hcju/proofwikiqa huggingface dataset:
By definition of upper closure of subset: :$\exists u \in \operatorname{fininfs}\left({F \cup X}\right): u \preceq x$ By definition of finite infima set: :there exists finite subset $Y$ of $F \cup X$: :$Y$ admits an infimum and $u = \inf Y$ We will prove that :$Y \setminus X \subseteq F$ Let $a \in Y \setminus X$. By definition of defference: :$a \in Y$ and $a \notin X$ By definition of subset: :$a \in F \cup X$ Thus by definition of union: :$a \in F$ {{qed|lemma}} Define $Z := Y \setminus X$. Case $Z = \varnothing$: By $F$ is non-empty: :$\exists b: b \in F$ By Meet Precedes Operands: :$b \wedge \inf X \preceq \inf X$ By Set Difference with Superset is Empty Set: :$Y \subseteq X$ By Existence of Non-Empty Finite Infima in Meet Semilattice: :$X$ admits an infimum. By Infimum of Subset: :$\inf X \preceq u$ Thus by definition of trantisivity: :$b \wedge \inf X \preceq x$ {{qed|lemma}} Case $Z \ne \varnothing$: By Set Difference is Subset: :$Z \subseteq Y$ By Subset of Finite Set is Finite: :$Z$ is finite. By Union of Finite Sets is Finite: :$Z \cup X$ is finite. By definitions of non-empty set and union: :$Z \cup X \ne \varnothing$ By Existence of Non-Empty Finite Infima in Meet Semilattice: :$X$ admits an infimum and :$Z \cup X$ admits an infimum. By Set is Subset of Union: :$Y \subseteq Y \cup X$ By Set Difference Union Second Set is Union: :$Y \subseteq Z \cup X$ By Infimum of Subset: :$\inf\left({Z \cup X}\right) \preceq u$ By Existence of Non-Empty Finite Infima in Meet Semilattice: :$Z$ admits an infimum By Infimum of Infima: :$\inf\left({Z \cup X}\right) = \left({\inf Z}\right) \wedge \inf X$ By Filtered in Meet Semilattice with Finite Infima: :$\inf Z \in F$ By definition of transitivity: :$\left({\inf Z}\right) \wedge \inf X \preceq x$ Thus :$\exists a \in S: a \in F \land x \succeq a \wedge \inf X$ {{qed}}
Since $\dfrac 1 n$ terminates after $m$ digits: :$\dfrac {10^m} n$ is an integer :$\dfrac {10^{m - 1}} n$ is not an integer From the first condition, we have $n = 2^p 5^q$ for some positive integers $p, q \le m$. This gives $m \ge \max \set {p, q}$. From the second condition, we cannot have both $p, q \le m - 1$. Therefore at least one of $p, q$ is equal to $m$. This gives $m \le \max \set {p, q}$. These results give $m = \max \set {p, q}$. {{qed}}
{{begin-eqn}} {{eqn | lo= \forall t \ge 1 :       | l = \size {e^{\psi t} }       | r = e^{a t}       | c = Modulus of Exponential is Exponential of Real Part }} {{eqn | o = <       | r = 2 e^{a t}       | c = Exponential of Real Number is Strictly Positive }} {{end-eqn}} The result follows from the definition of exponential order with $M = 1$, $K = 2$, and $a = a$. {{qed}} Category:Exponential Order o10hsbzpv7f6ttuwhf1ixrosismn783
{{begin-eqn}} {{eqn | lo= \forall t \ge 1:       | l = \size {\cos t}       | o = \le       | r = 1       | c = Real Cosine Function is Bounded }} {{eqn | ll= \leadsto       | l = \size {\cos t}       | o = <       | r = 2 }} {{eqn | r = 2 e^{0 t}       | c = Exponential of Zero }} {{end-eqn}} The result follows from the definition of exponential order with $M = 1$, $K = 2$, and $a = 0$. {{qed}}
The result follows from Real Cosine Function is Bounded and Bounded Function is of Exponential Order Zero. {{qed}}
{{begin-eqn}} {{eqn | l = \size {\sin t}       | o = \le       | r = 1       | c = Real Sine Function is Bounded }} {{eqn | ll =\leadsto       | l = \size {\sin t}       | o = <       | r = 2 }} {{eqn | r = 2 e^{0 t}       | c = Exponential of Zero }} {{end-eqn}} {{qed}}
If $\lambda = 0$, the theorem holds trivially. Let $\lambda \ne 0$. {{begin-eqn}} {{eqn | l = \size {\map f t}       | o = <       | r = K e^{a t}       | c = {{Defof|Exponential Order to Real Index}} }} {{eqn | ll= \leadsto       | l = \size \lambda \size {\map f t}       | o = <       | r = \size \lambda K e^{a t} }} {{eqn | ll= \leadsto       | l = \size {\lambda \, \map f t}       | o = <       | r = K' e^{a t}       | c = Modulus of Product, $K' = \size \lambda K$ }} {{end-eqn}} {{qed}} Category:Exponential Order mp7qgw2si88xgwoe25abb571f11yrtc
{{begin-eqn}} {{eqn | l = \size {\map f t}       | o = <       | r = K e^{a t}       | c = {{Defof|Exponential Order to Real Index}} }} {{eqn | ll= \leadsto       | l = \size {\map f {\lambda t} }       | o = <       | r = K e^{a \lambda t}       | c = replacing $t$ with $\lambda t$ }} {{end-eqn}} The result follows by the definition of exponential order. {{qed}} Category:Exponential Order ddyn9d7knsd6ra2jdgv0fe7a5nmip37
{{begin-eqn}} {{eqn | l = e^{\epsilon t}       | o = \ge       | r = 1 + \epsilon t       | c = Exponential of $t$ not less than $1 + t$ }} {{eqn | o = >       | r = \epsilon t }} {{eqn | ll= \leadsto       | l = K e^{\epsilon t}       | o = >       | r = t       | c = $K = \dfrac 1 \epsilon$ }} {{end-eqn}} {{qed}} Category:Exponential Order id26q54j2usaswssa7j8ydooc0fbkys
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval. By the definition of exponential order: {{begin-eqn}} {{eqn | l = \size {\map f t}       | o = <       | r = K_1 e^{a t} }} {{eqn | l = \size {\map g t}       | o = <       | r = K_2 e^{b t} }} {{eqn | ll= \leadsto       | l = \size {\map f t} \size {\map g t}       | o = <       | r = K_1 K_2 e^{a t} e^{b t}       | c = Positive Real Number Inequalities can be Multiplied }} {{eqn | ll= \leadsto       | l = \size {\map f t \, \map g t}        | o = <       | r = K' e^{\paren {a + b} t}       | c = Modulus of Product, Exponential of Sum, $K' = K_1 K_2$ }} {{end-eqn}} {{qed}} Category:Exponential Order niuixbh8q776zzq26qdxcjxctxjl0q1
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval. By the definition of exponential order: {{begin-eqn}} {{eqn | l = \size {\map f t}       | o = <       | r = K_1 e^{a t} }} {{eqn | l = \size {\map g t}       | o = <       | r = K_2 e^{b t} }} {{eqn | l = \size {\map f t} + \size {\map g t}       | o = <       | r = K_1 e^{a t} + K_2 e^{b t}       | c = Real Number Inequalities can be Added }} {{eqn | ll= \leadsto       | l = \size {\map f t + \map g t}       | o = <        | r = K_1 e^{\max \size {a, b} t} + K_2 e^{\max \size {a, b} t}       | c = Triangle Inequality for Real Numbers, Exponential is Strictly Increasing }} {{eqn | r = K' e^{\max \size {a, b} t}       | c = $K' = K_1 + K_2$ }} {{end-eqn}} {{qed}} Category:Exponential Order ev9cglofxub0yhotwdqt0tv8ec14ek9
Follows from: * Scalar Multiple of Function of Exponential Order * Sum of Functions of Exponential Order {{qed}} Category:Exponential Order l0rjg4v33skvpaty328lncl3s7n5zut
{{begin-eqn}} {{eqn | ll= \forall t \ge 1 :       | l = \size C       | o = <       | r = \size C + 1 }} {{eqn | r = \paren {\size C + 1} e^{0 t}       | c = Exponential of Zero }} {{end-eqn}} The result follows from the definition of exponential order, with $M = 1$, $K = \size C + 1$, and $a = 0$. {{qed}} Category:Exponential Order 21nfl3n3zd2qyw4fz7tm6yyetnnu7ry
If $P = 0$, the theorem holds trivially.  Let $P_n$ be a polynomial of degree $n$, where $n \ge 0$. The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial. === Basis for the Induction === Let $P_0$ be of degree zero. Then $P_0$ is a constant polynomial. By Constant Function is of Exponential Order Zero, $P_0 \in \mathcal E_0$. Therefore, by Raising Exponential Order, it is of exponential order $\epsilon$ as well. This is the basis for the induction. === Induction Hypothesis === Fix $n \in \N$ with $n \ge 0$. Assume: :$P_n \in \mathcal E_\epsilon$ That is: :$\size {\map {P_n} t} < K e^{\epsilon t}$ for some $K > 0$, for $\epsilon > 0$ arbitrarily small. This is our induction hypothesis. === Induction Step === Let $P_{n + 1}$ be of degree $n + 1$. By the definition of polynomial, :$P_{n + 1} = P_n + a_{n + 1} x^{n + 1}$ for some polynomials of degree $n$. $P_n$ is of exponential order $\epsilon$ by the induction hypothesis. Thus, by: :Sum of Functions of Exponential Order :Scalar Multiple of Function of Exponential Order :Natural Number Power is of Exponential Order Epsilon We have that $P_{n + 1}$ is of degree $\epsilon$. The result follows by the Principle of Mathematical Induction. {{qed}} Category:Exponential Order Category:Proofs by Induction jecyo3kibyn6re3xkbrhvioc4fg6hyj
{{begin-eqn}} {{eqn | l = 2^{n - 1} F_n       | r = \dfrac {2^n} {2 \sqrt 5} \left({\phi^n - \hat \phi^n}\right)       | c = Euler-Binet Formula }} {{eqn | r = \dfrac {\left({1 + \sqrt 5}\right)^n - \left({1 - \sqrt 5}\right)^n} {2 \sqrt 5}       | c = {{Defof|Golden Mean|index = 2}} }} {{eqn | r = \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \binom n j \sqrt 5^j - \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \left({-1}\right)^j \binom n j \sqrt 5^j       | c = Binomial Theorem }} {{eqn | r = \dfrac 1 {\sqrt 5} \sum_{\substack {0 \mathop \le j \mathop \le n \\ j \text { odd} } } \binom n j \sqrt 5^j       | c = even terms vanish, odd terms double up }} {{eqn | r = \dfrac 1 {\sqrt 5} \sum_{j \text { odd} } \binom n j 5^{j / 2}       | c = {{Defof|Binomial Coefficient}}: $\dbinom n j = 0$ for $j < 0$ and $j > n$ }} {{eqn | r = \sum_{j \text { odd} } \binom n j 5^{\left({j - 1}\right) / 2}       | c = gathering the spare $\sqrt 5$ into the index }} {{end-eqn}} Setting $j = 2 k + 1$ for $0 \le k \le \left({j - 1}\right) / 2$ gives: :$\displaystyle \sum_{k \mathop \ge 0} \binom n {2 k + 1} 5^k$ and the limits of the index of the summation are irrelevant, as $\dbinom n {2 k + 1} = 0$ for $j < 0$ and $j > n$. Hence the result.
The aliquot sum of an integer $n$ is the integer sum  of the aliquot parts of $n$. That is, the aliquot sum of $2016$ is the $\sigma$ function of $2016$ minus $2016$. Thus: {{begin-eqn}} {{eqn | l = \map \sigma {2016} - 2016       | r = 6552 - 2016       | c = {{SigmaLink|2016}} }} {{eqn | r = 4536       | c =  }} {{eqn | ll= \leadsto       | l = 4 \times \paren {\map \sigma {2016} - 2016}       | r = 18144       | c =  }} {{eqn | r = 9 \times 2016       | c =  }} {{end-eqn}} {{qed}}
From Fibonacci Number in terms of Smaller Fibonacci Numbers: :$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ Setting $m = n$: {{begin-eqn}} {{eqn | l = F_{2 n}       | r = F_{n - 1} F_n + F_n F_{n + 1}       | c =  }} {{eqn | r = F_n \paren {F_{n + 1} + F_{n - 1} }       | c =  }} {{eqn | r = \paren {F_{n + 1} - F_{n - 1} } \paren {F_{n + 1} + F_{n - 1} }       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = {F_{n + 1} }^2 - {F_{n - 1} }^2       | c = Difference of Two Squares }} {{end-eqn}} {{qed}}
From Fibonacci Number in terms of Smaller Fibonacci Numbers: :$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ Setting $m = 2 n$: {{begin-eqn}} {{eqn | l = F_{3 n}       | r = F_{2 n + n}       | c =  }} {{eqn | r = F_{2 n - 1} F_n + F_{2 n} F_{n + 1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = F_{n + \paren{n - 1} } F_n + F_{2 n} F_{n + 1}       | c =  }} {{eqn | r = \paren { {F_{n - 1} }^2 + {F_n}^2} F_n + F_{2 n} F_{n + 1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = \paren { {F_{n - 1} }^2 + {F_n}^2} F_n + \paren { {F_{n + 1} }^2 - {F_{n - 1} }^2} F_{n + 1}       | c = Fibonacci Number of Index 2n as Sum of Squares of Fibonacci Numbers }} {{eqn | r = {F_{n - 1} }^2 F_n + {F_n}^3 + {F_{n + 1} }^3 - {F_{n - 1} }^2 F_{n + 1}       | c = rearranging }} {{eqn | r = {F_{n - 1} }^2 F_n + {F_n}^3 + {F_{n + 1} }^3 - {F_{n - 1} }^2 \paren {F_n + F_{n - 1} }       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = {F_{n + 1} }^3 + {F_n}^3 - {F_{n - 1} }^3       | c = simplifying }} {{end-eqn}}
By definition of Fibonacci numbers: :$F_n = F_{n + 2} - F_{n + 1}$ and: :$F_{n + 3} = F_{n + 2} + F_{n + 1}$ Then $a$ can be expressed as: {{begin-eqn}} {{eqn | l = a       | r = \paren {F_{n + 2} - F_{n + 1} } \paren {F_{n + 2} + F_{n + 1} }       | c =  }} {{eqn | r = {F_{n + 2} }^2 - {F_{n + 1} }^2       | c =  }} {{end-eqn}} Now consider: {{begin-eqn}} {{eqn | r = F_{n + 2} F_{n + 3} - F_n F_{n + 1}       | o =        | c =  }} {{eqn | r = F_{n + 2} \paren {F_{n + 2} + F_{n + 1} } - \paren {F_{n + 2} - F_{n + 1} } F_{n + 1}       | c =  }} {{eqn | r = {F_{n + 2} }^2 + {F_{n + 1} }^2       | c = multiplying out }} {{end-eqn}} and it can be seen that the ordered triple: :$\paren {{F_{n + 2} }^2 - {F_{n + 1} }^2, 2 F_{n + 1} F_{n + 2}, {F_{n + 2} }^2 + {F_{n + 1} }^2}$ is in the form of Solutions of Pythagorean Equation. It remains to be shown that: :${F_{n + 2} }^2 + {F_{n + 1} }^2 = F_{2 n + 3}$ We have: {{begin-eqn}} {{eqn | l = F_{2 n + 3}       | r = F_{\paren {n + 1} + \paren {n + 2} }       | c =  }} {{eqn | r = F_{\paren {n + 2} } F_{\paren {n + 1} + 1} + F_{\paren {n + 2} - 1} F_{\paren {n + 1} }       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = {F_{n + 2} }^2 + {F_{n + 1} }^2       | c =  }} {{end-eqn}} and the proof is complete. {{qed|lemma}} The area is trivial: {{begin-eqn}} {{eqn | l = a       | r = F_n F_{n + 3}       | c =  }} {{eqn | l = b       | r = 2 F_{n + 1} F_{n + 2}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {a b} 2       | o = =       | r = \frac {F_n F_{n + 3} \times 2 F_{n + 1} F_{n + 2} } 2       | c =  }} {{eqn | r = F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}       | c =  }} {{end-eqn}} {{qed}}
Let: :$a = F_i, b = F_j, c = F_k, d = F_l$ where $F_n$ denotes the $n$th Fibonacci number. {{WLOG}}, further suppose that; :$a < b < c < d$ or equivalently: :$i < j < k < l$ Since $i, j, k, l$ are integers, the inequality could be written as: :$i \le j - 1 \le k - 2 \le l - 3$ Now consider: {{begin-eqn}} {{eqn | l = d - c       | r = F_l - F_k       | c =  }} {{eqn | o = \ge       | r = F_l - F_{l - 1}       | c = By assumption, $k - 2 \le l - 3$ }} {{eqn | r = F_{l - 2}       | c = {{Defof|Fibonacci Number}} }} {{eqn | o = \ge       | r = F_j       | c = By assumption, $j - 1 \le l - 3$ }} {{eqn | o = \ge       | r = F_j - F_i       | c =  }} {{eqn | r = b - a       | c =  }} {{end-eqn}} For $a, b, c, d$ be in arithmetic sequence: :$d - c = b - a$ This means that the all the inequalities above must be equalities: :$F_l - F_k = F_l - F_{l - 1}$ :$F_{l - 2} = F_j$ :$F_j = F_j - F_i$ So: :$F_i = 0$ and: :$F_k = F_{l - 1}$ :$F_j = F_{l - 2}$  The only Fibonacci numbers having different index but have the same value is $F_1 = F_2 = 1$. So one of the following is true: :$F_k = F_{l - 1} = 1$ :$F_j = F_{l - 2} = 1$  :$j - 1 = k - 2 = l - 3$ Suppose the third statement is true. Write $k = j + 1$, $l = j + 2$. Then: {{begin-eqn}} {{eqn | l = F_{j + 2} - F_{j + 1}       | r = F_{j + 1} - F_j       | c = $F_j, F_{j + 1}, F_{j + 2}$ form an arithmetic sequence }} {{eqn | ll = \leadsto       | l = F_j       | r = F_{j - 1}       | c = {{Defof|Fibonacci Number}} }} {{eqn | ll = \leadsto       | l = F_j - F_{j - 1}       | r = 0       | c =  }} {{eqn | ll = \leadsto       | l = F_{j - 2}       | r = 0       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} The only zero term of the Fibonacci numbers is $F_0$. This gives $j = 2$. Therefore the only arithmetic sequence among Fibonacci numbers satisfying the condition above is: :$F_0, F_2, F_3, F_4$ which corresponds to: :$0, 1, 2, 3$ Now suppose $F_j = 1$. Since $F_i, F_j, F_k, F_l$ form an arithmetic sequence: :$F_k = F_j + \paren {F_j - F_i} = 2$  :$F_l = F_k + \paren {F_j - F_i} = 3$  Which again gives the arithmetic sequence $0, 1, 2, 3$. Finally suppose $F_k = 1$.  Since $F_i, F_j, F_k$ form an arithmetic sequence: :$F_j = \dfrac 1 2 \paren {F_i + F_k} = \dfrac 1 2$ which is not an integer. So $F_k \ne 1$. All cases have been accounted for, and the only arithmetic sequence that can be formed is $0, 1, 2, 3$. {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 2} \paren {-1}^k \dfrac 1 {F_k F_{k + 1} }       | r = \sum_{k \mathop \ge 2} \paren {F_{k + 1} F_{k - 1} - F_k^2} \dfrac 1 {F_k F_{k + 1} }       | c = Cassini's Identity }} {{eqn | r = \sum_{k \mathop \ge 2} \paren {\dfrac {F_{k - 1} } {F_k} - \dfrac {F_k} {F_{k + 1} } }       | c =  }} {{eqn | r = \dfrac {F_1} {F_2} - \lim_{k \mathop \to \infty} \dfrac {F_k} {F_{k + 1} }       | c = Telescoping Series }} {{eqn | r = 1 - \dfrac 1 \phi       | c = Ratio of Consecutive Fibonacci Numbers }} {{eqn | r = \phi^{-1} \paren {\phi - 1}       | c =  }} {{eqn | r = \phi^{-2}       | c = {{Defof|Golden Mean}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k} F_{2 k + 2} }       | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} } - \dfrac 1 {F_{2 k + 2} } } \paren {\dfrac 1 {F_{2 k + 2} - F_{2k} } }       | c =  }} {{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} } - \dfrac 1 {F_{2 k + 2} } } \paren {\dfrac 1 {F_{2 k + 1} } }       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} F_{2 k + 1} } - \dfrac 1 {F_{2 k + 1} F_{2 k + 2} } }       | c =  }} {{eqn | r = \sum_{k \mathop \ge 2} \paren {-1}^k \dfrac 1 {F_k F_{k + 1} }       | c =  }} {{eqn | r = \phi^{-2}       | c = Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared }} {{end-eqn}} {{qed}}
First existence is demonstrated. Let $F_m \ge n$ such that $F_{m - 1} < n$. {{begin-eqn}} {{eqn | l = F_m       | r = F_{m - 1} + F_{m - 2}       | c = Definition of Fibonacci Numbers }} {{eqn | o = <       | r = 2 F_{m - 1}       | c = as $F_{m - 2} < F_{m - 1}$ }} {{eqn | o = <       | r = 2 n       | c = as $F_{m - 1} < n$ }} {{end-eqn}} This shows that the smallest Fibonacci number greater than $n$ is less than $2 n$. Thus there exists at least one Fibonacci number between $n$ and $2 n$. {{AimForCont}} there exist $3$ Fibonacci numbers between $n$ and $2 n$. Let $F_m \ge n$ be the smallest of those Fibonacci numbers. Then: {{begin-eqn}} {{eqn | l = F_{m + 2}       | r = F_m + F_{m + 1}       | c = Definition of Fibonacci Numbers }} {{eqn | o = >       | r = 2 F_m       | c = as $F_m < F_{m + 1}$ }} {{eqn | o = >       | r = 2 n       | c = as $F_m > n$ }} {{end-eqn}} But $F_{m + 2} < 2 n$ by hypothesis. Hence by Proof by Contradiction there can be no more than $2$ Fibonacci numbers between $n$ and $2 n$. Let $n = 2$. Then between $2$ and $4$ there exist $F_3 = 2$ and $F_4 = 3$. Let $n = 10$. Then between $10$ and $20$ there exists $F_7 = 13$ and no other Fibonacci numbers. Thus it has been demonstrated: :There always exists at least one Fibonacci number between $n$ and $2 n$ :There never exist more than $2$ Fibonacci number between $n$ and $2 n$ :There exist $n$ such that there exists exactly one Fibonacci number between $n$ and $2 n$ :There exist $n$ such that there exist exactly $2$ Fibonacci numbers between $n$ and $2 n$. The result is complete. {{qed}}
{{begin-eqn}} {{eqn | l = \sqrt t       | o = <       | r = K e^{a t}       | c = an Ansatz }} {{eqn | ll = \impliedby       | l = t       | o = <       | r = \paren {K e^{a t} }^2       | c = Square Root is Strictly Increasing }} {{eqn | r = K^2 e^{2 a t}       | c = Exponential of Product }} {{eqn | r =  K' e^{a' t}       | c = $K^2 = K', 2 a = a'$ }} {{end-eqn}} Recall from Identity is of Exponential Order Epsilon, $t < K'e^{a' t}$ for ''any'' $a' > 0$, arbitrarily small in magnitude. Therefore the inequality $\sqrt t < K e^{a t}$ has solutions of the same nature. {{qed}} Category:Exponential Order 3qz7nzmqw4ybbfwjlidamncckpjgvci
Let $F_m$ be an $n$-digit Fibonacci number. Then $F_m \ge 10^{n - 1} \ge 10$. We have: {{begin-eqn}} {{eqn | l = F_m       | r = F_{m - 1} + F_{m - 2}       | c = Definition of Fibonacci Numbers }} {{eqn | o = \le       | r = 2 F_{m - 1}       | c = as $F_{m - 2} \le F_{m - 1}$ }} {{end-eqn}} Thus: {{begin-eqn}} {{eqn | l = F_{m + 5}       | r = F_{m - 1} F_5 + F_m  F_6       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = 5 F_{m - 1} + 8 F_m       | c = }} {{eqn | o = >       | r = 2 F_m + 8 F_m       | c = $F_m \le 2 F_{m - 1}$ }} {{eqn | o = \ge       | r = 10^n       | c = }} {{end-eqn}} So $F_{m + 5}$ is a number of at least $n + 1$ digits. Therefore there is at most $5$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}, F_{m + 4}$. Similarly, $10 \le F_m < 10^n$. Suppose $F_m$ is the smallest Fibonacci numbers with $n$ digits. Then $F_{m - 1} < 10^{n - 1}$. Thus: {{begin-eqn}} {{eqn | l = F_{m + 3}       | r = F_{m - 1} F_3 + F_m F_4       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = 2 F_{m - 1} + 3 F_m       | c = }} {{eqn | o = <       | r = 8 F_{m - 1}       | c = $F_m \le 2 F_{m - 1}$ }} {{eqn | o = <       | r = 10^n       | c = }} {{end-eqn}} So there is at least $4$ Fibonacci numbers with $n$ digits: $F_m, F_{m + 1}, F_{m + 2}, F_{m + 3}$. Combining the arguments above, there are either $4$ or $5$ Fibonacci numbers with $n$ digits. {{qed}}
From Prime Number divides Fibonacci Number, either $F_{p - 1}$ or $F_{p + 1}$ is divisible by $p$. Thus: {{begin-eqn}} {{eqn | l = p       | o = \divides       | r = F_{p \pm 1}       | c = Prime Number divides Fibonacci Number }} {{eqn | lo= \forall n \in \Z_{>0}:       | l = F_{p \pm 1}       | o = \divides       | r = F_{n \paren {p \pm 1} }       | c = Divisibility of Fibonacci Number }} {{eqn | ll= \leadsto       | lo= \forall n \in \Z_{>0}:       | l = p       | o = \divides       | r = F_{n \paren {p \pm 1} }       | c =  }} {{end-eqn}} {{qed}}
It is worth noting the one case where $p = 5$: :$5 \divides F_5 = 5$ {{ProofWanted|Googling around suggests there is a proof based on the Law of Quadratic Reciprocity but I have not laid hands on it yet.}}
Let $F_n$ be a prime number. {{AimForCont}} $n$ is a composite number greater than $4$. Then $n = r s$ for some $1 < r, s < n$. Except for the case where $n = 4$, at least one of $r$ and $s$ is greater than $2$. From Divisibility of Fibonacci Number: :$F_r \divides F_n$ and: :$F_s \divides F_n$ where $\divides$ denotes divisibility. When $k > 2$ we have that $F_k > 1$. Thus when $n$ is composite such that $n > 4$, $F_n$ has at least one proper divisor. That is, $F_n$ is not prime. Thus by Proof by Contradiction, $n$ cannot be composite. The exception is when $n = 4$, as noted above in which case its only proper divisor is $2$. But $F_2 = 1$, and so divisibility by $F_2$ does not preclude primality. Indeed, $F_4 = 3$, which is prime. {{qed}}
Proof by Counterexample: :$F_{19} = 4181 = 37 \times 113$ {{qed}}
{{begin-eqn}} {{eqn | l = \int_a^b f \left({t}\right) \rd t       | r = \int_a^b f \left({\theta \left({t}\right)}\right) \theta' \left({t}\right) \rd t       | c = Complex Integration by Substitution: $\theta \left({t}\right) = t$, $\theta' \left({t}\right) = 1$ }} {{eqn | r = \int_\mathcal C f \left({t}\right) \rd t       | c = {{Defof|Complex Contour Integral}} }} {{end-eqn}} {{qed}} Category:Integral Calculus Category:Contour Integration nq6l8sekfh9v1ce0ulh7khwadykrppw
From Fibonacci Number in terms of Smaller Fibonacci Numbers, we have that: :$F_n = F_{k - 1} F_{n - k + 2} + F_{k - 2} F_{n - k + 1}$ for $2 \le k \le n$. {{AimForCont}} $F_n = F_m F_k$ for some $m, k \ge 3$. Then: {{begin-eqn}} {{eqn | l = F_m       | r = \dfrac {F_n} {F_k}       | c =  }} {{eqn | r = \dfrac {F_{k - 1} F_{n - k + 2} + F_{k - 2} F_{n - k + 1} } {F_k}       | c =  }} {{eqn | r = \paren {\dfrac {F_{k - 1} } {F_{k - 1} + F_{k - 2} } } F_{n - k + 2} + \paren {\dfrac {F_{k - 2} } {F_{k - 1} + F_{k - 2} } } F_{n - k + 1}        | c =  }} {{end-eqn}} The {{RHS}} is a weighted mean of $2$ consecutive Fibonacci numbers. Thus: :$F_{n - k + 2} < F_m < F_{n - k + 1}$ which cannot happen. The result follows by Proof by Contradiction. {{qed}}
{{begin-eqn}} {{eqn | l = \norm {\mathbf x^p}_q       | r = \paren {\sum_{n \mathop = 0}^\infty \size { {x_n}^p}^q}^{1 / q}       | c = {{Defof|P-Norm|$p$-Norm}} }} {{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^{p q} }^{1 / q}       | c = Power of Power }} {{eqn | r = \paren {\paren {\sum_{n \mathop = 0}^\infty \size {x_n}^{p q} }^{1 / p q} }^p       | c = Power of Power }} {{eqn | r = \norm {\mathbf x}_{p q}^p       | c = {{Defof|P-Norm|$p$-Norm}} }} {{end-eqn}} {{qed}} Category:Functional Analysis Category:Norm Theory Category:P-Sequence Metrics Category:P-Norms 7s0yq0l4umt2ll14z8pxdv54cwl87yc