avewright/proofwiki-math huggingface dataset:
Theorem: Euler-Binet Formula

The Fibonacci numbers have a closed-form solution:
:$F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1 / \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1}^n \phi^{-n} } {\sqrt 5} = \dfrac {\phi^n - \paren {1 - \phi}^n} {\phi - \paren {1 - \phi}}$
where $\phi$ is the golden mean.
Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
From {{Defof|Golden Mean|index = 2}}: $\phi = \dfrac {1 + \sqrt 5} 2$
Therefore, substituting $\sqrt 5 = 2\phi - 1 = \phi - \paren {1 - \phi} = \phi - \hat \phi$, the above can be written as:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\paren {\phi - \hat \phi}}$

Proof:
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is:
:$\map G z = \dfrac z {1 - z - z^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \map G z
      | r = \dfrac z {1 - z - z^2}
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }
      | c = Partial Fraction Expansion
}}
{{end-eqn}}
where:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
By Sum of Infinite Geometric Sequence:
:$\dfrac 1 {1 - \phi z} = 1 + \phi z + \phi^2 z^2 + \cdots$
and so:
:$\map G z = \dfrac 1 {\sqrt 5} \paren {1 + \phi z + \phi^2 z^2 + \cdots - 1 - \hat \phi z - \hat \phi^2 z^2 - \cdots}$
By definition, the coefficient of $z^n$ in $\map G z$ is exactly the $n$th Fibonacci number.
That is:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
{{qed}}
Theorem: Hölder's Inequality for Sums

Let $p, q \in \R_{>0}$ be strictly positive real numbers such that:
:$\dfrac 1 p + \dfrac 1 q = 1$
Let $\GF \in \set {\R, \C}$, that is, $\GF$ represents the set of either the real numbers or the complex numbers.
=== Formulation $1$ ===
{{:Hölder's Inequality for Sums/Formulation 1}}
=== Formulation $2$ ===
{{:Hölder's Inequality for Sums/Formulation 2}}

Proof:
Let $\sequence {x_k}_{k \mathop \in \N}$ and $\sequence {y_k}_{k \mathop \in \N}$ be infinite sequences in $\GF$ such that:
:$\forall m > n: x_m = y_m = 0$
Then we have:
{{begin-eqn}}
{{eqn | l = \sum \limits_{k \mathop = 1}^n \size {x_k y_k}
      | r = \sum \limits_{k \mathop \in \N} \size {x_k y_k}
      | c = {{hypothesis}}
}}
{{eqn | o = \le
      | r = \paren {\sum_{k \mathop \in \N} \size {x_k}^p}^{1 / p} \paren {\sum_{k \mathop \in \N} \size {y_k}^q}^{1 / q}
      | c = Hölder's Inequality for Sums
}}
{{eqn | r = \paren {\sum_{k \mathop = 1}^n \size {x_k}^p}^{1 / p} \paren {\sum_{k \mathop = 1}^n \size {y_k}^q}^{1 / q}
      | c = {{hypothesis}}
}}
{{end-eqn}}
{{qed}}
Theorem: Binet Form

Let $m \in \R$.
Define:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \sqrt {m^2 + 4}
}}
{{eqn | l = \alpha
      | r = \frac {m + \Delta} 2
}}
{{eqn | l = \beta
      | r = \frac {m - \Delta} 2
}}
{{end-eqn}}

Proof:
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
=== Basis for the Induction ===
$\map P 0$ is true, as this just says:
:$\dfrac {\phi^0 - \hat \phi^0} {\sqrt 5} = \dfrac {1 - 1} {\sqrt 5} = 0 = F_0$
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = \frac {\phi - \hat \phi} {\sqrt 5}
      | r = \frac {\paren {\frac {1 + \sqrt 5} 2} - \paren {\frac {1 - \sqrt 5} 2} } {\sqrt 5}
      | c = 
}}
{{eqn | r = \frac {\paren {1 - 1} + \paren {\sqrt 5 + \sqrt 5} } {2 \sqrt 5}
      | c = 
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = F_1
      | c = 
}}
{{end-eqn}}
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true for all $0 \le j \le k + 1$, then it logically follows that $\map P {k + 2}$ is true.
So this is our induction hypothesis:
:$\forall 0 \le j \le k + 1: F_j = \dfrac {\phi^j - \hat \phi^j} {\sqrt 5}$
Then we need to show:
:$F_{k + 2} = \dfrac {\phi^{k + 2} - \hat \phi^{k + 2} } {\sqrt 5}$
=== Induction Step ===
This is our induction step:
We observe that we have the following two identities:
:$\phi^2 = \paren {\dfrac {1 + \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 + 2 \sqrt 5} = \dfrac {3 + \sqrt 5} 2 = 1 + \phi$
:$\hat \phi^2 = \paren {\dfrac {1 - \sqrt 5} 2}^2 = \dfrac 1 4 \paren {6 - 2 \sqrt 5} = \dfrac {3 - \sqrt 5} 2 = 1 + \hat \phi$
This can also be deduced from the definition of the golden mean: the fact that $\phi$ and $\hat \phi$ are the solutions to the quadratic equation $x^2 = x + 1$.
Thus:
{{begin-eqn}}
{{eqn | l = \phi^{k + 2} - \hat \phi^{k + 2}
      | r = \paren {1 + \phi} \phi^k - \paren {1 + \hat \phi} \hat \phi^k
}}
{{eqn | r = \paren {\phi^k - \hat \phi^k} + \paren {\phi^{k + 1} - \hat \phi^{k + 1} }
}}
{{eqn | r = \sqrt 5 \paren {F_k + F_{k + 1} }
      | c = Induction Hypothesis
}}
{{eqn | r = \sqrt 5 F_{k + 2}
      | c = {{Defof|Fibonacci Numbers}}
}}
{{end-eqn}}
The result follows by the Second Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
{{qed}}
Theorem: Binet Form

Let $m \in \R$.
Define:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \sqrt {m^2 + 4}
}}
{{eqn | l = \alpha
      | r = \frac {m + \Delta} 2
}}
{{eqn | l = \beta
      | r = \frac {m - \Delta} 2
}}
{{end-eqn}}

Proof:
This follows as a direct application of the first Binet form:
:$U_n = m U_{n - 1} + U_{n - 2}$
where:
{{begin-eqn}}
{{eqn | l = U_0
      | r = 0
}}
{{eqn | l = U_1
      | r = 1
}}
{{end-eqn}}
has the closed-form solution:
:$U_n = \dfrac {\alpha^n - \beta^n} {\Delta}$
where:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \sqrt {m^2 + 4}
}}
{{eqn | l = \alpha
      | r = \frac {m + \Delta} 2
}}
{{eqn | l = \beta
      | r = \frac {m - \Delta} 2
}}
{{end-eqn}}
where $m = 1$.
{{qed}}
Theorem: Binet Form

Let $m \in \R$.
Define:
{{begin-eqn}}
{{eqn | l = \Delta
      | r = \sqrt {m^2 + 4}
}}
{{eqn | l = \alpha
      | r = \frac {m + \Delta} 2
}}
{{eqn | l = \beta
      | r = \frac {m - \Delta} 2
}}
{{end-eqn}}

Proof:
From Generating Function for Fibonacci Numbers, a generating function for the Fibonacci numbers is:
:$\map G z = \dfrac z {1 - z - z^2}$
Hence:
{{begin-eqn}}
{{eqn | l = \map G z
      | r = \dfrac z {1 - z - z^2}
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\dfrac 1 {1 - \phi z} - \dfrac 1 {1 - \hat \phi z} }
      | c = Partial Fraction Expansion
}}
{{end-eqn}}
where:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
By Sum of Infinite Geometric Sequence:
:$\dfrac 1 {1 - \phi z} = 1 + \phi z + \phi^2 z^2 + \cdots$
and so:
:$\map G z = \dfrac 1 {\sqrt 5} \paren {1 + \phi z + \phi^2 z^2 + \cdots - 1 - \hat \phi z - \hat \phi^2 z^2 - \cdots}$
By definition, the coefficient of $z^n$ in $\map G z$ is exactly the $n$th Fibonacci number.
That is:
:$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$
{{qed}}
Theorem: Binet-Cauchy Identity

:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
where all of the $a, b, c, d$ are elements of a commutative ring.
Thus the identity holds for $\Z, \Q, \R, \C$.

Proof:
Expanding the last term:
{{begin-eqn}}
{{eqn | o = 
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}
      | c = 
}}
{{eqn | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i c_i b_j d_j + a_j c_j b_i d_i}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i d_i b_j c_j + a_j d_j b_i c_i}
      | c = 
}}
{{eqn | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i c_i b_j d_j + a_j c_j b_i d_i} + \sum_{i \mathop = 1}^n a_i c_i b_i d_i
      | c = These new terms are the same
}}
{{eqn | o = 
      | ro= -
      | r = \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i d_i b_j c_j + a_j d_j b_i c_i} - \sum_{i \mathop = 1}^n a_i d_i b_i c_i
      | c = 
}}
{{eqn | r = \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_i c_i b_j d_j - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_i d_i b_j c_j
      | c = Completing the sums
}}
{{eqn | r = \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} - \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j}
      | c = Factoring terms indexed by $i$ and $j$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Binet-Cauchy Identity

:$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$
where all of the $a, b, c, d$ are elements of a commutative ring.
Thus the identity holds for $\Z, \Q, \R, \C$.

Proof:
This is a special case of the Cauchy-Binet Formula:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$
where:
:$\mathbf A$ is an $m \times n$ matrix
:$\mathbf B$ is an $n \times m$ matrix.
:For $1 \le j_1, j_2, \ldots, j_m \le n$:
::$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.
::$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.
In this case $m = 2$, giving:
:$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop \le n} \map \det {\mathbf A_{j_1 j_2} } \, \map \det {\mathbf B_{j_1 j_2} }$
{{finish}}
Theorem: Nicomachus's Theorem

{{begin-eqn}}
{{eqn | l = 1^3
      | r = 1
}}
{{eqn | l = 2^3
      | r = 3 + 5
}}
{{eqn | l = 3^3
      | r = 7 + 9 + 11
}}
{{eqn | l = 4^3
      | r = 13 + 15 + 17 + 19
}}
{{eqn | l = \vdots
      | o = 
}}
{{end-eqn}}
In general:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
In particular, the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $1^3 = 1$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$k^3 = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 + k - 1}$
Then we need to show:
:$\paren {k + 1}^3 = \paren {\paren {k + 1}^2 - \paren {k + 1} + 1} + \paren {\paren {k + 1}^2 - \paren {k + 1} + 3} + \dotsb + \paren {\paren {k + 1}^2 + \paren {k + 1} - 1}$
=== Induction Step ===
Let $T_k = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 + k - 1}$.
We can express this as:
:$T_k = \paren {k^2 - k + 1} + \paren {k^2 - k + 3} + \dotsb + \paren {k^2 - k + 2k - 1}$
We see that there are $k$ terms in $T_k$.
Let us consider the general term $\paren {\paren {k + 1}^2 - \paren {k + 1} + j}$ in $T_{k+1}$:
{{begin-eqn}}
{{eqn | l = \paren {k + 1}^2 - \paren {k + 1} + j
      | r = k^2 + 2 k + 1 - \paren {k + 1} + j
      | c = 
}}
{{eqn | r = k^2 - k + j + 2 k
      | c = 
}}
{{end-eqn}}
So, in $T_{k + 1}$, each of the terms is $2 k$ larger than the corresponding term for $T_k$.
So:
{{begin-eqn}}
{{eqn | l = T_{k + 1}
      | r = T_k + k \paren {2 k} + \paren {k + 1}^2 + \paren {k + 1} - 1
      | c = 
}}
{{eqn | r = k^3 + k \paren {2 k} + \paren {k + 1}^2 + \paren {k + 1} - 1
      | c = Induction Hypothesis
}}
{{eqn | r = k^3 + 2 k^2 + k^2 + 2 k + 1 + k + 1 - 1
      | c = 
}}
{{eqn | r = k^3 + 3 k^2 + 3 k + 1
      | c = 
}}
{{eqn | r = \paren {k + 1}^3
      | c = 
}}
{{end-eqn}}
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
Finally, note that the first term in the expansion for $\paren {n + 1}^3$ is $n^2 - n + 1 + 2 n = n^2 + n + 1$.
This is indeed two more than the last term in the expansion for $n^3$ .
{{qed}}
Theorem: Nicomachus's Theorem

{{begin-eqn}}
{{eqn | l = 1^3
      | r = 1
}}
{{eqn | l = 2^3
      | r = 3 + 5
}}
{{eqn | l = 3^3
      | r = 7 + 9 + 11
}}
{{eqn | l = 4^3
      | r = 13 + 15 + 17 + 19
}}
{{eqn | l = \vdots
      | o = 
}}
{{end-eqn}}
In general:
:$\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \dotsb + \paren {n^2 + n - 1}$
In particular, the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.

Proof:
From the definition:
:$\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$
can be written:
:$\paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}$
Writing this in sum notation:
{{begin-eqn}}
{{eqn | o = 
      | r = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 - n + 2 n - 1}
}}
{{eqn | r = \sum_{k \mathop = 1}^n \paren {n^2 - n + 2 k - 1}
}}
{{eqn | r = n \paren {n^2 - n} + \sum_{k \mathop = 1}^n \paren {2 k - 1}
}}
{{eqn | r = n^3 - n^2 + n^2
      | c = Odd Number Theorem
}}
{{eqn | r = n^3
}}
{{end-eqn}}
{{Qed}}
Theorem: Principle of Conservation of Angular Momentum

Newton's Laws of Motion imply the conservation of angular momentum in systems of masses in which no external force is acting.

Proof:
We start by stating Newton's Third Law of Motion in all its detail.
We consider a collection of massive bodies denoted by the subscripts $1$ to $N$.
These bodies interact with each other and exert forces on each other and these forces occur in equal and opposite pairs.
The force $F_{i j}$ exerted by body $i$ on body $j$ is related to the force exerted by body $j$ on body $i$ by:
:$(1): \quad \vec {F_{i j}} = -\vec {F_{j i} }$
The final part of Newton's Third Law of Motion is that these equal and opposite forces act through the line that connects the two bodies in question.
This can be stated thus:
:$(2): \quad \vec {F_{i j}} = a_{i j} \paren {\vec {r_j} - \vec {r_i} }$
where:
:$\vec{r_i}$ is the position of body $i$
:$a_{i j} $ is the magnitude of the force.
As a consequence of $(1)$:
:$a_{j i} = a_{i j}$
Let the total torque $\vec {\tau_{\operatorname {total} } }$ on the system be measured about an origin located at $\vec {r_0}$.
Thus:
{{begin-eqn}}
{{eqn | l = \vec {\tau_{\operatorname {total} } }
      | r = \sum_i \vec{\tau_i}
      | c = 
}}
{{eqn | r = \sum_i \paren {\paren {\vec {r_i} - \vec {r_0} } \times \sum_j \vec {F_{j i} } }
      | c = 
}}
{{eqn | r = \sum_{i, j} \paren {\paren {\paren {\vec {r_j} - \vec {r_0} } - \paren {\vec {r_j} - \vec {r_i} } } \times \vec {F_{j i} } }
      | c = 
}}
{{eqn | r = \sum_{i, j} \paren {\paren {\paren {\vec {r_j} - \vec {r_0} } - \paren {\vec {r_j} - \vec {r_i} } } \times a_{i j} \paren {\vec {r_j} - \vec {r_i} } }
      | c = 
}}
{{eqn | r = \sum_{i, j} \paren {\vec {r_j} - \vec {r_0} } \times  a_{i j} \paren {\vec {r_j} - \vec {r_i} }
      | c = 
}}
{{eqn | r = \sum_{i, j} \vec {r_j} \times  a_{i j} \paren {\vec {r_j} - \vec {r_i} } - \paren {\vec {r_0} \times \sum_{i, j} \vec {F_{i j} } }
      | c = 
}}
{{end-eqn}}
We have {{hypothesis}} that there is no external force.
Thus the second term disappears, and:
{{begin-eqn}}
{{eqn | l = \vec {\tau_{\operatorname {total} } }
      | r = \sum_{i, j} \vec {r_j} \times a_{i j} \paren {\vec {r_j} - \vec {r_i} }
      | c = 
}}
{{eqn | r = \sum_{i \mathop > j} \vec {r_j} \times a_{i j} \paren {\vec {r_j} - \vec {r_i} } + \vec {r_i} \times  a_{j i} \paren {\vec {r_j} - \vec {r_i} }
      | c = 
}}
{{eqn | r = \sum_{i \mathop > j} \vec {r_j} \times \vec {r_i} \paren {a_{j i} - a_{i j} }
      | c = 
}}
{{eqn | r = \vec 0
      | c = 
}}
{{end-eqn}}
In summary, in a system of masses in which there is no external force, the total torque on the system is equal to $0$.
This is because the pair of torque between two bodies must cancel out.
Since the rate of change of angular momentum is proportional to the torque, the angular momentum of a system is conserved when no external force is applied.
{{qed}}
Theorem: Lexicographic Order on Products of Well-Ordered Sets

Let $S$ be a set which is well-ordered by $\preccurlyeq$.
Let $\preccurlyeq_l$ be the lexicographic order on the set $T$ of all ordered tuples of $S$.
Then:
:$(1): \quad$ For a given $n \in \N: n > 0$, $\preccurlyeq_l$ is a well-ordering on the set $T_n$ of all ordered $n$-tuples of $S$;
:$(2): \quad \preccurlyeq_l$ is '''not''' a well-ordering on the set $T$ of ''all'' ordered tuples of $S$.

Proof:
It is straightforward to show that $\preccurlyeq_l$ is a total ordering on both $T_n$ and $T$.
{{handwaving|there ought to be a result here that should be linked to}}
It remains to investigate the nature of $\preccurlyeq_l$ as to whether or not it is a well-ordering.
Theorem: Basis Expansion of Irrational Number

A basis expansion of an irrational number never terminates and does not recur.

Proof:
We use a Proof by Contraposition.
Thus, we show that if a basis expansion of a (real) number terminates or recurs, then that number is rational.
{{AimForCont}} $x \in \R$ were to terminate in some number base $b$.
Then (using the notation of that definition):
:$\exists k \in \N: f_k = 0$
and so we can express $x$ precisely as:
:$x = \sqbrk {s . d_1 d_2 d_3 \ldots d_{k - 1} }_b$
This means:
:$x = s + \dfrac {d_1} b + \dfrac {d_2} {b^2} + \dfrac {d_3} {b^3} + \cdots + \dfrac {d_{k - 1} } {b^{k - 1} }$
This is the same as:
:$x = \dfrac {s b^{k - 1} + d_1 b^{k - 2} + d_2 b^{k - 3} + \cdots + d_{k - 1} } {b^{k - 1} }$
Both numerator and denominator are integers and so $x$ is a rational number.
Now suppose $x \in \R$ were to recur in some number base $b$.
Let $y$ be the recurring part of $x$.
That is, $y$ is obtained from $x$ by first subtracting $\floor x$ and then subtracting the non-recurring part of the basis expansion (if any).
Then we can express $y$ precisely as:
:$y = \sqbrk {0 . 0 0 \ldots 0 d_1 d_2 \ldots d_{k - 1} d_1 d_2 \ldots d_{k - 1} \ldots}_b$
for some number $m$ (possibly $m = 0$) of zeroes following the point.
We define:
:$z:= \sqbrk {0 . d_1 d_2 \ldots d_{k - 1} }_b = \dfrac {d_1} b + \dfrac {d_2} {b^2} + \dfrac {d_3} {b^3} + \cdots + \dfrac {d_{k - 1} } {b^{k - 1} }$
Then $z$ is a rational number.
Furthermore:
:$y = \dfrac z {b^m} + \dfrac z {b^{m + k} } + \dfrac z {b^{m + 2 k} } + \cdots = \dfrac z {b^m} \paren {1 + \dfrac 1 {b^k} + \dfrac 1 {b^{2 k} } + \cdots} = \dfrac z {b^m} \cdot \dfrac 1 {1 - \dfrac 1 {b^k} }$
so $y$ is a rational number as well.
Thus as $x - y$ is rational, $x$ is also rational.
{{Qed}}
Theorem: Determinant of Combinatorial Matrix

Let $C_n$ be the combinatorial matrix of order $n$ given by:
:<nowiki>$C_n = \begin{bmatrix}
x + y & y & \cdots & y \\
y & x + y & \cdots & y \\
\vdots & \vdots & \ddots & \vdots \\
y & y & \cdots & x + y
\end{bmatrix}$</nowiki>
Then the determinant of $C_n$ is given by:
:$\map \det {C_n} = x^{n - 1} \paren {x + n y}$

Proof:
Take the determinant $\map \det {C_n}$:
:<nowiki>$\map \det {C_n} = \begin{vmatrix}
x + y & y & y & \cdots & y \\
y & x + y & y & \cdots & y \\
y & y & x + y & \cdots & y \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & y & y & \cdots & x + y
\end{vmatrix}$</nowiki>
Subtract column $1$ from columns $2$ to $n$.
From Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant:
:<nowiki>$\map \det {C_n} = \begin{vmatrix}
x + y & -x & -x & \cdots & -x \\
y & x & 0 & \cdots & 0 \\
y & 0 & x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & 0 & 0 & \cdots & x
\end{vmatrix}$</nowiki>
Add rows $2$ to $n$ to row $1$.
Again, from Multiple of Row Added to Row of Determinant this will have no effect on the value of the determinant:
:<nowiki>$\map \det {C_n} = \begin{vmatrix}
x + n y & 0 & 0 & \cdots & 0 \\
y & x & 0 & \cdots & 0 \\
y & 0 & x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
y & 0 & 0 & \cdots & x
\end{vmatrix}$</nowiki>
This is now the determinant of a (lower) triangular matrix.
From Determinant of Triangular Matrix, it follows immediately that:
:$\map \det {C_n} = x^{n - 1} \paren {x + n y}$
{{qed}}
Theorem: Power Set is Closed under Union

Let $S$ be a set.
Let $\powerset S$ be the power set of $S$.
Then:
:$\forall A, B \in \powerset S: A \cup B \in \powerset S$

Proof:
Let $A, B \in \powerset S$.
Then by the definition of power set, $A \subseteq S$ and $B \subseteq S$.
We also have $A \cup B \subseteq S \iff A \subseteq S \land B \subseteq S$ from Union is Smallest Superset.
Thus $A \cup B \in \powerset S$, and closure is proved.
{{qed}}
Theorem: Matrix Product with Adjugate Matrix

Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Let $\adj {\mathbf A}$ be its adjugate matrix.
Then:
{{begin-eqn}}
{{eqn | l = \mathbf A \cdot \adj {\mathbf A}
      | r = \map \det {\mathbf A} \cdot \mathbf I_n
}}
{{eqn | l = \adj {\mathbf A} \cdot \mathbf A
      | r = \map \det {\mathbf A} \cdot \mathbf I_n
}}
{{end-eqn}}
where $\map \det {\mathbf A}$ is the determinant of $\mathbf A$, and $\mathbf I_n$ denotes the unit matrix of order $n$.

Proof:
Let $\mathbf A = \paren {a_{i j} }$.
Let $A_{i j}$ denote the cofactor of $a_{i j} \in \mathbf A$.
Theorem: Square of Ones Matrix

Let $\mathbf J = \sqbrk 1_n$ be a square ones matrix of order $n$.
Then $\mathbf J^2 = n \mathbf J$.
That is:
:<nowiki>$\begin{bmatrix}
1 & 1 & \cdots & 1 \\
1 & 1 & \cdots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \cdots & 1
\end{bmatrix}^2 = \begin{bmatrix}
n & n & \cdots & n \\
n & n & \cdots & n \\
\vdots & \vdots & \ddots & \vdots \\
n & n & \cdots & n
\end{bmatrix}$</nowiki>

Proof:
Follows directly from the definition of matrix multiplication:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
In this case, $m = n$ and $a_{i k} = b_{k j} = 1$.
Hence:
:$\ds c_{i j} = \sum_{k \mathop = 1}^n 1 \times 1 = n$
{{qed}}
Category:Matrix Algebra
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