nyuuzyou/wikis huggingface dataset:
Find A Suitable Nail Design Shop With an Ease

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(A cap C) cup (B cap Complement C) = Empty iff B subset C subset Complement A

== Theorem ==
Let $A$, $B$ and $C$ be subsets of a universe $\Bbb U$.
Then:
:$\paren {A \cap C} \cup \paren {B \cap \map \complement C} = \O \iff B \subseteq C \subseteq \map \complement A$
where $\map \complement C$ denotes the complement of $C$ in $\Bbb U$.
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $11 \ \text{(e)}$
Category:Set Union
Category:Set Intersection
Category:Set Complement
Category:Empty Set
Category:Subsets
*-Algebra Homomorphism between C*-Algebras is Norm-Decreasing

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ and $\struct {B, \dagger, \norm {\, \cdot \,}_B}$ be $\text C^\ast$-algebras.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Then:
:$\norm {\map \phi x}_B \le \norm x_A$ for each $x \in A$.
== Proof ==
Let $x \in A$.
Then:
From Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism: Corollary, we have:
:$\map {\sigma_B} {\map \phi {x x^\ast} } \cup \set 0 \subseteq \map {\sigma_A} {x x^\ast} \cup \set 0$ 
Hence we have:
Hence we obtain:
:$\norm {\map \phi x}_B^2 \le \norm x^2_A$
So:
:$\norm {\map \phi x}_B \le \norm x_A$ for each $x \in A$. 
Category:C*-Algebras
*-Algebra Homomorphism preserves Canonical Preordering of C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,}_A}$ and $\struct {B, \square, \norm {\, \cdot \,}_B}$ be $\text C^\ast$-algebras.
Let $\phi : A \to B$ be a $\ast$-algebra homomorphism.
Let $\le_A$ and $\le_B$ be the canonical preorderings on $A$ and $B$ respectively.
Let $x, y, z \in A$ be such that:
:$x \le_A y \le_A z$
Then we have:
:$\map \phi x \le_B \map \phi y \le_B \map \phi z$
== Proof ==
Since $\phi$ is an algebra homomorphism, we aim to show that:
:$\map \phi {y - x}$ is positive
and:
:$\map \phi {z - y}$ is positive.
Since $y - x$ and $z - y$ are Hermitian and $\phi$ is a $\ast$-algebra homomorphism, we have that:
:$\map \phi {y - x}$ and $\map \phi {z - y}$ are Hermitian.
Since $y - x$ are $z - y$ are positive, we have:
:$\map {\sigma_A} {y - x} \subseteq \hointr 0 \infty$
and:
:$\map {\sigma_A} {z - y} \subseteq \hointr 0 \infty$
Hence from Spectrum of Image of Element of Unital Algebra under Unital Algebra Homomorphism: Corollary, we have:
:$\map {\sigma_B} {\map \phi {y - x} } \subseteq \map {\sigma_A} {y - x} \cup \set 0 \subseteq \hointr 0 \infty$
and:
:$\map {\sigma_B} {\map \phi {z - y} } \subseteq \map {\sigma_A} {z - y} \cup \set 0 \subseteq \hointr 0 \infty$ 
Hence $\map \phi {y - x}$ and $\map \phi {z - y}$ are positive.
Hence $\map \phi y - \map \phi x$ and $\map \phi z - \map \phi y$ are positive.
Hence:
:$\map \phi x \le_B \map \phi y \le_B \map \phi z$
Category:Canonical Preorderings of C*-Algebras
*-Algebra obtains Banach *-Algebra Structure through *-Algebra Isomorphism

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $\struct {B, \square, \norm {\, \cdot \,}_B}$ be a Banach $\ast$-algebra.
Let $\phi : A \to B$ be a $\ast$-algebra isomorphism. 
Define $\norm {\, \cdot \,}_A : A \to \R$ by:
:$\norm a_A = \norm {\map \phi a}_B$
Then $\struct {A, \ast, \norm {\, \cdot \,}_A}$ is a Banach $\ast$-algebra.
Further if $\struct {B, \square, \norm {\, \cdot \,}_B}$ is a $\text C^\ast$-algebra, then so is $\struct {A, \ast, \norm {\, \cdot \,}_A}$.
== Proof ==
From Algebra obtains Norm Structure through Algebra Isomorphism:
:$\norm {\, \cdot \,}_A$ is an algebra norm.
We want to show that:
:$\norm {a^\ast}_A = \norm a_A$
for each $a \in A$. 
Let $a \in A$.
We have:
By definition, $\phi : A \to B$ is an isometric isomorphism.
Hence from Inverse of Isometric Isomorphism between Normed Vector Spaces is Isometric Isomorphism, $\phi^{-1} : B \to A$ is an isometric isomorphism.
From Metric Space Completeness is Preserved by Isometry, the completeness of $B$ implies the completeness of $A$.
So $\struct {A, \norm {\, \cdot \,}_A}$ is a Banach algebra.
Hence $\struct {A, \ast, \norm {\, \cdot \,}_A}$ is a Banach $\ast$-algebra.
Now suppose that $\struct {B, \square, \norm {\, \cdot \,}_B}$ is a $\ast$-algebra so that:
:$\norm {b b^\square}_B = \norm b^2_B$
for each $b \in B$.
We want to show that:
:$\norm {a a^\ast}_A = \norm a^2_A$
for each $a \in A$.
We have:
Category:Banach *-Algebras
Category:C*-Algebras
-1

== Number ==
$-1$ (minus one) is:
:The largest strictly negative integer
:The $1$st integer $n$ such that $\dbinom n 0 + \dbinom n 1 + \dbinom n 2 + \dbinom n 3 = m^2$ for integer $m$:
::$\dbinom {-1} 0 + \dbinom {-1} 1 + \dbinom {-1} 2 + \dbinom {-1} 3 = 0^2$
== Also see ==
=== Next in sequence: $0$ ===
* 
Category:Specific Numbers
Category:-1
-1^n by -n choose k-1 equals -1^k by -k choose n-1

== Theorem ==
Let $n, k \in \Z_{\ge 0}$.
Then:
:$\paren {-1}^n \dbinom {-n} {k - 1} = \paren {-1}^k \dbinom {-k} {n - 1}$
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $16$
Category:Binomial Coefficients
0.999...=1/Proof 1

== Theorem ==
== Proof ==
By Sum of Infinite Geometric Sequence:
:$0.999 \ldots = \dfrac a {1 - r}$
where $a = \dfrac 9 {10}$ and $r = \dfrac 1 {10}$.
Since our ratio is less than $1$, then we know that $\ds \sum_{n \mathop = 0}^\infty \frac 9 {10} \paren {\frac 1 {10} }^n$ must converge to:
:$\dfrac a {1 - r} = \dfrac {\frac 9 {10} } {1 - \frac 1 {10} } = \dfrac {\frac 9 {10} } {\frac 9 {10} } = 1$
Category:0.999...=1
0.999...=1/Proof 4

== Theorem ==
== Proof ==
We begin with the knowledge that:
Now we divide $9$ by $9$ using the standard process of long division, only instead of stating that $90$ divided by $9$ is $10$, we say that it is "$9$ remainder $9$," yielding the following result:
 0.9999...
 -----------
 9)9.0000...
 8.1
 ---
 90
 81
 --
 90
 81
 --
 9...
Thus, we are compelled to believe that:
:$0.999\ldots = \dfrac 9 9 = 1$
Category:0.999...=1
0 in B-Algebra is Left Cancellable Element

== Theorem ==
Let $\struct {X, \circ}$ be a $B$-Algebra.
Then:
:$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$
== Proof ==
Let $x, y \in X$ and let $0 \circ x = 0 \circ y$.
Then:
So we have shown:
:$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x \circ y = 0$
From $B$-Algebra Identity: $x \circ y = 0 \iff x = y$:
:$\forall x, y \in X: 0 \circ x = 0 \circ y \implies x = y$
Hence the result.
Category:B-Algebras
1+1 = 2

== Theorem ==
Define $0$ as the unique element in the set $P \setminus \map s P$, where:
:$P$ is the Peano Structure
:$\map s P$ is the image of the mapping $s$ defined in Peano structure
:$\setminus$ denotes the set difference.
Then:
:$1 + 1 = 2$
where:
:$1 := \map s 0$
:$2 := \map s 1 = \map s {\map s 0}$
:$+$ denotes addition
:$=$ denotes equality
:$\map s n$ denotes the successor mapping.
== Proof 1 ==
== Proof 2 ==
Category:Mathematical Logic
Category:1+1 = 2
1+1 = 2/Proof 1

== Theorem ==
== Proof ==
$1$ is defined as $\map s 0$ and $2$ as $\map s {\map s 0}$.
Hence the statement to be proven becomes:
:$\map s 0 + \map s 0 = \map s {\map s 0}$
Thus:
Category:1+1 = 2
1+1 = 2/Proof 2

== Theorem ==
== Proof ==
Defining $1$ as $\map s 0$ and $2$ as $\map s {\map s 0}$, the statement to be proven becomes:
:$\map s 0 + \map s 0 = \map s {\map s 0}$
By the definition of addition:
:$\forall m \in P: \forall n \in P: m + \map s n = \map s {m + n}$
Letting $m = \map s 0$ and $n = 0$:
By the definition of addition:
:$\forall m: m + 0 = m$
Letting $m = \map s 0$:
:$\map s 0 + 0 = \map s 0$
Taking the successor of both sides: 
Applying Equality is Transitive to $(1)$ and $(2)$ we have:
:$\map s 0 + \map s 0 = \map s {\map s 0}$
Hence the result.
Category:1+1 = 2
1+2+...+n+(n-1)+...+1 = n^2

== Theorem ==
:$\forall n \in \N: 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 = n^2$
== Direct Proof 1 ==
== Direct Proof 2 ==
== Proof by Induction ==
== Proof 4 ==
== Illustration ==
== Sources ==
* : $25$
* : $25$
Category:Sums of Sequences
Category:Square Numbers
Category:1+2+...+n+(n-1)+...+1 = n^2
1+2+...+n+(n-1)+...+1 = n^2/Proof 3

== Theorem ==
== Proof ==
Proof by induction:
=== Basis for the Induction ===
$n = 1$ holds trivially.
Just to make sure, we try $n = 2$:
:$1 + 2 + 1 = 4$
Likewise $n^2 = 2^2 = 4$.
So shown for basis for the induction.
=== Induction Hypothesis ===
This is the induction hypothesis:
:$1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1 = k^2$
Now we need to show true for $n = k + 1$:
:$1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1 = \paren {k + 1}^2$
=== Induction Step ===
This is the induction step:
The result follows by induction.
Category:Proofs by Induction
Category:1+2+...+n+(n-1)+...+1 = n^2
1+2+...+n+(n-1)+...+1 = n^2/Proof 4

== Theorem ==
== Proof ==
Let $T_n = 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1$.
We have $T_1 = 1$
and
Thus we have:
Category:1+2+...+n+(n-1)+...+1 = n^2
1,000,000

== Number ==
$1 \, 000 \, 000$ (one million) is:
:$2^6 \times 5^6$
:The $6$th power of $10$ after $(1)$, $10$, $100$, $1000$, $10 \, 000$, $100 \, 000$:
::$1 \, 000 \, 000 = 10^6$
:The $10$th sixth power after $1$, $64$, $729$, $4096$, $15 \, 625$, $46 \, 656$, $117 \, 649$, $262 \, 144$, $531 \, 441$:
::$1 \, 000 \, 000 = 10 \times 10 \times 10 \times 10 \times 10 \times 10$
:The $100$th cube number:
::$1 \, 000 \, 000 = 100 \times 100 \times 100$
:The $1000$th square number:
::$1 \, 000 \, 000 = 1000 \times 1000$
== Also see ==
* 
* 
* 
* 
== Sources ==
* : $1,000,000$
* : $1,000,000$
Category:Specific Numbers
Category:1,000,000
1,000,000,000,061

== Number ==
$1 \, 000 \, 000 \, 000 \, 061$ is:
:A prime number
:The smaller of a pair of twin primes with $1 \, 000 \, 000 \, 000 \, 063$
== Also see ==
* 
* 
== Sources ==
* : $1,000,000,000,061$
* : $1,000,000,000,061$
Category:Specific Numbers
Category:1,000,000,000,061
1,000,000,000,063

== Number ==
$1 \, 000 \, 000 \, 000 \, 063$ is:
:A prime number
:The larger of a pair of twin primes with $1 \, 000 \, 000 \, 000 \, 061$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1,000,000,000,063
1,002,000,000,000/Historical Note

== Historical Note on $1 \, 002 \, 000 \, 000 \, 000$ ==
According to , $1 \, 002 \, 000 \, 000 \, 000$ was the number of possible syllables that calculated as the number of syllables that could be made from the letters of the Greek alphabet.
If this is true, then it forms the first attempt on record at solving a significant problem of combinatorics.
== Sources ==
* : $1,002,000,000,000$
* : $1,002,000,000,000$
Category:Historical Notes
1,003,003,001

== Number ==
$1 \, 003 \, 003 \, 001$ is:
:$7^3 \times 11^3 \times 13^3$
:The $1001$st cube number:
::$1 \, 003 \, 003 \, 001 = 1001 \times 1001 \times 1001$
:The $7$th palindromic cube after $1$, $8$, $343$, $1331$, $1 \, 030 \, 301$, $1 \, 367 \, 631$:
::$1 \, 003 \, 003 \, 001 = 1001^3$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1,003,003,001
1,026,753,849

== Number ==
$1 \, 026 \, 753 \, 849$ is:
:$3^2 \times 11^2 \times 971^2$
:The $32 \, 043$rd square number:
::$1 \, 026 \, 753 \, 849 = 32 \, 043 \times 32 \, 043$
:The smallst pandigital square (in the sense where pandigital excludes the zero)
== Also see ==
* Smallest Pandigital Square
* 
== Sources ==
* : $1,026,753,849$
Category:Specific Numbers
Category:1,026,753,849
1,030,301

== Number ==
$1 \, 030 \, 301$ (one million, thirty thousand, three hundred and one) is:
:$101^3$
:The $5$th palindromic cube after $1$, $8$, $343$, $1331$:
::$1 \, 030 \, 301 = 101^3$
:The $101$st cube number:
::$1 \, 030 \, 301 = 101 \times 101 \times 101$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1,030,301
1,031,223,314

== Number ==
$1 \, 031 \, 223 \, 314$ is:
:$2 \times 11 \times 46 \, 873 \, 787$
:A fixed point in the ascending pea pattern:
::$1031223314$ is one $0$, three $1$s, two $2$s, three $3$s, one $4$
== Also see ==
* Ascending Pea Pattern on $1 \, 031 \, 223 \, 314$
== Sources ==
* : $1,031,223,314$
Category:Specific Numbers
Category:1,031,223,314
1,048,576

== Number ==
$1 \, 048 \, 576$ (one million, forty-eight thousand, five hundred and seventy-six) is:
:$2^{20}$
:The $1$st of a sequence of $7$ numbers which are square and anagrams of each other
:The number of different binary operations with an identity element that can be applied to a set with $4$ elements
:The number of different commutative binary operations that can be applied to a set with $4$ elements
:The $5$th power of $16$ after $(1)$, $16$, $256$, $4096$, $65 \, 536$:
::$1 \, 048 \, 576 = 16^5$
:The $10$th power of $4$ after $(1)$, $4$, $16$, $64$, $256$, $1024$, $4096$, $16 \, 384$, $65 \, 536$, $262 \, 144$:
::$1 \, 048 \, 576 = 4^{10}$
:The $16$th fifth power after $1$, $32$, $243$, $1024$, $3125$, $7776$, $16 \, 807$, $32 \, 768$, $59 \, 049$, $100 \, 000$, $161 \, 051$, $248 \, 832$, $371 \, 293$, $537 \, 824$, $759 \, 375$:
::$1 \, 048 \, 576 = 16 \times 16 \times 16 \times 16 \times 16$
:Hence $100 \, 000$ in hexadecimal.
:The $20$th power of $2$ after $(1)$, $2$, $4$, $8$, $16$, $\ldots$, $16 \, 384$, $32 \, 768$, $65 \, 536$, $131 \, 072$, $262 \, 144$, $524 \, 288$:
::$1 \, 048 \, 576 = 2^{20}$
:The $21$st almost perfect number after $1$, $2$, $4$, $8$, $16$, $\ldots$, $16 \, 384$, $32 \, 768$, $65 \, 536$, $131 \, 072$, $262 \, 144$, $524 \, 288$:
::$\map {\sigma_1} {1 \, 048 \, 576} = 2 \, 097 \, 151 = 2 \times 524 \, 288 - 1$
:The $32$nd fourth power:
::$1 \, 048 \, 576 = 32 \times 32 \times 32 \times 32$
:The $1024$th square number:
::$1 \, 048 \, 576 = 1024 \times 1024$
== Also see ==
* 
* 
* 
* 
* 
* 
* 
* 
* 
* 
== Sources ==
* : $1,048,576$
* : $1,048,576$
Category:Specific Numbers
Category:1,048,576
1,056,784

== Number ==
$1 \, 056 \, 784$ (one million, fifty-six thousand, seven hundred and eighty-four) is:
:$2^4 \times 257^2$
:The $2$nd after $1 \, 048 \, 576$ of a sequence of $7$ numbers which are square and anagrams of each other
:The $1028$th square number:
::$1 \, 056 \, 784 = 1028 \times 1028$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1,056,784
1,059,683,225,053,915,111,058,165,141,686,995

== Number ==
$1 \, 059 \, 683 \, 225 \, 053 \, 915 \, 111 \, 058 \, 165 \, 141 \, 686 \, 995$ is:
:$3 \times 5 \times 84 \, 089 \times 73 \, 919 \, 059 \times 150 \, 031 \, 897 \times 75 \, 754 \, 002 \, 239$
:The $2$nd term of a general Fibonacci sequence whose first $2$ terms are coprime such that all its terms were mistakenly believed at one time to be composite:
::$1 \, 786 \, 772 \, 701 \, 928 \, 802 \, 632 \, 268 \, 715 \, 130 \, 455 \, 793$, $1 \, 059 \, 683 \, 225 \, 053 \, 915 \, 111 \, 058 \, 165 \, 141 \, 686 \, 995$, $2 \, 846 \, 455 \, 926 \, 982 \, 717 \, 743 \, 326 \, 880 \, 272 \, 142 \, 788$, $\ldots$
== Also see ==
* 
Category:Specific Numbers
Category:1,059,683,225,053,915,111,058,165,141,686,995
1,085,764

== Number ==
$1 \, 085 \, 764$ (one million, eighty-five thousand, seven hundred and sixty-four) is:
:$2^2 \times 521^2$
:The $3$rd after $1 \, 048 \, 576$, $1 \, 056 \, 784$ of a sequence of $7$ numbers which are square and anagrams of each other
:The $1042$nd square number:
::$1 \, 085 \, 764 = 1042 \times 1042$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1,085,764
1,098,765,432

== Number ==
$1 \, 098 \, 765 \, 432$ is:
:$2^3 \times 3^3 \times 827 \times 6151$
:Remains pandigital when multiplied by $2$, $4$, $5$, $7$ and $8$
== Also see ==
* Pandigital Integers remaining Pandigital on Multiplication
== Sources ==
* : $123,456,789$
* : $123,456,789$
Category:Specific Numbers
Category:1,098,765,432
1,111,111

== Number ==
$1 \, 111 \, 111$ (one million, one hundred and eleven thousand, one hundred and eleven) is:
:$239 \times 4649$
:The $7$th repuint after $1$, $11$, $111$, $1111$, $11 \, 111$, $111 \, 111$
:The $38$th palindromic integer whose square is also palindromic integer:
::$1 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 654 \, 321$
== Also see ==
* 
* 
Category:Specific Numbers
Category:1,111,111
1,111,111,111

== Number ==
$1 \, 111 \, 111 \, 111$ is:
:$11 \times 41 \times 271 \times 9091$
:The $10$th repunit after $1$, $11$, $111$, $1111$, $\ldots$, $111 \, 111 \, 111$
 
:The $103$rd Kaprekar number, and the smallest with $10$ digits:
::$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321 \to 123 \, 456 \, 790 + 0 \, 987 \, 654 \, 321 = 1 \, 111 \, 111 \, 111$
== Also see ==
* 
* 
== Sources ==
* : $297$
* : $1,111,111,111$
* : $297$
* : $1,111,111,111$
Category:Specific Numbers
Category:1,111,111,111
1,111,111,111,111

== Number ==
$1 \, 111 \, 111 \, 111 \, 111$ is:
:$53 \times 79 \times 265 \, 371 \, 653$
:The $13$th repunit
 
== Also see ==
* 
Category:Specific Numbers
Category:1,111,111,111,111
1,111,111,111,111,111

== Number ==
$1 \, 111 \, 111 \, 111 \, 111 \, 111$ is:
:$11 \times 17 \times 73 \times 101 \times 137 \times 5 \, 882 \, 353$
:The $16$th repunit
 
== Also see ==
* 
Category:Specific Numbers
Category:1,111,111,111,111,111
1,111,111,111,111,111,111

== Number ==
$1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$ is:
:A prime number
:The $2$nd repunit prime
:The $13$th unique period prime after $3$, $11$, $37$, $101$, $9091$, $9901$, $333 \, 667$, $909 \, 091$, $99 \, 990 \, 001$, $999 \, 999 \, 000 \, 001$, $9 \, 999 \, 999 \, 900 \, 000 \, 001$, $909 \, 090 \, 909 \, 090 \, 909 \, 091$: its period is $19$:
::$\dfrac 1 {1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111} = 0 \cdotp \dot 00000 \, 00000 \, 00000 \, 000 \dot 9$
:The $19$th repunit
:The $23$rd permutable prime after $2$, $3$, $5$, $7$, $11$, $13$, $17$, $31$, $37$, $71$, $73$, $79$, $97$, $113$, $131$, $199$, $311$, $337$, $373$, $733$, $919$, $991$
== Also see ==
* Repunit 19 is Unique Period Prime with Period 19
* 
* 
* 
* 
* 
== Historical Note ==
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Specific Numbers
Category:1,111,111,111,111,111,111
1,111,111,111,111,111,111,111

== Number ==
$1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$ is:
:$11^2 \times 23 \times 4093 \times 8779 \times 21649 \times 513239$
:The $22$nd repunit
== Also see ==
* 
Category:Specific Numbers
Category:1,111,111,111,111,111,111,111
1,111,111,111,111,111,111/Historical Note

== Historical Note on $1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$ ==
The prime nature of $1 \, 111 \, 111 \, 111 \, 111 \, 111 \, 111$ was discovered by a reader of one of 's newspaper puzzle columns in $1918$.
== Sources ==
* : $1,111,111,111,111,111,111$
* : $1,111,111,111,111,111,111$
Category:Historical Notes
1,111,111,111/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$297$
== Mistake ==
:So is $1,111,111,111$ [a Kaprekar number ], the smallest Kaprekar number of $10$ digits whose square is $12345678900987654321$.
== Correction ==
The square is incorrect. It should be:
:$1234567900987654321$
The mistake is not repeated under the section for $1,111,111,111$ itself, in which the square is given correctly.
In of $1997$, the number $1,111,111,111^2$ is again given incorrectly in Section $297$, but differently:
:$12, 345, 671, 900, 987, 654, 321$
== Sources ==
* : $297$
Category:Mistakes/Curious and Interesting Numbers
1,111,111,111/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$297$
== Mistake ==
:So is $1,111,111,111$ [a Kaprekar number ], the smallest Kaprekar number of $10$ digits whose square is $12, 345, 671, 900, 987, 654, 321$.
== Correction ==
The square is incorrect. It should be:
:$1,234,567,900,987,654,321$
The mistake is not repeated under the section for $1,111,111,111$ itself, in which the square is given correctly.
In of $1986$, the number $1,111,111,111^2$ was again given incorrectly in Section $297$, but differently:
:$12345678900987654321$
== Sources ==
* : $297$
Category:Mistakes/Curious and Interesting Numbers
1,122,659

== Number ==
$1 \, 122 \, 659$ (one million, one hundred and twenty-two thousand, six hundred and fifty-nine) is:
:The $87 \, 359$th prime number
:The $1$st term of the smallest Cunningham chain of the first kind of length $7$:
::$1 \, 122 \, 659$, $2 \, 245 \, 319$, $4 \, 490 \, 639$, $8 \, 981 \, 279$, $17 \, 962 \, 559$, $35 \, 925 \, 119$, $71 \, 850 \, 239$
== Also see ==
* 
* 
== Sources ==
* : $1,122,659$
* : $1,122,659$
Category:Specific Numbers
Category:1,122,659
1,175,265

== Number ==
$1 \, 175 \, 265$ (one million, one hundred and seventy-five thousand, two hundred and sixty-five) is:
:$3^2 \times 5 \times 7^2 \times 13 \times 41$
:The smaller of the $9$th odd amicable pair, with $1 \, 438 \, 983$:
::$\map {\sigma_1} {1 \, 175 \, 265} = \map {\sigma_1} {1 \, 438 \, 983} = 2 \, 614 \, 248 = 1 \, 175 \, 265 + 1 \, 438 \, 983$
== Also see ==
* 
== Sources ==
* : $1,175,265$
* : $1,175,265$
Category:Specific Numbers
Category:1,175,265
1,234,321

== Number ==
$1 \, 234 \, 321$ (one million, two hundred and thirty-four thousand, three hundred and twenty-one) is:
:$11^2 \times 101^2$
:The $4$th Wonderful Demlo number after $1$, $121$, $12 \, 321$
:The $1111$th square number:
::$1 \, 234 \, 321 = 1111 \times 1111$
:A palindromic square of a palindromic number
== Also see ==
* Square of Small-Digit Palindromic Number is Palindromic/Examples/1111
* 
* 
== Sources ==
* : $1,234,321$
* : $1,234,321$
Category:Specific Numbers
Category:1,234,321
1,234,567,891

== Number ==
$1 \, 234 \, 567 \, 891$ is:
:The $1$st of the only $3$ known prime numbers consisting of strings of consecutive ascending digits starting from $1$ (allowing either $0$ or $1$ to follow $9$)
== Also see ==
* 
* 
== Sources ==
* : $1,234,567,891$
* : $1,234,567,891$
Category:Specific Numbers
Category:1,234,567,891