hcju/proofwikiqa huggingface dataset:
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$. By definition: :$\forall \epsilon \in \R_{>0}: \exists N \in \N: \forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$ Let $\epsilon < 1$, say: $\epsilon = \dfrac 1 2$. By the definition of $d$: :$\forall m, n \in \N: x_m \ne x_n \implies \map d {x_m, x_n} \ge 1$ So the only possible way for: :$\forall m, n \in \N: m, n \ge N: \map d {x_n, x_m} < \epsilon$ is for $x_m = x_n$. Hence the result. {{qed}}
Let $\sequence {x_n}$ be a Cauchy sequence in $\struct {\Z_{>0}, d}$. From Cauchy Sequence in Positive Integers under Usual Metric is eventually Constant: :$\sequence {x_n}$ is a convergent sequence to some $n \in \Z_{>0}$. Hence the result by definition of complete metric space. {{qed}}
For a general $x_m, x_n \in \sequence {x_n}$ as defined: {{begin-eqn}} {{eqn | l = \map \delta {x, y}       | r = \frac {\size {x_m - x_n} } {x_m x_n}       | c = Definition of $\delta$ }} {{eqn | r = \size {\frac 1 {x_m} - \frac 1 {x_n} }       | c = algebra }} {{eqn | n = 1       | r = \size {\frac 1 m - \dfrac 1 n}       | c = Definition of $\sequence {x_n}$ }} {{end-eqn}} Let $\epsilon \in \R_{>0}$. Then by the Archimedean Principle: :$\exists N \in \N: N > \dfrac 1 \epsilon$ from which it follows that: :$\epsilon > \dfrac 1 N$ Thus: {{begin-eqn}} {{eqn | lo= \forall m, n \in \N:       | l = m, n       | o = >       | r = N       | c =  }} {{eqn | ll= \leadsto       | l = \map \delta {x_m, x_n}       | r = \size {\frac 1 m - \frac 1 n}       | c = from $(1)$ above }} {{eqn | o = <       | r = \max \set {\frac 1 m, \frac 1 n}       | c =  }} {{eqn | o = <       | r = \frac 1 N       | c =  }} {{eqn | o = <       | r = \epsilon       | c =  }} {{end-eqn}} Therefore $\sequence {x_n}$ is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$. {{qed}}
Consider the sequence $\sequence {x_n}$ in $\Z_{>0}$ defined as: :$\forall n \in \N: x_n = n$ From Cauchy Sequence in Positive Integers under Scaled Euclidean Metric: :$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Z_{>0}, \delta}$. But $\sequence {x_n}$ is not convergent to any $m \in \Z_{>0}$. Hence the result, by definition of complete metric space. {{qed}}
By definition, $\tau_d$ is the topology induced by the Euclidean metric $d$. Consider the complement of $A$ in $\R^2$: :$A' := \R^2 \setminus A$ Thus: :$A := \set {\tuple {x, y} \in \R^2: x y < 1}$ Let $a = \tuple {x_a, y_a} \in A^2$. Let $\epsilon = \size {1 - x_a y_a}$. Then the open $\epsilon$-ball of $a$ in $\R^2$ lies entirely in $A'$. As $a$ is arbitrary, it follows that any such $a$ has an open $\epsilon$-ball of $a$ in $\R^2$ which lies entirely in $A'$. Thus, by definition, $A'$ is open in $\R^2$. So, also by definition, $A$ is closed in $\R^2$. {{qed}}
By definition, the real number plane with the usual (Euclidean) topology on $\R^2$ is the product space of $\struct {\R, d}$ with $\struct {\R, d}$, where $\struct {\R, d}$ is the real number line with the usual (Euclidean) topology The result follows from Projection from Product Topology is Open. {{qed}}
Consider the set $A \subseteq R^2$ of all points defined as: :$A := \set {\tuple {x, y} \in \R^2: x y \ge 1}$ By Subset of Euclidean Plane whose Product of Coordinates are Greater Than or Equal to 1 is Closed: :$A$ is a closed set in $\struct {\R^2, d}$. By inspection, it can be seen that the image of $A$ under $\rho$ is: :$\rho \sqbrk A = \openint \gets 0 \cup \openint 0 \to$ which by Union of Open Sets of Metric Space is Open is open. Hence the result by definition of closed mapping. {{qed}}
Let $\tuple {a, b} \in A$. Consider any point $\tuple {x_1, y_1} \in A$ whose coordinates are both irrational. By definition, either $a$ or $b$ is irrational. {{WLOG}} suppose $a$ is irrational. Then the union of the straight lines $x = a, y = y_1$ is an arc-connected subset of $A$ connecting $\tuple {x_1, y_1}$ to $\tuple {a, b}$. Hence any point in $A$ can be connected to $\tuple {x_1, y_1}$ by an arc. Hence the result, by definition of arc-connected. {{qed}}
Recall the definition of compact subspace: :$\left({\varnothing, \tau_\varnothing}\right)$ is '''compact in $T$''' {{iff}} every open cover $\mathcal C \subseteq \tau_\varnothing$ for $\varnothing$ has a finite subcover. The only open cover for $\varnothing$ that is contained in $\varnothing$ is $\left\{{\varnothing}\right\}$ itself. This has only one finite subcover, and that is $\left\{{\varnothing}\right\}$. This is a finite subcover. Hence the result, by definition of compact subspace. {{qed}} Category:Compact Spaces Category:Empty Set 40suo9x8c2qi1cy4gxxnfxno2z386bq
Let $\tau_K$ be the subspace topology on $K$. Let $T_K = \left({K, \tau_K}\right)$ be the topological subspace determined by $K$. By Closed Set in Topological Subspace, $H \cap K$ is closed in $T_K$. By Closed Subspace of Compact Space is Compact, $H \cap K$ is compact in $T_K$. By Compact in Subspace is Compact in Topological Space, $H \cap K$ is compact in $T$. {{qed}}
Let $\left\langle{U_\alpha}\right\rangle$ be an open cover of $H \cap K$: :$\displaystyle H \cap K \subseteq \bigcup_\alpha U_\alpha$ Then: :$\displaystyle K \subseteq \bigcup_\alpha U_\alpha \cup \left({S \setminus H}\right)$ Since $H$ is closed in $T$, $\left({S \setminus H}\right)$ is open in $T$. Hence $\left\langle{U_\alpha}\right\rangle \cup S \setminus H$ is an open cover of $H$. We have that $K$ is compact in $T$. It follows by definition that a finite subcover: :$\left\{{U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}, S \setminus H}\right\}$ of $H$ exists. Thus: :$H \cap K \subseteq \left\{{U_{\alpha_1}, U_{\alpha_2}, \ldots, U_{\alpha_n}}\right\}$ and $H \cap K$ is compact in $T$. {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\mathcal U_n := \displaystyle \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$. $P \left({0}\right)$ is the case: :$\mathcal U_0 := \displaystyle \bigcup_{i \mathop = 1}^0 U_i$ From Union of Empty Set: :$\displaystyle \bigcup_{i \mathop = 1}^0 U_i = \varnothing$ From Empty Set is Compact Space it follows that: :$\mathcal U_0$ is compact in $T$. $P \left({1}\right)$ is true, as this just says: :$U_1$ is compact in $T$. === Basis for the Induction === $P \left({2}\right)$ is the case: :$U_1 \cup U_2$ is compact in $T$. which is proved as follows: Let $\mathcal C$ be an open cover of $U_1 \cup U_2$. Then $\mathcal C$ is an open cover of both $U_1$ and $U_2$. As $U_1$ and $U_2$ are both compact in $T$: :$U_1$ has a finite subcover $C_1$ of $\mathcal C$ :$U_2$ has a finite subcover $C_2$ of $\mathcal C$. Their union $C_1 \cup C_2$ is a finite subcover of $\mathcal C$ for $U_1 \cup U_2$. From Union of Finite Sets is Finite it follows that $C_1 \cup C_2$ is finite. As $\mathcal C$ is arbitrary, it follows by definition that $U_1$ and $U_2$ is compact in $T$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\mathcal U_k := \displaystyle \bigcup_{i \mathop = 1}^k U_i$ is compact in $T$. Then we need to show: :$\mathcal U_{k+1} := \displaystyle \bigcup_{i \mathop = 1}^{k+1} U_i$ is compact in $T$. === Induction Step === This is our induction step: We have that: :$\displaystyle \bigcup_{i \mathop = 1}^{k+1} U_i = \left({\bigcup_{i \mathop = 1}^k U_i}\right) \cup U_{k+1}$ By the induction hypothesis: :$\mathcal U_k$ is compact in $T$. By the basis for the induction: :$\mathcal U_k \cup U_{k+1}$ is compact in $T$. So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N: \bigcup_{i \mathop = 1}^n U_i$ is compact in $T$. {{qed}} Category:Compact Spaces Category:Set Union izenikmugsid70js3fh0qx26zm046b3
Let $\Z_{\ge n_0}$ denote the set: :$S = \set {n \in \Z: n \ge n_0}$ Let $S$ be the set of integers defined as: :$S = \set {n \in \Z_{\ge n_0}: \map P n}$ That is, the set of all integers for which $n \ge n_0$ and for which $\map P n$ holds. From Subset of Set with Propositional Function we have that: :$S \subseteq \Z_{\ge n_0}$ From $(1)$ we have that $\map P {n_0}$. Hence $n_0 \in S$. Let $k \in S$. Then $\map P k$ holds. But by $(2)$, $\map P {k + 1}$ also holds. This implies $k + 1 \in S$. So as: :$S \subseteq \Z_{\ge n_0}$ and: :$S$ satisfies $(1)$ and $(2)$ it follows by the Principle of Finite Induction that $S = \Z_{\ge n_0}$. Hence for all $n \ge n_0$, $\map P n$ holds. {{qed}}
Let $\Z_{\ge n_0} := \set {n \in \Z: n \ge n_0}$. {{AimForCont}} $S \ne \Z_{\ge n_0}$. Let $S' = \Z_{\ge n_0} \setminus S$. Because $S \ne \Z_{\ge n_0}$ and $S \subseteq \Z_{\ge n_0}$, we have that $S' \ne \O$. By definition, $\Z_{\ge n_0}$ is bounded below by $n_0$. From Set of Integers Bounded Below by Integer has Smallest Element, $S'$ has a minimal element. Let $k$ be this minimal element of $S'$. By $(1)$ we have that: :$n_0 \in S$ and so: :$n_0 \notin S'$ Hence: :$k \ne n_0$ and so: :$k > n_0$ It follows that: :$k - 1 \le n_0$ Because $k$ is the minimal element of $S'$: :$k - 1 \notin S'$ and so: :$k - 1 \in S$ But by $(2)$: :$\paren {k - 1} + 1 = k \in S$ So we have: :$k \in S$ and: :$k \notin S$ Hence by Proof by Contradiction $S = \Z_{\ge n_0}$. {{qed}}
{{questionable|This only takes on board a subset of $\N$, where we need a subset of $\Z$}} Consider $\N$ defined as a naturally ordered semigroup. The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result. {{qed}}
Let $A \subseteq P$ be defined by: :$A := \set {n \in P: \map Q n}$ From $(1)$, $0 \in A$. From $(2)$: :$\forall n \in P: n \in A \implies \map s n \in A$ As this holds for all $n \in P$, it holds a fortiori for all $n \in A$. Thus the condition: :$n \in A \implies \map s n \in A$ is satisfied. So by Axiom $(\text P 5)$ of the Peano Axioms: :$A = P$ That is: :$\forall n \in P: \map Q n$ {{qed}}
This is nothing but a reformulation of Axiom $(P5)$ of the Peano Axioms. {{qed}} Category:Peano's Axioms o33monttem5erz7x2wmjus9jyovdugs
{{AimForCont}} that $T \subsetneq S$. That is, $T$ is a proper subset of $S$: : $T \ne S$ Let $T' = S \setminus T$. Then by Set Difference with Proper Subset: :$T' \ne \O$ By axiom $NO1$, $S$ is well-ordered. By definition of well-ordered set, it follows that $T'$ has a smallest element $x$. By definition of $T$: :$0 \in T$ and so by  definition of $T'$: :$0 \notin T'$ so: :$0 \prec x$ By Sum with One is Immediate Successor in Naturally Ordered Semigroup: :$1 \preceq x$ By the definition of a naturally ordered semigroup: :$\exists y \in S: y \circ 1 = x$ Again by Sum with One is Immediate Successor in Naturally Ordered Semigroup: :$y \prec x$ We have that $x$ is the smallest element of $T'$ and $y \prec x$. Therefore: :$y \notin T'$ and so :$y \in T$ But from the definition of $T$: :$y \in T \implies y \circ 1 = x \in T$ But then by the definition of $T'$: :$x \in T' \implies x \notin T$ From this contradiction, it follows that: :$T = S$ {{qed}}
From the definition of Fibonacci numbers: :$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$ Proof by induction: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :$\gcd \set {F_n, F_{n + 1} } = 1$ === Basis for the Induction === $\map P 2$ is the case: :$\gcd \set {F_2, F_3} = \gcd \set {2, 3} = 1$ Thus $\map P 2$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\gcd \set {F_k, F_{k + 1} } = 1$ Then we need to show: :$\gcd \set {F_{k + 1}, F_{k + 2} } = 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \gcd \set {F_{k + 1}, F_{k + 2} }       | r = \gcd \set {F_{k + 1}, F_{k + 2} - F_{k + 1} }       | c = Common Divisor Divides Integer Combination }} {{eqn | r = \gcd \set {F_{k + 1}, F_k}       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \gcd \set {F_k, F_{k + 1} }       | c =  }} {{eqn | r = 1       | c = Induction Hypothesis }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \ge 2: \gcd \set {F_n, F_{n + 1} } = 1$ {{qed}}
From the initial definition of Fibonacci numbers: :$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$ Let $n = k m - r$ where $0 \le r < m$ We have: :$m \divides n \iff r = 0$ The proof proceeds by induction on $k$. For all $k \in \N_{>0}$, let $\map P k$ be the proposition: :$r = 0 \iff F_m \divides F_{k m - r}$ === Basis for the Induction === $\map P 1$ is the case: :$r = 0 \iff F_m \divides F_{m - r}$ which holds because $F_{m - r} < F_m$ unless $r = 0$. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$r = 0 \iff F_m \divides F_{k m - r}$ Then we need to show: :$r = 0 \iff F_m \divides F_{k m + m - r}$ === Induction Step === This is our induction step: Let $F_{k m - r} = a F_m + b$ where $0 \le b < F_m$. We have: :$b = 0 \iff F_m \divides F_{k m - r} \iff r = 0$ by the induction hypothesis. {{begin-eqn}} {{eqn | l = F_{k m + m - r}       | r = F_{m - 1} F_{k m - r} + F_m F_{k m - r + 1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = a F_m F_{m - 1} + b F_{m - 1} + F_m F_{k m - r + 1}       | c =  }} {{eqn | r = F_m \paren {a F_{m - 1} + F_{k m - r + 1} } + b F_{m - 1}       | c =  }} {{end-eqn}} We have that $F_{m - 1}$ and $F_m$ are coprime by Consecutive Fibonacci Numbers are Coprime. Let $F_m \divides b F_{m - 1}$. Then there exists an integer $k$ such that $k F_m \divides b F_{m - 1}$, by the definition of divisibility. Then: :$\dfrac k b = \dfrac {F_{m - 1} } {F_m}$ We have that $F_{m - 1}$ and $F_m$ are coprime. Thus by Coprime Numbers form Fraction in Lowest Terms: :$\dfrac {F_{m - 1} } {F_m}$ is in canonical form. Then by Ratios of Fractions in Lowest Terms :$F_m \divides b$ Because $0 \le b < F_m$, the only case is when $b = 0$. Therefore: :$F_m \divides b F_{m - 1} \iff b = 0$ Therefore: :$F_m \divides F_{k m + m - r} \iff F_m \divides b F_{m - 1} \iff b = 0 \iff r = 0$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall m, n > 2 : m \divides n \iff F_m \divides F_n$ {{qed}}
From the initial definition of Fibonacci numbers, we have: :$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$ Proof by induction: For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle \forall m \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{m + 1}       | r = F_{m - 1} + F_m       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{m - 1} \times 1 + F_m \times 1       | c =  }} {{eqn | r = F_{m - 1} F_1 + F_m F_2       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{m - 1} F_n + F_m F_{n + 1}       | c = for $n = 1$ }} {{end-eqn}} and so $P \left({1}\right)$ is seen to hold. $P \left({2}\right)$ is the case: {{begin-eqn}} {{begin-eqn}} {{eqn | l = F_{m + 2}       | r = F_{m + 1} + F_m       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{m - 1} + F_m + F_m       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{m - 1} \times 1 + F_m \times 2       | c =  }} {{eqn | r = F_{m - 1} F_2 + F_m F_3       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{m - 1} F_n + F_m F_{n + 1}       | c = for $n = 2$ }} {{end-eqn}} and so $P \left({2}\right)$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ and $P \left({k-1}\right)$ are true, where $k > 1$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle F_{m + k} = F_{m - 1} F_k + F_m F_{k + 1}$ and: :$\displaystyle F_{m + k - 1} = F_{m - 1} F_{k - 1} + F_m F_k$ from which it is to be shown: :$\displaystyle F_{m + k + 1} = F_{m - 1} F_{k + 1} + F_m F_{k + 2}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = F_{m + k + 1}       | r = F_{m + k} + F_{m + k - 1}       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{m - 1} F_k + F_m F_{k + 1} + F_{m - 1} F_{k - 1} + F_m F_k       | c = Induction Hypothesis }} {{eqn | r = F_{m - 1} \left({F_k + F_{k - 1} }\right) + F_m \left({F_{k + 1} + F_k}\right)       | c =  }} {{eqn | r = F_{m - 1} F_{k + 1} + F_m F_{k + 2}       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $P \left({k}\right) \land P \left({k - 1}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall m, n \in \Z_{>0} : F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$ {{qed}}
{{begin-eqn}} {{eqn | r = F_{m - 1} F_n + F_m F_{n + 1}       | o =  }} {{eqn | r = \dfrac {\phi^{m - 1} - \hat \phi^{m - 1} } {\sqrt 5} \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} + \dfrac {\phi^m - \hat \phi^m} {\sqrt 5} \dfrac {\phi^{n + 1} - \hat \phi^{n + 1} } {\sqrt 5}       | c = Euler-Binet Formula }} {{eqn | r = \dfrac {\phi^{m + n - 1} - \phi^{m - 1} \hat \phi^n - \phi^n \hat \phi^{m - 1} + \hat \phi^{m + n - 1} + \phi^{m + n + 1} - \phi^m \hat \phi^{n + 1} - \phi^{n + 1} \hat \phi^m + \hat \phi^{m + n + 1} } 5       | c =  }} {{eqn | r = \dfrac {\phi^{m + n - 1} \left({1 + \phi^2}\right) + \hat \phi^{m + n - 1} \left({1 + \hat \phi^2}\right) - \phi^{m - 1} \hat \phi^n \left({1 + \phi \hat \phi}\right) -\phi^n \hat \phi^{m - 1} \left({1 + \phi \hat \phi}\right)} 5       | c =  }} {{eqn | r = \dfrac {\phi^{m + n - 1} \left({1 + \phi^2}\right) + \hat \phi^{m + n - 1} \left({1 + \hat \phi^2}\right)} 5       | c = as $\phi \hat \phi = -1$ }} {{eqn | r = \dfrac {\phi^{m + n - 1} \left({2 + \phi}\right) + \hat \phi^{m + n - 1} \left({2 + \hat \phi}\right)} 5       | c = as both $\phi$ and $\hat \phi$ satisfy $x^2 = x + 1$ }} {{eqn | r = \dfrac {\phi^{m + n - 1} \left({2 + \dfrac {1 + \sqrt 5} 2}\right) + \hat \phi^{m + n - 1} \left({2 + \dfrac {1 - \sqrt 5} 2}\right) } 5       | c = Definition of $\phi$ and $\hat \phi$ }} {{eqn | r = \dfrac {\phi^{m + n - 1} \left({\dfrac {5 + \sqrt 5} 2}\right) + \hat \phi^{m + n - 1} \left({\dfrac{ 5 - \sqrt 5} 2}\right)} 5       | c =  }} {{eqn | r = \dfrac {\phi^{m + n - 1} \left({\dfrac{1 + \sqrt 5} 2}\right) - \hat \phi^{m + n - 1} \left({\dfrac {1 - \sqrt 5} 2}\right)} {\sqrt 5}       | c = dividing numerator and denominator by $\sqrt 5$ }} {{eqn | r = \dfrac {\phi^{m + n} - \hat \phi^{m + n} } {\sqrt 5}       | c = Definition of $\phi$ and $\hat \phi$ }} {{eqn | r = F_{m + n}       | c = Euler-Binet Formula }} {{end-eqn}} {{qed}}
From the initial definition of Fibonacci numbers, we have: :$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$ {{WLOG}}, let $m \le n$. Let $h$ be $\gcd \set {m, n}$. Let $a$ and $b$ be integers such that $m = h a$ and $n = \map h {a + b}$. $a$ and $a + b$ are coprime by Integers Divided by GCD are Coprime. Therefore, $a$ and $b$ are coprime by Integer Combination of Coprime Integers. {{begin-eqn}} {{eqn | l = \gcd \set {F_m, F_n}       | r = \gcd \set {F_{h a}, F_{h a - 1} F_{h b} + F_{h a} F_{h b + 1} }       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = \gcd \set {F_{h a}, F_{h a - 1} F_{h b} }       | c = GCD with Remainder }} {{end-eqn}} Let $u$ and $v$ be integers such that $F_{h a} = u F_h$ and $F_{h b} = v F_h$, whose existence is proved by Divisibility of Fibonacci Number. We have that $F_{h a}$ and $F_{h a - 1}$ are coprime by Consecutive Fibonacci Numbers are Coprime. Therefore, $u$ and $F_{h a - 1}$ are coprime by Divisor of One of Coprime Numbers is Coprime to Other. {{begin-eqn}} {{eqn | l = \gcd \set {F_{h a}, F_{h a - 1} F_{h b} }       | r = F_h \gcd \set {u, v F_{h a - 1} }       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = F_h \gcd \set {u, v}       | c = Solution of Linear Diophantine Equation }} {{eqn | r = \gcd \set {F_m, F_{n - m} }       | c = }} {{end-eqn}} Therefore: :$\forall m, n \in \Z_{>2} : \gcd \set {F_m, F_n} = \gcd \set {F_m, F_{n - m} }$ This can be done recurrently to produce the result, in a fashion similar to the Euclidean Algorithm.  Since $a$ and $b$ are coprime, the result would be $\gcd \set {F_h, F_h}$. Therefore: :$\forall m, n > 2 : \gcd \set {F_m, F_n} = F_{\gcd \set {m, n} }$ {{qed}}
From the initial definition of Fibonacci numbers, we have: :$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$ By definition of the extension of the Fibonacci numbers to negative integers: :$F_n = F_{n + 2} - F_{n - 1}$ The proof proceeds by induction. For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :$F_{-n} = \left({-1}\right)^n F_n$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{-1}       | r = F_1 - F_0       | c =  }} {{eqn | r = 1 - 0       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = \left({-1}\right)^{1 + 1} F_1       | c =  }} {{end-eqn}} So $P(1)$ is seen to hold. $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_{-2}       | r = F_0 - F_{-1}       | c =  }} {{eqn | r = 0 - 1       | c =  }} {{eqn | r = -1       | c =  }} {{eqn | r = \left({-1}\right)^{2 + 1} F_2       | c =  }} {{end-eqn}} So $P(2)$ is seen to hold. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ and $P \left({k-1}\right)$ are true, where $k > 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$F_{-\left({k - 1}\right)} = \left({-1}\right)^k F_{k - 1}$ :$F_{-k} = \left({-1}\right)^{k + 1} F_k$ Then we need to show: :$F_{-\left({k + 1}\right)} = \left({-1}\right)^{k + 2} F_{k + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = F_{- \left({k + 1}\right)}       | r = F_{-\left({k - 1}\right)} - F_{-k}       | c = {{Defof|Fibonacci Number for Negative Index}} }} {{eqn | r = \left({-1}\right)^k F_{k - 1} - \left({-1}\right)^{k + 1} F_k       | c = Induction Hypothesis }} {{eqn | r = \left({-1}\right)^k F_{k - 1} + \left({-1}\right)^k F_k       | c =  }} {{eqn | r = \left({-1}\right)^k \left({F_{k - 1} + F_k}\right)       | c =  }} {{eqn | r = \left({-1}\right)^k \left({F_{k + 1} }\right)       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = \left({-1}\right)^{k + 2} \left({F_{k + 1} }\right)       | c =  }} {{end-eqn}} So $P \left({k}\right) \land P \left({k-1}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0} : F_{-n} = \left({-1}\right)^{n + 1} F_n$ {{qed}}
{{begin-eqn}} {{eqn | l = F_{m - n}       | r = F_{m + \left({-n}\right)}       | c = {{Defof|Integer Subtraction}} }} {{eqn | r = F_{m - 1} F_{-n} + F_m F_{-n + 1}       | c = Fibonacci Number in terms of Smaller Fibonacci Numbers }} {{eqn | r = \left({-1}\right)^{n + 1} F_{m - 1} F_n + \left({-1}\right)^n F_m F_{n - 1}       | c = Fibonacci Number with Negative Index }} {{end-eqn}} {{qed}} Category:Fibonacci Numbers dm87q8ycvy4zkzxcxa3zjfjjtpfmnvt