nyuuzyou/wikis huggingface dataset:
Connected Graph with only Even Vertices has no Bridge

== Theorem ==
Let $G$ be a connected graph whose vertices are all even.
Then no edge of $G$ is a bridge.
== Proof ==
Let the vertices of $G$ all be even.
Then by Characteristics of Eulerian Graph, $G$ is Eulerian.
By definition of Eulerian, $G$ therefore contains a Eulerian circuit.
Thus every edge of $G$ lies on a circuit of $G$.
From Condition for Edge to be Bridge, if an edge $e$ of $G$ is a bridge, then it does not lie on a circuit.
Hence no edge of $G$ is a bridge.
== Sources ==
* : $\S 2.4$: Cut-Vertices and Bridges: Problem $48$
Category:Graph Theory
Connected Open Subset of Euclidean Space is Path-Connected

== Theorem ==
Let $\R^n$ be a Euclidean $n$-space.
Let $U$ be a connected open subset of $\R^n$.
Then $U$ is path-connected.
== Proof ==
Let $a \in U$.
Let $H \subseteq U$ be the subset of points in $U$ which can be joined to $a$ by a path in $U$.
Let $K = U \setminus H$.
Let $x \in H$.
Then:
:$\exists \epsilon > 0: \map {B_\epsilon} x \subseteq U$
where $\map {B_\epsilon} x$ is the open $\epsilon$-ball of $x$.
Given any $y \in \map {B_\epsilon} x$, there is a (straight line) path $g$ in $\map {B_\epsilon} x \subseteq U$ connecting $x$ and $y$.
But since $x \in H$, there is a path $f$ in $U$ joining $a$ to $x$.
From Joining Paths makes Another Path, traversing $f$, and then $g$ forms a path from $a$ to $y$.
It follows that $y \in H$, and therefore $\map {B_\epsilon} x \subseteq H$.
Thus $H$ is open.
By a similar argument, $K$ is also shown to be open:
If $x \in K$, then $\map {B_\epsilon} x \subseteq U$ for some $\epsilon > 0$.
If any point in $\map {B_\epsilon} x$ can be joined to $a$ by a path in $U$, then so could $x$.
It is clear that $H \cap K = \O$ and $H \cup K = U$ by definition of set difference.
As, trivially, $a \in H$, we have $H \ne \O$.
Knowing that $U$ is connected, it follows that $K = \O$, and $H = U$.
Hence the result.
== Sources ==
* : $6.4$: Comparison of Definitions: Proposition $6.4.2$
Category:Connected Spaces
Category:Path-Connected Spaces
Category:Euclidean Space
Connected Riemannian Manifold with Restricted Exponential Map defined on Whole Tangent Space admits Minimizing Geodesic Segment

== Theorem ==
Let $\struct {M, g}$ be a connected Riemannian manifold.
Let $T_p M$ be the tangent space at $p \in M$.
Let $\exp_p$ be the restricted exponential map defined on the whole $T_p M$.
Then for all $q \in M$ there is a minimizing geodesic segment from $p$ to $q$.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Connected Riemannian Manifold with Restricted Exponential Map defined on Whole Tangent Space is Metrically Complete

== Theorem ==
Let $\struct {M, g}$ be a connected Riemannian manifold.
Let $T_p M$ be the tangent space at $p \in M$.
Let $\exp_p$ be the restricted exponential map defined on the whole $T_p M$.
Then $M$ is metrically complete.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Connected Riemannian Manifolds with Local Isometry

== Theorem ==
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be connected Riemannian manifolds.
Let $\pi : \tilde M \to M$ be a local isometry.
Then $M$ is complete, and $\pi$ is a Riemannian covering map.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Connected Riemannian Manifolds with Local Isometry/Corollary

== Theorem ==
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be connected Riemannian manifolds.
Let $\pi : \tilde M \to M$ be a Riemannian covering map.
Then $M$ is complete $\tilde M$ complete.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Completeness
Category:Riemannian Manifolds
Connected Set in Subspace

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq B \subseteq S$.
Let $T_B = \struct {B, \tau_B}$ be the topological space where $\tau_B$ is the subspace topology on $B$.
Then
:$A$ is connected in $T_B$ $A$ is connected in $T$.
== Proof ==
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.
Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$.
By the definition of a connected set, $A$ is connected in $T$ $\struct {A, \tau_A}$ is a connected topological space.
Similarly, $A$ is connected in $T_B$ $\struct {A, \tau'_A}$ is a connected topological space.
From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$.
Hence $A$ is connected in $T_B$ $A$ is connected in $T$.
Category:Connected Spaces
Connected Space is Connected Between Two Points

== Theorem ==
Let $T$ be a topological space which is connected.
Then $T$ is connected between two points.
== Proof ==
By definition of connected space, $T$ admits no separation.
Therefore, vacuously, every partition has one open containing $t_1, t_2 \in T$, for all $t_1, t_2 \in T$.
That is, for all $t_1, t_2 \in T$, $T$ is connected between $t_1$ and $t_2$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Connectedness Between Two Points
Category:Connected Spaces
Connected Space is not necessarily Locally Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is connected.
Then it is not necessarily the case that $T$ is also a locally connected space.
== Proof ==
Let $C$ be the closed topologist's sine curve embedded in the real Euclidean plane.
From Closed Topologist's Sine Curve is Connected, $C$ is connected in $T$
From Closed Topologist's Sine Curve is not Locally Connected, $C$ is not locally connected.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Locally Connected Spaces
Category:Connected Spaces
Connected Subgraph of Tree is Tree

== Theorem ==
Let $T$ be a tree.
Let $S$ be a subgraph of $T$ such that $S$ is connected.
Then $S$ is also a tree.
== Proof ==
Follows directly from the fact that by definition, $T$ has no circuits.
As $T$ has no circuits, then nor can $S$ have.
Hence $S$ is a connected simple graph with no circuits.
Thus by definition, $S$ is a tree.
Category:Tree Theory
Connected Subset of Union of Disjoint Open Sets

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $A$ be a connected set of $T$.
Let $U, V$ be disjoint open sets.
Let $A \subseteq U \cup V$.
Then
:either $A \subseteq U$ or $A \subseteq V$.
== Proof ==
Let $U' = A \cap U$ and $V' = A \cap V$.
By definition $U'$ and $V'$ are open sets in the subspace $\struct{A, \tau_A}$. 
From Intersection is Empty Implies Intersection of Subsets is Empty $U'$ and $V'$ are disjoint.
Hence $U'$ and $V'$ are separated sets by definition.
Now
Since $A$ is connected then one of $U'$ or $V'$ is empty.
 assume that $V' = \O$.
Then
== Sources ==
* : $\S 4$ Connectedness and Compactness, Proposition $4.9 \ \text {(a)}$
Category:Connected Spaces
Connected Subspace Lie in One Component of Separation

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A_1 \mid A_2$ be a separation of $T$.
Let $X$ be a connected set of $T$.
Then there exist $i, j \in \set { 1, 2 }$ with $i \ne j$ such that $X \subseteq A_i$, and $X \cap A_j = \O$.
== Proof ==
By definition of separation, $A_1$ and $A_2$ are disjoint.
Define $B_i = X \cap A_i$ for $i \in \set {1, 2}$.
From Intersection is Subset, $B_i \subseteq A_i$ for $i \in \set {1, 2}$.
From Subsets of Disjoint Sets are Disjoint, $B_1$ and $B_2$ are disjoint.
The union of $B_1$ and $B_2$ is:
By definition of separation, $A_1$ and $A_2$ are open in $T$.
Let $S_X = \struct {X, \tau_X}$ be the topological subspace of $T$.
Then $B_1$ and $B_2$ are open in $S_X$.
If both $B_1$ and $B_2$ are non-empty, they would form a separation of $S_X$, which contradicts the assumption that $X$ is connected.
Hence either $B_1 = \O$, or $B_2 = \O$.
== Sources ==
* : $3$: Connectedness and Compactness: $\S 23$: Connected Spaces
Category:Connected Sets
Category:Separations
Connected Subspace of Linearly Ordered Space

== Theorem ==
Let $\struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq S$.
Then $Y$ is connected in $\struct {S, \tau}$ both of the following hold:
:$(1): \quad Y$ is convex in $S$
:$(2): \quad \struct {Y, \preceq \restriction_Y}$ is a linear continuum, where $\restriction$ denotes restriction.
== Proof ==
=== Necessary Conditions ===
Let $Y$ be connected in $\struct {S, \tau}$.
 $Y$ is not convex in $S$.
Then there exist $a, b, c \in S$ such that:
:$a \prec b \prec c$
:$a, c \in Y$ but $b \notin Y$
Recall that:
:$b^\prec$ denotes the (strict) lower closure of $b$: $b^\prec = \set {u \in S: u \prec b}$
:$b^\succ$ denotes the (strict) upper closure of $b$: $b^\succ = \set {u \in S: b \prec u}$
We have that $Y \cap b^\prec$ and $Y \cap b^\succ$ are separated in $Y$.
That is, by definition, $Y$ is disconnected.
Hence by Proof by Contradiction it follows that $Y$ is convex in $S$.
 $Y$ is convex in $S$, but not a linear continuum.
Then by Order Topology on Convex Subset is Subspace Topology, the subspace topology on $Y$ is the same as the order topology on $Y$.
Thus by Linearly Ordered Space is Connected iff Linear Continuum, $Y$ is disconnected.
Hence by Proof by Contradiction it follows that $Y$ is a linear continuum.
=== Sufficient Condition ===
Suppose that $Y$ is convex in $S$ and a linear continuum.
Then the result follows from:
:Order Topology on Convex Subset is Subspace Topology
:Linearly Ordered Space is Connected iff Linear Continuum.
== Also see ==
* Subset of Real Numbers is Interval iff Connected
* Compact Subspace of Linearly Ordered Space
Category:Linearly Ordered Spaces
Category:Connected Spaces
Connected Vertices are Connected by Path

== Theorem ==
Let $G = \struct {V, E}$ be a simple graph.
Let $x, y \in V$.
Let there exist a walk $w: \N_n \to V$ from $x$ to $y$.
Then there exists a subsequence $z_n$ of $w$ which is a path from $x$ to $y$.
== Proof ==
We represent a walk as a sequence of vertices.
However, the same argument will work for the representation as an alternating sequence of vertices and edges.
The proof proceeds by induction on the length of $w$.
If the length of $w$ is $0$, then it is trivially a path.
Suppose that every walk of length less than $n$ has a subsequence forming a path between its endpoints.
Let $w$ have length $n$.
If $w$ is not a path, then it must have a cycle.
That is, for some $j, k \in \Dom w$, $j < k$ and $w_j = w_k$.
Then removing $w_{j + 1}, \dots, w_k$ from $w$ will form a new walk $w'$, from $x$ to $y$ which is strictly shorter than $w$.
By the inductive hypothesis, $w'$ has as subsequence which is a path from $x$ to $y$.
Category:Graph Theory
Connected and Locally Path-Connected Implies Path Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a connected and locally path-connected topological space.
Then $T$ is path-connected.
== Proof ==
By:
:Path Component of Locally Path-Connected Space is Closed
:Path Component of Locally Path-Connected Space is Open
the path components of $T$ are clopen.
Because $T$ is connected, every path component equals $S$.
That is, $T$ is path-connected.
Category:Connected Spaces
Category:Path-Connected Spaces
Category:Locally Path-Connected Spaces
Connected iff no Proper Clopen Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is connected
:
:there exists no non-empty proper subset of $S$ which is clopen in $T$.
== Proof ==
=== Sufficient Condition ===
Let $T$ be connected.
 there exists $H \subset S$ such that:
:$H$ is clopen in $T$
:$H$ is a non-empty proper subset of $S$, that is:
::$\O \ne H \ne S$
Then $H$ and $\relcomp S H$ are open sets whose union is $S$.
Thus $\set {H \mid \relcomp S H}$ form a partition of $S$.
Hence by definition, $T$ is not connected.
This contradicts our assumption.
Hence, by Proof by Contradiction, such $H$ does not exist.
=== Necessary Condition ===
Let there exist no non-empty proper subset of $S$ which is clopen in $T$.
 $T$ is not connected (disconnected).
By definition, there is a partition $\set {A \mid B}$ of $S$.
Then $\relcomp S A = B$ is open.
Hence by definition $A$ is closed.
Thus $A$ is clopen such that:
:$\O \ne A \ne S$
But then by definition $A$ is a non-empty proper subset of $S$ which is clopen in $T$.
This contradicts our assumption.
Hence, by Proof by Contradiction, $T$ is connected.
Category:Connected Spaces
Connectedness Between Two Points is an Equivalence Relation

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff \exists S \subseteq T: S$ is connected between the two points $a$ and $b$
where $a, b \in X$.
Then $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
Trivially, any separation of $T$ is a partition of $S$.
So any $a \in S$ will be in exactly one of the open sets forming the partition.
So $a \sim a$ and $\sim$ is seen to be reflexive.
=== Symmetry ===
If $a \sim b$ then each separation of $T$ includes a single open set $U \ne \O$ which contains both $a$ and $b$.
Trivially it follows that $b \sim a$ and so $\sim$ is seen to be symmetric.
=== Transitivity ===
Let $a \sim b$ and $b \sim c$.
Then by definition:
:Each separation of $T$ includes a single open set $U_1 \ne \O$ which contains both $a$ and $b$.
:Each separation of $T$ includes a single open set $U_2 \ne \O$ which contains both $b$ and $c$.
So $b \in U_1$ and $b \in U_2$ and so $U_1 \cap U_2 \ne \O$.
So by the definition of partition it follows that $U_1 = U_2$ and so $a \sim c$.
So $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Examples of Equivalence Relations
Category:Connectedness Between Two Points
Connectedness of Points is Equivalence Relation

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $a \sim b $ denote the relation:
:$a \sim b \iff a$ is connected to $b$
where $a, b \in S$.
Then $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
We have that $\set a \subseteq S$ is a (degenerate) connected set of $S$.
So $a$ is in the same connected set as itself and so $a \sim a$.
So $\sim$ has been shown to be reflexive.
=== Symmetry ===
If $x \sim y$ then $x$ is in the same connected set as $y$ by definition.
Trivially it follows that $y$ is in the same connected set as $x$ and so $y \sim x$.
So $\sim$ has been shown to be symmetric.
=== Transitivity ===
Let $x \sim y$ and $y \sim z$.
Then by definition:
:$x$ is in the same connected set as $y$
:$y$ is in the same connected set as $z$
By Union of Connected Sets with Non-Empty Intersections is Connected it follows that $x$ is in the same connected set as $z$.
Thus so $x \sim z$.
So $\sim$ has been shown to be transitive.
$\sim$ has been shown to be reflexive, symmetric and transitive.
Hence by definition it is an equivalence relation.
== Sources ==
* : $6.5$: Components
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Examples of Equivalence Relations
Category:Connected Spaces
Connecting Homomorphism is Functorial

== Theorem ==
Let $A$ be a ring with unity.
Let:
:$\begin{xy}\xymatrix{
&&& M_1 \ar@{->}[rr]
 \ar@{->}[dl]^{f_1}
 \ar@{->}[dd]^{\phi_1}|!{[d];[d]}\hole
&& M_2 %
 \ar@{->}[rr]
 \ar@{->}[dl]^{f_2}
 \ar@{->}[dd]^{\phi_2}|!{[d];[d]}\hole
&& M_3 \ar@{->}[dl]^{f_3}
 \ar@{->}[dd]^{\phi_3}|!{[d];[d]}\hole
 \ar@{->}[rr]
&& 0
\\
&& M_1' \ar@{->}[rr]
 \ar@{->}[dd]^{\phi_1'}
&& M_2' \ar@{->}[rr]
 \ar@{->}[dd]^{\phi_2'}
&& M_3' \ar@{->}[dd]^{\phi_3'}
 \ar@{->}[rr]
&& 0
\\
& 0 \ar@{->}[rr]|!{[r];[r]}\hole
&& N_1 \ar@{->}[rr]|!{[r];[r]}\hole
 \ar@{->}[dl]_{g_1}
&& N_2 \ar@{->}[rr]|!{[r];[r]}\hole
 \ar@{->}[dl]_{g_2}
&& N_3 \ar@{->}[dl]_{g_3}
\\
%
0 \ar@{->}[rr]
&&
N_1' \ar@{->}^(.65){e’}[rr]
&&
N_2' \ar@{->}[rr]
&&
N_3'
}\end{xy}$
be a commutative diagram of $A$-modules.
Suppose that the rows are exact.
Let 
:$\delta : \map \ker {\phi_3} \to \map {\operatorname {coker} } {\phi_1}$ 
:$\delta' : \map \ker {\phi_3'} \to \map {\operatorname {coker} } {\phi_1'}$ 
be the boundary homomorphisms coming from Snake Lemma applied to the front diagram and the back diagram.
Then the diagram:
:$\begin{xy}\xymatrix{
\map \ker {\phi_3} \ar[r]^{\delta} \ar[d] & \map {\operatorname{coker} } {\phi_1} \ar[d] \\
\map \ker {\phi_3'} \ar[r]^{\delta'} & \map {\operatorname{coker} } {\phi_1'} \\
}
\end{xy}$
is commutative.
Here the vertical arrows in the commutative square are induced by $\tuple {f_3, g_3}$ and $\tuple {f_1, g_1}$.
== Proof ==
Category:Homological Algebra
Consecutive Fibonacci Numbers are Coprime

== Theorem ==
Let $F_k$ be the $k$th Fibonacci number.
Then:
:$\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$
where $\gcd \set {a, b}$ denotes the greatest common divisor of $a$ and $b$.
That is, a Fibonacci number and the one next to it are coprime.
== Proof ==
From the definition of Fibonacci numbers:
:$F_1 = 1, F_2 = 1, F_3 = 2$
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\gcd \set {F_n, F_{n + 1} } = 1$
=== Basis for the Induction ===
$\map P 2$ is the case:
:$\gcd \set {F_2, F_3} = \gcd \set {1, 2} = 1$
Thus $\map P 2$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\gcd \set {F_k, F_{k + 1} } = 1$
Then we need to show:
:$\gcd \set {F_{k + 1}, F_{k + 2} } = 1$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$
== Sources ==
* : $5$
* : $5$
* 
* 
Category:Fibonacci Numbers
Category:Coprime Integers
Category:Proofs by Induction
Consecutive Integers are Coprime

== Theorem ==
$\forall h \in \Z$, $h$ and $h + 1$ have only two common factors: $1$ and $-1$.
That is, consecutive integers are always coprime.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Consecutive Integers are Coprime
Category:Coprime Integers
Consecutive Integers are Coprime/Proof 2

== Theorem ==
== Proof ==
Let $k \in \Z: k \divides h$.
Also assume $k \divides \paren {h + 1}$.
Thus:
:$\exists a, b \in \N: a k = h, b k = \paren {h + 1}$
Then:
:$\paren {h + 1} - h = b k - a k$
and so:
:$1 = \paren {b - a} k$
Since the integers form an integral domain, $\paren {b - a} \in \Z$.
Thus either $k = 1$ and $b - a = 1$, or $k = -1$ and $b - a = -1$.
Therefore, only $1$ and $-1$ can be factors of both $h$ and $\paren {h + 1}$.
Category:Consecutive Integers are Coprime
Consecutive Integers are Coprime/Proof 3

== Theorem ==
== Proof ==
A direct application of GCD of Integer with Integer + $n$:
:$\gcd \set {a, a + n} \divides n$
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Problems $2.2$: $12$
Category:Consecutive Integers are Coprime
Consecutive Integers which are Powers of 2 or 3

== Theorem ==
The only pairs of consecutive positive integers which are powers of $2$ or $3$ are:
:$\tuple {1, 2}$, $\tuple {2, 3}$, $\tuple {3, 4}$, $\tuple {8, 9}$
== Proof ==
Let $a$ and $b$ be two arbitrary consecutive positive integers.
=== Both Integers Powers of $2$ ===
The only powers of $2$ that differ by $1$ are $2^0$ and $2^1$, which gives us the pair $\tuple {1, 2}$.
=== Both Integers Powers of $3$ ===
The powers of $3$ are:
:$1, 3, 9, \ldots$
and so trivially there are no two consecutive positive integers which are powers of $3$.
=== The Remaining Case ===
There is no need to consider the case where $a$ or $b$ is $1$, as that has already been investigated.
So, let $a = 2^m$ and $b = 3^n$ where both $m$ and $n$ are greater than $0$.
From Powers of 3 Modulo 8, $b \equiv 1 \pmod 8$ or $b \equiv 3 \pmod 8$.
For $m \ge 3$ we have that:
:$2^m \equiv 0 \pmod 8$
while:
:$2^1 \equiv 2 \pmod 8$
:$2^2 \equiv 4 \pmod 8$
Let $a = b + 1$.
Then we have the cases:
:$b = 1, a = 2$
which has already been covered, and:
:$b = 3, a = 4$
which gives us the pairs $\tuple {3, 4}$.
Otherwise, for $a \ge 8$, we have:
and we have demonstrated that there are no such $b$, as $b \equiv 1 \pmod 8$ or $b \equiv 3 \pmod 8$
Let $b = a + 1$.
Let $b = 3^n$ where $n$ is odd.
Then $b \equiv 3 \pmod 8$
which means:
:$b \equiv 2 \pmod 8$
leading to the pair $\tuple {2, 3}$.
No further pairs are possible for $b = 3^n$ with $n$ odd.
Let $b = 3^n$ where $n$ is even.
Thus:
The only such $p$ and $q$ are $2^p = 2$ and $2^q = 4$
This means that the only possible pair fulfilling this condition is $\tuple {8, 9}$.
Thus all such possible pairs have been found.
== Historical Note ==
== Sources ==
* 
Category:Powers of 2
Category:Powers of 3
Consecutive Integers whose Product is Primorial

== Theorem ==
The following primorials can be expressed as the product of consecutive integers:
:$2, 6, 30, 210, 510 \, 510$
No others are known.
The corresponding indices of those primorials are:
:$2, 3, 5, 7, 17$
The corresponding values of $n$ such that $p\# = \paren {n - 1} n$ are:
:$2, 3, 6, 15, 715$
== Proof ==
== Sources ==
* 
* : $714$
* : $714$
Category:Primorials
Category:Recreational Mathematics
Consecutive Integers whose Product is Primorial/Mistake/First Edition

== Source Work ==
:
:The Dictionary
::$714$
== Mistake ==
:They discovered on computer that only primorial $1$, $2$, $3$, $4$ and $7$ can be represented as the product of consecutive numbers, up to primorial $3049$.
== Correction ==
There are three problems here:
:$(1): \quad 4$ is wrong -- it should be $5$. (This has been corrected in of $1997$.)
:$(2): \quad$ The primorial of $17$, which the section discusses, is omitted from this sentence. It perhaps ought to start:
::::Apart from primorial $17$, ...
:$(3): \quad$ The primorial of $1$ is generally accepted as being $1$. There are no two consecutive numbers whose product is $1$: $0 \times 1 = 0$, $1 \times 2 = 2$.
== Sources ==
* : $714$
Category:Mistakes/Curious and Interesting Numbers
Consecutive Integers whose Product is Primorial/Mistake/Second Edition

== Source Work ==
:
:The Dictionary
::$714$
== Mistake ==
:They discovered on computer that only primorial $1$, $2$, $3$, $5$ and $7$ can be represented as the product of consecutive numbers, up to primorial $3049$.
== Correction ==
There are two problems here:
:$(1): \quad$ The primorial of $17$, which the section discusses, is omitted from this sentence. It perhaps ought to start:
::::Apart from primorial $17$, ...
:$(2): \quad$ The primorial of $1$ is generally accepted as being $1$. There are no two consecutive numbers whose product is $1$: $0 \times 1 = 0$, $1 \times 2 = 2$.
== Sources ==
* : $714$
Category:Mistakes/Curious and Interesting Numbers