nyuuzyou/wikis huggingface dataset:
Sum of Antecedents of Proportion

== Theorem ==
== Proof ==
Let a first magnitude $AB$ have to a second $C$ the same ratio as a third $DE$ has to a fourth $F$.
Let also a fifth $BG$ have to the second $C$ the same ratio as a sixth $EF$ has to a fourth $F$.
:
We need to show that $AG : C = DH : F$.
We have that $BG : C = EH : F$ and so $C : BG = F : EH$.
We also have that $AB : C = DE : F$.
So from Equality of Ratios Ex Aequali $AB : BG = DE : EH$.
From Magnitudes Proportional Separated are Proportional Compounded it follows that $AG : GB = DH : HE$.
But $BG : C = EH : F$.
So from Equality of Ratios Ex Aequali, $AG : C = DH : F$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Ratios
Sum of Arccotangents

== Theorem ==
:$\arccot a + \arccot b = \arccot \dfrac {a b - 1} {a + b}$
where $\arccot$ denotes the arccotangent.
== Proof ==
Let $x = \arccot a$ and $y = \arccot b$.
Then:
== Also see ==
* Difference of Arccotangents
Category:Arccotangent Function
Sum of Arcsecant and Arccosecant

== Theorem ==
Let $x \in \R$ be a real number such that $\size x \ge 1$.
Then:
: $\arcsec x + \arccsc x = \dfrac \pi 2$
where $\arcsec$ and $\arccsc$ denote arcsecant and arccosecant respectively.
== Proof ==
Let $y \in \R$ such that:
:$\exists x \in \R: \size x \ge 1$ and $x = \map \csc {y + \dfrac \pi 2}$
Then:
Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
Then we can write:
:$-y = \arccsc x$
But then:
:$\map \csc {y + \dfrac \pi 2} = x$
Now because $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that:
:$0 \le y + \dfrac \pi 2 \le \pi$
Hence:
:$y + \dfrac \pi 2 = \arcsec x$
That is:
:$\dfrac \pi 2 = \arcsec x + \arccsc x$
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.76$
Category:Arcsecant Function
Category:Arccosecant Function
Sum of Arcsine and Arccosine

== Theorem ==
Let $x \in \R$ be a real number such that $-1 \le x \le 1$.
Then:
:$\arcsin x + \arccos x = \dfrac \pi 2$
where $\arcsin$ and $\arccos$ denote arcsine and arccosine respectively.
== Proof ==
Let $y \in \R$ such that:
:$\exists x \in \closedint {-1} 1: x = \map \cos {y + \dfrac \pi 2}$
Then:
Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
Then we can write:
:$-y = \arcsin x$
But then:
:$\map \cos {y + \dfrac \pi 2} = x$
Now since $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that:
:$0 \le y + \dfrac \pi 2 \le \pi$
Hence:
:$y + \dfrac \pi 2 = \arccos x$
That is:
:$\dfrac \pi 2 = \arccos x + \arcsin x$
== Note ==
Note that from Derivative of Arcsine Function and Derivative of Arccosine Function, we have:
:$\map {D_x} {\arcsin x + \arccos x} = \dfrac 1 {\sqrt {1 - x^2} } + \dfrac {-1} {\sqrt {1 - x^2}} = 0$
which is what (from Derivative of Constant) we would expect.
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.74$
* : $\S 16.5 \ (3)$
Category:Arcsine Function
Category:Arccosine Function
Sum of Arctangent and Arccotangent

== Theorem ==
Let $x \in \R$ be a real number.
Then:
:$\arctan x + \arccot x = \dfrac \pi 2$
where $\arctan$ and $\arccot$ denote arctangent and arccotangent respectively.
== Proof ==
Let $y \in \R$ such that:
:$\exists x \in \R: x = \map \cot {y + \dfrac \pi 2}$
Then:
Suppose $-\dfrac \pi 2 \le y \le \dfrac \pi 2$.
Then we can write:
:$-y = \arctan x$
But then:
:$\map \cot {y + \dfrac \pi 2} = x$
Now because $-\dfrac \pi 2 \le y \le \dfrac \pi 2$ it follows that:
:$0 \le y + \dfrac \pi 2 \le \pi$
Hence:
:$y + \dfrac \pi 2 = \arccot x$
That is:
:$\dfrac \pi 2 = \arccot x + \arctan x$
== Sources ==
* : $\S 5$: Trigonometric Functions: $5.75$
Category:Arctangent Function
Category:Arccotangent Function
Sum of Arctangents

== Theorem ==
Let $\arctan a + \arctan b \in \openint {-\dfrac \pi 2} {\dfrac \pi 2}$ 
Then:
:$\arctan a + \arctan b = \map \arctan {\dfrac {a + b} {1 - a b} }$
where $\arctan$ denotes the arctangent.
== Corollary ==
== Proof ==
== Also see ==
* Difference of Arctangents
== Sources ==
* : $\text V$. Trigonometry: Formulae $(39)$
Category:Sum of Arctangents
Category:Arctangent Function
Sum of Arctangents/Corollary

== Theorem == 
Let $n \in \N_{>0}$ where $\N_{>0}$ denotes the non-zero natural numbers.
Then: 
:$\ds \map \arctan {\frac 1 n} + \map \arctan {\frac {n - 1} {n + 1} } = \frac \pi 4$
where $\arctan$ denotes the arctangent.
== Proof ==
Category:Sum of Arctangents
Category:Arctangent Function
Sum of Arctangents/Mistake

== Source Work ==
:
:$\text V$. Trigonometry: Formulae $(39)$
== Mistake ==
:$\tan^{-1} a + \tan^{-1} b = \tan^{-1} \dfrac {a + b} {1 - a b}$
== Correction ==
This is valid only if $-\dfrac \pi 2 < \tan^{-1} a + \tan^{-1} b < \dfrac \pi 2$.
Outside that range, either:
:$\size {\tan^{-1} a + \tan^{-1} b} = \dfrac \pi 2$ in which case $\tan^{-1} \dfrac {a + b} {1 - a b}$ is undefined
or:
:$\size {\tan^{-1} a + \tan^{-1} b} > \dfrac \pi 2$ in which case $\tan^{-1} \dfrac {a + b} {1 - a b}$ is either $\tan^{-1} a + \tan^{-1} b + \pi$ or $\tan^{-1} a + \tan^{-1} b - \pi$ according to whether positive or negative.
== Sources ==
* : $\text V$. Trigonometry: Formulae $(39)$
Category:Mistakes/A Note Book in Pure Mathematics
Sum of Arithmetic-Geometric Sequence

== Theorem ==
Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sum of Infinite Arithmetic-Geometric Sequence
== Linguistic Note ==
== Sources ==
* : $\S 19$: Arithmetic-Geometric Series: $19.6$
* : $\S 21$: Series of Constants: Arithmetic-Geometric Series: $21.6.$
Category:Sums of Sequences
Category:Sum of Arithmetic-Geometric Sequence
Category:Arithmetic-Geometric Sequences
Sum of Arithmetic-Geometric Sequence/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
=== Basis for the Induction ===
$\map P 1$ is the case:
demonstrating that $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P m$ is true, where $m \ge 1$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{k \mathop = 0}^{m - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^m} } {1 - r} + \frac {r d \paren {1 - m r^{m - 1} + \paren {m - 1} r^m} } {\paren {1 - r}^2}$
Then we need to show:
:$\ds \sum_{k \mathop = 0}^m \paren {a + k d} r^k = \frac {a \paren {1 - r^{m + 1} } } {1 - r} + \frac {r d \paren {1 - \paren {m + 1} r^m + m r^{m + 1} } } {\paren {1 - r}^2}$
=== Induction Step ===
This is our induction step:
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \N_{> 0}: \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
Category:Sum of Arithmetic-Geometric Sequence
Sum of Arithmetic Sequence

== Theorem ==
Let $\sequence {a_k}$ be an arithmetic sequence defined as:
:$a_k = a + k d$ for $k = 0, 1, 2, \ldots, n - 1$
Then its closed-form expression is:
== Proof ==
We have that:
:$\ds \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} = a + \paren {a + d} + \paren {a + 2 d} + \dotsb + \paren {a + \paren {n - 1} d}$
Then:
So:
Hence the result.
== Also presented as ==
== Examples ==
== Historical Note ==
== Linguistic Note ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Sum of Arithmetic Progression to $n$ Terms: $3.1.9$
* : $\S 19$: Arithmetic Series: $19.1$
* 
* : $\S 1.2.3$: Sums and Products: Example $4$. $(15)$
* 
* 
* : $\S 21$: Series of Constants: Arithmetic Series: $21.1.$
Category:Sums of Sequences
Category:Arithmetic Sequences
Category:Sum of Arithmetic Sequence
Sum of Arithmetic Sequence/Also presented as

== Sum of Arithmetic Sequence: Also presented as ==
The Sum of Arithmetic Sequence can also be seen presented in the forms:
Some present it as:
:$\ds \sum_{k \mathop = 0}^n \paren {a + k d} = a \paren {n + 1} + \dfrac 1 2 d n \paren {n + 1}$
The reality is that this is a messy result that cannot be presented elegantly.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Sum of Arithmetic Progression to $n$ Terms: $3.1.9$
* : $\S 19$: Arithmetic Series: $19.1$
* 
* : $\S 1.2.3$: Sums and Products: Example $4$. $(15)$
* 
* 
* : $\S 21$: Series of Constants: Arithmetic Series: $21.1.$
Category:Sum of Arithmetic Sequence
Sum of Arithmetic Sequence/Examples/a0 = i, d = 2+2i

== Example of Sum of Arithmetic Sequence ==
Let $A_n$ be the arithmetic sequence of $n$ terms defined as:
Then:
:$A_n = n \paren {n - 1} + n^2 i$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $3$
Category:Examples of Sum of Arithmetic Sequence
Sum of Bernoulli Numbers by Binomial Coefficients Vanishes

== Theorem ==
:$\forall n \in \Z_{>1}: \ds \sum_{k \mathop = 0}^{n - 1} \binom n k B_k = 0$
where $B_k$ denotes the $k$th Bernoulli number.
== Proof ==
Take the definition of Bernoulli numbers:
:$\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
From the definition of the exponential function:
Thus:
By Product of Absolutely Convergent Series, we will let:
Then:
Multiplying $c_n$ through by $\paren {n + 1 }!$ gives:
But those coefficients are the binomial coefficients:
Hence the result.
== Examples ==
== Also see ==
* Definition:Bernoulli Numbers/Recurrence Relation
* Sum of Euler Numbers by Binomial Coefficients Vanishes
== Sources ==
* : Chapter $\text {B}.20$: The Bernoulli Numbers and some Wonderful Discoveries of Euler
Category:Bernoulli Numbers
Category:Binomial Coefficients
Category:Sum of Bernoulli Numbers by Binomial Coefficients Vanishes
Sum of Bernoulli Numbers by Binomial Coefficients Vanishes/Examples

== Example of Use of Sum of Bernoulli Numbers by Binomial Coefficients Vanishes ==
:$\begin{array}{r|cccccccccc}
B_k & \dbinom n 0 & & \dbinom n 1 & & \dbinom n 2 & & \dbinom n 3 & & \dbinom n 4 & & \dbinom n 5 \\
\hline
B_0 = 1 & 1 B_0 & & & & & & & & & & & = 1 \\
B_1 = -\frac 1 2 & 1 B_0 & + & 2 B_1 & & & & & & & & & = 0 \\
B_2 = +\frac 1 6 & 1 B_0 & + & 3 B_1 & + & 3 B_2 & & & & & & & = 0 \\
B_3 = 0 & 1 B_0 & + & 4 B_1 & + & 6 B_2 & + & 4 B_3 & & & & & = 0 \\
B_4 = -\frac 1 {30} & 1 B_0 & + & 5 B_1 & + & 10 B_2 & + & 10 B_3 & + & 5 B_4 & & & = 0 \\
B_5 = 0 & 1 B_0 & + & 6 B_1 & + & 15 B_2 & + & 20 B_3 & + & 15 B_4 & + & 6 B_5 & = 0 \\
\end{array}$
Category:Sum of Bernoulli Numbers by Binomial Coefficients Vanishes
Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
where $B_n$ denotes the $n$th Bernoulli number.
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
== Sources ==
* : $\S 21$: Relationships of Bernoulli and Euler Numbers: $21.5$
* : $\S 23$: Bernoulli and Euler Numbers: Relationships of Bernoulli and Euler Numbers: $23.5.$
Category:Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient
Category:Bernoulli Numbers
Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient/Also presented as

== Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient: Also presented as == 
Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient can also be presented using the archaic form of the Bernoulli numbers:
== Sources ==
* : $\S 21$: Relationships of Bernoulli and Euler Numbers: $21.5$
* : $\S 23$: Bernoulli and Euler Numbers: Relationships of Bernoulli and Euler Numbers: $23.5.$
Category:Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient
Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient/Proof 1

== Theorem ==
== Proof ==
Let $B_k$ denote the kth Bernoulli number
Let $B_k \left(x\right)$ denote the kth Bernoulli polynomial
By Value of Odd Bernoulli Polynomial at One Half:
By Odd Bernoulli Numbers Vanish:
Category:Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient
Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 2$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{k \mathop = 1}^r \dbinom {2 r + 1} {2 k} 2^{2 k} B_{2 k} = 2 r$
from which it is to be shown that:
:$\ds \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \paren {r + 1} + 1} {2 k} 2^{2 k} B_{2 k} = 2 \paren {r + 1}$
=== Induction Step ===
This is the induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall n \in \Z_{>0}: \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$
Category:Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient
Sum of Big-O Estimates/Real Analysis

== Theorem ==
Let $c$ be a real number.
Let $f, g : \hointr c \infty \to \R$ be real functions.
Let $\OO$ denote big-$\OO$ notation.
Let $R_1 : \hointr c \infty \to \R$ be a real function such that $f = \map \OO {R_1}$.
Let $R_2 : \hointr c \infty \to \R$ be a real function such that $g = \map \OO {R_2}$.
Then: 
:$f + g = \map \OO {\size {R_1} + \size {R_2} }$
== Proof ==
Since: 
:$f = \map \OO {R_1}$
there exists $x_1 \in \hointr c \infty$ and a real number $C_1$ such that: 
:$\size {\map f x} \le C_1 \size {\map {R_1} x}$
for $x \ge x_1$.
Similarly, since:
:$g = \map \OO {R_2}$
there exists $x_2 \in \hointr c \infty$ and a real number $C_2$ such that: 
:$\size {\map g x} \le C_2 \size {\map {R_2} x}$
Set:
:$x_0 = \max \set {x_1, x_2}$
and:
:$C = \max \set {C_1, C_2}$
Then, for $x \ge x_0$ we have:
That is, by the definition of big-$\OO$ notation, we have: 
:$f + g = \map \OO {\size {R_1} + \size {R_2} }$
Category:Big-O Notation
Sum of Big-O Estimates/Sequences

== Theorem ==
Let $\sequence {a_n},\sequence {b_n},\sequence {c_n},\sequence {d_n}$ be sequences of real or complex numbers.
Let:
:$a_n = \map \OO {b_n}$
:$c_n = \map \OO {d_n}$
where $\OO$ denotes big-$\OO$ notation.
Then:
:$a_n + c_n = \map \OO {\size {b_n} + \size {d_n} }$
== Proof ==
Since:
:$a_n = \map \OO {b_n}$
there exists a positive real number $C_1$ and natural number $N_1$ such that: 
:$\size {a_n} \le C_1 \size {b_n}$
for all $n \ge N_1$. 
Similarly, since: 
:$c_n = \map \OO {d_n}$
there exists a positive real number $C_2$ and natural number $N_2$ such that: 
:$\size {c_n} \le C_2 \size {d_n}$
for all $n \ge N_2$.
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n \ge N$ we have: 
Let:
:$C = \max \set {C_1, C_2}$
Then:
So:
:$\size {a_n + c_n} \le C \size {\size {b_n} + \size {d_n} }$
for $n \ge N$. 
So:
:$a_n + c_n = \map \OO {\size {b_n} + \size {d_n} }$
Category:Big-O Notation
Sum of Binomial Coefficients over Lower Index

== Theorem ==
:$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$
where $\dbinom n i$ is a binomial coefficient.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.6$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Properties of Binomial Coefficients: $3.7$
* : $\text {3-1}$ Permutations and Combinations: Exercise $2$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.32$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $36$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: Properties of Binomial Coefficients: $3.7.$
Category:Sum of Binomial Coefficients over Lower Index
Category:Binomial Coefficients
Sum of Binomial Coefficients over Lower Index/Corollary

== Theorem ==
:$\ds \forall n \in \Z_{\ge 0}: \sum_{i \mathop \in \Z} \binom n i = 2^n$
where $\dbinom n i$ is a binomial coefficient.
== Proof ==
From the definition of the binomial coefficient, when $i < 0$ and $i > n$ we have $\dbinom n i = 0$.
The result follows directly from Sum of Binomial Coefficients over Lower Index.
Category:Binomial Coefficients
Category:Sum of Binomial Coefficients over Lower Index
Sum of Binomial Coefficients over Lower Index/Proof 1

== Theorem ==
== Proof ==
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 0}^n \binom n i = 2^n$
$\map P 0$ is true, as this just says $\dbinom 0 0 = 1$.
This holds by definition.
=== Basis for the Induction ===
$\map P 1$ is true, as this just says $\dbinom 1 0 + \dbinom 1 1 = 2$.
This holds by Binomial Coefficient with Zero and Binomial Coefficient with One (or Binomial Coefficient with Self).
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:
:$\ds \sum_{i \mathop = 0}^k \binom k i = 2^k$
Then we need to show:
:$\ds \sum_{i \mathop = 0}^{k + 1} \binom {k + 1} i = 2^{k + 1}$
=== 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:
:$\ds \forall n \in \N: \sum_{i \mathop = 0}^n \binom n i = 2^n$
Category:Sum of Binomial Coefficients over Lower Index
Sum of Binomial Coefficients over Lower Index/Proof 2

== Theorem ==
== Proof ==
Let $S$ be a set with $n$ elements.
From the definition of $r$-combination, $\ds \sum_{i \mathop = 0}^n \binom n i$ is the total number of subsets of $S$.
Hence $\ds \sum_{i \mathop = 0}^n \binom n i$ is equal to the cardinality of the power set of $S$.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.9$
Category:Sum of Binomial Coefficients over Lower Index
Sum of Binomial Coefficients over Lower Index/Proof 3

== Theorem ==
== Proof ==
From the Binomial Theorem, we have that:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{i \mathop = 0}^n \binom n i x^{n - i} y^i$
Putting $x = y = 1$ we get:
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Relations between coeffficients
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(a)}$
Category:Sum of Binomial Coefficients over Lower Index
Sum of Binomial Coefficients over Upper Index

== Theorem ==
Let $m \in \Z$ be an integer such that $m \ge 0$.
Then:
where $\dbinom j m$ denotes a binomial coefficient.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.34$
* : $\S 1.2.6$: Binomial Coefficients: $\text{E} \ (11)$
Category:Binomial Coefficients
Category:Sum of Binomial Coefficients over Upper Index
Sum of Binomial Coefficients over Upper Index/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\ds \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$
=== Basis for the Induction ===
$\map P 0$ says:
:$\dbinom 0 m = \dbinom 1 {m + 1}$
When $m = 0$ we have by definition:
:$\dbinom 0 0 = 1 = \dbinom 1 1$
When $m > 0$ we also have by definition:
:$\dbinom 0 m = 0 = \dbinom 1 {m + 1}$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P r$ is true, where $r \ge 0$, then it logically follows that $\map P {r + 1}$ is true.
So this is our induction hypothesis:
:$\ds \sum_{k \mathop = 0}^r \binom k m = \binom {r + 1} {m + 1}$
Then we need to show:
:$\ds \sum_{k \mathop = 0}^{r + 1} \binom k m = \binom {r + 2} {m + 1}$
=== Induction Step ===
This is our induction step:
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\ds \forall m, n \in \Z, m \ge 0, n \ge 0: \sum_{k \mathop = 0}^n \binom k m = \binom {n + 1} {m + 1}$
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $\text{E} \ (11)$
Category:Sum of Binomial Coefficients over Upper Index
Sum of Bounded Linear Transformations is Bounded Linear Transformation

== Theorem ==
Let $\mathbb F \in \set {\R, \C}$. 
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\mathbb F$. 
Let $A, B : \HH \to \KK$ be bounded linear transformations.
Let $\norm \cdot$ be the norm on the space of bounded linear transformations.
Then: 
:$A + B$ is a bounded linear transformation
with: 
:$\norm {A + B} \le \norm A + \norm B$
== Proof ==
From Addition of Linear Transformations, we have that: 
:$A + B$ is a linear transformation.
It remains to show that $A + B$ is bounded.
Let $\norm \cdot_\HH$ be the inner product norm on $\HH$. 
Let $\norm \cdot_\KK$ be the inner product norm on $\KK$.
Since $A$ is a bounded linear transformation, from Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\norm {A x}_\KK \le \norm A \norm x_\HH$
for all $x \in \HH$. 
Similarly, since $B$ is a bounded linear transformation we have: 
:$\norm {B x}_\KK \le \norm B \norm x_\HH$ 
for all $x \in \HH$. 
Let $x \in \HH$. 
Then, we have: 
So, taking $c = \norm A +\norm B$, we have: 
:$\norm {\paren {A + B} x}_\KK \le c \norm x_\HH$
for all $x \in \HH$.
So:
:$A + B$ is a bounded linear transformation.
Note that:
:$\norm A + \norm B \in \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$
while, by the definition of the norm, we have: 
:$\norm {A + B} = \inf \set {c > 0: \forall h \in \HH: \norm {\paren {A + B} h}_\KK \le c \norm h_\HH}$
So, by the definition of infimum:
:$\norm {A + B} \le \norm A + \norm B$
Category:Linear Transformations on Hilbert Spaces
Sum of Cardinals is Associative

== Theorem ==
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.
Then:
: $\mathbf a + \paren {\mathbf b + \mathbf c} = \paren {\mathbf a + \mathbf b} + \mathbf c$
where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
== Proof ==
Let $\mathbf a = \card A, \mathbf b = \card B$ and $\mathbf c = \card C$ for some sets $A$, $B$ and $C$.
Let $A, B, C$ be pairwise disjoint, that is:
:$A \cap B = \O$
:$B \cap C = \O$
:$A \cap C = \O$
Then we can define:
:$A \sqcup B := A \cup B$
:$B \sqcup C := B \cup C$
:$A \sqcup C := A \cup C$
where $A \sqcup B$ denotes the disjoint union of $A$ and $B$.
Then we have:
:$\mathbf a + \mathbf b = \card {A \sqcup B} = \card {A \cup B}$
:$\mathbf b + \mathbf c = \card {B \sqcup C} = \card {B \cup C}$
Then:
Then:
Similarly:
Then:
Finally note that from Union is Associative:
:$A \cup \paren {B \cup C} = \paren {A \cup B} \cup C$
== Sources ==
* : $\S 8$: Theorem $8.5: \ (2)$
Category:Cardinals
Sum of Cardinals is Commutative

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be cardinals.
Then:
:$\mathbf a + \mathbf b = \mathbf b + \mathbf a$
where $\mathbf a + \mathbf b$ denotes the sum of $\mathbf a$ and $\mathbf b$.
== Proof ==
Let $\mathbf a = \map \Card A$ and $\mathbf b = \map \Card B$ for some sets $A$ and $B$ such that $A \cap B = \O$.
Then:
== Sources ==
* : $\S 8$: Theorem $8.5: \ (1)$
Category:Cardinals