nyuuzyou/wikis huggingface dataset:
One-Sided Limit of Real Function/Examples/Reciprocal of 1 + e to the Reciprocal of x

== Examples of One-Sided Limits of Real Functions ==
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = \dfrac 1 {1 + e^{1 / x} }$
Then:
== Proof ==
:
We have:
:$\dfrac 1 x \to +\infty$ as $x \to 0^+$
and so:
:$1 + e^{1/x} \to +\infty$ as $x \to 0^+$
Thus:
:$\dfrac 1 {1 + e^{1 / x} } \to 0$ as $x \to 0^+$
and remains positive.
Then we have:
:$\dfrac 1 x \to -\infty$ as $x \to 0^-$
and so:
:$-e^{1/x} \to 0$ as $x \to 0^-$
Thus:
:$\dfrac 1 {1 + e^{1 / x} } \to 1$ as $x \to 0^-$
and remains less than $1$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity: Example $\text C$
Category:Examples of Limits of Real Functions
One-Sided Limit of Real Function/Examples/e^-1 over x at 0 from Left

== Example of One-Sided Limit of Real Functions ==
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = e^{-1 / x}$
Then:
== Proof ==
:
By definition of the limit from the left:
:$\ds \lim_{x \mathop \to a^-} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: a - \delta < x < a \implies \size {\map f x - L} < \epsilon$
In this case we are interested in the situation where $a = 0$, and we wish to demonstrate that $L \to +\infty$ at that point.
Hence the condition we need to ascertain is:
:$\lnot \paren {\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: -\delta < x < 0 \implies \size {e^{-1 / x} } < \epsilon}$
That is:
:$\exists \epsilon \in \R_{>0}: \nexists \delta \in \R_{>0}: \forall x \in \Bbb I: -\delta < x < 0 \implies \size {e^{-1 / x} } < \epsilon$
or:
:$\exists \epsilon \in \R_{>0}: \forall \delta \in \R_{>0}: \forall x \in \Bbb I: -\delta < x < 0 \implies \size {e^{-1 / x} } > \epsilon$
Let $\epsilon \in \R_{>0}$ be chosen arbitrarily such that $\epsilon < 1$.
Let $x < 0$.
Let $z = -x$.
Thus:
:$z > 0$
Then we have:
So, having been given an arbitrary $\epsilon \in \R_{>0}$, let $\delta = \dfrac 1 {\ln \epsilon}$.
Then:
:$-\delta < -x < 0 \implies \size {e^{-1 / x} } > \epsilon$
Hence by definition of limit from the left:
:$\ds \lim_{x \mathop \to 0^-} e^{-1 / x} = +\infty$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $1 \, \text {(c)}$
Category:Examples of Limits of Real Functions
One-Sided Limit of Real Function/Examples/e^-1 over x at 0 from Right

== Example of One-Sided Limit of Real Functions ==
Let $f: \R \to \R$ be the real function defined as:
:$\map f x = e^{-1 / x}$
Then:
== Proof ==
:
By definition of the limit from the right:
:$\ds \lim_{x \mathop \to a^+} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: a < x < a + \delta \implies \size {\map f x - L} < \epsilon$
In this case we are interested in the situation where $a = 0$, and we wish to demonstrate that $L = 0$ at that point.
Hence the condition we need to ascertain is:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \Bbb I: 0 < x < \delta \implies \size {e^{-1 / x} } < \epsilon$
Let $\epsilon \in \R_{>0}$ be chosen arbitrarily.
Let $x > 0$.
We have that:
:$e^{-1 / x} > 0$
and so:
:$\size {e^{-1 / x} } = e^{-1 / x}$
Then we have:
So, having been given an arbitrary $\epsilon \in \R_{>0}$, let $\delta = \dfrac 1 {\map \ln {1 / \epsilon} }$.
Then:
:$0 < x < \delta \implies \size {e^{-1 / x} } < \epsilon$
Hence by definition of limit from the right:
:$\ds \lim_{x \mathop \to 0^+} e^{-1 / x} = 0$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $1 \, \text {(b)}$
Category:Examples of Limits of Real Functions
One-Step Subgroup Test

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subset of $G$.
Then $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$ :
:$(1): \quad H \ne \O$, that is, $H$ is non-empty
:$(2): \quad \forall a, b \in H: a \circ b^{-1} \in H$.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Let $H$ be a subset of $G$ that fulfils the conditions given.
It is noted that the fact that $H$ is non-empty is one of the conditions.
It is also noted that the group operation of $\struct {H, \circ}$ is the same as that for $\struct {G, \circ}$, that is, $\circ$.
So it remains to show that $\struct {H, \circ}$ is a group.
We check the four group axioms:
==== ====
From Restriction of Associative Operation is Associative, associativity is inherited by $\struct {H, \circ}$ from $\struct {G, \circ}$.
==== ====
Let $e$ be the identity of $\struct {G, \circ}$.
Since $H$ is non-empty, $\exists x \in H$.
If we take $a = x$ and $b = x$, then $a \circ b^{-1} = x \circ x^{-1} = e \in H$, where $e$ is the identity element.
==== ====
If we take $a = e$ and $b = x$, then $a \circ b^{-1} = e \circ x^{-1} = x^{-1} \in H$.
Thus every element of $H$ has an inverse also in $H$.
==== ====
Let $x, y \in H$.
Then $y^{-1} \in H$, so we may take $a = x$ and $b = y^{-1}$.
So:
:$a \circ b^{-1} = x \circ \paren {y^{-1} }^{-1} = x \circ y \in H$
Thus, $H$ is closed.
Therefore, $\struct {H, \circ}$ satisfies all the group axioms, and is therefore a group.
Therefore $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.
=== Necessary Condition ===
Now suppose $\struct {H, \circ}$ is a subgroup of $\struct {G, \circ}$.
:$(1): \quad H \le G \implies H \ne \O$ from the fact that $H$ is a group and therefore can not be empty.
:$(2): \quad$ As $\struct {H, \circ}$ is a group, it is closed and every element has an inverse. So it follows that $\forall a, b \in H: a \circ b^{-1} \in H$.
== Also see ==
* One-Step Subgroup Test using Subset Product
* Two-Step Subgroup Test
== Linguistic Note ==
== Sources ==
* : $\S 5.2$. Subgroups
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.4: \ 1^\circ - 3^\circ$
* : $\S 1.9$: Subgroups: Lemma $8$
* : Chapter $2$: Subgroups and Cosets: $\S 35 \alpha$
* : $\S 1.2$
* : $0$: Some Conventions and some Basic Facts
* : $\S 36.4$: Subgroups
* : $\S 3.2$: Groups; the axioms
* : Chapter $4$: Subgroups: Proposition $4.2$
Category:Named Theorems
Category:Subgroups
One-Step Subgroup Test/Linguistic Note

== Linguistic Note on One-Step Subgroup Test ==
The One-Step Subgroup Test is so called despite the fact that, on the face of it, there are two steps to the test.
This is because the fact that the subset must be non-empty is frequently assumed as one of the "givens", and is then not specifically included as one of the tests to be made.
Category:Linguistic Notes
One-Step Subgroup Test using Subset Product

== Theorem ==
Let $G$ be a group.
Let $\O \subset H \subseteq G$ be a non-empty subset of $G$.
Then $H$ is a subgroup of $G$ :
:$H H^{-1} \subseteq H$
where:
:$H^{-1}$ is the inverse of $H$
:$H H ^{-1}$ is the product of $H$ with $H^{-1}$.
== Proof ==
This is a reformulation of the One-Step Subgroup Test in terms of subset product.
=== Necessary Condition ===
Let $H$ be a subgroup of $G$.
Let $x, y \in H$.
Then by the definition of subset product:
:$x y^{-1} \in H H^{-1}$
As $H \le G$, from the One-Step Subgroup Test, $x y^{-1} \in H$.
Thus $H H^{-1} \subseteq H$.
=== Sufficient Condition ===
Let:
:$H H^{-1} \subseteq H$
From the definition of subset product:
:$\forall x, y \in H: x y^{-1} \in H$
So by the One-Step Subgroup Test, $H$ is a subgroup of $G$.
== Also presented as ==
This result can also be presented as:
$H$ is a subgroup of $G$ :
:$H^{-1} H \subseteq H$
and the same argument applies.
== Also see ==
* Two-Step Subgroup Test using Subset Product
== Linguistic Note ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Theorem $9.1: \ 3^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets: Exercise $9.10 \ \text {(b)}$
* : Chapter $\text{II}$: Groups: Theorem $2$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $6 \ \text{(iii)}$
* : $\S 3.2$: Groups; the axioms
Category:Subgroups
Category:Subset Products
One-Step Vector Subspace Test

== Theorem ==
Let $V$ be a vector space over a division ring $K$ whose unity is $1_K$.
Let $U \subseteq V$ be a non-empty subset of $V$ such that:
:$\forall u, v \in U: \forall \lambda \in K: u + \lambda v \in U$
Then $U$ is a subspace of $V$.
== Proof ==
We need to verify the vector space axioms for $U$.
We start with observing that the properties for a unitary module:
:
:
:
:
are true for all elements of $V$.
Hence, since $U \subseteq V$, they hold for all elements of $U$ as well.
The same holds for the axioms:
:
:
From Vector Inverse is Negative Vector, we have for all $u \in U$:
:$u + \paren {-1_K} u = 0_V$
which is an element of $U$.
Since $U$ is non-empty, this means $0_V \in U$.
Hence it is seen that axioms:
:
:
are satisfied.
The last axiom that remains is .
From and :
:$\forall u, v \in U: u + v = u + 1_K v \in U$
Having verified all the vector space axioms, we conclude that $U$ is a subspace of $V$.
== Also see ==
* Vector Subspace of Real Vector Space
* Two-Step Vector Subspace Test
Category:Linear Algebra
Category:Vector Subspaces
One-Time Pad is Only Known Unbreakable Encryption Method

== Open Question ==
The only known method of encryption which is (technically) unbreakable is the one-time pad.
It is an open question as to whether there exist any others.
== Discussion ==
It is known that One-Time Pad is Unbreakable.
However, it has not been proved that there do not exist other methods of encryption that are also (technically) unbreakable.
We just have not actually found any yet.
== Sources ==
* 
Category:One-Time Pads
Category:Open Questions
One-Way Classification/Examples/Arbitrary Example 1

== Example of Use of One-Way Classification ==
This is an example of the use of one-way classification: 
$3$ treatments $A$, $B$ and $C$ may be allocated to $13$ experimental units.
:$5$ of those experimental units are chosen at random to receive treatment $A$
:$4$ of those experimental units are chosen at random to receive treatment $B$
:$4$ of those experimental units are chosen at random to receive treatment $C$.
== Sources ==
* 
* 
Category:Examples of One-Way Classification
One-to-Many Image of Set Difference

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation.
Let $A$ and $B$ be subsets of $S$.
Then:
:$(1): \quad \RR \sqbrk A \setminus \RR \sqbrk B = \RR \sqbrk {A \setminus B}$
 $\RR$ is one-to-many.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
=== Sufficient Condition ===
First, to show that $(1)$ holds if $\RR$ is one-to-many.
From Image of Set Difference under Relation, we already have:
:$\RR \sqbrk A \setminus \RR \sqbrk B \subseteq \RR \sqbrk {A \setminus B}$
So we just need to show:
:$\RR \sqbrk {A \setminus B} \subseteq \RR \sqbrk A \setminus \RR \sqbrk B$
Let $t \notin \RR \sqbrk A \setminus \RR \sqbrk B$.
Then by De Morgan's Laws:
:$t \notin \RR \sqbrk A \lor t \in \RR \sqbrk B$
Suppose $t \notin \RR \sqbrk A$.
Then by definition of a relation:
:$\neg \exists s \in A: \tuple {s, t} \in \RR$
By Image of Subset is Subset of Image:
:$\RR \sqbrk {A \setminus B} \subseteq \RR \sqbrk A$
Thus, by definition of subset and Rule of Transposition:
:$t \notin \RR \sqbrk A \implies t \notin \RR \sqbrk {A \setminus B}$
Now suppose $t \in \RR \sqbrk B$.
Then:
:$\exists s \in B: \tuple {s, t} \in \RR$
Because $\RR$ is one-to-many:
:$\forall x \in S: \tuple {x, t} \in \RR \implies x = s$
and thus:
:$x \in B$
Thus:
:$x \notin A \setminus B$
and hence:
:$t \notin \RR \sqbrk {A \setminus B}$
So by Proof by Cases:
:$t \notin \RR \sqbrk A \setminus \RR \sqbrk B \implies t \notin \RR \sqbrk {A \setminus B}$
The result follows from Set Complement inverts Subsets:
:$S \subseteq T \iff \map \complement T \subseteq \map \complement S$
=== Necessary Condition ===
Now for the converse: If $(1)$ holds, it is to be shown that $\RR$ is one-to-many.
Let $s, t \in S$ be distinct.
That is, $s \ne t$.
Then in particular:
:$\set s \setminus \set t = \set s$
Applying $(1)$ to these two sets, it follows that:
:$\RR \sqbrk {\set s} \setminus \RR \sqbrk {\set t} = \RR \sqbrk {\set s}$
By Set Difference with Disjoint Set, this implies that:
:$\RR \sqbrk {\set s} \cap \RR \sqbrk {\set t} = \O$
It follows that every element of $T$ can be related to at most one element of $S$.
That is, $\RR$ is one-to-many.
Category:Set Difference
Category:One-to-Many Relations
Category:One-to-Many Image of Set Difference
One-to-Many Image of Set Difference/Corollary 1

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation which is one-to-many.
Let $A \subseteq B \subseteq S$.
Then:
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$
where $\complement$ (in this context) denotes relative complement.
== Proof ==
We have that $A \subseteq B$.
Then by definition of relative complement:
:$\relcomp B A = B \setminus A$
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk B \setminus \RR \sqbrk A$
Hence, when $A \subseteq B$:
:$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$
means exactly the same thing as:
:$\RR \sqbrk B \setminus \RR \sqbrk A = \RR \sqbrk {B \setminus A}$
Hence the result from One-to-Many Image of Set Difference.
Category:Relative Complement
Category:One-to-Many Image of Set Difference
One-to-Many Image of Set Difference/Corollary 2

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation which is one-to-many.
Let $A$ be a subset of $S$.
Then:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
where $\complement$ (in this context) denotes relative complement.
In the language of direct image mappings this can be presented as:
:$\forall A \in \powerset S: \map {\paren {\complement_{\Img \RR} \circ \RR^\to} } A = \map {\paren {\RR^\to \circ \complement_S} } A$
== Proof ==
By definition of the image of $\RR$:
:$\Img \RR = \RR \sqbrk S$
So, when $B = S$ in One-to-Many Image of Set Difference: Corollary 1:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \relcomp {\RR \sqbrk S} {\RR \sqbrk A}$
Hence:
:$\relcomp {\Img \RR} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
means exactly the same thing as:
:$\relcomp {\RR \sqbrk S} {\RR \sqbrk A} = \RR \sqbrk {\relcomp S A}$
that is:
:$\RR \sqbrk S \setminus \RR \sqbrk A = \RR \sqbrk {S \setminus A}$
Hence the result from One-to-Many Image of Set Difference.
Category:Set Difference
Category:Relative Complement
Category:One-to-Many Image of Set Difference
One-to-Many Relation/Examples/Inverse Sine

== Examples of One-to-Many Relations ==
Let $\RR \subseteq \R \times \R$ be the relation on $\R$ defined as:
:$\forall \tuple {x, y} \in \R \times \R: \tuple {x, y} \in \RR \iff x = \sin y$
Then $\RR$ is a one-to-many relation.
== Proof ==
Let:
:$x_1, x_2 \in \R: \tuple {x_1, y} \in \RR \land \tuple {x_2, y} \in \RR$
Then:
:$x_1 = \sin y = x_2$
and so:
:$x_1 = x_2$
Hence the result.
== Sources ==
* 
Category:Examples of One-to-Many Relations
One-to-Many Relation Composite with Inverse is Coreflexive

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation which is one-to-many.
Then the composite of $\RR$ with its inverse is a coreflexive relation:
:$\RR^{-1} \circ \RR \subseteq \Delta_X$
That is, by Relation is Symmetric and Antisymmetric iff Coreflexive, $\RR^{-1} \circ \RR$ is both symmetric and antisymmetric.
== Proof ==
As $\RR$ is one-to-many, it follows from Inverse of Many-to-One Relation is One-to-Many that $\RR^{-1}$ is many-to-one.
Let $\tuple {x, z} \in \RR^{-1} \circ \RR$.
Then:
:$\exists y \in S: \tuple {x, y} \in \RR, \tuple {y, z} \in \RR^{-1}$
As $\tuple {x, y} \in \RR$, from the definition of inverse relation:
:$\tuple {y, x} \in \RR^{-1}$
As $\RR^{-1}$ is many-to-one:
:$\tuple {y, z} = \tuple {y, x}$
Thus if $\tuple {x, z} \in \RR^{-1} \circ \RR$, it follows that:
:$\tuple {x, z} = \tuple {x, x}$
Thus:
:$\RR^{-1} \circ \RR \subseteq \Delta_X$
The result follows by definition of coreflexive relation.
Category:Coreflexive Relations
One (Category) is Terminal Object

== Theorem ==
Let $\mathbf{Cat}$ be the category of categories.
Let $\mathbf 1$ be the category one.
Then $\mathbf 1$ is a terminal object of $\mathbf{Cat}$.
== Proof ==
Let $\mathbf C$ be an object of $\mathbf{Cat}$, i.e. a small category.
From Singleton is Terminal Object of Category of Sets, there exist unique mappings:
:$F_0: \mathbf C_0 \to \mathbf 1_0 = \left\{{*}\right\}$
:$F_1: \mathbf C_1 \to \mathbf 1_1 = \left\{{\operatorname{id}_*}\right\}$
since the latter sets are singletons.
It remains to verify that $F: \mathbf C \to \mathbf 1$ so defined, in fact is a functor.
Trivially, $F$ preserves identity morphisms.
That $F$ has the morphism property follows from:
:$\operatorname{id}_* \circ \operatorname{id}_* = \operatorname{id}_*$
Hence $F: \mathbf C \to \mathbf 1$ constitutes a functor.
The result follows by definition of terminal object.
== Sources ==
* : $\S 2.2$: Example $2.11$: $2$
Category:Category of Categories
One Choose n

== Theorem ==
:$\dbinom 1 n = \begin{cases} 1 & : n \in \set {0, 1} \\ 0 & : \text {otherwise} \end{cases}$
where $\dbinom 1 n$ denotes a binomial coefficient.
== Proof ==
By definition of binomial coefficient:
:$\dbinom m n = \begin{cases}
\dfrac {m!} {n! \paren {m - n}!} & : 0 \le n \le m \\
& \\
0 & : \text { otherwise } \end{cases}$
Thus when $n > 1$:
:$\dbinom 1 n = 0$
and when $n < 0$:
:$\dbinom 1 n = 0$
Then:
Category:Examples of Binomial Coefficients‎
One Diagonal of Kite Bisects the Other

== Theorem ==
Let $ABCD$ be a kite such that:
:$AC$ and $BD$ are its diagonals
:$AB = BC$
:$AD = DC$
Then $BD$ is the perpendicular bisector of $AC$.
== Proof ==
:
Let $AC$ and $BD$ meet at $E$.
From Diagonals of Kite are Perpendicular, $AC$ and $BD$ are perpendicular.
That is:
:$\angle AEB = \angle CEB$
both being right angles.
Consider the triangles $\triangle ABE$ and $\triangle CBE$.
We have that:
:$\angle AEB = \angle CEB$ are both right angles.
:$AB = BC$
:$BE$ is common
Hence by Triangle Right-Angle-Hypotenuse-Side Congruence, $\triangle ABE$ and $\triangle CBE$ are congruent.
Thus:
:$AE = CE$
and it has already been established that $BD$ is perpendicular to $AC$.
Hence the result by definition of perpendicular bisector.
== Sources ==
* 
Category:Kites
One Eighth as Pandigital Fraction

== Theorem ==
There are $46$ ways $\dfrac 1 8$ can be expressed as a pandigital fraction:
:$\dfrac 1 8 = \dfrac {3187} {25496}$
:$\dfrac 1 8 = \dfrac {4589} {36712}$
:$\dfrac 1 8 = \dfrac {4591} {36728}$
:$\dfrac 1 8 = \dfrac {4689} {37512}$
:$\dfrac 1 8 = \dfrac {4691} {37528}$
:$\dfrac 1 8 = \dfrac {4769} {38152}$
:$\dfrac 1 8 = \dfrac {5237} {41896}$
:$\dfrac 1 8 = \dfrac {5371} {42968}$
:$\dfrac 1 8 = \dfrac {5789} {46312}$
:$\dfrac 1 8 = \dfrac {5791} {46328}$
:$\dfrac 1 8 = \dfrac {5839} {46712}$
:$\dfrac 1 8 = \dfrac {5892} {47136}$
:$\dfrac 1 8 = \dfrac {5916} {47328}$
:$\dfrac 1 8 = \dfrac {5921} {47368}$
:$\dfrac 1 8 = \dfrac {6479} {51832}$
:$\dfrac 1 8 = \dfrac {6741} {53928}$
:$\dfrac 1 8 = \dfrac {6789} {54312}$
:$\dfrac 1 8 = \dfrac {6791} {54328}$
:$\dfrac 1 8 = \dfrac {6839} {54712}$
:$\dfrac 1 8 = \dfrac {7123} {56984}$
:$\dfrac 1 8 = \dfrac {7312} {58496}$
:$\dfrac 1 8 = \dfrac {7364} {58912}$
:$\dfrac 1 8 = \dfrac {7416} {59328}$
:$\dfrac 1 8 = \dfrac {7421} {59368}$
:$\dfrac 1 8 = \dfrac {7894} {63152}$
:$\dfrac 1 8 = \dfrac {7941} {63528}$
:$\dfrac 1 8 = \dfrac {8174} {65392}$
:$\dfrac 1 8 = \dfrac {8179} {65432}$
:$\dfrac 1 8 = \dfrac {8394} {67152}$
:$\dfrac 1 8 = \dfrac {8419} {67352}$
:$\dfrac 1 8 = \dfrac {8439} {67512}$
:$\dfrac 1 8 = \dfrac {8932} {71456}$
:$\dfrac 1 8 = \dfrac {8942} {71536}$
:$\dfrac 1 8 = \dfrac {8953} {71624}$
:$\dfrac 1 8 = \dfrac {8954} {71632}$
:$\dfrac 1 8 = \dfrac {9156} {73248}$
:$\dfrac 1 8 = \dfrac {9158} {73264}$
:$\dfrac 1 8 = \dfrac {9182} {73456}$
:$\dfrac 1 8 = \dfrac {9316} {74528}$
:$\dfrac 1 8 = \dfrac {9321} {74568}$
:$\dfrac 1 8 = \dfrac {9352} {74816}$
:$\dfrac 1 8 = \dfrac {9416} {75328}$
:$\dfrac 1 8 = \dfrac {9421} {75368}$
:$\dfrac 1 8 = \dfrac {9523} {76184}$
:$\dfrac 1 8 = \dfrac {9531} {76248}$
:$\dfrac 1 8 = \dfrac {9541} {76328}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Third as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Ninth as Pandigital Fraction
Category:One Eighth
Category:Pandigital Fractions
One Equals Minus One

== Paradox ==
== Resolution ==
This is a falsidical paradox arising from incorrect reasoning about the nature of square roots.
=== Explanation ===
The property:
:$\sqrt a \times \sqrt b = \sqrt {a b}$ 
can only be used when: 
:$a \ge 0$ and $b \ge 0$
It is specifically false when both $a$ and $b$ are negative.
Since $-1 < 0$, this property of square roots cannot be applied to this statement.
This is how the analysis is to go:
and all is well with the world.
== Also see ==
* Definition:Complex Number
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $128$
* 
Category:Complex Analysis
Category:Mistakes
Category:Falsidical Paradoxes
One Fifth as Pandigital Fraction

== Theorem ==
There are $12$ ways $\dfrac 1 5$ can be expressed as a pandigital fraction:
:$\dfrac 1 5 = \dfrac {2697} {13485}$
:$\dfrac 1 5 = \dfrac {2769} {13845}$
:$\dfrac 1 5 = \dfrac {2937} {14685}$
:$\dfrac 1 5 = \dfrac {2967} {14835}$
:$\dfrac 1 5 = \dfrac {2973} {14865}$
:$\dfrac 1 5 = \dfrac {3297} {16485}$
:$\dfrac 1 5 = \dfrac {3729} {18645}$
:$\dfrac 1 5 = \dfrac {6297} {31485}$
:$\dfrac 1 5 = \dfrac {7629} {38145}$
:$\dfrac 1 5 = \dfrac {9237} {46185}$
:$\dfrac 1 5 = \dfrac {9627} {48135}$
:$\dfrac 1 5 = \dfrac {9723} {48615}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Half as Pandigital Fraction
* One Third as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Eighth as Pandigital Fraction
* One Ninth as Pandigital Fraction
Category:One Fifth
Category:Pandigital Fractions
One Half as Pandigital Fraction

== Theorem ==
There are $12$ ways $\dfrac 1 2$ can be expressed as a pandigital fraction:
:$\dfrac 1 2 = \dfrac {6729} {13 \, 458}$
:$\dfrac 1 2 = \dfrac {6792} {13 \, 584}$
:$\dfrac 1 2 = \dfrac {6927} {13 \, 854}$
:$\dfrac 1 2 = \dfrac {7269} {14 \, 538}$
:$\dfrac 1 2 = \dfrac {7293} {14 \, 586}$
:$\dfrac 1 2 = \dfrac {7329} {14 \, 658}$
:$\dfrac 1 2 = \dfrac {7692} {15 \, 384}$
:$\dfrac 1 2 = \dfrac {7923} {15 \, 846}$
:$\dfrac 1 2 = \dfrac {7932} {15 \, 864}$
:$\dfrac 1 2 = \dfrac {9267} {18 \, 534}$
:$\dfrac 1 2 = \dfrac {9273} {18 \, 546}$
:$\dfrac 1 2 = \dfrac {9327} {18 \, 654}$
== Proof ==
Can be verified by brute force.
== Also see ==
* One Third as Pandigital Fraction
* One Quarter as Pandigital Fraction
* One Fifth as Pandigital Fraction
* One Sixth as Pandigital Fraction
* One Seventh as Pandigital Fraction
* One Eighth as Pandigital Fraction
* One Ninth as Pandigital Fraction
== Historical Note ==
== Sources ==
* : $0 \cdotp 5$
* : $0 \cdotp 5$
Category:One Half
Category:Pandigital Fractions
One Hundred Fowls

== Classic Problem ==
$100$ fowls were sold for $100$ shillings.
:The cockerels were sold for $5$ shillings each.
:The hens were sold for $3$ shillings each.
:The chicks were sold for $\frac 1 3$ of a shilling each.
How many of each were sold?
== Solution ==
There were either:
:$0$ cockerels, $25$ hens and $75$ chicks;
:$4$ cockerels, $18$ hens and $78$ chicks;
:$8$ cockerels, $11$ hens and $81$ chicks;
:$12$ cockerels, $4$ hens and $84$ chicks.
== Proof ==
Let $x$, $y$ and $z$ denote the number of cockerels, hens and chicks respectively.
We are to solve for $x, y, z \in \N$:
$7 x$ must be divisible by $4$.
Hence so must $x$.
Write $x = 4 n$.
Then the problem is reduced to:
:$7 n + y = 25$
Hence $0 \le n \le 3$.
We list the solutions $\tuple {n, x, y, z}$ for each $n$:
:$\tuple {0, 0, 25, 75}, \tuple {1, 4, 18, 78}, \tuple {2, 8, 11, 81}, \tuple {3, 12, 4, 84}$
== Version from Bakhshali Manuscript ==
== Historical Note ==
== Sources ==
* 
* 
* : Sun Tsu Suan Ching: $74$
Category:Classic Problems
Category:Examples of Diophantine Problems
Category:One Hundred Fowls Problems