 \, 088$ is the smallest even integer whose Euler $\phi$ value is shared by no odd integer.
== Sources ==
* 
* 
* : $33,817,088$
Category:33,817,088
Category:Euler Phi Function
Smallest Fermat Pseudoprime to Bases 2, 3, 5 and 7

== Theorem ==
The smallest Fermat pseudoprime to bases $2$, $3$, $5$ and $7$ is $29 \, 341$.
== Proof ==
== Sources ==
* 
* : $29,351$
Category:Fermat Pseudoprimes
Category:29,341
Smallest Fermat Pseudoprime to Bases 2, 3, 5 and 7/Mistake

== Source Work ==
:
:The Dictionary
::$29,351$
== Mistake ==
:[$29,351$ is] Simultaneously a pseudoprime in bases $2$, $3$, $5$ and $7$.
This appears to be a misprint for $29,341$.
It remains to be seen whether the mistake came originally from , or whether it was transcribed incorrectly from there into 
== Sources ==
* 
* : $29,351$
Category:Mistakes/Curious and Interesting Numbers
Smallest Field/Cayley Tables

== Cayley Tables for the Smallest Field ==
The smallest field can be completely described by showing its Cayley tables.
In purely abstract form as $\struct {\set {0_R, 1_R}, +, \circ}$:
:$\begin{array} {r|rr}
\struct {\set {0_R, 1_R}, +} & 0_R & 1_R \\
\hline
0_R & 0_R & 1_R \\
1_R & 1_R & 0_R \\
\end{array}
\qquad
\begin{array} {r|rr}
\struct {\set {0_R, 1_R}, \circ} & 0_R & 1_R \\
\hline
0_R & 0_R & 0_R \\
1_R & 0_R & 1_R \\
\end{array}$
=== Ring of Integers Modulo $2$ ===
It can also be expressed as the ring of integers modulo $2$ $\struct {\Z_2, +_2, \times_2}$:
=== Parity Ring ===
It can also be expressed in terms of integer parity:
Category:Examples of Cayley Tables
Smallest Field containing Subfield and Complex Number/Examples

== Examples of Smallest Field containing Subfield and Complex Number ==
=== Complex Numbers ===
=== Numbers of Type $a + b \sqrt 2: a, b \in \Q$ ===
=== Smallest Field Containing $\Q$, $\sqrt 2$ and $\sqrt 3$ ===
=== Numbers of Type $a + b \sqrt [3] 2: a, b \in \Q$ ===
Category:Examples of Field Extensions
Smallest Field containing Subfield and Complex Number/Examples/Field Containing Rationals, Root 2 and Root 3

== Example of Smallest Field containing Subfield and Complex Number ==
The smallest field containing $\Q$, $\sqrt 2$ and $\sqrt 3$ is:
:$\set {a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6: a, b, c, d \in \Q}$
This forms a vector space of dimension $4$ which has basis $\set {1, \sqrt 2, \sqrt 3, \sqrt 6}$.
== Proof ==
Let $K := \Q \sqbrk {\sqrt 2}$ denote the set:
:$K := \set {a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.
It is immediately apparent that $\sqrt 3 \notin K$.
Let $K'$ be a field containing $K$ and $\sqrt 3$.
Then $K'$ must contain all real numbers of the form:
:$\paren {a + b \sqrt 2} \paren {c + d \sqrt 2} \sqrt 3$
where $a, b, c, d \in \Q$.
This can be written:
:$\set {a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6}$
and real numbers of this form do actually form a field.
Let $L := \Q \sqbrk {\sqrt 3}$ denote the set:
:$L := \set {a + b \sqrt 3: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 3$ where $a$ and $b$ are rational numbers.
Let $L'$ be a field containing $L$ and $\sqrt 2$.
Then $L'$ must contain all real numbers of the form:
:$\paren {a + b \sqrt 3} \paren {c + d \sqrt 3} \sqrt 2$
where $a, b, c, d \in \Q$.
This can be written:
:$\set {a + b \sqrt 3 + c \sqrt 2 + d \sqrt 6}$
which is the same set as before.
That is:
:$K \sqbrk {\sqrt 2} = L \sqbrk {\sqrt 3}$
But a field containing $\sqrt 2$ and $\sqrt 3$ must also contain $\sqrt 6$.
Thus such a field contains all numbers of the form $a + b \sqrt 2 + c \sqrt 3 + d \sqrt 6$.
As these form a field, this is the smallest field containing $\Q$, $\sqrt 2$ and $\sqrt 3$.
Let this be denoted $\Q \sqbrk {\sqrt 2, \sqrt 3}$.
So:
:$\Q \sqbrk {\sqrt 2, \sqrt 3} = K \sqbrk {\sqrt 3}$
We have that:
:$\index {K \sqrt 3} K = 2$
and from Smallest Field containing Subfield and Complex Number: Numbers of Type $a + b \sqrt 2: a, b \in \Q$:
:$\index K Q = 2$
So by Degree of Field Extensions is Multiplicative:
:$\index {\Q \sqbrk {\sqrt 2, \sqrt 3} } \Q = 4$
and it is seen that $\set {1, \sqrt 2, \sqrt 3, \sqrt 6}$ forms a basis for $\Q \sqbrk {\sqrt 2, \sqrt 3}$.
By definition of basis and dimension, it is seen that $\map \dim {\Q \sqbrk {\sqrt 2, \sqrt 3} } = 4$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $73$
Category:Examples of Field Extensions
Smallest Field containing Subfield and Complex Number/Examples/Numbers of Type Rational a plus b cube root 2

== Example of Smallest Field containing Subfield and Complex Number ==
Let $\Q \sqbrk {\sqrt [3] 2}$ denote the set:
:$\Q \sqbrk {\sqrt [3] 2} := \set {a + b \sqrt [3] 2 + c \sqrt [3] {2^2}: a, b, c \in \Q}$
Then:
:$\Q \sqbrk {\sqrt [3] 2}$ is the smallest field containing $\Q$ and $\sqrt [3] 2$
and:
:$\index {\Q \sqbrk {\sqrt [3] 2} } \Q = 3$
== Sources ==
* : Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $74$
Category:Examples of Field Extensions
Smallest Field containing Subfield and Complex Number/Examples/Numbers of Type Rational a plus b root 2

== Example of Smallest Field containing Subfield and Complex Number ==
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.
Then $\Q \sqbrk {\sqrt 2}$ is the smallest field containing $\Q$ and $\sqrt 2$.
Formally, $\Q \sqbrk {\sqrt 2}$ is the field extension of $\Q$ for the minimal polynomial of $\sqrt 2$, the second-degree polynomial $x^2 - 2$.
Therefore, $\Q \sqbrk {\sqrt 2}$ is the vector space of dimension $2$ isomorphic to $\Q \sqbrk x / \gen {x^2 - 2}$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $72$
* : Chapter $20$: Extension Fields: $\S 1$. Splitting Fields
Category:Examples of Field Extensions