nyuuzyou/wikis huggingface dataset:
Taxicab Metric on Real Number Plane is not Rotation Invariant

== Theorem ==
Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.
Let $d_1$ denote the taxicab metric on $\R^2$.
Then it is not necessarily the case that:
:$\forall x, y \in \R^2: \map {d_1} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_1} {x, y}$
== Proof ==
Proof by Counterexample:
Let $x = \tuple {0, 0}$ and $y = \tuple {0, 1}$ be arbitrary points in $\R^2$.
Then:
Now let $\alpha = \dfrac \pi 4 = 45 \degrees$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 22$
Category:Geometric Rotations
Category:Taxicab Metric
Taxicab Metric on Real Vector Space is Metric

== Theorem ==
The taxicab metric on the real vector space $\R^n$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $2$
Category:Taxicab Metric
Category:Taxicab Metric on Real Vector Space is Metric
Taxicab Metric on Real Vector Space is Metric/Proof 2

== Theorem ==
== Proof ==
The taxicab metric on $\R^n$ is:
:$\ds \map {d_1} {x, y} = \sum_{i \mathop = 1}^n \size {x_i - y_i}$
for $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \R^n$.
=== Proof of ===
So holds for $d_1$.
=== Proof of ===
So holds for $d_1$.
=== Proof of ===
So holds for $d_1$.
=== Proof of ===
So holds for $d_1$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Examples $2.2.3$
Category:Taxicab Metric on Real Vector Space is Metric
Taxicab Norm is Norm

== Theorem ==
The taxicab norm is a norm on the real and complex numbers.
== Proof ==
By P-Norm is Norm, $\norm {\, \cdot \,}_p$ is a norm.
By definition, the taxicab norm is $\norm {\, \cdot \,}_1$.
Therefore, the taxicab norm is a norm.
== Also see ==
* P-Norm is Norm
* Euclidean Space is Normed Vector Space
Category:Taxicab Norm
Taylor's Theorem/One Variable

== Theorem ==
== Proof ==
=== Integral Version ===
=== Proof using Cauchy Mean Value Theorem ===
An alternative proof, which holds under milder technical assumptions on the function $f$, can be supplied using the Cauchy Mean Value Theorem.
=== Proof using Rolle's Theorem directly ===
Yet another proof for Lagrange Form of the Remainder can be constructed applying Rolle's theorem directly $n$ times; this proof might be easier to visualize geometrically.
== Also see ==
* Mean Value Theorem: Taylor's Theorem when $n = 0$
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(iii)}$ Taylor's expansion
* 
* 
* 
Category:Taylor's Theorem
Taylor's Theorem/One Variable/Integral Version

== Theorem ==
== Proof ==
This proof requires $f^{\paren n}$ to be absolutely continuous on $\closedint a x$, so that the Fundamental Theorem of Calculus holds.
Except at the end when the Mean Value Theorem is invoked, differentiability of $f^{\paren n}$ need not be assumed, since absolute continuity implies:
:differentiability almost everywhere
:the validity of the Fundamental Theorem of Calculus
provided the integrals involved are understood as Lebesgue integrals.
Consequently, the integral form of the remainder holds with this particular weakening of the assumptions on $f$.
We first prove Taylor's Theorem with the integral remainder term.
The Fundamental Theorem of Calculus states that:
:$\ds \int_a^x \map {f'} t \rd t = \map f x - \map f a$
which can be rearranged to:
:$\ds \map f x = \map f a + \int_a^x \map {f'} t \rd t$
Now we can see that an application of Integration by Parts yields:
Another application yields:
:$\ds \map f x = \map f a + \paren {x - a} \map {f'} a + \frac 1 2 \paren {x - a}^2 \map {f' '} a + \frac 1 2 \int_a^x \paren {x - t}^2 \map {f' ' '} t \rd t$
By repeating this process, we may derive Taylor's theorem for higher values of $n$.
This can be formalized by applying the technique of Principle of Mathematical Induction.
So, suppose that Taylor's theorem holds for a $n$, that is, suppose that:
We can rewrite the integral using Integration by Parts.
A primitive of $\paren {x - t}^n$ as a function of $t$ is given by $\dfrac {-\paren {x - t}^{n + 1} } {n + 1}$.
So:
The last integral can be solved immediately, which leads to
:$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$
Category:Taylor's Theorem
Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem

== Theorem ==
== Proof ==
Let $G$ be a real-valued function continuous on $\closedint a x$ and differentiable with non-vanishing derivative on $\openint a x$.
Let:
:$\map F t = \map f t + \dfrac {\map {f'} t} {1!} \paren {x - t} + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$
By the Cauchy Mean Value Theorem:
:$(1): \quad \dfrac {\map {F'} x} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$
for some $\xi \in \openint a x$.
Note that the numerator:
:$\map F x - \map F a = R_n$
is the remainder of the Taylor polynomial for $\map f x$.
On the other hand, computing $\map {F'} \xi$:
:$\map {F'} \xi = \map {f'} \xi - \map {f'} \xi + \dfrac {\map {f' '} \xi} {1!} \paren {x - \xi} - \dfrac {\map {f' '} \xi} {1!} \paren {x - \xi} + \dotsb + \dfrac {\map {f^{\paren {n + 1} } } t} {n!} \paren {x - \xi}^n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n$
Putting these two facts together and rearranging the terms of $(1)$ yields:
:$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {n!} \paren {x - \xi}^n \dfrac {\map G x - \map G a} {\map {G'} \xi}$
which was to be shown.
Note that the Lagrange Form of the Remainder comes from taking $\map G t = \paren {x - t}^{n + 1}$ and the given Cauchy Form of the Remainder comes from taking $\map G t = t - a$.
Category:Taylor's Theorem
Taylor's Theorem/One Variable/Proof by Rolle's Theorem

== Theorem ==
== Proof ==
Let the function $g$ be defined as:
:$\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$
Then:
:$\map {g^{\paren k} } a = 0$
for $k = 0, \dotsc, n$, and $\map g x = 0$.
Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$.
Then there exist:
:$\xi_1, \ldots, \xi_{n + 1}$
between $a$ and $x$ such that:
:$\map {g'} {\xi_1} = 0, \map {g' '} {\xi_2} = 0, \ldots, \map {g^{\paren {n + 1} } } {\xi_{n + 1} } = 0$
Let $\xi = \xi_{n + 1}$.
Then:
:$0 = \map {g^{\paren {n + 1} } } \xi = \map {f^{\paren {n + 1} } } \xi - \dfrac {\paren {n + 1}!} {\paren {x - a}^{n + 1} } \map {R_n} x$
and the formula for $\map {R_n} x$ follows.
== Sources ==
* : $\S 11.10$
Category:Taylor's Theorem
Taylor's Theorem/One Variable/Statement of Theorem

== Theorem ==
Let $f$ be a real function which is:
:of differentiability class $C^n$ on the closed interval $\closedint a x$
and:
:at least $n + 1$ times differentiable on the open interval $\openint a x$.
Then:
where $R_n$ (sometimes denoted $E_n$) is known as the error term or remainder, and can be presented in one of $2$ forms:
;Lagrange Form
:$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi} {\paren {n + 1}!} \paren {x - a}^{n + 1}$
for some $\xi \in \openint a x$.
;Cauchy Form
:$R_n = \dfrac {\map {f^{\paren {n + 1} } } \xi \paren {x - \xi}^n} {n!} \paren {x - a}$
for some $\xi \in \openint a x$.
=== Taylor Series Expansion ===
The expression:
:$\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {\paren {x - a}^n} {n!} \map {f^{\paren n} } a$
where $n$ is taken to the limit, is known as the Taylor series expansion of $f$ about $a$.
== Also presented as ==
== Sources ==
* : Chapter $53$: Maclaurin's and Taylor's Formulas with Remainders
* : $\S 9.7$
Category:Taylor's Theorem
Taylor's Theorem/One Variable/Statement of Theorem/Also presented as

== Taylor's Theorem in One Variable Also presented as ==
Taylor's Theorem in One Variable can also be presented in a form like this or similar:
Let $f$ be a real function which is at least $n + 1$ times differentiable on the open interval $\openint a b$.
Let $\xi$ be a real number in $\openint a b$.
Then for a given $x \in \openint a b$:
where $E_n$ satisfies:
:$E_n = \dfrac 1 {\paren {n + 1}!} \paren {x - \xi}^{n + 1} \map {f^{\paren {n + 1} } } \eta$
for some $\eta$ between $x$ and $\xi$.
== Sources ==
* : $\S 11.10$
Category:Taylor's Theorem
Taylor's Theorem/One Variable with Two Functions

== Theorem ==
Let $f$ and $g$ be real functions satisfying following conditions:
:$(1): \quad f$ is $n + 1$ times differentiable on the open interval $\openint a x$
:$(2): \quad f$ is of differentiability class $C^n$ on the closed interval $\closedint a x$
:$(3): \quad g$ is $k + 1$ times differentiable on the open interval $\openint a x$
:$(4): \quad g$ is of differentiability class $C^k$ on the closed interval $\closedint a x$
:$(5): \quad \map {g^{\paren {k + 1}}} t \ne 0$ for any $t \in \openint a x$
Then the following equation holds for some real number $\xi \in \openint a x$:
:$\dfrac {\map {f^{\paren {n + 1} } } \xi /n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k} } a} {k!} \paren {x - a}^k}$
or equivalently:
== Proof ==
We define $F$ and $G$ as follows:
Then $F$ and $G$ are continuous on $\closedint a x$ and differentiable on $\openint a x$.
Differentiating $t$:
By condition $(5)$ of the statement of the theorem, it follows that $G$ does not vanish on $\openint a x$.
By Cauchy Mean Value Theorem, there exists a real number $\xi \in \openint a x$ such that:
:$\dfrac {\map {F'} \xi} {\map {G'} \xi} = \dfrac {\map F x - \map F a} {\map G x - \map G a}$
That is:
:$\dfrac {\map {f^{\paren {n + 1} } } \xi / n!} {\map {g^{\paren {k + 1} } } \xi /k!} \paren {x - \xi}^{n - k} = \dfrac {\map f x - \map f a - \map {f'} a \paren {x - a} - \dfrac {\map {f^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {f^{\paren n} } a} {n!} \paren {x - a}^n} {\map g x - \map g a - \map {g'} a \paren {x - a} - \dfrac {\map {g^{\prime\prime} } a} {2!} \paren {x - a}^2 - \dotsb - \dfrac {\map {g^{\paren k}} a} {k!} \paren {x - a}^k}$
== Also see ==
* Definition:Lagrange Form of Remainder of Taylor Series: taking $k = n$ and $\map g x = \paren {x - a}^{n + 1}$. 
* Definition:Cauchy Form of Remainder of Taylor Series: taking $k = 0$ and $\map g x = x - a$. 
Category:Taylor's Theorem
Taylor's Theorem/Two Variables

== Theorem ==
Let $f: \R^2 \to \R$ be a real-valued function which is appropriately differentiable in a neighborhood of the point $\tuple {a, b}$.
Then for $\sqrt {h^2 + k^2} < R$ for some $R \in \R$:
== Proof ==
== Sources ==
* : $4.$ Mathematics: $4.4$ Differential calculus: $\text {(iv)}$ Taylor's series
Category:Taylor's Theorem
Taylor Expansion for Polynomials/Order 1

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $f(X)\in R[X]$ be a polynomial.
Let $a\in R$.
Then there exists a polynomial $g(X)\in R[X]$ such that:
:$f(X+a) = f(X) + a f'(X) +a^2 g(X)$
where $f'$ denotes the formal derivative of $f$.
== Proof ==
By linearity, it suffices to prove this for $f(X)=X^n$.
This is now a direct consequence of the Binomial Theorem.
Category:Polynomial Theory
Taylor Series of Analytic Function has infinite Radius of Convergence

== Theorem ==
Let $F$ be a complex function.
Let $F$ be analytic everywhere.
Let the restriction of $F$ to $\R \to \C$ be a real function $f$.
This means:
:$\forall x \in \R: \map f x = \map \Re {\map F {x, 0} }, 0 = \map \Im {\map F {x, 0} }$
where $\tuple {x, 0}$ denotes the complex number with real part $x$ and imaginary part $0$.
Let $x_0$ be a point in $\R$.
Then:
:the Taylor series of $f$ about $x_0$ converges to $f$ at every point in $\R$.
== Proof ==
The result follows by Convergence of Taylor Series of Function Analytic on Disk for the case $R = \infty$.
== Also see ==
* Taylor Series reaches closest Singularity
Category:Taylor Series
Category:Real Analysis
Taylor Series of Holomorphic Function

== Theorem == 
Let $a \in \C$ be a complex number.
Let $r > 0$ be a real number.
Let $f$ be a function holomorphic on some open ball, $D = B \paren {a, r}$. 
Then: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty \frac {\map {f^{\paren n} } a} {n!} \paren {z - a}^n$ 
 
for all $z \in D$. 
== Proof == 
In Holomorphic Function is Analytic, it is shown that: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t} \paren {z - a}^n$ 
for all $z \in D$.
From Cauchy's Integral Formula for Derivatives, we have: 
:$\ds \frac 1 {2 \pi i} \int_{\partial D} \frac {\map f t} {\paren {t - a}^{n + 1} } \rd t = \frac {\map {f^{\paren n} } a} {n!}$
Hence the result. 
Category:Holomorphic Functions
Category:Taylor Series
Taylor Series of Logarithm of Gamma Function

== Theorem ==
Let $\gamma$ denote the Euler-Mascheroni constant.
Let $\map \zeta s$ denote the Riemann zeta function.
Let $\map \Gamma z$ denote the gamma function.
Let $\Log$ denote the natural logarithm.
Then $\map \Log {\map \Gamma z}$ has the power series expansion:
which is valid for all $z \in \C$ such that $\cmod {z - 1} < 1$.
== Proof ==
Hence:
Therefore:
Setting:
:$z = \paren {x - \floor x} \leadsto \d z = \d x$
:$M = \floor x + 1$:
we have:
:$z + M = x + 1$
Hence:
Recall Stirling's Formula for Gamma Function:
We now have:
:$\ds \frac \d {\d z} \map \Log {\map \Gamma {z + 1} } = \map \Log z + \map \OO {\frac 1 z}$
Therefore:
We now have:
:$\ds \lim_{M \mathop \to \infty} \frac \d {\d z} \map \Log {\map \Gamma {z + M} } = \lim_{M \mathop \to \infty} \map \Log {z + M} $
Setting $z = 1$ into $(2)$:
Also:
:$\ds \frac {\d^{1 + k} } {\d z^{1 + k}} \map \Log {\map \Gamma {z + 1} } = \map \OO {\frac 1 z}$
shows that:
:$\ds \lim_{M \mathop \to \infty} \frac {\d^{1 + k} } {\d z^{1 + k} } \map \Log {\map \Gamma {M + 1} } = 0$
thus for $n > 1$:
Thus by definition of Taylor series:
From Zeroes of Gamma Function, we see that $\map \Gamma z$ is non-zero everywhere.
Thus $\map \Log {\map \Gamma z}$ has poles only where $\Gamma$ does, that is, the negative integers.
Since the radius of convergence of a power series is equal to the distance of its center to the closest point where the function is not analytic:
The radius of convergence of $\map \Log {\map \Gamma z}$ is $\cmod {1 - 0} = 1$.
Category:Gamma Function
Category:Natural Logarithms
Taylor Series reaches closest Singularity

== Theorem ==
Let $F$ be a complex function.
Let $F$ be analytic everywhere except at a finite number of singularities.
Let a singularity of $F$ be one of the following:
:a pole
:an essential singularity
:a branch point
In the latter case $F$ is a restriction of a multifunction to one of its branches.
Let $x_0$ be a real number.
Let $F$ be analytic at the complex number $\tuple {x_0, 0}$.
Let $R \in \R_{>0}$ be the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$.
Let the restriction of $F$ to $\R \to \C$ be a real function $f$.
This means:
:$\forall x \in \R: \map f x = \map \Re {\map F {x, 0} }, 0 = \map \Im {\map F {x, 0} }$
where $\tuple {x, 0}$ denotes the complex number with real part $x$ and imaginary part $0$.
Then:
:the Taylor series of $f$ about $x_0$ converges to $f$ at every point $x \in \R$ satisfying $\size {x - x_0} < R$
== Proof ==
We have that $F$ is analytic everywhere except at its singularities.
Also, the distance from the complex number $\tuple {x_0, 0}$ to the closest singularity of $F$ is $R$.
Therefore:
:$F$ is analytic at every point $z \in \C$ satisfying $\size {z - \tuple {x_0, 0} } < R$
where $\tuple {x_0 , 0}$ denotes the complex number with real part $x_0$ and imaginary part $0$.
The result follows by Convergence of Taylor Series of Function Analytic on Disk.
== Also see ==
* Taylor Series of Analytic Function has infinite Radius of Convergence
Category:Taylor Series
Category:Real Analysis
Telescoping Series/Example 1

== Theorem ==
Let $\sequence {b_n}$ be a sequence in $\R$.
Let $\sequence {a_n}$ be a sequence whose terms are defined as:
:$a_k = b_k - b_{k + 1}$
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = b_1 - b_{n + 1}$
If $\sequence {b_n}$ converges to zero, then:
:$\ds \sum_{k \mathop = 1}^\infty a_k = b_1$
== Proof ==
If $\sequence {b_k}$ converges to zero, then $b_{n + 1} \to 0$ as $n \to \infty$.
Thus:
:$\ds \lim_{n \mathop \to \infty} a_n = b_1 - 0 = b_1$
So:
:$\ds \sum_{k \mathop = 1}^\infty a_k = b_1$
== Linguistic Note ==
Category:Telescoping Series
Telescoping Series/Example 2

== Theorem ==
Let $\sequence {b_n}$ be a sequence in $\R$.
Let $\sequence {a_n}$ be a sequence whose terms are defined as:
:$a_k = b_k - b_{k - 1}$
Then:
:$\ds \sum_{k \mathop = m}^n a_k = b_n - b_{m - 1}$
== Proof ==
== Linguistic Note ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $19$
Category:Telescoping Series
Temperature of Body under Newton's Law of Cooling

== Theorem ==
Let $B$ be a body in an environment whose ambient temperature is $H_a$.
Let $H$ be the temperature of $B$ at time $t$.
Let $H_0$ be the temperature of $B$ at time $t = 0$.
Then:
:$H = H_a - \paren {H_0 - H_a} e^{-k t}$
where $k$ is some positive constant.
== Proof ==
By Newton's Law of Cooling:
:The rate at which a hot body loses heat is proportional to the difference in temperature between it and its surroundings.
We have the differential equation:
:$\dfrac {\d H} {\d t} \propto - \paren {H - H_a}$
That is:
:$\dfrac {\d H} {\d t} = - k \paren {H - H_a}$
where $k$ is some positive constant.
This is an instance of the Decay Equation, and so has a solution:
:$H = H_a + \paren {H_0 - H_a} e^{-k t}$
== Historical Note ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 4$: Growth, Decay and Chemical Reactions: Problem $5$
* : Chapter $1$: Mathematical Models: $\S 1.2$: Mathematical Models: Example $\S 1.2$
:::Beware - the expression given is incorrect.
Category:Thermodynamics
Tempered Distribution Space is Proper Subset of Distribution Space

== Theorem ==
Let $\map {\DD'} \R$ be the distribution space.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then $\map {\SS'} \R$ is a proper subset of $\map {\DD'} \R$:
:$\map {\SS'} \R \subsetneqq \map {\DD'} \R$
== Proof ==
By Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space we have that $\map {\SS'} \R \subseteq \map {\DD'} \R$.
Consider the real function $\map f x = e^{x^2}$.
We have that:
:Real Power Function for Positive Integer Power is Continuous
:Exponential Function is Continuous/Real Numbers
:Composite of Continuous Mappings is Continuous
Thus, $f$ is a continuous real function.
Also:
:$\forall x \in \R : e^{x^2} < \infty$
Hence, $f$ is locally integrable.
By Locally Integrable Function defines Distribution, $T_f \in \map {\DD'} \R$.
 $T_f$ is a tempered distribution.
We have that $e^{-x^2}$ is a Schwartz test function.
Then:
Hence, $\map {T_f} {e^{-x^2} } \notin \R$.
This is a contradiction.
Therefore, $T_f \notin \map {\SS'} \R$ while at the same time $T_f \in \map {\DD'} \R$.
== Sources ==
* : Chapter $\S 6.5$: A glimpse of distribution theory. Fourier transform of (tempered) distributions
Category:Tempered Distributions
Tensor Product is Module

== Theorem ==
Let $R$ be a ring.
Let $M$ be a $R$-right module.
Let $N$ be a $R$-left module.
Then:
:$\ds T = \bigoplus_{s \mathop \in M \times N} R s$
is a left module.
== Proof ==
=== Axiom 1 ===
Let $x, y \in T$ with $x = \family {s_i}_{i \mathop \in I}$ and $y = (t_i)_{i \mathop \in I}$.
Let $\lambda\in R$.
Then:
=== Axiom 2 ===
Let $x \in T$ with $x = \family {s_i}_{i \mathop \in I}$
Let $\lambda, \mu \in R$.
Then:
=== Axiom 3 ===
Let $x\in T$ with $x = \family {s_i}_{i \mathop \in I}$.
Let $\lambda, \mu \in R$.
Then:
Category:Tensor Algebra
Category:Module Theory
Tensor Product of Free Modules is Free

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $F$ and $F'$ be free $A$-modules.
Then the tensor product $F \otimes_A F'$ is a free $A$-module.
== Proof ==
By Free Module is Isomorphic to Free Module on Set there are sets $I$ and $I'$ and isomorphisms $\Psi : A^{\paren I} \to F$ and $\Psi' : A^{\paren {I'} } \to F'$.
By Tensor Product Distributes over Direct Sum, there is an isomorphism:
:$\ds A^{\paren I} \otimes_A A^{\paren {I'} } \cong \bigoplus_{i \mathop \in I} \bigoplus_{i' \mathop \in I'} A$
By Direct Sum of Direct Sums is Direct Sum, there is an isomorphism:
:$\ds \bigoplus_{i \mathop \in I} \bigoplus_{i' \mathop \in I'} A \cong \bigoplus_{\tuple {i, i'} \mathop \in I \times I'} A$
Hence $A^{\paren I} \otimes_A A^{\paren {I'} } \cong A^{\paren {I \times I'} }$ by definition of $A^{\paren {I \times I'} }$.
By Free Module on Set is Free $A^{\paren {I \times I'} }$ is free.
By Module Isomorphic to Free Module is Free $A^{\paren I} \otimes_A A^{\paren {I'} }$ is free.
Category:Commutative Algebra
Category:Free Modules
Category:Homological Algebra
Category:Module Theory
Tensor Product of Projective Modules is Projective

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $P$ and $Q$ be projective $A$-modules.
Then the tensor product $P \otimes_A Q$ is a projective $A$-module.
== Proof ==
By Projective iff Direct Summand of Free Module, there exist $A$-modules $P'$ and $Q'$, such that $P \oplus P'$ and $Q \oplus Q'$ are free.
By Tensor Product Distributes over Direct Sum, there is an isomorphism
: $\paren {P \oplus P'} \otimes_A \paren {Q \oplus Q'} \cong \paren {P \otimes_A Q} \oplus \paren {P' \otimes_A Q} \oplus \paren {P \otimes_A Q'} \oplus \paren {P' \otimes_A Q'}$
By Tensor Product of Free Modules is Free the is free.
Hence $P \otimes_A Q$ is a direct summand of a free module.
By Projective iff Direct Summand of Free Module $P \otimes_A Q$ is projective.
Category:Commutative Algebra
Tensor with Zero Element is Zero in Tensor

== Theorem ==
Let $R$ be a ring.
Let $M$ be a right $R$-module.
Let $N$ be a left $R$-module.
Let $M \otimes_R N$ denote their tensor product.
Then:
:$0\otimes_R n = m \otimes_R 0 = 0 \otimes_R 0$
is the zero in $M \otimes_R N$.
== Proof ==
Let $m \in M$ and $n \in N$
Then
Hence $0 \otimes_R n$, $m \otimes_R 0$ and $0 \otimes_R 0$ must all be identity elements for $M \otimes_R N$ as a left module.
Category:Tensor Algebra
Category:Homological Algebra
Term of Computable Real Sequence is Computable

== Theorem ==
Let $\sequence {x_i}$ be a computable real sequence.
Then, for each $i \in \N$:
:$x_i$ is a computable real number.
== Proof ==
By definition of computable real sequence, there exists a total recursive function $f : \N^2 \to \N$ such that:
:For every $m,n \in \N$, $\map f {m, n}$ codes an integer $k$ such that:
::$\dfrac {k - 1} {n + 1} < x_m < \dfrac {k + 1} {n + 1}$
Let $g : \N \to \N$ be defined as:
:$\map g n = \map f {i, n}$
By:
* Constant Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
it follows that $g$ is a total recursive function.
Additionally, for every $n \in \N$, $\map g n$ codes an integer $k$ such that:
:$\dfrac {k - 1} {n + 1} < x_i < \dfrac {k + 1} {n + 1}$
Thus, by definition, $x_i$ is a computable real number.
Category:Computability Theory
Termial on Real Numbers is Extension of Integers

== Theorem ==
The termial function as defined on the real numbers is an extension of its definition on the integers $\Z$.
== Proof ==
From the definition of the termial function on the integers:
:$\ds n? = \sum_{k \mathop = 1}^n k = 1 + 2 + \cdots + n$
From Closed Form for Triangular Numbers, we have that:
:$\ds \forall n \in \Z_{> 0}: \sum_{k \mathop = 1}^n k = \dfrac {n \paren {n + 1} } 2$
This agrees with the definition of the termial function on the real numbers.
Hence the result, by definition of extension.
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: $(10)$
Category:Termial Function
Terminal Object as Limit

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $\mathbf C$ have a terminal object $1$.
Then $1$ is the limit of the unique diagram $D: \mathbf 0 \to \mathbf C$, where $\mathbf 0$ is the zero category.
== Proof ==
== Sources ==
* : $\S 5.4$: Example $5.19$
Category:Category Theory