nyuuzyou/wikis huggingface dataset:
Dilogarithm Function/Examples

== Examples of Dilogarithm Function ==
=== Example: $\map {\Li_2} {-\phi}$ ===
=== Example: $\map {\Li_2} {-1}$ ===
=== Example: $\map {\Li_2} {-\dfrac 1 \phi}$ ===
=== Example: $\map {\Li_2} 0$ ===
=== Example: $\map {\Li_2} {1 - \dfrac 1 \phi}$ ===
=== Example: $\map {\Li_2} {\dfrac 1 2 }$ ===
=== Example: $\map {\Li_2} {\dfrac 1 \phi}$ ===
=== Example: $\map {\Li_2} 1$ ===
Category:Examples of Dilogarithm Function
Category:Spence's Function
Dilogarithm Reflection Formula

== Theorem ==
:$\map {\Li_2} z + \map {\Li_2} {1 - z} = \map \zeta 2 - \map \ln z \map \ln {1 - z}$
where:
:$\map {\Li_2} z$ is the Dilogarithm function of $z$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$.
== Proof ==
From the definition of the dilogarithm function:
:$\ds \map {\Li_2} z = -\int_0^z \dfrac {\map \ln {1 - x} } x \rd x$
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
== Sources ==
* : Chapter $\text {2}$. The Hypergeometric Functions
Category:Spence's Function
Category:Reflection Formulas
Dilogarithm of Minus Golden Mean

== Theorem ==
:$\map {\Li_2} {-\phi} = -\dfrac 3 5 \map \zeta 2 - \paren {\map \ln \phi}^2$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\phi$ denotes the golden mean.
== Proof ==
We now note:
== Sources ==
* : Chapter $1$. Dilogarithm
Category:Spence's Function
Category:Golden Mean
Dilogarithm of Minus One

== Theorem ==
:$\map {\Li_2} {-1} = -\dfrac 1 2 \map \zeta 2$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$.
== Proof ==
Category:Spence's Function
Dilogarithm of Minus Reciprocal of Golden Mean

== Theorem ==
:$\map {\Li_2} {-\dfrac 1 \phi} = -\dfrac 2 5 \map \zeta 2 + \dfrac 1 2 \paren {\map \ln \phi}^2$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\phi$ denotes the golden mean.
== Proof ==
We now note:
== Sources ==
* : Chapter $1$. Dilogarithm
Category:Spence's Function
Category:Golden Mean
Dilogarithm of Minus Z Plus Dilogarithm of Minus Reciprocal of Z

== Theorem ==
:$\map {\Li_2} {-z} + \map {\Li_2} {-\dfrac 1 z} = -\map \zeta 2 - \dfrac 1 2 \map {\ln^2} z$
where:
:$\map {\Li_2} z$ is the dilogarithm function of $z$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$.
== Proof ==
From the definition of the dilogarithm function:
:$\ds \map {\Li_2} z = -\int_0^z \dfrac {\map \ln {1 - x} } x \rd x$
Taking the derivative of both sides at $-\dfrac 1 z$
Now integrating both sides $z$, we obtain:
We now solve for $C$ by setting $z = 1$
Therefore:
:$\map {\Li_2} {-z} + \map {\Li_2} {-\dfrac 1 z} = -\map \zeta 2 - \dfrac 1 2 \map {\ln^2} z$
== Sources ==
* : Chapter $\text {1}$. Dilogarithm
Category:Spence's Function
Dilogarithm of One

== Theorem ==
:$\map {\Li_2} 1 = \map \zeta 2$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$.
== Proof ==
Category:Spence's Function
Dilogarithm of One Half

== Theorem ==
:$\map {\Li_2} {\dfrac 1 2} = \dfrac 1 2 \paren {\map \zeta 2 - \paren {\map \ln 2}^2}$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$.
== Proof ==
Category:Spence's Function
Dilogarithm of One Minus Reciprocal of Golden Mean

== Theorem ==
:$\map {\Li_2} {1 - \dfrac 1 \phi} = \dfrac 2 5 \map \zeta 2 - \paren {\map \ln \phi}^2$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\phi$ denotes the golden mean.
== Proof ==
We first note the following:
We now note:
== Sources ==
* : Chapter $\text {1}$. Dilogarithm
Category:Spence's Function
Category:Golden Mean
Dilogarithm of One Minus Z Plus Dilogarithm of One Minus Reciprocal of Z

== Theorem ==
:$\map {\Li_2} {1 - z} + \map {\Li_2} {1 - \dfrac 1 z} = -\dfrac 1 2 \map {\ln^2} z$
where:
:$\map {\Li_2} z$ is the Dilogarithm function of $z$
:$z \in \C$ and $z < 1$.
== Proof ==
From the definition of the dilogarithm function:
:$\ds \map {\Li_2} z = -\int_0^z \dfrac {\map \ln {1 - x} } x \rd x$
Taking the derivative of both sides at $-\dfrac z {1 - z}$
Now integrating both sides $z$, we obtain:
We now solve for $C$ by setting $z = 0$
Therefore:
== Sources ==
* : Chapter $\text {1}$. Dilogarithm
Category:Spence's Function
Dilogarithm of Reciprocal of Golden Mean

== Theorem ==
:$\map {\Li_2} {\dfrac 1 \phi} = \dfrac 3 5 \map \zeta 2 - \paren {\map \ln \phi}^2$
where:
:$\map {\Li_2} x$ is the dilogarithm function of $x$
:$\map \zeta 2$ is the Riemann $\zeta$ function of $2$
:$\phi$ denotes the golden mean.
== Proof ==
We first note the following:
We now note:
== Sources ==
* : Chapter $1$. Dilogarithm
Category:Spence's Function
Category:Golden Mean
Dilogarithm of Square

== Theorem ==
:$\map {\Li_2} z + \map {\Li_2} {-z} = \dfrac 1 2 \map {\Li_2} {z^2}$
where $\Li_2$ denotes the dilogarithm function.
== Proof 1 ==
== Proof 2 ==
Category:Dilogarithm of Square
Category:Spence's Function
Dimension of Affine Algebraic Set is Dimension of Affine Coordinate Ring

== Theorem ==
Let $k$ be a field.
Let $Y \subseteq k^n$ be an affine algebraic set.
Let $\map A Y$ be the affine coordinate ring.
Then:
:$\map \dim Y = \map \dim {\map A Y}$
where:
:$\map \dim Y$ is the Krull dimension of $Y$ with respect to Zariski topology
:$\map \dim {\map A Y}$ is the Krull dimension of $\map A Y$
== Proof ==
== Sources ==
* $\text{I}.1$ Affine Varieties
Category:Algebraic Geometry
Category:Affine Geometry
Dimension of Algebraic Dual

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be an $n$-dimensional $R$-module.
Let $G^*$ be the algebraic dual of $G$.
Then $G^*$ is also $n$-dimensional.
=== Dimension of Double Dual ===
== Proof ==
Let $\struct {R, +_R, \circ}_R$ denote the $R$-module $R$.
From Dimension of $R$-Module $R$ is $1$ we have that the dimension of $\struct {R, +_R, \circ}_R$ is $1$.
By definition, the algebraic dual of $G$ is the $R$-module $\map {\LL_R} {G, R}$ of all linear forms on $G$.
From Dimension of Set of Linear Transformations the dimension of $G^*$ is the product of the dimension of $G$ with the dimension of $\struct {R, +_R, \circ}_R$.
That is:
:$\map \dim {G^*} = n \times 1$
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Algebraic Duals
Dimension of Annihilator on Algebraic Dual

== Theorem ==
Let $G$ be an $n$-dimensional vector space over a field.
Let $G^*$ be the algebraic dual of $G$.
Let $M$ be an $m$-dimensional subspace of $G$.
Let $M^\circ$ be the annihilator of $M$.
Then:
:$M^\circ$ is an $\paren {n - m}$-dimensional subspace of $G^*$.
== Proof ==
Let $\sequence {a_n}$ be an ordered basis of $G$ such that $\sequence {a_m}$ is an ordered basis of $M$.
Let $\sequence { {a_n}'}$ be the ordered dual basis of $G^*$.
Let $\ds t = \sum_{k \mathop = 1}^n \lambda_k {a_k}' \in M^\circ$.
Then:
So $t$ is a linear combination of $\set { {a_k}': m + 1 \le k \le n}$.
But ${a_k}'$ clearly belongs to $M^\circ$ for each $k \in \closedint {m + 1} n$.
Therefore $M^\circ$ has dimension $n - m$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$
Category:Annihilators
Category:Vector Subspaces
Dimension of Bohr Radius

== Theorem ==
The Bohr radius has the dimension $\mathsf L$.
== Proof ==
By definition, the Bohr radius is:
:$a_0 = \dfrac {\varepsilon_0 h^2} {\pi \E^2 m_\E}$
where:
:$\varepsilon_0$ denotes the vacuum permittivity
:$h$ denotes Planck's constant
:$\E$ denotes the elementary charge
:$m_\E$ denotes the electron rest mass
We have:
Category:Bohr Radius
Category:Dimensions of Measurement
Dimension of Double Dual

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be an $n$-dimensional $R$-module.
Let $G^{**}$ be the double dual of $G$.
Then $G^{**}$ is also $n$-dimensional.
== Proof ==
By definition, the double dual of $G$ is the algebraic dual of the algebraic dual $G^*$ of $G$.
From Dimension of Algebraic Dual:
:$\map \dim {G^**} = \map \dim {G^*}$
Also from Dimension of Algebraic Dual::
:$\map \dim {G^*} = \map \dim G$
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Algebraic Duals
Dimension of Free Vector Space on Set

== Theorem ==
Let $k$ be a division ring.
Let $X$ be a set.
Let $k^{\paren X}$ be the free vector space on $X$.
The vector space $k^{\paren X}$ has dimension the cardinality of $X$.
== Proof ==
Follows from:
:Canonical Basis of Free Module on Set is Basis
:Cardinality of Canonical Basis of Free Module on Set
Category:Vector Spaces
Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $T : X \to Y$ be a linear transformation.
Then: 
:$\dim T \sqbrk X \le \dim X$
== Proof ==
From Vector Space has Basis, there exists a basis $\BB$ for $X$.
By Image of Generating Set of Vector Space under Linear Transformation is Generating Set of Image, $T \sqbrk \BB$ is a generator for $T \sqbrk X$.
From Generator of Vector Space Contains Basis, there exists a basis $\BB'$ for $T \sqbrk X$ such that $\BB' \subseteq T \sqbrk \BB$.
From Cardinality of Image of Mapping not greater than Cardinality of Domain, we have $\card {T \sqbrk \BB} \le \card \BB$. 
Also, since $\BB' \subseteq T \sqbrk \BB$, we have: 
:$\card {\BB'} \le \card {T \sqbrk \BB}$
Since $\BB'$ is a basis for $T \sqbrk X$, we have: 
:$\card {\BB'} = \dim T \sqbrk X$
and since $\BB$ is a basis for $X$ we have: 
:$\card \BB = \dim X$
So we obtain: 
Category:Linear Transformations
Dimension of Measurement/Examples

== Examples of Dimensions of Measurement ==
* Displacement has dimension $\mathsf L$.
* Velocity has dimension $\mathsf {L T}^{-1}$ (change in displacement, that is length traveled, per unit of time).
* Acceleration has dimension $\mathsf {L T}^{-2}$ (change in velocity per unit of time).
* Force has dimension $\mathsf {M L T}^{-2}$ (mass times acceleration, from Newton's Second Law of Motion).
Category:Dimensions of Measurement
Dimension of Orthogonal Complement With Respect to Bilinear Form

== Theorem ==
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension.
Let $f$ be a nondegenerate bilinear form on $V$.
Let $U\subset V$ be a subspace.
Let $U^\perp$ be its orthogonal complement.
Then:
:$\map \dim U + \map \dim U^\perp = \map \dim V$
== Proof ==
Category:Bilinear Forms (Linear Algebra)
Dimension of Preimage under Evaluation Isomorphism of Annihilator on Algebraic Dual

Let $G$ be an $n$-dimensional vector space over a field.
Let $J: G \to G^{**}$ be the evaluation isomorphism.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Let $N$ be a $p$-dimensional subspace of $G^*$.
Then:
:$\map {J^\gets} {N^\circ}$ is an $\paren {n - p}$-dimensional subspace of $G$
where 
== Proof ==
We have that $N$ is a $p$-dimensional subspace of $G^*$.
From Dimension of Annihilator on Algebraic Dual we have that $N^\circ$ has dimension $n - p$.
From Inverse Evaluation Isomorphism of Annihilator we also have that $J^{-1} \sqbrk {N^\circ}$ also has dimension $n - p$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$
Category:Annihilators
Category:Vector Subspaces
Dimension of Proper Subspace is Less Than its Superspace

== Theorem ==
Let $G$ be a vector space whose dimension is $n$.
Let $H$ be a subspace of $G$.
Then $H$ is finite dimensional, and $\map \dim H \le \map \dim G$.
If $H$ is a proper subspace of $G$, then $\map \dim H < \map \dim G$.
== Proof ==
Let $H$ be a subspace of $G$.
Every linearly independent subset of the vector space $H$ is a linearly independent subset of the vector space $G$.
Therefore, it has no more than $n$ elements by Size of Linearly Independent Subset is at Most Size of Finite Generator.
So the set of all natural numbers $k$ such that $H$ has a linearly independent subset of $k$ vectors has a largest element $m$, and $m \le n$.
Now, let $B$ be a linearly independent subset of $H$ having $m$ vectors.
If the subspace generated by $B$ were not $H$, then $H$ would contain a linearly independent subset of $m + 1$ vectors.
This follows by Linearly Independent Subset also Independent in Generated Subspace.
This would contradict the definition of $m$.
Hence $B$ is a generator for $H$ and is thus a basis for $H$.
Thus $H$ is finite dimensional and $\map \dim H \le \map \dim G$.
Now, if $\map \dim H = \map \dim G$, then a basis of $H$ is a basis of $G$ by Sufficient Conditions for Basis of Finite Dimensional Vector Space, and therefore $H = G$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.13$
Category:Dimension of Proper Subspace
Dimension of R-Module R is 1

== Theorem ==
Let $\struct {R, +, \times}$ be a ring whose unity is $1_R$.
Let $\struct {R, +_R, \circ}_R$ denote the $R$-module $R$.
Then the dimension of $\struct {R, +_R, \circ}_R$ is $1$.
== Proof ==
We have by definition that the $R$-module $R$ is the special case of the $R$-module $R^n$ where $n = 1$.
From $R$-module $R^n$ is $n$-Dimensional it follows that $\struct {R, +_R, \circ}_R$ is $1$-dimensional.
Category:Module on Cartesian Product
Dimension of Radical of Bilinear Form

== Theorem ==
Let $\mathbb K$ be a field.
Let $V$ be a vector space over $\mathbb K$ of finite dimension $n > 0$.
Let $f$ be a bilinear form on $V$.
Let $\map \Rad V$ denote the radical of $V$.
Let $\map {\operatorname {rk} } f$ be the rank of $f$.
Then:
:$\map \dim {\map \Rad V} = n - \map {\operatorname {rk} } f$
where $\dim$ denotes dimension.
== Proof ==
Category:Bilinear Forms (Linear Algebra)
Dimension of Rydberg Constant

== Theorem ==
The Rydberg constant has the dimension $\mathsf {L^{-1} }$.
== Proof ==
By definition, the Rydberg constant is:
:$R_\infty = \dfrac {m_\E \E^4} {8 \varepsilon_0^2 h^3 c}$
where:
:$m_\E$ denotes the electron rest mass
:$\E$ denotes the elementary charge
:$\varepsilon_0$ denotes the vacuum permittivity
:$h$ denotes Planck's constant
:$c$ denotes the speed of light.
We have:
Category:Rydberg Constant
Category:Dimensions of Measurement
Dimension of Set of Linear Transformations

== Theorem ==
Let $R$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module such that $\map \dim G = n$.
Let $\struct {H, +_H, \circ}_R$ be a unitary $R$-module such that $\map \dim H = m$.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Then:
:$\map \dim {\map {\LL_R} {G, H} } = n m$
where $\map \dim {\map {\LL_R} {G, H} }$ denotes the dimension of $\map {\LL_R} {G, H}$.
== Proof ==
Consider the set:
:$B = \set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$
From Basis for Set of Linear Transformations, $B$ is a basis for $\map {\LL_R} {G, H}$.
It is seen that by construction, $B$ has $n m$ elements.
The result follows by definition of dimension.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.8$
Category:Linear Transformations
Dimension of Spring Force Constant

== Theorem ==
The dimension of a spring force constant is $\mathsf {M T}^{-2}$.
== Proof ==
From Hooke's Law, we have:
:$\mathbf F = -k \mathbf x$
where:
:$\mathbf F$ is a force, of dimension $\mathsf {M L T}^{-2}$
:$k$ is the spring force constant
:$\mathbf x$ is a displacement, of dimension $\mathsf L$.
Let the dimension of $k$ be $D$.
Then we have:
:$\mathsf {M L T}^{-2} = D \mathsf L$
from which the result follows immediately:
:$D = \mathsf {M T}^{-2}$
Category:Dimensional Analysis
Dimension of Universal Gravitational Constant

== Theorem ==
The dimension of the universal gravitational constant $G$ is $M^{-1} L^3 T^{-2}$.
== Proof ==
From Newton's Law of Universal Gravitation:
:$\mathbf F = \dfrac {G m_1 m_2 \mathbf r} {r^3}$
We have that:
: The dimension of force is $M L T^{-2}$
: The dimension of displacement is $L$
: The dimension of mass is $M$.
Let $x$ be the dimension of $G$.
Then we have:
:$M L T^{-2} = x \dfrac {M^2 L}{L^3}$
Hence, after algebra:
:$x = M^{-1} L^3 T^{-2}$
Category:Universal Gravitational Constant
Category:Dimensional Analysis
Dimension of Vector Space of Polynomial Functions

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose unity is $1_F$.
Let $F_n \sqbrk X$ be the ring of polynomials over $F$ whose degree is less than $n$.
Then the dimension of the vector space $F_n \sqbrk X$ is $n$.
== Proof ==
Let $B$ be the set of all the identity functions $I^k$ on $F_n \sqbrk X$ where $n \in \Z_{\ge 0}$.
By definition, every element of $F_n \sqbrk X$ is a linear combination of $B$.
Suppose:
:$\ds \sum_{k \mathop = 0}^m \alpha_k I^k = 0, \alpha_m \ne 0$
Then by differentiating $m$ times, we obtain from Nth Derivative of Nth Power:
:$m! \alpha_m = 0$
whence $\alpha_m = 0$ which is a contradiction.
Hence $B$ is linearly independent and therefore is a basis for $F_n \sqbrk X$.
The result follows from Linearly Independent Set is Basis iff of Same Cardinality as Dimension.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $67$
Category:Dimension of Vector Space
Category:Polynomial Theory
Dimension of Vector Space on Cartesian Product

== Theorem ==
Let $\struct {K, +, \circ}$ be a division ring.
Let $n \in \N_{>0}$.
Let $\mathbf V := \struct {K^n, +, \times}_K$ be the $K$-vector space $K^n$.
Then the dimension of $\mathbf V$ is $n$.
== Proof ==
Let the unity of $K$ be $1$, and the zero of $K$ be $0$.
Consider the vectors:
Thus $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
From Standard Ordered Basis is Basis, $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is a basis of $\mathbf V$.
The result follows from Linearly Independent Set is Basis iff of Same Cardinality as Dimension.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $66$
Category:Vector Space on Cartesian Product
Category:Dimension of Vector Space
Dini's Theorem

== Theorem ==
Let $K \subseteq \R$ be compact.
Let $\sequence {f_n}$ be a sequence of continuous real functions defined on $K$.
Let $\sequence {f_n}$ converge pointwise to a continuous function $f$.
Suppose that:
:$\forall x \in K : \sequence {\map {f_n} x}$ is monotone.
Then the convergence of $\sequence {f_n}$ to $f$ is uniform.
== Proof ==
For each $n$, define $d_n : K \to \R$ by:
:$d_n := \size {f_n - f}$
We have:
:$(1): \quad \sequence {d_n}$ converge pointwise to $0$
since $\sequence {f_n}$ converges pointwise to $f$.
We also have:
:$(2): \quad \forall x \in K : \sequence {\map {d_n} x}$ is monotonically decreasing
since $\sequence {\map {f_n} x}$ is monotone.
Let $\epsilon > 0$.
Consider $\sequence {A_n}$ defined by:
:$A_n := \set {x \in K : \map {d_n} x < \epsilon}$
As $d_n$ are continuous, $A_n$ are open sets.
In view of $(1)$, we have:
:$K = \bigcup A_n$
Since $K$ is compact, there exists a finite subcover:
:$K = A_{n_1} \cup \cdots \cup A_{n_m}$
for some $m \in \N_{>0}$.
Let $N := \max \set {n_1, \ldots, n_m}$ be the maximum element among $n_1, \ldots, n_m$.
Then, since $\sequence {A_n}$ is increasing in view of $(2)$, we have:
:$\forall n \ge N : K = A_n$
That is:
:$\forall n \ge N : \forall x \in K :\quad \map {d_n} x = \size{\map {f_n} x - \map f x} < \epsilon$
Category:Real Analysis
Diophantine Equation y cubed equals x squared plus 2

== Theorem ==
The indeterminate Diophantine equation:
:$y^3 = x^2 + 2$
has only one solution in the Natural Numbers:
:$x = 5, y = 3$
== Proof ==
Assume that $x$ is even:
Therefore, the is $2 \paren {2 k^2 + 1} \equiv 2 \pmod 4$
If $y$ is odd, then the will be odd:
and if $y$ is even, then the will be $\equiv 0 \pmod 4$
Therefore, $x$ and $y$ must both be odd since the can never be $\equiv 2 \pmod 4$
Let us rewrite $x$ as $x = y + a$
We now demonstrate that $a$ must be even:
Therefore:
From the , we know that $y$ is odd and therefore $y - 1$ is even and the only even term on the is $2$.
Therefore, our only solution is: $y - 1 = 2 \leadsto y = 3$ and $x = 5$
== Historical Note ==
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
Category:Diophantine Equations
Diophantine Equation y cubed equals x squared plus 2/Historical Note

== Historical Note on Diophantine Equation $y^3 = x^2 + 2$ ==
The Diophantine equation $y^3 = x^2 + 2$ was proved to have only the solution $x = 5, y = 3$ by by use of the Method of Infinite Descent.
He submitted it, without proof, along with a number of others, to in a letter dated $14$ August $1659$.
== Sources ==
* : Chapter $\text{IV}$: The Prince of Amateurs
Category:Historical Notes
Diophantus Updated/Solution

== Solution to Diophantus Updated ==
=== Solution ===
Inside Amanda.
== Proof ==
Let $M$ be the age in years of Amanda now.
Let $C$ be the age in years of John now.
We have:
So John is $-\dfrac 3 4$ years old, that is, $-9$ months.
That is, John will be born in $9$ months time.
So, right now, John's father is inside Amanda.
Diophantus of Alexandria/Arithmetica/Book 1

== Problems by : Book $\text I$ ==
=== Problem $1$ ===
=== Problem $2$ ===
=== Problem $3$ ===
=== Problem $4$ ===
=== Problem $5$ ===
=== Problem $6$ ===
=== Problem $7$ ===
=== Problem $8$ ===
=== Problem $9$ ===
=== Problem $10$ ===
=== Problem $11$ ===
=== Problem $12$ ===
=== Problem $13$ ===
=== Problem $14$ ===
=== Problem $15$ ===
=== Problem $16$ ===
=== Problem $17$ ===
=== Problem $18$ ===
=== Problem $19$ ===
=== Problem $20$ ===
=== Problem $21$ ===
=== Problem $22$ ===
=== Problem $23$ ===
=== Problem $24$ ===
=== Problem $25$ ===
=== Problem $26$ ===
=== Problem $27$ ===
=== Problem $28$ ===
=== Problem $29$ ===
=== Problem $30$ ===
=== Problem $31$ ===
=== Problem $32$ ===
=== Problem $33$ ===
=== Problem $34$ ===
=== Problem $35$ ===
=== Problem $36$ ===
=== Problem $37$ ===
=== Problem $38$ ===
=== Problem $39$ ===
Category:Diophantus of Alexandria/Arithmetica
Category:Examples of Algebra Problems
Diophantus of Alexandria/Arithmetica/Book 1/Problem 1

== Problem ==
:To divide a given number into two having a given difference.
== Proof ==
Let $N$ be the given number.
Let $d$ be the given difference.
Let $x$ be the smaller of the two numbers into which $N$ is to be divided.
Thus the greater of the two numbers is $x + d$.
We also have that the greater of the two numbers is $N - x$.
Hence the smaller of the two numbers is found by:
=== Example: $N = 100$, $d = 40$ ===
== Sources ==
* : Book $\text I$: Problem $1$
* : The Arithmetica: Book $\text {I}$: Problems: $1.$
Category:Diophantus of Alexandria/Arithmetica
Diophantus of Alexandria/Arithmetica/Book 1/Problem 1/100, 40

== Problem ==
:Given number $100$, given difference $40$.
== Solution ==
The required numbers are $30$ and $70$.
== Proof ==
Let $N$ be the given number.
Let $d$ be the given difference.
Let $x$ be the smaller of the two numbers into which $N$ is to be divided.
From the exposition of Problem $1$:
Here we have:
== Sources ==
* : Book $\text I$: Problem $1$
* : The Arithmetica: Book $\text {I}$: Problems: $1.$
Category:Diophantus of Alexandria/Arithmetica
Diophantus of Alexandria/Arithmetica/Book 1/Problem 15

== Problem ==
:Two numbers are such that:
::if the first receives $30$ from the second, they are in the ratio $2 : 1$
::if the second receives $50$ from the first, they are in the ratio $1 : 3$.
What are these numbers?
== Solution ==
:The first number is $98$.
:The second number is $94$.
== Proof ==
Let $x$ and $y$ be the numbers requested.
Then we have:
== Sources ==
* : Book $\text I$: Problem $15$
* : The First Pure Number Puzzles: $24$
Category:Examples of Algebra Problems
Diophantus of Alexandria/Arithmetica/Book 1/Problem 17

== Problem ==
:The sums of $4$ numbers, omitting each of the numbers in turn, are $22$, $24$, $27$ and $20$ respectively
What are the numbers?
== Solution ==
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Book $\text I$: Problem $17$
* : The Arithmetica: Book $\text {I}$: $17$
* : The First Pure Number Puzzles: $25$
Category:Examples of Algebra Problems
Category:Diophantus of Alexandria/Arithmetica/Book 1/Problem 17