hcju/proofwikips huggingface dataset:
Magic Square/Examples/Order 3
== Example of Order $3$ Magic Square == Order $3$ magic square:  :$\begin{array}{|c|c|c|} \hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end{array}$  == Also known as == {{:Magic Square/Examples/Order 3/Also known as}} == Also see == * Smallest Magic Square is of Order 3 * Magic Constant of Order 3 Magic Square == Historical Note == {{:Magic Square/Examples/Order 3/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Hilbert-Waring Theorem/Particular Cases/4/Historical Note|next = Magic Constant of Order 3 Magic Square}}: $9$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Hilbert-Waring Theorem/Particular Cases/4/Historical Note|next = Magic Constant of Order 3 Magic Square}}: $9$ * {{BookReference|The Penguin Dictionary of Mathematics|1998|David Nelson|ed = 2nd|edpage = Second Edition|prev = Definition:Magic Square|next = Magic Square/Examples/Order 3/Also known as|entry = magic square}} * {{BookReference|The Penguin Dictionary of Mathematics|2008|David Nelson|ed = 4th|edpage = Fourth Edition|prev = Definition:Magic Square|next = Magic Square/Examples/Order 3/Also known as|entry = magic square}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Definition:Magic Square|next = Magic Square/Examples/Order 4/Dürer|entry = magic square}}
Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary
== Corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions == Let $f$ be a real function which is Darboux integrable on any closed interval $\mathbb I$.  Let $a_0, a_1, \ldots, a_n$ be real numbers, where $n \in \N$ and $n \ge 2$. Then: :$\displaystyle \int_{a_0}^{a_n} \map f t \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$  == Proof == Proof by induction: === Basis for the Induction === According to Sum of Integrals on Adjacent Intervals for Integrable Functions, $n = 2$ holds. This is the basis for the induction. === Induction Hypothesis === This is our induction hypothesis: :$\displaystyle \int_{a_0}^{a_k} \map f t \rd x = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$ Now we need to show true for $n = k + 1$: :$\displaystyle \int_{a_0}^{a_{k + 1} } \map f t \rd t = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \int_{a_0}^{a_{k + 1} } \map f t \rd t       | r = \int_{a_0}^{a_k} \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t + \int_{a_k}^{a_{k + 1} } \map f t \rd t       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t       | c =  }} {{end-eqn}} The result follows by induction.  == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Piecewise Continuous Function with One-Sided Limits is Darboux Integrable|next = Definition:Real Right-Hand Derivative}}: Chapter Two: $\S 1$. Piecewise-Continuous Functions: $(5)$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Improper Integral on Open Below Interval|next = Definite Integral of Constant Multiple of Real Function}}: $\S 15$: General Formulas involving Definite Integrals: $15.7$
Integration by Substitution/Corollary
== Corollary to Integration by Substitution ==  Let $f : \R \to \R$ be a real function. Let $f$ be integrable. Let $a$, $b$, and $c$ be real numbers. Then: :$\displaystyle \int_{a - c}^{b - c} \map f t \rd t = \int_a^b \map f {t - c} \rd t$  == Proof == Let $\map \phi u = u - c$. By Sum Rule for Derivatives, Derivative of Identity Function, and Derivative of Constant, we have: :$\map {\phi'} u = 1$ By Integration by Substitution: {{begin-eqn}} {{eqn | l = \int_{\map \phi a}^{\map \phi b} \map f t \rd t       | r = \int_a^b \map f {\map \phi u} \map {\phi'} u \rd u       | c =  }} {{eqn | ll = \leadsto       | l = \int_{a - c}^{b - c} \map f t \rd t       | r = \int_a^b \map f {u - c} \paren 1 \rd u       | c =  }} {{eqn | r = \int_a^b \map f {u - c} \rd u       | c =  }} {{eqn | r = \int_a^b \map f {t - c} \rd t       | c =  }} {{end-eqn}}
Finite Ordinal Times Ordinal/Lemma
Finite Ordinal Times Ordinal/Lemma 0 51259 275440 2016-11-06T09:54:20Z Kc kennylau 2331 Created page with "== Lemma ==  Let $m$ be a finite ordinal.  Let $m \ne 0$, where $0$ is the zero ordinal.   Then:  : $m \times \omeg..." wikitext text/x-wiki == Lemma == Let $m$ be a finite ordinal. Let $m \ne 0$, where $0$ is the zero ordinal. Then: : $m \times \omega = \omega$ where $\omega$ denotes the minimal infinite successor set. == Proof == {{begin-eqn}} {{eqn | ll = \forall n \in \omega       | l = m \times n       | o = \in       | r = \omega       | c = Natural Number Multiplication is Closed }} {{eqn | l = \bigcup_{n \mathop \in \omega} \left({ m \times n }\right)       | o = \le       | r = \omega       | c = Supremum Inequality for Ordinals }} {{eqn | ll = \implies       | l = m \times \omega       | o = \le       | r = \omega       | c = Definition of Ordinal Multiplication }} {{end-eqn}} Also, $\omega \le \left({ m \times \omega }\right)$ by Subset is Right Compatible with Ordinal Multiplication. The lemma follows from the definition of equality. {{qed|lemma}}
Binary Logarithm/Examples/10
== Example of Binary Logarithm ==  The binary logarithm of $10$ is: :$\log_2 10 \approx 3 \cdotp 32192 \, 80948 \, 87362 \, 34787 \, 0319 \ldots$  {{OEIS|A020862}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Square Root/Examples/10|next = Number of Binary Digits in Power of 10}}: $3 \cdotp 321 \, 928 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Square Root/Examples/10|next = Number of Binary Digits in Power of 10}}: $3 \cdotp 32192 \, 8 \ldots$
Bound for Analytic Function and Derivatives
Bound for Analytic Function and Derivatives 0 51357 435939 435768 2019-11-21T10:51:19Z Ivar Sand 2302 Refactor template removed wikitext text/x-wiki == Lemma == Let $f$ be a complex function. Let $z_0$ be a point in $\C$. Let $r$ be a real number in $\R_{>0}$. Let $\Gamma$ be a circle in $\C$ with center at $z_0$ and radius $r$. Let $f$ be analytic on $\Gamma$ and its interior. Let $t \in \C$ be such that $\cmod {t - z_0} < r$. Then a real number $M$ exists such that, for every $n \in \N$: :$\displaystyle \cmod {\map {f^{\paren n} } t} \le \frac {M r \, n!} {\paren {r - \cmod {t - z_0} }^\paren {n + 1} }$ == Proof == === Lemma (Analytic Function Bounded on Circle) === {{:Bound for Analytic Function and Derivatives/Analytic Function Bounded on Circle}}{{qed|lemma}} We have: :$f$ is analytic on $\Gamma$ and its interior :$t$ is in the interior of $\Gamma$ Therefore: :$\displaystyle \map {f^{\paren n} } t = \frac {n!} {2 \pi i} \int_\Gamma \frac {\map f z} {\paren {z - t}^{\paren {n + 1} } } \rd z$ by Cauchy's Integral Formula for Derivatives where $\Gamma$ is traversed counterclockwise. We have that $f$ is bounded on $\Gamma$ by Lemma (Analytic Function Bounded on Circle). Therefore, there is a positive real number $M$ that satisfies: :$M \ge \cmod {\map f z}$ for every $z$ on $\Gamma$ We have $\cmod {t - z_0} < r$. Therefore: :$0 < r - \cmod {t - z_0}$ We observe that $r - \cmod {t - z_0}$ is the minimum distance between $t$ and $\Gamma$. Therefore: :$\paren {r - \cmod {t - z_0} } \le \cmod {z - t}$ for every $z$ on $\Gamma$ We get: {{begin-eqn}} {{eqn | l = \cmod {\map {f^{\paren n} } t}       | r = \cmod {\frac {n!} {2 \pi i} \int_\Gamma \frac {\map f z} {\paren {z - t}^{\paren {n + 1} } } \rd z} }} {{eqn | o = \le       | r = \frac {n!} {2 \pi} \int_\Gamma \frac {\cmod {\map f z} } {\cmod {z - t}^{\paren {n + 1} } } \cmod {\d z} }} {{eqn | o = \le       | r = \frac {n!} {2 \pi} \int_\Gamma \frac M {\cmod {z - t}^{\paren {n + 1} } } \cmod {\d z}       | c = as $M \ge \cmod {\map f z}$ for every $z$ on $\Gamma$ }} {{eqn | o = \le       | r = \frac {n!} {2 \pi} \int_\Gamma \frac M {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } \cmod {\d z}       | c = as $0 < \paren {r - \cmod {t - z_0} } \le \cmod {z - t}$ for every $z$ on $\Gamma$ }} {{eqn | r = \frac {n!} {2 \pi} \frac M {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } \int_\Gamma \cmod {\d z} }} {{eqn | r = \frac {n!} {2 \pi} \frac M {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } 2 \pi r }} {{eqn | r = \frac {M r \, n!} {\paren {r - \cmod {t - z_0} }^{\paren {n + 1} } } }} {{end-eqn}}  {{explain|The notation $\cmod {\d z}$ has not been raised before on  -- we need to do something about that}}
Pythagorean Triangle/Examples/3-4-5
== Example of Primitive Pythagorean Triangle ==  The triangle whose sides are of length $3$, $4$ and $5$ is a primitive Pythagorean triangle.  :300px == Proof == {{begin-eqn}} {{eqn | l = 3^2 + 4^2       | r = 9 + 16       | c =  }} {{eqn | r = 25       | c =  }} {{eqn | r = 5^2       | c =  }} {{end-eqn}} It follows by Pythagoras's Theorem that $3$, $4$ and $5$ form a Pythagorean triple. Note that $3$ and $4$ are coprime. Hence, by definition, $3$, $4$ and $5$ form a primitive Pythagorean triple. The result follows by definition of a primitive Pythagorean triangle.  == Also see == * Smallest Pythagorean Triangle is 3-4-5 * Pythagorean Triangle whose Area is Half Perimeter: its area and semiperimeter are both $6$ * Pythagorean Triangle with Sides in Arithmetic Sequence == Historical Note == {{:Pythagorean Triangle/Examples/3-4-5/Historical Note}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Definition:Pythagorean Triangle|next = Pythagoras's Theorem}}: $5$ * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = 6/Historical Note|next = Definition:Perfect Number}}: $6$ * {{BookReference|Curious and Interesting Puzzles|1992|David Wells|prev = Definition:Pythagorean Triple|next = Babylonian Mathematics/Examples/Sliding Ladder}}: Pythagorean Triples * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Definition:Pythagorean Triangle|next = Pythagoras's Theorem}}: $5$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = 6/Historical Note|next = Definition:Perfect Number}}: $6$
Volume of Unit Hypersphere/Sequence
== Sequence of Volumes of Unit Hyperspheres ==  The sequence of volumes of the unit sphere in $n$-dimensional space begins as follows: {{begin-eqn}} {{eqn | l = n = 1       | o = :       | r = \map V 1 = 2 }} {{eqn | l = n = 2       | o = :       | r = \map V 2 = 3.1 }} {{eqn | l = n = 3       | o = :       | r = \map V 3 = 4.2 }} {{eqn | l = n = 4       | o = :       | r = \map V 4 = 4.9 }} {{eqn | l = n = 5       | o = :       | r = \map V 5 = 5.264 }} {{eqn | l = n = 6       | o = :       | r = \map V 6 = 5.2 }} {{eqn | l = n = 7       | o = :       | r = \map V 7 = 4.7 }} {{end-eqn}}  == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Maximum Volume of Unit Radius Sphere in Fractional Dimensions|next = 6}}: $5 \cdotp 256 \, 946 \, 404 \, 860 \ldots$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Maximum Volume of Unit Radius Sphere in Fractional Dimensions|next = 6}}: $5 \cdotp 25694 \, 64048 \, 60 \ldots$
X to the x is not of Exponential Order/Lemma
== Lemma ==  Let $f: \R \to \R$ be defined on $\left [{0 \,.\,.\, \to} \right)$ with $f \left({x}\right) = x^x$. Suppose there exist strictly positive real constants $M, K, a \in \R_{> 0}$ such that: :$\forall t \ge M: \left\vert {f \left({t}\right)} \right \vert < K e^{a t}$ Then there exists a constant $C$ such that: :$\forall t > C: \left\vert {f \left({t}\right)} \right \vert > K e^{a t}$  == Proof == By the definition of power: :$f \left({t}\right) = \exp \left({t \ln t}\right)$  By Exponential of Real Number is Strictly Positive, we can reduce the lemma into the existence of $C$ such that: :$\forall t > C: f \left({t}\right) > K e^{a t}$ We will divide into two cases. === Case 1: $K > 1$ === Assume that $t > K e^a$. {{begin-eqn}} {{eqn | l = a       | o = >       | r = 0       | c =  }} {{eqn | l = e^a       | o = >       | r = e^0       | c = Exponential is Strictly Increasing }} {{eqn | l = e^a       | o = >       | r = 1       | c = Exponential of Zero }} {{eqn | l = K e^a       | o = >       | r = 1       | c = As $K > 1$ }} {{eqn | l = t       | o = >       | r = 1       | c = As $t > K e^a$ }} {{eqn | n = 1       | l = t \ln K       | o = >       | r = \ln K       | c =  }} {{eqn | l = t       | o = >       | r = K e^a       | c = Assumption }} {{eqn | l = \ln t       | o = >       | r = \ln \left({K e^a}\right)       | c = Logarithm is Strictly Increasing }} {{eqn | l = \ln t       | o = >       | r = a + \ln K       | c =  }} {{eqn | l = t \ln t       | o = >       | r = a t + t \ln K       | c =  }} {{eqn | l = t \ln t       | o = >       | r = a t + \ln K       | c = from $(1)$ }} {{eqn | l = \exp \left({t \ln t}\right)       | o = >       | r = \exp \left({a t + \ln K}\right)       | c = Exponential is Strictly Increasing }} {{eqn | l = f \left({t}\right)       | o = >       | r = K e^{a t}       | c = Exponential is Strictly Increasing }} {{end-eqn}} Here, $C = K e^a$. === Case 2: $K \le 1$ === Assume that $t > e^a$. {{begin-eqn}} {{eqn | l = \ln K       | o = \le       | r = \ln 1       | c = Logarithm is Strictly Increasing }} {{eqn | n = 1       | l = \ln K       | o = \le       | r = 0       | c = Logarithm of 1 is 0 }} {{eqn | l = t       | o = >       | r = e^a       | c = by assumption }} {{eqn | l = \ln t       | o = >       | r = a       | c = Logarithm is Strictly Increasing }} {{eqn | l = t \ln t       | o = >       | r = a t       | c =  }} {{eqn | l = t \ln t       | o = >       | r = a t + \ln k       | c = from $(1)$ }} {{eqn | l = \exp \left({t \ln t}\right)       | o = >       | r = \exp \left({a t + \ln K}\right)       | c = Exponential is Strictly Increasing }} {{eqn | l = f \left({t}\right)       | o = >       | r = K e^{a t}       | c = Exponential is Strictly Increasing }} {{end-eqn}} Here, $C = e^a$.
Schanuel's Conjecture
== Conjecture == Let $z_1, \cdots, z_n$ be complex numbers that are linearly independent over the rational numbers $\Q$. Then: :the extension field $\map \Q {z_1, \cdots, z_n, e^{z_1}, \cdots, e^{z_n} }$ has transcendence degree at least $n$ over $\Q$ where $e^z$ is the complex exponential of $z$. {{Namedfor|Stephen Hoel Schanuel|cat = Schanuel}}
Characterization of Cosine Integral Function
== Definition == Let $\Ci: \R_{>0}: \R$ denote the cosine integral function: :$\map \Ci x = \displaystyle \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {\cos t} t \rd t$ Then: :$\map \Ci x = -\gamma - \ln x + \displaystyle \int_{t \mathop \to 0}^{t \mathop = x} \frac{1 - \cos t} t \rd t$ where $\gamma$ is the Euler-Macheroni constant. == Proof == {{proof wanted|Laplace transform to the one, then invert to the other?}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Definition:Cosine Integral Function|next = Power Series Expansion for Cosine Integral Function plus Logarithm}}: $\S 35$: Miscellaneous Special Functions: Cosine Integral: $35.14$ * {{MathWorld|Cosine Integral|CosineIntegral}}
Sum of Terms of Magic Square/Sequence
== Sequence of Sums of Terms of Magic Squares ==  The sequence of the sum totals of all the entries in magic squares of order $n$ begins: :$1, \paren {10,} \, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10 \, 440, 14 \, 365, 19 \, 306, 25 \, 425, 32 \, 896, \ldots$ However, note that while $10 = \dfrac {2^2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist.  {{OEIS|A037270}}
Magic Constant of Magic Square/Sequence
== Sequence of Magic Constants of Magic Squares ==  The sequence of the magic constants of magic squares of order $n$ begins: :$1, (5,) \, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, \ldots$ However, note that while $5 = \dfrac {2 \paren {2^2 + 1} } 2$, a magic square of order $2$ does not actually exist.  {{OEIS|A006003}}
Schanuel's Conjecture Implies Transcendence of Pi to the power of Euler's Number/Lemma
== Lemma == Let Schanuel's Conjecture be true. Let $z_1 = \ln \ln \pi$, $z_2 = 1 + \ln \ln \pi$, $z_3 = \ln \pi$, $z_4 = e \ln \pi$, and $z_5 = i \pi$. Then, $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$. == Proof == Assume the truth of Schanuel's Conjecture. Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. Equivalently, they are linearly independent over the integers $\Z$. Let $a, b \in \Z$ such that: :$a z_1 + b z_3 = 0$ Substituting: :$a \ln \ln \pi + b \ln \pi = 0$ Applying the exponential function to both sides: :$\paren {\ln \pi}^a \pi^b = 1$ By Schanuel's Conjecture Implies Algebraic Independence of Pi and Log of Pi over the Rationals, the above equation only has solution when $a = b = 0$. Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. Since $z_5$ is wholly imaginary, $z_1$, $z_3$, and $z_5$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_3, z_5, e^{z_1}, e^{z_3}, e^{z_5} }$ has transcendence degree at least $3$ over the rational numbers $\Q$. That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, \ln \pi, \pi, e^{i \pi} }$ has transcendence degree at least $3$ over $\Q$. However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic. Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$. Therefore, $\ln \ln \pi$, $\ln \pi$, and $i \pi$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, $z_5$, and $1$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_3, z_5, 1, e^{z_1}, e^{z_3}, e^{z_5}, e}$ has transcendence degree at least $4$ over the rational numbers $\Q$. That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, 1, \ln \pi, \pi, e^{i \pi}, e}$ has transcendence degree at least $4$ over $\Q$. However, $1$ is algebraic. Moreover, by Euler's Identity, $e^{i \pi} = -1$ is algebraic. Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$. Therefore, $\ln \ln \pi$, $\ln \pi$, $i \pi$, and $e$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$, $z_5$, and $e$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, $e z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$. That is, $z_1$, $z_3$, $z_4$, and $z_5$ must be algebraically independent over $\Q$. Therefore, $z_1$, $1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$. Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$. That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$. {{qed|lemma}}
Schanuel's Conjecture Implies Transcendence of 2 to the power of Euler's Number/Lemma
{{MissingLinks|or rather, correct the existing links}} == Lemma == Let: :$z_1 = \ln \ln 2$ :$z_2 = 1 + \ln \ln 2$ :$z_3 = \ln 2$ :$z_4 = e \ln 2$ Let Schanuel's Conjecture be true. Then $z_1$, $z_2$, $z_3$ and $z_4$ are linearly independent over the rational numbers $\Q$. == Proof == Assume the truth of Schanuel's Conjecture. $2$ is algebraic. Hence, by the Corollary of the weaker Hermite-Lindemann-Weierstrass theorem, $\ln 2$ is transcendental. Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. Equivalently, by Linearly Independent over the Rational Numbers iff Linearly Independent over the Integers, they are linearly independent over the integers $\Z$. Let $a, b \in \Z$ such that: :$a z_1 + b z_3 = 0$ Substituting: :$a \ln \ln 2 + b \ln 2 = 0$ Applying the exponential function to both sides: :$\left({\ln 2}\right)^a 2^b = 1$ Since $\ln 2$ is transcendental, the above equation only has solution when $a = b = 0$. Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, e^{z_1}, e^{z_3}}\right)$ has transcendence degree at least $2$ over the rational numbers $\Q$. That is, the extension field $\Q \left({\ln \ln 2, \ln 2, \ln 2, 2}\right)$ has transcendence degree at least $2$ over $\Q$. However, $2$ is algebraic. Therefore, $\ln \ln 2$ and $\ln 2$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, and $1$ are linearly independent over the rational numbers $\Q$. By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, 1, e^{z_1}, e^{z_3}, e}\right)$ has transcendence degree at least $3$ over the rational numbers $\Q$. That is, the extension field $\Q \left({\ln \ln 2, \ln 2, 1, \ln 2, 2, e}\right)$ has transcendence degree at least $3$ over $\Q$. However, $1$ and $2$ are algebraic. Therefore, $\ln \ln 2$, $\ln 2$, $e$ must be algebraically independent over $\Q$. That is, $z_1$, $z_3$, and $e$ must be algebraically independent over rational numbers $\Q$. It follows that $z_1$, $z_3$, $e z_3$ must be algebraically independent over rational numbers $\Q$. That is, $z_1$, $z_3$, $z_4$ must be algebraically independent over $\Q$. Therefore, $z_1$, $1$, $z_3$, $z_4$ must be linearly independent over $\Q$. Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$ must be linearly independent over $\Q$. That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$ are linearly independent over the rational numbers $\Q$. {{qed|lemma}}
Divisor Counting Function/Examples/12
Divisor Counting Function/Examples/12 0 52622 451779 292229 2020-03-02T09:43:28Z Prime.mover 59 Prime.mover moved page Tau Function/Examples/12 to Divisor Counting Function/Examples/12 wikitext text/x-wiki == Example of Use of $\tau$ Function ==  :$\tau \left({12}\right) = 6$  where $\tau$ denotes the $\tau$ Function. == Proof == From Tau Function from Prime Decomposition: :$\displaystyle \tau \left({n}\right) = \prod_{j \mathop = 1}^r \left({k_j + 1}\right)$ where: :$r$ denotes the number of distinct prime factors in the prime decomposition of $n$ :$k_j$ denotes the multiplicity of the $j$th prime in the prime decomposition of $n$. We have that: :$12 = 2^2 \times 3$ Thus: :$\tau \left({12}\right) = \tau \left({2^2 \times 3^1}\right) = \left({2 + 1}\right) \left({1 + 1}\right) = 6$ The divisors of $12$ can be enumerated as: :$1, 2, 3, 4, 6, 12$