nyuuzyou/wikis huggingface dataset:
Irrational Number is Limit of Unique Simple Infinite Continued Fraction

== Theorem ==
Let $x$ be an irrational number.
Then the continued fraction expansion of $x$ is the unique simple infinite continued fraction that converges to $x$.
== Proof ==
Follows from:
* Continued Fraction Expansion of Irrational Number Converges to Number Itself
* Simple Infinite Continued Fraction is Uniquely Determined by Limit.
== Also see ==
* Correspondence between Irrational Numbers and Simple Infinite Continued Fractions, a more precise statement
== Sources ==
* 
Category:Simple Continued Fractions
Category:Irrational Numbers
Irrational Numbers are Uncountably Infinite

== Theorem ==
The set $\R \setminus \Q$ of irrational numbers is uncountable.
== Proof ==
From Real Numbers are Uncountable, $\R$ is an uncountable set.
From Rational Numbers are Countably Infinite $\Q$ is countable.
The result follows from Uncountable Set less Countable Set is Uncountable.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $156$
* 
* 
* 
* 
Category:Irrational Numbers
Category:Uncountable Sets
Irrational Numbers form G-Delta Set in Reals

== Theorem ==
Let $\R \setminus \Q$ denote the set of irrational numbers.
Let $\struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Then $\R \setminus \Q$ forms a $G_\delta$ set in $\R$.
== Proof ==
The result follows from Rational Numbers are Countably Infinite.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $2$
Category:Irrational Numbers
Category:Real Number Line with Euclidean Topology
Category:G-Delta Sets
Irrational Numbers form Metric Space

== Theorem ==
Let $\mathbb I = \R \setminus \Q$ be the set of all irrational numbers.
Let $d: \mathbb I \times \mathbb I \to \R$ be defined as:
:$\map d {x_1, x_2} = \size {x_1 - x_2}$
where $\size x$ is the absolute value of $x$.
Then $d$ is a metric on $\mathbb I$ and so $\struct {\mathbb I, d}$ is a metric space.
== Proof ==
From the definition of absolute value:
:$\size {x_1 - x_2} = \sqrt {\paren {x_1 - x_2}^2}$
This is the Euclidean metric.
in Euclidean Metric on Real Vector Space is Metric this is shown to be a metric.
== Sources ==
* : Part $\text {II}$: Counterexamples: $31$. The Irrational Numbers: $4$
Category:Irrational Number Space
Irrationality of Logarithm

== Theorem ==
Let $a, b \in \N_{>0}$ such that both $\nexists m, n \in \N_{>0}: a^m = b^n$.
Then $\log_b a$ is irrational.
== Proof ==
 $\log_b a$ is rational.
Then:
:$\exists p, q \in \N_{>0} : \log_b a = \dfrac p q$
where $p \perp q$.
Then:
which contradicts the initial assumption:
:$\nexists m, n \in \N: a^m = b^n$
Hence the result by Proof by Contradiction.
Category:Number Theory
Category:Irrationality Proofs
Category:Logarithms
Irrationals are Everywhere Dense in Reals/Normed Vector Space

== Theorem ==
Let $\struct {\R, \size {\, \cdot \,} }$ be the normed vector space of real numbers.
Let $\R \setminus \Q$ be the set of irrational numbers.
Then $\R \setminus \Q$ is everywhere dense in $\struct {\R, \size {\, \cdot \,} }$
== Proof ==
Let $x \in \R$.
Let $\epsilon \in \R_{\mathop > 0}$
Either $x \in \Q$ or $x \in \R \setminus \Q$.
Suppose $x \in \R \setminus \Q$.
Let $y := x$.
Then:
:$\size {x - y} < \epsilon$
Suppose $x \in \Q$.
Let $n \in \N : n > \dfrac {\sqrt 2} \epsilon$
Let $y := x + \dfrac {\sqrt 2} n$
Then $y \in \R \setminus \Q$.
Furthermore:
In both cases $x$ was arbitrary.
Hence:
:$\forall x \in \R : \exists \epsilon \in \R_{\mathop > 0} : \exists y \in \R \setminus \Q : \size {x - y} < \epsilon$
By definition, $\R \setminus \Q$ is dense in $\R$.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Real Analysis
Category:Normed Vector Spaces
Category:Irrational Number Space
Category:Denseness
Irrationals are Everywhere Dense in Reals/Topology

== Theorem ==
Let $T = \struct {\R, \tau}$ denote the real number line with the usual (Euclidean) topology.
Let $\R \setminus \Q$ be the set of irrational numbers.
Then $\R \setminus \Q$ is everywhere dense in $T$.
== Proof ==
Let $x \in \R$.
Let $U \subseteq \R$ be an open set of $T$ such that $x \in U$.
From Basis for Euclidean Topology on Real Number Line, there exists an open interval $V_0 = \openint {x - \epsilon} {x + \epsilon} \subseteq U$ for some $\epsilon > 0$ such that $x \in V_0$.
From Between two Real Numbers exists Rational Number:
:$\exists p \in \Q: p \in \openint {x - \epsilon} {x + \epsilon}$
Thus, we can define the open interval $V_1 = \openint x p \subseteq V_0$.
Similarly:
:$\exists q \in \Q: q \in \openint x p$
We can then define an open interval $V_2 = \openint q p \subseteq V_1$.
We have $V_2 \subseteq V_1$, $V_1 \subseteq V_0$ and $V_0 \subseteq U$.
By successively applying Subset Relation is Transitive, it follows that $V_2 \subseteq U$.
Note that $x \notin V_2$, since $x < q < p < x + \epsilon$.
From Between two Rational Numbers exists Irrational Number, there exists $y \in \R \setminus \Q: y \in \openint q p = V_2$.
As $x \notin V_2$, it must be the case that $x \ne y$.
Since $V_2 \subseteq U$, $U$ is an open set of $T$ containing $x$ which also contains an element of $\R \setminus \Q$ other than $x$.
As $U$ is arbitrary, it follows that every open set of $T$ containing $x$ also contains an element of $\R \setminus \Q$ other than $x$.
That is, $x$ is by definition a limit point of $\R \setminus \Q$.
As $x$ is arbitrary, it follows that all elements of $\R$ are limit points of $\R \setminus \Q$.
The result follows from the definition of everywhere dense.
Category:Real Analysis
Category:Real Number Line with Euclidean Topology
Category:Irrational Number Space
Category:Denseness
Irreducible Component is Closed

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $Y$ be an irreducible component of $T$.
Then $Y$ is closed in $T$.
== Proof ==
By Closure of Irreducible Subspace is Irreducible, the closure $Y^-$ of $Y$ is irreducible.
By Set is Subset of its Topological Closure, $Y \subseteq Y^-$.
Because $Y$ is an irreducible component, we must have $Y = Y^-$.
By Set is Closed iff Equals Topological Closure, $Y$ is closed in $T$.
Category:Irreducible Spaces
Irreducible Components of Hausdorff Space are Points

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-empty Hausdorff space.
Then the irreducible components of $T$ are the singleton sets.
== Proof ==
By Subspace of Hausdorff Space is Hausdorff, the irreducible components of $T$ are also Hausdorff.
By Irreducible Hausdorff Space is Singleton, they can only be singletons.
By Trivial Topological Space is Irreducible, every singleton of $X$ is indeed irreducible.
Category:Irreducible Spaces
Irreducible Elements of 5th Cyclotomic Ring

== Theorem ==
Let $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ denote the $5$th cyclotomic ring.
The following elements of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$ are irreducible:
:$2$
:$3$
:$1 + i \sqrt 5$
:$1 - i \sqrt 5$
== Proof ==
Let $z = x + i y$ be an element of $\Z \sqbrk {i \sqrt 5}$ in the set $S$, where:
:$S := \set {2, 3, 1 + i \sqrt 5, 1 - i \sqrt 5}$
Let $z$ have a non-trivial factorization:
:$z = z_1 z_2$
where neither $z_1$ nor $z_2$ are units of $\Z \sqbrk {i \sqrt 5}$.
Let $\map N z$ denote the field norm of $z \in \Z \sqbrk {i \sqrt 5}$.
Then:
Then we have:
From Elements of 5th Cyclotomic Ring with Field Norm 1, the only elements of $\Z \sqbrk {i \sqrt 5}$ whose field norm is $1$ are the units of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$: $1$ and $-1$.
From 5th Cyclotomic Ring has no Elements with Field Norm of 2 or 3, none of $4$, $6$ and $9$ have proper divisors which are field norms of elements of $\Z \sqbrk {i \sqrt 5}$.
Thus either $z_1$ or $z_2$ is a unit of $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
So none of the elements of $S$ has a non-trivial factorization in $\struct {\Z \sqbrk {i \sqrt 5}, +, \times}$.
Hence the result, by definition of irreducible.
== Sources ==
* : Chapter $9$: Rings: Exercise $19 \ \text {(iii)}$
Category:Cyclotomic Rings
Irreducible Elements of Ring of Integers

== Theorem ==
Let $\struct {\Z, +, \times}$ be the ring of integers.
The irreducible elements of $\struct {\Z, +, \times}$ are the prime numbers and their negatives.
== Proof ==
We have that Integers form Integral Domain.
Therefore the concept of irreducible is defined.
Let $p$ be a prime number.
By definition, the only divisors of $p$ are $1, -1, p, -p$.
From Units of Ring of Integers, $1$ and $-1$ are (the only) units of $\Z$.
From Associates are Unit Multiples, $p$ and $-p$ are (the only) associates of each other.
Hence the result, from the definition of irreducible.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Example $57$
Category:Integers
Irreducible Hausdorff Space is Singleton

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-empty topological space which is irreducible and Hausdorff.
Then $S$ is a singleton.
== Proof ==
Suppose $S$ has exactly one element.
Then by definition $T = \struct {S, \tau}$ is the trivial topological space.
Hence, from Trivial Topological Space is Irreducible, $S$ is irreducible.
Suppose $S$ has at least two distinct elements:
:$x, y \in S, x \ne y$
By definition of irreducible space:
:there are no two disjoint open sets $T$ such that $x$ is in one and $y$ is in the other:
:$\nexists U_1, U_2 \in \tau: U_1 \cap U_2 = \O, x \in U_1, y \in U_2$
This contradicts the fact that $T$ is Hausdorff.
Thus $S$ has only one element.
Category:Irreducible Spaces
Category:Hausdorff Spaces
Irreducible Polynomial/Examples

== Examples of Irreducible Polynomials ==
=== $x^2 - 2$ in Ring of Polynomials over Reals ===
=== $x^2 - 2$ in Ring of Polynomials over Rationals ===
=== $X^2 + 1$ in Ring of Polynomials over Reals ===
=== $x^2 + 1$ in Ring of Polynomials over Real Numbers ===
=== $x^2 + 1$ in Ring of Polynomials over Complex Numbers ===
=== $8 x^3 - 6 x - 1$ in Ring of Polynomials over Rationals ===
=== $x^3 + x^2 - 2 x - 1$ in Ring of Polynomials over Rationals ===
Category:Examples of Irreducible Polynomials
Irreducible Polynomial/Examples/8 x^3 - 6 x - 1 in Rationals

== Examples of Irreducible Polynomials ==
Consider the polynomial:
:$\map P x = 8 x^3 - 6 x - 1$
over the ring of polynomials $\Q \sqbrk X$ over the rational numbers.
Then $\map P x$ is irreducible.
== Proof ==
 $\map P x$ has proper factors.
Then one of these has to be of degree $1$.
Thus from Factors of Polynomial with Integer Coefficients have Integer Coefficients we have:
:$8 x^3 - 6 x - 1 = \paren {a x + b} \paren {c^2 + d x + e}$
for some $a, b, c, d, e \in \Z$.
Hence:
The only possible degree $1$ factors with integer coefficients are:
:$x \pm 1, 2 x \pm 1, 4 x \pm 1, 8 x \pm 1$
By trying each of these possibilities, it is determined that no integer value of $d$ gives the correct values.
Hence the result.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 31$. Polynomials with Integer Coefficients: Example $59$
Category:Examples of Irreducible Polynomials
Irreducible Polynomial/Examples/x^2 + 1 in Complex Numbers

== Examples of Irreducible Polynomials ==
Consider the polynomial:
:$\map P x = x^2 + 1$
over the ring of polynomials $\C \sqbrk X$ over the complex numbers.
Then $\map P x$ is not irreducible, as:
:$x^2 + 1 \equiv \paren {x + i} \paren {x - i}$
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Example $58$
Category:Examples of Irreducible Polynomials
Irreducible Polynomial/Examples/x^2 + 1 in Real Numbers

== Examples of Irreducible Polynomials ==
Consider the polynomial:
:$\map P x = x^2 + 1$
over the ring of polynomials $\R \sqbrk X$ over the complex numbers.
Then $\map P x$ is irreducible, as its factors:
:$x^2 + 1 \equiv \paren {x + i} \paren {x - i}$
are complex.
== Sources ==
* 
* 
Category:Examples of Irreducible Polynomials
Irreducible Polynomial/Examples/x^2 - 2 in Rationals

== Examples of Irreducible Polynomials ==
Consider the polynomial:
:$\map P x = x^2 - 2$
over the ring of polynomials $\Q \sqbrk X$ over the rational numbers.
Then $\map P x$ is irreducible.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Example $58$
Category:Examples of Irreducible Polynomials
Irreducible Polynomial/Examples/x^2 - 2 in Reals

== Examples of Irreducible Polynomials ==
Consider the polynomial:
:$\map P x = x^2 - 2$
over the ring of polynomials $\R \sqbrk X$ over the real numbers.
Then $\map P x$ is not irreducible, as from Difference of Two Squares:
:$x^2 - 2 \equiv \paren {x + \sqrt 2} \paren {x - \sqrt 2} $
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 29$. Irreducible elements: Example $58$
Category:Examples of Irreducible Polynomials
Irreducible Representations of Abelian Group

== Theorem ==
Let $\struct {G, \cdot}$ be a finite abelian group.
Let $V$ be a non-null vector space over an algebraically closed field $K$.
Let $\rho: G \to \GL V$ be a linear representation.
Then $\rho$ is irreducible $\map {\dim_K} V = 1$, where, $\dim_K$ denotes dimension.
== Proof ==
=== Sufficient Condition ===
Suppose that $\map {\dim_K} V = 1$.
That $\rho$ is irreducible is shown on Representation of Degree One is Irreducible.
=== Necessary Condition ===
Suppose that $\rho$ is an irreducible linear representation.
Let $g \in G$ be arbitrary. Now, for all $h \in G$, have:
 
Now, combining Commutative Linear Transformation is G-Module Homomorphism and Corollary to Schur's Lemma (Representation Theory) yields that:
:$\exists \lambda_g \in K: \map \rho g = \lambda_g \operatorname{Id}_V$
That is, there is a $\lambda_g \in K$ such that $\map \rho g$ is the linear mapping of multiplying by $\lambda_g$.
Hence, $\forall v \in V: \map {\map \rho g} v = \lambda_g v$.
It follows that any vector subspace of $V$ of dimension $1$ is invariant. 
So, had $V$ any proper vector subspace of dimension $1$, $\rho$ would not be irreducible.
Since $V$ is non-null, it follows from Trivial Vector Space iff Zero Dimension that $\map {\dim_K} V > 0$.
Hence necessarily $\map {\dim_K} V = 1$.
Category:Representation Theory
Irreducible Space is Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is connected.
== Proof ==
Let $T = \struct {S, \tau}$ be irreducible.
Then:
:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
So trivially there are no two open sets that can form a separation of $T$.
The result follows from definition of connected.
== Also see ==
* Irreducible Component is Contained in Connected Component
* Space is Irreducible iff Open Sets are Connected
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Irreducible Spaces
Category:Connected Spaces
Irreducible Space is Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is locally connected.
== Proof ==
Let $T = \struct {S, \tau}$ be irreducible.
Then:
:$\forall U_1, U_2 \in \tau: U_1, U_2 \ne \O \implies U_1 \cap U_2 \ne \O$
So trivially there are no two open sets that can form a separation of $T$.
As a basis consists of open sets, this applies to all sets in a basis for $T$.
The result follows from definition of locally connected.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Irreducible Spaces
Category:Locally Connected Spaces
Irreducible Space is Pseudocompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is irreducible.
Then $T$ is pseudocompact.
== Proof ==
We have that Continuous Real-Valued Function on Irreducible Space is Constant.
A constant mapping is trivially bounded.
Hence the result by definition of pseudocompact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Pseudocompact Spaces
Category:Irreducible Spaces
Irreducible Space is not necessarily Path-Connected

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space which is irreducible.
Then $T$ is not necessarily path-connected.
== Proof ==
Let $T$ be a countable finite complement space.
From Finite Complement Space is Irreducible, $T$ is an irreducible space.
From Countable Finite Complement Space is not Path-Connected, $T$ is not path-connected.
Hence the result.
== Also see ==
* Finite Irreducible Space is Path-Connected
* Irreducible Space with Finitely Many Open Sets is Path-Connected
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Irreducible Spaces
Category:Path-Connected Spaces
Irreducible Space with Finitely Many Open Sets is Path-Connected

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a irreducible topological space.
Let its topology $\tau$ be finite.
Then $T$ is path-connected.
== Proof ==
Follows immediately from:
* Irreducible Space with Finitely Many Open Sets has Generic Point
* Topological Space with Generic Point is Path-Connected
== Also see ==
* Irreducible Space is not necessarily Path-Connected
Category:Irreducible Spaces
Category:Path-Connected Spaces
Irreducible Subspace is Contained in Irreducible Component

== Theorem ==
Let $T = \struct {S_1, \tau}$ be a non-empty topological space.
Let $S_2$ be an irreducible subset of $S_1$.
Then there exists an irreducible component $S_3$ of $T$ such that $S_2 \subseteq S_3$.
== Outline of Proof ==
We apply Zorn's Lemma to the set of irreducible subspaces, ordered by the subset relation.
== Proof ==
By definition, an irreducible component of $X$ is an irreducible subspace that is maximal among the irreducible subspaces, ordered by the subset relation.
Consider the ordered set $\struct {\AA, \subseteq}$, where:
:$\AA := \leftset {S : S}$ is an irreducible subset of $S_1$ such that $\rightset {S_2 \subseteq S}$
We need to show that $\AA$ has a maximal element.
As $S_2 \in \AA$, we have $\AA \ne \O$.
Thus, in view of Zorn's Lemma, it suffices to show:
:each non-empty chain in $\AA$ has an upper bound in $\AA$.
Let $C$ be a non-empty chain in $\AA$.
Let:
:$M := \bigcup C$
We now show that $M$ is an upper bound of $C$.
By construction of $M$, it suffices to show:
:$M \in \AA$
As $C \subseteq \AA$, we have:
:$S_2 \subseteq M$
So, if remains to show that $M$ is an irreducible subset of $S_1$.
Let $U, V$ are non-empty open sets of $\struct {M, \tau_M}$.
In view of the definition of irreducible space, we need to show:
:$U \cap V \ne \O$
By definition of $\tau_M$, there are $U_1, V_1 \in \tau$ such that:
:$U = U_1 \cap M$
and:
:$V = V_1 \cap M$
Since:
:$\bigcup _{N \in C} \paren {U_1 \cap N} = U_1 \cap M \ne \O$
there exists $N_U \in C$ such that:
:$U_1 \cap N_U \ne \O$
Similarly, there exists $N_V \in C$ such that:
:$V_1 \cap N_V \ne \O$
As $C$ is a chain, we have either:
:$N_U \subseteq N_V$ or $N_V \subseteq N_U$
, suppose:
:$N_U \subseteq N_V$
Let:
:$U' := U_1 \cap N_V$
and:
:$V' := V_1 \cap N_V$
Then $U', V'$ are non-empty open sets of $N_V$.
Since $N_V$ is irreducible, we have:
:$U' \cap V' \ne \O$
This implies:
:$U \cap V \ne \O$
== Also see ==
* Point is Contained in Irreducible Component
Category:Irreducible Spaces
Is 2 to the power of Euler's Number Rational?

== Open Question ==
It is not known whether $2$ to the power of Euler's number $e$:
:$2^e$
is rational or irrational.
== Progress ==
See Schanuel's Conjecture Implies Transcendence of 2 to the power of Euler's Number.
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Open Questions
Is Apéry's Constant Transcendental?

== Open Question ==
It is not known whether Apéry's constant $\map \zeta 3$ is transcendental or not.
== Also see ==
* Apéry's Theorem: $\map \zeta 3$ is irrational
== Sources ==
* : $1 \cdotp 202 \, 056 \ldots$
* : $1 \cdotp 20205 \, 6 \ldots$
* 
Category:Open Questions
Category:Apéry's Constant
Is Euler's Number Normal?

== Open Question ==
It is not known whether Euler's Number $e$ is normal.
== Sources ==
* : $0 \cdotp 12345 67891 01112 13141 51617 18192 02122 \ldots$
* : $0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
Category:Open Questions
Category:Normal Numbers
Is Euler-Mascheroni Constant Irrational?

== Open Question ==
It is not even known whether the Euler-Mascheroni constant $\gamma$ is irrational, let alone whether it is transcendental or not.
== Progress ==
In $1977$ it was established by that if $\gamma$ is a rational number expressible as the ratio of two integers $\dfrac a b$, it would be necessary for $b$ to be greater than $10^{10 \, 000}$.
By $1980$ that lower limit had been raised, according to in collaboration with , to $10^{15 \, 000}$.
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* 
* 
* : $0,57721 56649 \ldots$
* : $0 \cdotp 577 \, 215 \, 664 \, 901 \, 532 \, 860 \, 606 \, 512 \, 090 \, 082 \, 402 \, 431 \ldots$
* : $0 \cdotp 57721 \, 56649 \, 01532 \, 86060 \, 65120 \, 90082 \, 40243 \, 1 \ldots$
Category:Open Questions
Category:Euler-Mascheroni Constant
Is Pi Normal?

== Open Question ==
It is not known whether $\pi$ (pi) is normal.
== Historical Note ==
== Sources ==
* : $0 \cdotp 12345 67891 01112 13141 51617 18192 02122 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $0 \cdotp 12345 \, 67891 \, 01112 \, 13141 \, 51617 \, 18192 \, 02122 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Open Questions
Category:Normal Numbers
Is Pi Normal?/Historical Note

== Open Question ==
The question as to the normality or not of $\pi$ (pi) arose when investigated the $707$-digit approximation to $\pi$ that had been published by in $1853$.
 noticed that the digit $7$ occurred less frequently than the other digits: only $44$ times against the expected $61$.
He determined that the probability of this low frequency was $45$ to $1$ against.
However, as discovered by in around $1945$, had made a mistake in the $528$th digit. The correct calculation did not show that bias against $7$.
Since that time, all tests for normality that have been done on all known digits of $\pi$ have shown no suggestion that $\pi$ is not normal.
Work continues.
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Historical Notes
Is Pi multiplied by Euler's Number Rational?

== Open Question ==
It is not known whether the product of $\pi$ (pi) and Euler's number $e$:
:$\pi \times e$
is rational or irrational.
== Progress ==
By:
: Transcendence of Sum or Product of Transcendentals
: Euler's Number is Transcendental
: Pi is Transcendental
at least one of $\pi + e$ and $\pi e$ is transcendental.
Also, see Schanuel's Conjecture Implies Transcendence of Pi by Euler's Number.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $47$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Open Questions
Category:Pi
Category:Euler's Number
Is Pi plus Euler's Number Rational?

== Open Question ==
It is not known whether the sum of $\pi$ (pi) and Euler's number $e$:
:$\pi + e$
is rational or irrational.
== Progress ==
By:
: Transcendence of Sum or Product of Transcendentals
: Euler's Number is Transcendental
: Pi is Transcendental
at least one of $\pi + e$ and $\pi e$ is transcendental.
Also, see Schanuel's Conjecture Implies Transcendence of Pi plus Euler's Number.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $47$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Open Questions
Category:Pi
Category:Euler's Number