hcju/proofwikiqa huggingface dataset:
From definition of$\bmod 1$, the statement above is equivalent to the statement: :$F_{2 n + 1} \phi - \phi^{-2 n - 1}$ is an integer We have: {{begin-eqn}} {{eqn | l = \phi^2 - \phi \sqrt 5       | r = \paren {\frac {1 + \sqrt 5} 2}^2 - \paren {\frac {1 + \sqrt 5} 2} \sqrt 5 }} {{eqn | r = \frac {6 + 2 \sqrt 5} 4 - \frac {5 + \sqrt 5} 2 }} {{eqn | r = -1 }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = F_{2 n + 1} \phi - \phi^{-2 n - 1}       | r = \frac {\phi^{2 n + 1} - \paren {-1}^{2 n + 1} \phi^{-2 n - 1} } {\sqrt 5} \phi - \phi^{-2 n - 1}       | c = Euler-Binet Formula }} {{eqn | r = \frac {\phi^{2 n + 2} + \phi^{-2 n} - \phi^{-2 n - 1} \sqrt 5} {\sqrt 5}  }} {{eqn | r = \frac {\phi^{2 n + 2} + \phi^{-2 n - 2} \paren {\phi^2 - \phi \sqrt 5} } {\sqrt 5}  }} {{eqn | r = \frac {\phi^{2 n + 2} - \phi^{-2 n - 2} } {\sqrt 5}       | c = as $\phi^2 - \phi \sqrt 5 = -1$ }} {{eqn | r = \frac {\phi^{2 n + 2} - \paren {-1}^{-2 n - 2} \phi^{-2 n - 2} } {\sqrt 5} }} {{eqn | r = F_{2 n + 2}       | c = Euler-Binet Formula }} {{end-eqn}} which is an integer. {{qed}}
Follows directly from Residue of Fibonacci Number Modulo Fibonacci Number. {{qed}}
{{begin-eqn}} {{eqn | l = \sin n z       | r = \map \sin {\dfrac {n \pi} 2 + i n \ln \phi}       | c =  }} {{eqn | r = \frac {e^{i \paren {\paren {n \pi / 2} + i n \ln \phi} } - e^{-i \paren {\paren {n \pi / 2} + i n \ln \phi} } } {2 i}       | c = Sine Exponential Formulation }} {{eqn | r = \frac {e^{i n \pi / 2} e^{-n \ln \phi} - e^{-i n \pi / 2} e^{n \ln \phi} } {2 i}       | c =  }} {{eqn | r = \frac {e^{-n \ln \phi} \paren {\cos \frac {n \pi} 2 + i \sin \frac {n \pi} 2} - e^{n \ln \phi} \paren {\cos \paren {-\frac {n \pi} 2} + i \map \sin {-\frac {n \pi} 2} } } {2 i}       | c = Euler's Formula and Corollary }} {{eqn | r = \frac {e^{-n \ln \phi} \paren {i \sin \frac {n \pi} 2} - e^{n \ln \phi} \paren {i \map \sin {-\frac {n \pi} 2} } } {2 i}       | c = Cosine of Half-Integer Multiple of Pi }} {{eqn | r = \frac {i^n e^{-n \ln \phi} - \paren {-i}^n e^{n \ln \phi} } {2 i}       | c = Sine of Half-Integer Multiple of Pi and simplification }} {{eqn | r = \frac {i^{n - 1} \paren {e^{-n \ln \phi} + e^{n \ln \phi} } } 2       | c = simplification }} {{eqn | r = i^{n - 1} \frac {\phi^n + \frac 1 {\phi^n} } 2       | c = Exponential of Natural Logarithm }} {{eqn | r = i^{n - 1} \frac {\phi^n - \paren {-\frac 1 {\phi^n} } } 2       | c =  }} {{eqn | r = i^{n - 1} \frac {\phi^n - \hat \phi^n} 2       | c = Reciprocal Form of One Minus Golden Mean }} {{end-eqn}} Setting $n = 1$: :$\sin z = i^0 \frac {\phi^1 - \hat \phi^1} 2 = \frac {\phi - \hat \phi} 2$ Thus: {{begin-eqn}} {{eqn | l = \dfrac {\sin n z} {\sin z}       | r = \dfrac {i^{n - 1} \frac {\phi^n - \hat \phi^n} 2} {\frac {\phi - \hat \phi} 2}       | c =  }} {{eqn | r = i^{n - 1} \dfrac {\phi^n - \hat \phi^n } {\phi - \hat \phi}       | c =  }} {{eqn | r = i^{n - 1} \dfrac {\frac {\phi^n - \hat \phi^n} {\sqrt 5} } {\frac {\phi - \hat \phi} {\sqrt 5} }       | c =  }} {{eqn | r = i^{n - 1} \dfrac {F_n} {F_1}       | c = Euler-Binet Formula }} {{eqn | r = i^{n - 1} F_n       | c = {{Defof|Fibonacci Number}}: $F_1 = 1$ }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \cos z       | r = \cos \left({\dfrac {\pi} 2 + i \ln \phi}\right)       | c =  }} {{eqn | r = \frac {e^{i \left({\left({\pi / 2}\right) + i \ln \phi}\right)} + e^{-i \left({\left({\pi / 2}\right) + i \ln \phi}\right)} } 2       | c = Cosine Exponential Formulation }} {{eqn | r = \frac {e^{i \pi / 2} e^{-\ln \phi} + e^{-i \pi / 2} e^{\ln \phi} } 2       | c =  }} {{eqn | r = \frac {e^{-\ln \phi} \left({\cos \frac \pi 2 + i \sin \frac \pi 2}\right) + e^{\ln \phi} \left({\cos \left({-\frac \pi 2}\right) + i \sin \left({-\frac \pi 2}\right)}\right)} 2       | c = Euler's Formula and Corollary }} {{eqn | r = \frac {e^{-\ln \phi} \left({i \sin \frac \pi 2}\right) + e^{\ln \phi} \left({i \sin \left({-\frac \pi 2}\right)}\right)} 2       | c = Cosine of Half-Integer Multiple of Pi }} {{eqn | r = \frac {i e^{-\ln \phi} - i e^{\ln \phi} } 2       | c = Sine of Half-Integer Multiple of Pi and simplification }} {{eqn | r = -i \frac {\phi - \frac 1 {\phi} } 2       | c = Exponential of Natural Logarithm }} {{eqn | r = -i \frac {\phi^2 - 1} {2 \phi}       | c =  }} {{eqn | r = -i \frac {\phi} {2 \phi}       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \frac {-i} 2       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \sin \left({n + 1}\right) z + \sin \left({n - 1}\right) z       | r = 2 \sin n z \cos z       | c = Simpson's Formula for Sine by Cosine }} {{eqn | r = -i \sin n z       | c =  }} {{end-eqn}} {{finish|Where to from here?}}
We have that: :$\phi^0 = 1$ and so $1$ has the representation $S_1$ as: :$S_1 := \left[{1 \cdotp 00}\right]_\phi$ By inspection it is seen that $S_1$ is the simplest form of $1$. Expanding $S_1$: :$S_2 := \left[{0 \cdotp 11}\right]_\phi$ which can then be expressed as: :$S_2 := \left[{0 \cdotp 1100}\right]_\phi$ Expanding $S_2$: :$S_3 := \left[{0 \cdotp 1011}\right]_\phi$ and so: :$S_4 := \left[{0 \cdotp 101011}\right]_\phi$ Continuing in this way we obtain an infinite sequence $\left\langle{S_n}\right\rangle$ all of which are a representation of $1$. {{qed}}
By Solution to Quadratic Equation: {{begin-eqn}} {{eqn | l = x       | r = \dfrac {-\left({-1}\right) \pm \sqrt {\left({-1}\right)^2 - 4 \times 1 \times \left({-1}\right)} } {2 \times 1}       | c = Solution to Quadratic Equation }} {{eqn | r = \dfrac {1 \pm \sqrt 5} 2       | c =  }} {{end-eqn}} Thus :$x = \dfrac {1 + \sqrt 5} 2$ and: :$x = \dfrac {1 - \sqrt 5} 2$ The result follows: By definition of golden mean: :$\phi = \dfrac {1 + \sqrt 5} 2$ Then: {{begin-eqn}} {{eqn | r = \dfrac {1 - \sqrt 5} 2       | c =  }} {{eqn | r = \dfrac {2 - 1 - \sqrt 5} 2       | c =  }} {{eqn | r = \dfrac {2 - \left({1 + \sqrt 5}\right)} 2       | c =  }} {{eqn | r = 1 - \dfrac {1 + \sqrt 5} 2       | c =  }} {{eqn | r = 1 - \phi       | c =  }} {{end-eqn}} {{qed}} Category:Golden Mean siaxqz46gj94fkii1bfrxi78oivfvr4
Define the set: :$Y = \left\{ { y \in T_1: S_y \text{ is an initial segment of } \left({T_2,\preccurlyeq}\right) } \right\}$. Then: {{begin-eqn}} {{eqn | l = S_x \left({T_1}\right)       | r =  \left\{ {b \in T_1, x \in T_1: b \prec x}\right\}       | c = {{Defof|Initial Segment}} }} {{eqn | r = \left\{ {b \in T_2, x \in T_2: b \prec x}\right\}       | c = {{Defof|Proper Subtower in Set}}, as $T_1 \subseteq T_2$ }} {{eqn | r = S_x \left({T_2}\right)       | c = {{Defof|Initial Segment}} }} {{end-eqn}} By induction on well-ordered sets, $Y = T_1$. That is, $\left({T_1,\preccurlyeq}\right)$ is an initial segment in $\left({T_2,\preccurlyeq}\right)$. {{qed}}
Let $z \in \C$. Let $w \in \C$ such that $w + 2 = z$. Then: {{begin-eqn}} {{eqn | l = \phi^z       | r = \phi^{w + 2}       | c =  }} {{eqn | r = \phi^w \phi^2       | c =  }} {{eqn | r = \phi^w \left({\phi + 1}\right)       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \phi^{w + 1} + \phi^w       | c =  }} {{eqn | r = \phi^{z - 1} + \phi^{z - 2}       | c =  }} {{end-eqn}} {{qed}}
Let $100$ appear anywhere within $x$. Then: :$x = \phi^r + \displaystyle \sum_{c \mathop \in C} \phi^c$ where: :$C \subset \Z$ :$r \in \Z$ :$r \notin C, r - 1 \notin C, r - 2 \notin C$ That is, the instance of $100$ corresponds to the indices $r, r - 1, r - 2$. From Power of Golden Mean as Sum of Smaller Powers: :$\phi^r = \phi^{r - 1} + \phi^{r - 2}$ and so: :$x = \phi^{r - 1} + \phi^{r - 2} + \displaystyle \sum_{c \mathop \in C} \phi^c$ That is, the indices instance of $100$ corresponds to the indices $r, r - 1, r - 2$ now correspond to the string $011$. {{qed}}
Note that step $(2)$ is an instance of a simplification of $S$. From 100 in Golden Mean Number System is Equivalent to 011, it has been established that $011$ is equivalent to $100$. The following need to be established: :$010101 \ldots$ is equivalent to $100$ :Replacing the leftmost $011$ with $100$ reduces the overall number of instances of $011$. {{questionable|Except it doesn't, necessarily -- for example it converts $01011$ to $01100$. Something more sophisticated is needed.}} Let the first $1$ of $010101 \ldots$ represent the instance of $\phi^n$ for some $n \in \Z$. It follows that the first $0$ of $010101 \ldots$ represents the instance of $\phi^{n + 1}$ for some $n \in \Z$. Thus $010101 \ldots$ represents the real number $y$ where: :$y = \phi^n + \phi^{n - 2} + \phi^{n - 4} + \cdots$ and so: {{begin-eqn}} {{eqn | l = y       | r = \phi^n \displaystyle \sum_{k \mathop \ge 0} \phi^{-2 k}       | c =  }} {{eqn | r = \phi^n \dfrac 1 {1 - \phi^{-2} }       | c = Sum of Infinite Geometric Sequence }} {{eqn | r = \phi^n \dfrac {\phi^2} {\phi^2 - 1}       | c =  }} {{eqn | r = \phi^n \dfrac {\phi^2} {\paren {\phi + 1} - 1}       | c = Square of Golden Mean equals One plus Golden Mean }} {{eqn | r = \phi^{n + 1}       | c = after simplification }} {{end-eqn}} This can represented in the golden mean number system by $100$, where the $1$ corresponds to the first instance of $\phi^{n + 1}$. Hence $010101 \ldots$ is equivalent to $100$. {{qed|lemma}} {{finish|so far so good, but I'm going to the beach.}}
The proof proceeds by strong induction. For all $n \in \Z_{>0}$, let $P \left({n}\right)$ be the proposition: :$\len \left({S_n}\right) = F_n$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \len \left({S_1}\right)       | r = \len \left({\text a}\right)       | c = {{Defof|Fibonacci String}} }} {{eqn | r = 1       | c = {{Defof|Length of String}} }} {{eqn | r = F_1       | c = {{Defof|Fibonacci Number}}: $F_1 = 1$ }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = \len \left({S_2}\right)       | r = \len \left({\text b}\right)       | c = {{Defof|Fibonacci String}} }} {{eqn | r = 1       | c = {{Defof|Length of String}} }} {{eqn | r = F_2       | c = {{Defof|Fibonacci Number}}: $F_2 = 1$ }} {{end-eqn}} Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $1 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true. This is the induction hypothesis: :$\len \left({S_k}\right) = F_k$ and: :$\len \left({S_{k - 1} }\right) = F_{k - 1}$ from which it is to be shown that: :$\len \left({S_{k + 1} }\right) = F_{k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \len \left({S_{k + 1} }\right)       | r = \len \left({S_k S_{k - 1} }\right)       | c = {{Defof|Fibonacci String}} }} {{eqn | r = \len \left({S_k}\right) + \len \left({S_{k - 1} }\right)       | c = {{Defof|Length of String}} }} {{eqn | r = F_k + F_{k - 1}       | c = Induction Hypothesis }} {{eqn | r = F_{k + 1}       | c = {{Defof|Fibonacci Number}} }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>0}: \len \left({S_n}\right) = F_n$ {{qed}}
The proof proceeds by strong induction. For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition: :There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_n$. $\map P 1$ is the case: :$S_1 = \text a$ === Basis for the Induction === $\map P 2$ is the case: :$S_2 = \text b$ $\map P 3$ is the case: :$S_3 = \text {ba}$ This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $2 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. This is the induction hypothesis: :There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_k$. and: :There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_{k - 1}$. from which it is to be shown that: :There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_{k + 1}$. === Induction Step === This is the induction step: By definition of Fibonacci string: :$S_{k + 1} = S_k S_{k - 1}$ concatenated. By the induction hypothesis, neither $S_k$ nor $S_{k - 1}$ have any double $\text a$'s or triple $\text b$'s. The only way for $S_{k + 1}$ to have a double $\text a$ is for $S_n$ to end with $\text a$ and for $S_{k - 1}$ to begin with $\text a$. But from Fibonacci String Begins with ba, for all $k > 2$, $S_{k - 1}$ begins with $\text b$. When $k = 2$, $S_{k - 1}$ does begin with $\text a$, but then $S_{k + 1} = \text {ba}$, which has no double $\text a$. Similarly, the only way for $S_{k + 1}$ to have a triple $\text b$, either $S_k$ must end with double $\text b$ or $S_{k - 1}$ must begin with double $\text b$. From Fibonacci String Begins with ba, $S_{k - 1}$ does not begin with double $\text b$. From Fibonacci String Ends with ab or ba, $S_k$ does not begin with double $\text b$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :For all $n \in \Z_{>0}$, there are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_n$. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition: :$S_n$ begins with $\text{ba}$ We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these begin with $\text{ba}$. === Basis for the Induction === $P \left({3}\right)$ is the case: :$S_3 = \text{ba}$ Thus $P \left({3}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$S_k$ begins with $\text{ba}$ from which it is to be shown that: :$S_{k + 1}$ begins with $\text{ba}$ === Induction Step === This is the induction step: By definition of Fibonacci string: :$S_{k + 1} = S_k S_{k - 1}$ concatenated. So $S_{k + 1}$ begins with $S_k$. By the induction hypothesis, $S_k$ begins with $\text{ba}$. Thus $S_{k + 1}$ likewise begins with $\text{ba}$. So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :for all $n \in \Z$ such that $n \ge 3$, $S_n$ begins with $\text{ba}$. Category:Fibonacci Strings m1amfujyneiyn3x1ok7in380usedwui
The proof proceeds by induction. For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition: :$S_n$ ends either with $\text{ba}$ or with $\text{ab}$ We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these ends either with $\text{ba}$ or with $\text{ab}$. === Basis for the Induction === $P \left({3}\right)$ is the case: :$S_3 = \text{ba}$ Thus $P \left({3}\right)$ is seen to hold. $P \left({4}\right)$ is the case: :$S_4 = \text{bab}$ Thus $P \left({4}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$S_k$ ends either with $\text{ba}$ or with $\text{ab}$ and: :$S_{k - 1}$ ends either with $\text{ba}$ or with $\text{ab}$ from which it is to be shown that: :$S_{k + 1}$ ends either with $\text{ba}$ or with $\text{ab}$ === Induction Step === This is the induction step: By definition of Fibonacci string: :$S_{k + 1} = S_k S_{k - 1}$ concatenated. So $S_{k + 1}$ end with $S_{k - 1}$. By the induction hypothesis, $S_k$ ends either with $\text{ba}$ or with $\text{ab}$. Thus $S_{k + 1}$ likewise ends either with $\text{ba}$ or with $\text{ab}$. So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :for all $n \in \Z$ such that $n \ge 3$, $S_n$ends either with $\text{ba}$ or with $\text{ab}$. Category:Fibonacci Strings fa27blmnmughxk5hw6jd7wiavesx2th
The proof proceeds by strong induction. For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition: :$S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$. === Basis for the Induction === $\map P 3$ is the case: :$S_n = \text {ba}$ It can be seen that $S_n$ has $F_1 = 1$ instance of $\text a$ and $F_2 = 1$ instance of $\text b$. Thus $\map P 3$ is seen to hold. $\map P 4$ is the case: :$S_n = \text {bab}$ It can be seen that $S_n$ has $F_2 = 1$ instance of $\text a$ and $F_3 = 2$ instances of $\text b$. Thus $\map P 4$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $\map P j$ is true, for all $j$ such that $4 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. This is the induction hypothesis: :$S_k$ has $F_{k - 2}$ instances of $\text a$ and $F_{k - 1}$ instances of $\text b$ and: :$S_{k - 1}$ has $F_{k - 3}$ instances of $\text a$ and $F_{k - 2}$ instances of $\text b$ from which it is to be shown that: :$S_{k + 1}$ has $F_{k - 1}$ instances of $\text a$ and $F_k$ instances of $\text b$. === Induction Step === This is the induction step: By definition of Fibonacci string: :$S_{k + 1} = S_k S_{k - 1}$ concatenated. By the induction hypothesis: :$S_k$ has $F_{k - 2}$ instances of $\text a$ and $F_{k - 1}$ instances of $\text b$ and: :$S_{k - 1}$ has $F_{k - 3}$ instances of $\text a$ and $F_{k - 2}$ instances of $\text b$ So: :$S_{k + 1}$ has $F_{k - 2} + F_{k - 3}$ instances of $\text a$ and so by definition of Fibonacci numbers: :$S_{k + 1}$ has $F_{k - 1}$ instances of $\text a$ and: :$S_{k + 1}$ has $F_{k - 1} + F_{k - 2}$ instances of $\text b$ and so by definition of Fibonacci numbers: :$S_{k + 1}$ has $F_k$ instances of $\text b$. So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :for all $n \in \Z$ such that $n \ge 3$: $S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$. {{qed}}
The proof proceeds by strong induction. For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition: :for all $m \in \Z$ such that $1 < m \le n$, the initial part of $S_n$ of length $F_m$ is $S_m$. === Basis for the Induction === $P \left({3}\right)$ is the case: :$S_3 = \text{ba}$ For $m = 2$, $S_3$ starts with $S_2 = \text{b}$, which is the initial part of $S_3$ of length $F_2 = 1$. Thus $P \left({3}\right)$ is seen to hold. $P \left({4}\right)$ is the case: :$S_4 = \text{bab}$ For $m = 2$, $S_4$ starts with $S_2 = \text{b}$, which is the initial part of $S_4$ of length $F_2 = 1$. For $m = 3$, $S_4$ starts with $S_3 = \text{ba}$, which is the initial part of $S_4$ of length $F_3 = 2$. Thus $P \left({4}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $4 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true. This is the induction hypothesis: :for all $m \in \Z$ such that $1 < m \le k$, the initial part of $S_k$ of length $F_m$ is $S_m$. from which it is to be shown that: :for all $m \in \Z$ such that $1 < m \le k + 1$, the initial part of $S_{k + 1}$ of length $F_m$ is $S_m$. === Induction Step === This is the induction step: By definition of Fibonacci string: :$S_{k + 1} = S_k S_{k - 1}$ concatenated. Thus the initial part of $S_{k + 1}$ of length $F_k$ is $S_k$. By the induction hypothesis, for all $m \in \Z$ such that $1 < m < k - 1$, the initial part of $S_k$ of length $F_m$ is $S_m$. But we also have that the initial part of $S_{k + 1}$ of length $F_k$ is $S_k$.  So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :for all $m \in \Z$ such that $1 < m \le n$, the initial part of $S_n$ of length $F_m$ is $S_m$. {{qed}}
The proof proceeds by strong induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$ === Basis for the Induction === $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = \dfrac {3^0 - \left({-2}\right)^0} 5       | r = \dfrac {1 - 1} 5       | c =  }} {{eqn | r = 0       | c =  }} {{eqn | r = a_0       | c =  }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = \dfrac {3^1 - \left({-2}\right)^1} 5       | r = \dfrac {3 - \left({-2}\right)} 5       | c =  }} {{eqn | r = \dfrac {3 + 2} 5       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = a_1       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({j}\right)$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $P \left({k + 1}\right)$ is true. This is the induction hypothesis: :$a_k = \dfrac {3^k - \left({-2}\right)^k} 5$ and: :$a_{k - 1} = \dfrac {3^{k - 1} - \left({-2}\right)^{k - 1} } 5$ from which it is to be shown that: :$a_{k + 1} = \dfrac {3^{k + 1} - \left({-2}\right)^{k + 1} } 5$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = a_{k + 1}       | r = a_k + 6 a_{k - 1}       | c = Definition of $a_n$ }} {{eqn | r = \dfrac {3^k - \left({-2}\right)^k} 5 + 6 \times \dfrac {3^{k - 1} - \left({-2}\right)^{k - 1} } 5       | c =  }} {{eqn | r = \dfrac {3^k - \left({-2}\right)^k + 6 \times 3^{k - 1} - 6 \left({-2}\right)^{k - 1} } 5       | c =  }} {{eqn | r = \dfrac {3^{k - 1} \left({3 + 6}\right) - \left({-2}\right)^{k - 1} \left({-2 + 6}\right)} 5       | c =  }} {{eqn | r = \dfrac {3^{k - 1} \times 9 - \left({-2}\right)^{k - 1} \times 4} 5       | c =  }} {{eqn | r = \dfrac {3^{k - 1} \times 3^2 - \left({-2}\right)^{k - 1} \times \left({-2}\right)^2} 5       | c =  }} {{eqn | r = \dfrac {3^{k + 1} - \left({-2}\right)^{k + 1} } 5       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: a_n = \dfrac {3^n - \left({-2}\right)^n} 5$ {{qed}} {{improve|Given that this result was given as a question in a chapter on the Fibonacci numbers, one presumes there is a way to solve this by using Fibonacci numbers, or a Fibonacci-like approach. Maybe use generating functions.}}
The proof proceeds by induction. For all $n \in \Z_{\ge 1}$, let $P \left({n}\right)$ be the proposition: :$f \left({n}\right) = m$ for $F_m < n \le F_{m + 1}$ === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = 0       | c =  }} {{eqn | l = F_0       | o = <       | m = 1       | mo= \le       | r = F_{0 + 1}       | c =  }} {{eqn | r = 1       | c = {{Defof|Fibonacci Number}}: $F_1 = 1$ }} {{eqn | ll= \leadsto       | l = m       | r = 0       | c =  }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$f \left({r}\right) = m$ for $F_m < r \le F_{m + 1}$ from which it is to be shown that: :$f \left({r + 1}\right) = m$ for $F_m < {r + 1} \le F_{m + 1}$ === Induction Step === This is the induction step: {{finish}} {{begin-eqn}} {{eqn | l =        | o = =       | r =        | c =  }} {{eqn | ll= \leadsto       | l =        | o = =       | r =        | c =  }} {{end-eqn}} So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 1}: f \left({n}\right) = m$ for $F_m < n \le F_{m + 1}$ {{qed}}