We slightly refine the Collatz sequence as follows:  $$f(n) =  \begin{cases}        \text{Od}(n) &amp; \text{if } n \equiv 0 \pmod 2\\       3n + 1 &amp; \text{if } n \equiv 1 \pmod 2\\    \end{cases} $$  Where Od(n) is the  of n. And   $$a_i(n) =  \begin{cases}        n &amp; \text{for } i = 0\\       f(a_{i-1}(n)) &amp; \text{for } i &gt; 0\\  \end{cases} $$  We wish to classify numbers by their refined Collatz stopping times, $C(n) = s$.  $$C(n) = s \iff a_s = 1$$  And we wish to describe sets of natural numbers grouped by stopping time class: $$C_s = \{n\in\mathbb{N} \big| C(n) = s\}$$  Note: \begin{align} C_0 &amp;= \{1\}\\ C_1 &amp;= \{n,a \in \mathbb{N} | n = 2^a\}\\ C_2 &amp;= \bigg\{n,a \in \mathbb{N}, a&gt;1 \bigg|n = \frac{4^a - 1}{3}\bigg\}\\ C_3 &amp;= \{n\in\mathbb{N}, c_1\in C_1, c_2\in C_2\bigg|n=c_1 \times c_2\}\\ C_4 &amp;= \bigg\{n\in \mathbb{N},a_1,a_2 \in \mathbb{Z}^*, r_1\in \{1,2\}, 3a_1+r_1\geq2\bigg|n=\frac{1}{3}\bigg(2^{3 - r_1}\times4^{a_2}\times\frac{4^{3a_1 + r_1} - 1}{3} - 1\bigg)\bigg\}\\ C_5 &amp;= \{n\in\mathbb{N}, c_1\in C_1, c_4\in C_4\bigg|n=c_1 \times c_4\} \end{align}  This seems to be quickly getting out hand. What is $C_6$? What texts might have similar analysis?
123693
Yes, there is no one who doesn't know this problem.My question is only about curiosity.     $$C(n) = \begin{cases} n/2 &amp;\text{if } n \equiv 0 \pmod{2}\\ 3n+1 &amp; \text{if } n\equiv 1 \pmod{2} .\end{cases}$$   On this problem, I caught something like this.I'm sure, We all realized that.  For example, $n=19$, we have $6$ odd steps.  We know that, even steps are not important, because each even number is converted to an odd number.     $19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$   Then, for $n=77$, We have also $6$ odd steps.     $77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1$   For $n=9$      $9\Longrightarrow 7 \Longrightarrow   11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1$   Again we have $k=6$ odd steps.     I want to know / learn / ask, for $k=6$, (Generalized: for any number $k$ ) can we produce a formula(s) to catch all such numbers, which gives the result  $1$?    Thank you!
53291
Problem  The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:   $a_1=\sqrt2$, $a_{n+1}=\sqrt{2+a_n}$.   Evaluate the limit $\lim_{n\to\infty} a_n$.  Solution   Show that the sequence $(a_n)_{n=1}^\infty$ is monotonic. The statement $$V(n): a_n &lt; a_{n+1}$$ holds for $n = 1$, that is $\sqrt2 &lt; \sqrt{2+\sqrt2}$. Let us assume the statement holds for $n$ and show that $V(n) \implies V(n+1)$. We have that $$a_n &lt; a_{n+1}.$$ Adding 2 to both sides and taking square roots, we have that $$\sqrt{2+a_n} &lt; \sqrt{2+a_{n+1}},$$ that is $a_{n+1} &lt; a_{n+2}$ by definition. Find bounds for $a_n$. The statement $$W(n): 0 &lt; a_n &lt; 2$$ holds for $n=1$, that is $0 &lt; \sqrt2 &lt; 2$. Let us assume the statement holds for $n$ and show that $W(n) \implies W(n+1)$. We have that $$0 &lt; a_n &lt; 2.$$ Adding two and taking square roots, we have that $$0 &lt; \sqrt2 &lt; \sqrt{2+a_n} &lt; \sqrt4 = 2.$$ The limit $\lim_{n\to\infty} a_n$ exists, because $(a_n)_{n=1}^\infty$ is a bounded monotonic sequence. Let $A = \lim_{n\to\infty} a_n$. Therefore the limit $\lim_{n \to\infty} a_{n+1}$ exists as well and $\lim_{n \to\infty} a_{n+1} = A$. (For $(n_k)_{k=1}^\infty = (2,3,4, \dots)$, we have that $(a_{n_k})_{k=1}^\infty$ is a subsequence of $(a_n)_{n=1}^\infty$, from which the statement follows.) We have that $a_{n+1} = f(a_n)$. That means that $A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}$. Solving the equation $A = \sqrt{2 + A}$, we get $A = -1 \lor A = 2$. Putting it all together, we get that $A = 2$, because the terms of the sequence are increasing and $a_1 &gt; 0$.   Is my solution correct?
96091
The Collatz Conjecture is a famous conjecture in mathematics that has lasted for over 70 years. It goes as follows:  Define $f(n)$ to be as a function on the natural numbers by:  $f(n) = n/2$ if $n$ is even and $f(n) = 3n+1$ if $n$ is odd  The conjecture is that for all $n \in \mathbb{N}$, $n$ eventually converges under iteration by $f$ to $1$.  I was wondering if the "5n+1" problem has been solved. This problem is the same as the Collatz problem except that in the above one replaces $3n+1$ with $5n+1$.
1110
I'm currently working my way through Harvard's online abstract algebra lectures (if you're interested, you can find them ).  The lectures come complete with notes and homework problems.  Of course, since I don't actually go to Harvard, I can't hand in the homework assignments to find out if I'm doing them correctly.  So I've decided to try to crowd source the grading of my solutions.  I'll post individual questions as I finish them and wait for comments.  I would like to get critiques of not just my reasoning but the style of my write ups as well.  Also, any alternative approaches to the problem would be welcome.  I've looked for forums dedicated to these kinds of OCW courses but have not been able to find any.  This surprises me.  It seems like, with the advent of these free online educational resources, an online meeting place for those who take advantage of them would be a natural offshoot.  So this might be a bit of an experiment.  Unless I'm the 873rd person to post something like this here.  If that's the case, sorry for being so long winded.  Anyway, on with the question.  This one is assigned in the 4th lecture.     Let $V$ denote the Klein 4-group.  Show that $\operatorname{Aut}(V)$ is isomorphic to $S_3$.   (I've moved my solution below to keep this question from showing up in the unanswered list.)  Thanks...
109902
I'm reading the proof of the multiplicative property of $$s_k(n)=\sum_{d|(n,k)}f(d)g\bigg( \frac kd\bigg)$$ The book wrote that in order to understand the proof, we need to know if $a,b,k,m$ are integers such that $(a,k)=(b,m)=1$, then $$(ab,km)=(a,m)(b,k).$$  I have written proof, but I don't know is it consistent, and this seems used more tools than it necessarily has to.   My proof: Let $$S_1=\{d: d|(a,m), d&gt;0\}\\S_2=\{d: d|(b,k), d&gt;0\}\\S=\{d:d|(ab,km), d&gt;0\}.$$ I will prove the statement by showing that the function $\phi:S_1\times S_2\to S$ defined as $$\phi(d_1,d_2)=d_1d_2$$ is bijective, and complete the proof by connecting the maximal elements from these partial ordering sets.  Obviously, for $d_1\in S_1, d_2\in S_2$, this tells us $d_1 $ is both divisor of $a$ and $m$, and $d_2$ is both divisor of $b$ and $k$, hence $d_1d_2$ is both divisor of $ab$ and $km$, so we can tell $\phi(S_1\times S_2)\subseteq S$.  To show one-to-one, if we have $c_1,d_1\in S_1$ and $c_2,d_2\in S_2$ such that $$\phi(c_1,c_2)=\phi(d_1,d_2)$$ then $$c_1c_2=d_1d_2$$  In particular, $c_1|d_1d_2$, we need to show $c_1|d_1$. Indeed, if $(c_1,d_2)&gt;1$, this implies $a,k$ have common factor larger than $1$, contradicts the assumption. Therefore $c_1|d_1$. A similar argument implies $d_1|c_1$, therefore $c_1=d_1$. Again a similar argument implies $c_2=d_2$, which shows one-to-one.   To show onto, let $d\in S$, since $d|ab$, we can split $d$ into $d_1d_2$ such that $d_1|a$ and $d_2|b$. Since $(a,k)=1$, we see $(d_1,k)=1$, by property of $S$ we see also $d_1|m$, therefore $d_1\in S_1$, a similar argument implies $d_2\in S_2$. So every number in $S$ has a preimage in $S_1\times S_2$  This completes the proof. Is there any mistake in the proof?     Edit: A proof about the statement: If $d|ab$ then there are $d_1,d_2$ such that $d=d_1d_2$ and $d_1|a, d_2|b$.  Let $ab=p_1^{e_1}\cdots p_r^{e_r}$ for distinct prime factors $p_1,\cdots,p_r$, then we can write $$a=p_1^{x_1}\cdots p_r^{x_r}\\b=p_1^{y_1}\cdots p_r^{y_r}$$ where some of $x_i, y_i$ can be zero, and $x_i+y_i=e_i$ for all $i=1,\cdots,r$.  Since $d|ab$, we can write $$d=p_1^{f_1}\cdots p_r^{f_r}$$ with $f_i\leq e_i$ for all $i=1,\cdots,r$. Then since $$f_i\leq x_i+y_i$$ Split into two cases:   Case 1: $f_i&gt; x_i$, then we can choose $g_i=x_i$ and $h_i=f_i-x_i$.  Case 2: $f_i\leq x_i$, then we can choose $g_i=f_i$ and $h_i=0$.  Then the choice $$d_1=\prod_{i=1}^r p_i^{g_i}, \quad d_2=\prod_{i=1}^r p_i^{h_i}$$ will prove the statement.
42409
Came across the following exercise in Bartle's Elements of Real Analysis. This is the solution I came up with. Would be grateful if someone could verify it for me and maybe suggest better/alternate solutions.  I also looked up these related questions - , ,  - but was not happy with proofs given there. I seem to need some help understanding these. Any such help is appreciated.      Show that every sequence in $\Bbb R$ either has a monotone increasing sub-sequence or a monotone decreasing sub-sequence.   Let $(x_n)$ be a sequence in $\Bbb R$. Suppose $(x_n)$ is not bounded. Without loss of generality we may assume that $(x_n)$ is not bounded above. Therefore given any real number there is a member of the sequence which is greater. Let $x_{n_1}$ be any member of the sequence.   There is $x_{n_2} \gt \sup\{x_1, x_2, ..., x_{n_1} \}$. For $i \gt 1$ let $x_{n_i} = \{x_1, x_2, ..., x_{n_{i - 1}}\}$ then $(x_{n_k})$ forms a monotone subsequence of $(x_n)$.   Now suppose instead that $(x_n)$ is bounded. By the Bolzano-Weierstrass Theorem there is a subsequence $(y_n)$ of $(x_n)$ which converges to a limit $y$. Without loss of generality there are infinitely many distinct values in $(y_n)$ that are unequal to $y$. Let $y_{k1}$ be the first such element. Let $y_{k2}$ be any element in $\{ y' \in (y_n) \ \ | \ \ |y' - y | \lt |y - y_{k1}|  \}$.   For $i \gt 1$ let $y_{ki} \in \{ y' \in (y_n)  \ \ | \ \ |y' - y | \lt |y - y_{k \ i - 1}|  \}$. Such $y_{ki}$ exists for every $i \in  \Bbb N$ since $ \lim (y_n) = y $. Now let $(y_{kn})$ be the sub-sequence of $(y_n)$ thus formed. At least one of the two following sets must contain infinitely many elements.    $\{ y \in (y_{kn}) \ \ | \ \ y \gt x\}$ $\{ y \in (y_{kn}) \ \ | \ \ y \lt x\}$   The one which does forms a monotone subsequence.
3499
1.(a) Consider the recurrence relation $a_{n+2}a_n = a^2_{n+1} + 2$ with $a_1 = a_2 = 1$.  (i) Assume that all $a_n$ are integers. Prove that they are all odd and the integers $a_n$ and $a_{n+1}$ are coprime for $n \in \mathbb{N}$.  (ii) Assume that the set $\left\{ a_n,\ a_{n+1},\ a_{n+2}\right\}$ is pairwise coprime for $n \in \mathbb{N}$. Prove that all $a_n$ are integers by induction.  (b) Consider the recurrence relation $a_{n+2}a_n = a^2_{n+1} + 1$ with $a_ 1= 1,\ a_2 = 2$ and compare this sequence to the Fibonacci numbers. What do you find? Formulate it as a mathematical statement and prove it.  I have no idea where to start with 1(ai) but with 1(aii) i have started with:  Given that $\left\{a_n,\ a_{n+1},\ a_{n+2}\right\}$ is pairwise coprime you get $gcd(a_n,a_{n+1})=1,\  gcd(a_n,a_{n+2})=1,\  gcd(a_{n+1},a_{n+2})=1$  Using the initial terms, you can do base induction on $a_{n+2} = \frac{a^2_{n+1}+2}{a_n}$ to prove whether the next terms will be integers.  $a_3 = \frac{1^2+2}{1}$ $a_4 = \frac{3^2+2}{1}$ $a_5 = \frac{11^2+2}{3}$  gives:  $a_1=1,\ a_2 = 1,\ a_3 = 3,\ a_4 = 11,\ a_5 = 41$  which are all integers. So the base induction is correct.  Now for the inductive step $n=k, k \rightarrow k+1$  $a_{k+3} = \frac{a^2_{k+2}+2}{a_{k+1}}$. I am not sure how to prove this for the inductive step.  Also for (b) how do you formulate a mathematical statement and prove it?
3816
I have a sequence $(a_n)$ where for each natural number $n$,   $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})$$ and I want to find its limit as $n\to\infty$.   I obviously couldn't prove it and after several futile attempts decided to post it here.   Here is a list of a few observations which I got from those attempts:   The sequence $(a_n)$ is a strictly increasing sequence. To prove this, I rewrote each element as $$a_n = (1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})= \frac{(n+1)\cdots(n+n)}{n^n}= \frac{(2n)!}{n!n^n}.$$ Then $$\frac{a_{n+1}}{a_n}= \frac{2(n+1)!}{(n+1)!(n+1)^{n+1}}\frac{n!n^n}{(2n)!}=\frac{(2n+1)(2n+2)}{(n+1)^2(1+\frac{1}{n})^n} \to \frac{4}{e}$$ as $n\to\infty$. Since $\frac{4}{e}&gt;1$ we have $a_{n+1}&gt;a_n$ eventually. The limit of this sequence is bounded below by $e$. By replacing $1,2, \ldots, n$ with $1$ in the expression of $a_n$, we get $a_n \geq (1+\frac{1}{n})^n$. And thus $\lim{(a_n)}\geq e$.  $\lim{(a_n)}\geq e^2$ and $\lim{(a_n)}\geq e^3$. The first assertion follows from the fact that $$a_n\geq(1+\frac{1}{n})(1+\frac{2}{n})^{n-1}= \frac{(1+\frac{1}{n})(1+\frac{2}{n})^{n}}{(1+\frac{2}{n})} \to e^2.$$ And the last one follows the same way because $$a_n\geq (1+\frac{1}{n})(1+\frac{2}{n})(1+\frac{3}{n})^{n-2}.$$   Now I have a gut feeling that for any natural number $k$, one can show that for all sufficiently large natural number $n$, $$a_n\geq (1+\frac{1}{n})\cdots(1+\frac{k-1}{n})(1+\frac{k}{n})^{n-(k-1)}.$$ And therefore for all $k \in \mathbb{N}$, $\lim{(a_n)}\geq e^k$ making the sequence divergent. But I'm really not sure about this approach and I'll appreciate any help towards this end. Thank you.  [Note: As this sequence is quite common, there may be other posts on math.SE asking the same question. I didn't search for them because I just don't know how to search for an expression this big. Though a link related to any previous question concerning this particular sequence will be good enough, I will greatly appreciate if someone takes the trouble to look into my approach/observations and point out where  I'm going wrong.]
61026
Motivating question:    I'm aware that one can construct a hierarchy of number systems via the Cayley–Dickson process:  $$\mathbb{R} \subset \mathbb{C} \subset \mathbb{H} \subset \mathbb{O} \subset \mathbb{S} \subset \ldots $$  "Reals" $\subset$ "Complex" $\subset$ "Quaternions" $\subset$ "Octonions" $\subset$ "Sedenions" $\subset$ $\ldots$  and that at each step you're given a multiplication table that tell how the elements interact. As you move up the ladder, certain "nice" properties are lost: ordering,  commutativity, associativity, multiplicative normedness, etc... Given the multiplication table, you can show that these properties don't hold.   Eric Naslund noted that "the first 4 are very special as they are the unique 4 normed division algebras over ℝ", no surprise then that these $2^n$-ions have found quite a bit of use. I'm interested in the sequence itself however, irrespective of how useful a $2^{256}$-ion might be (ducenti-quinquaginta-sex-ion?).   I feel like something deeper is going on here though that I don't understand. Why are these particular properties lost at each step? Is it possible to quantify the process such that, at the $2^n$-ion you can say something about the symmetry of the multiplication table*?  * I'm making an ansatz that there is a connection between the symmetry of the multiplication table and these "nice" properties.
21836
Let $M_n(\mathbb{F})$ be the set of all $n\times n$ with entries in $\mathbb{F}$ and let $\exp:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be defined by  $$ \exp(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!},$$ for all $A\in M_n(\mathbb{C}).$  I want to prove that $\exp$ is a surjective map from $M_n(\mathbb{C})$ to $GL(n,\mathbb{C})=\left\{A\in M_n(\mathbb{C})\,\middle| \det(A)\neq0\right\}$, how do I go about that?   I mean saying that $\exp:M_n(\mathbb{R})\to GL(n,\mathbb{R})$ is an analogous to saying $\exp:\mathbb{R}\to \mathbb{R}_{&gt;0}$ and this is also pretty intuitive, since, in analogy with the case of numbers, $A^0=I\;\forall A$, so $\exp(0)=I+0+\frac{0^2}{2!}+\dots=I$, so even for $A=0$ we get $\det\left(\exp(A)\right)\neq0$ and so because of the first term we can never get a zero determinant. But I have no idea how to prove the subjectiveness.  Thanks in advance.
72019
Show that the sequence $\{a_n\}_{n\in \mathbb{N}} = \{x: x=$ the nth decimal digit of Champernowne's constant$\}$ is not periodic.  For those who don't know what Champernowne's constant is, it's the following number:  $C_{10} = 0.123456789101112131415161718192021...$  Well, before someone says that $\{a_n\}_{n\in \mathbb{N}}$ is definitely aperiodic because Champernowne's constant is irrational(and even transcendental), I have to say that I'm well aware of the fact that a number is rational if and only if the sequence of its digits is periodic, but I'm looking for another proof that the sequence of the digits of $C_{10}$ is aperiodic that doesn't use anything but a smart way of showing that $a_{n+k} = a_n$ can't hold for any $k\in \mathbb{N}$.  I've observed that by grouping the terms of a sequence in a clever way we can show that no such $k$ can exist for some sequences. For example I created the following sequence:  $1,0,0,1,1,1,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,1,1,1,1,1,1,\cdots$  The sequence is created this way: First you put one $1$, then two $0$'s, and in the nth step you add either $1$'s or $0$'s n times depending on whether n is odd or even respectively.   It's easy to show this sequence is not periodic, because no matter how large $k$ is you can always find $n$ such that $a_n = 1$ but $a_{n+k}=0$ or $a_n=0$ but $a_{n+k}=1$. That proves the sequence is not periodic and as a corollary it' proved that the real number $0.1001110000111110000001111111\cdots$ is irrational.  After showing that I've been trying to apply this technique of grouping to the sequence in problem without much success so far.  I think if the periodicity of a sequence $\{a_n\}$ is denoted by $\Omega(\{a_n\})$ then if we define a new sequence $\bar{a}_m(n) := \bar{a}_{m,n} = [a_n]_m$ then $\Omega(\{\bar{a}_{m,n}\}) \mid \Omega({a_n})$. This shouldn't be very hard to prove  I think.  Now, to show that   $1,2,3,4,5,6,7,8,9,1,0,1,1,1,2,1,3,1,4,1,5,1,6,1,7,1,8,1,9,2,0,2,1,\cdots$  is not periodic it's sufficient to prove that the sequence   $1,0,1,0,1,0,1,0,1,1,0,1,1,1,0,1,0,1,0,1,\cdots$  is not periodic. But I haven't progressed much from here  :/
65683
I have a repeated composition of functions ${T_n}(z) = {\tau _0} \circ {\tau _1} \circ {\tau _2} \circ  \cdots  \circ {\tau _n}(z)$  By analogy with $\sum\limits_{i = 1}^n {} ,\prod\limits_{i = 1}^n {} ,\bigcup\limits_{i = 1}^n {} ,\bigcap\limits_{i = 1}^n {} ,$ I want to write ${T_n}(z) = \left( {\mathop  \circ \limits_{i = 0}^n {\tau _i}} \right)(z)$ or even ${T_n}(z) = {\mathop  \circ \limits_{i = 0}^n {\tau _i}} (z)$. Can I do this?
74716
I was looking at the Fibonacci sequence:  $$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,\ldots, F_n : F_n + F_{n - 1} = F_{n + 1} \land F_0 = 0$$  Here I noticed something; $\qquad F_n^2 + F_{n\pm1}^2 = F_{2n\pm1} : n\in \mathbb{N}$  For example; $$\ \ F_3^2 + F_{3 + 1}^2 = F_3^2 + F_4^2 = 2^2 + 3^2 = F_{2\times3 + 1} = F_{6 + 1} = F_7 = 13$$ $$F_3^2 + F_{3 - 1} = F_3^2 + F_2^2 = 2^2 + 1^2 = F_{2\times3 - 1} = F_{6 - 1} = F_5 = 5$$ However, no matter how true this can seem, I cannot seem to prove/disprove it.    My Attempt: $$\because (a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab : (a,b)\in \mathbb{Z}$$ $$\implies F_n^2 + F_{n\pm 1}^2 = (F_n + F_{n\pm 1})^2 - 2F_nF_{n\pm 1}$$ $$\implies F_n^2 + F_{n - 1}^2 = F_{n + 1}^2 - 2F_nF_{n - 1} : F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_n^2 + F_{n + 1}^2 = F_{n + 2}^2 - 2F_nF_{n + 1} : F_n + F_{n + 1} = F_{n + 2} \iff F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_{2n\pm 1} + 2F_nF_{n\pm 1} = F_{n + 2\lor1}^2 : F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1} \tag{+2 if +, +1 if -}$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1} - F_{n\pm 1}^2}\sqrt{F_{2n\pm 1} - F_n^2}= F_{n + 2\lor1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{(F_{2n\pm 1} - F_{n\pm 1}^2)(F_{2n\pm 1} - F_n^2)} = F_{n + 2\lor1}^2$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1}(F_{2n\pm 1} - F_n^2 - F_{n\pm 1}^2) - (F_nF_{n\pm 1})^2} = F_{n + 2\lor1}^2$$.  $$\because F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{-(F_nF_{n\pm 1})^2} = F_{n + 2\lor1} \neq F_{2n\pm 1} + 2\sqrt{\big(-(F_nF_{n\pm 1})\big)^2}$$ $$\therefore F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} : i = \sqrt{-1}$$  I have looked over every step and cannot see where I am wrong. The only part that must be wrong is that:  $$F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} \iff F_n^2 + F_{n\pm1}^2 = F_{2n\pm 1}$$  I can seperate $F_{2n\pm 1}$ as $F_n^2 + F_{n\pm 1}^2$ (if true) so I turn the equation into:  $$F_n^2 + F_{n\pm 1}^2 + 2F_nF_{n\pm 1}i = F_{n + 2\lor1}$$ $$\therefore (F_n + F_{n\pm 1})\bigg(\frac{F_n}{F_{n\pm 1}} + \frac{F_{n\pm 1}}{F_n} + 2i\bigg) = F_{n + 2\lor1}$$ Now I know that the greater the value of $n$, one of the fractions in the above equation (depending on which sign is used in $F_{n\pm 1}$) will get closer to the golden ratio $φ = \frac{1 + \sqrt{5}}{2}$ but $i = \sqrt{-1}$ is not a "real" number, although $F_n$ is.  I then searched for other formulae for finding $F_n$ and came across the one below (being Binet's Formula):  $$F_n = \bigg\{\frac{φ^n - ψ^n}{φ - ψ} : ψ = φ - \frac{\sqrt{20}}{2} = \frac{1 - \sqrt{5}}{2} \ \land \ φ - ψ = \sqrt{5}\bigg\}$$  But I do not know what how to turn that into $F_{n\pm 1}$. Could you please show me where I went wrong (because I must have gone wrong somewhere), and also if you wish, could you please show me how to turn the above fraction into $F_{n\pm 1}$ (if possible).  I only used the $(a + b)^2$ expansion rule because my skill level in mathematics is not particularly high compared to most of the users on this site (especially when logging into the MSE and finding fancy symbols like $\nabla$ and $\partial$ and $\sum$ which all look Greek to me, however the only "fancy" thing I know is that $i = i^5 = \cdots = \sqrt{-1}$) so if you used any other techniques in trying to prove this formula, could you please show me step by step?  Much appreciated. Thank you in advance.    Edit: I believe this question is a duplicate of this other question $\longrightarrow$   and I can go here to serve as a little hint $\longrightarrow$
72678
Where $n \in \mathbb{N}$   and  $$ F_n = \begin{cases} 0                 &amp; \text{ if } n = 0 \\ 1                 &amp; \text{ if } n = 1 \\ F_{n-1} + F_{n-2} &amp; \text{ if } n &gt; 1 \end{cases} $$  This is basically describing the famous Fibonacci sequence.  If we try the base case with 1, it works (for $0$ I am not sure...)  When $n = 1$  $2 \cdot0 - 1^2 = (-1)^1 = -1$  For the inductive hypothesis, we assume this $F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n$ is true.  Now, for the inductive step, we try to prove for $n+1$, so for  $F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}$.  Since $n$ is always a natural number, and it will be always or even or odd, the $-1$ raised to $n$ will be always either $-1$ (when $n$ is odd) or $1$ (when $n$ is even).   Thus, $F_{n+1} \cdot F_{n-1} - F^2_n$ = -($F_{n+2} \cdot F_{n} - F^2_{n+1}$).  Or simply: $$ (-1)^n = -(-1)^{n+1} $$ $$ 1 \cdot (-1)^n = -(-1)\cdot(-1)^n $$ $$ 1 = -(-1) $$ Which is true.  I don't know if this is sufficient... I can arrive at the second step and say that we know this $(-1)^n$ is equals to $F_{n+1} \cdot F_{n-1} - F^2_n$, but I am not sure...
100946
The  (ESC), states that for every natural number $n \geq 2$, there exists a set of natural numbers $a, b, c$ , such that the following equation is satisfied:  $$\frac{4}{n}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\tag{1}$$  The basic approach to solving this problem outlined by Mordell [Ref1] is described below  By defining $t$ and $m$ as positive integers greater than zero and $q$ a positive integer greater than one we can observe that  a) There is always a solution for even $n$, since if $n=2^qt$ we have the trivial solution $$\frac{4}{4t}=\frac{1}{t}$$  In the remaining case $n=2(2t+1)$, a solution in the form of two Egyptian fractions can always be found  e.g. $$\frac{4}{2(2t+1)}=\frac{2}{2t+1}=\frac{1}{t+1}+\frac{1}{(t+1)(2t+1)}$$  b) If $(1)$ is a solution for some particular prime $n$ then all composite numbers $mn$ divisible by $n$ are also solutions, thus  $$\frac{4}{mn}=\frac{1}{ma}+\frac{1}{mb}+\frac{1}{mc}$$  will also be a solution. This means that we can simplify the analysis to the cases where $n$ is a prime greater than 2.  Using Mordell's approach we have just shown that we only need to consider the cases where $n$ is prime and where $n \equiv 1 \pmod{2} \;\;[meaning \;\;n=2t+1]$  The argument continues...  Mordell goes on to show in turn that the search can be reduced further to the cases when $$n \equiv 1 \pmod{4} \;\;[meaning \;numbers \;\;n=4t+1]$$ $$n \equiv 1 \pmod{8} \;\;[meaning \;numbers \;\;n=8t+1]$$ $$n \equiv 1 \pmod{3} \;\;[meaning \;numbers \;\;n=3t+1]$$ $$n \equiv 1,2,4 \pmod{7} \;\;[meaning \;numbers \;\;n=7t+1,n=7t+2 \;or\;n=7t+4 ]$$ $$n \equiv 1,4 \pmod{5}  \;\;[meaning \;numbers \;\; n=5t+1 \;or\;n=5t+4]$$  Assembling these results together, Mordell showed that the conjecture can be proved in this context except for the cases when $$n \equiv 1,11^2,13^2,17^2,19^2,23^2 \pmod{840}$$  Mordell stated that since the first prime meeting this condition is 1009, this is proof that the conjecture holds for $n&lt;1009$.  This basic approach can be pursued further. Other workers have shown that the conjecture holds for much higher values of $n$ using similar methods as can be seen on the above Wikipedia page.   Note that other intermediate results can be constructed from the above congruence's, e.g. $n \equiv 1 \pmod{24}$.  The question is:  Are there any other elementary approaches to solving this problem than the one outlined by Mordell (and described above)?  [Ref1] Louis J. Mordell (1969) Diophantine Equations, Academic Press, London, pp. 287-290.
90614
By naming an equivalence class in the domain that is assigned at least two different values prove that the following is not a well defined function.     $$f : \Bbb Z_{3} \to \Bbb Z_{6} \;\;\;\text{ given by } f(\overline x) = [x]  $$      In this case we represent an element of the domain as an $\bar x$ and use the notation $[x]$ for equivalence classes in the co-domain.   $f(\overline0) =  [0] \;,$  $ \Bbb Z_{3}  \quad (3x+0)\;\; \overline 0 = \{ ...-6,-3,0,3,6...  \}, \; \Bbb Z_{6}\;  (6x+0)\; \overline0 =\{ ...-12,-6,0,6,12...\}$   $f(\overline1) = [1], $ $\qquad   \; (3x+1) \; \;\;\;\overline   1 = \{ ...-5,-2,1,4,7 ... \},\; \;  (6x+1)\;\overline1 =\{...-11,-5,1,7,13.. \}$  $f(\overline2) = [2], $ $\qquad  \qquad \qquad \;\overline 2 = \{ ...-4,-3,2,5,8 ... \},\;\;\overline 2 = \{ ...-10,-4,2,8,14  ...\},\;$  $f(\overline3) = [3] ,$  $\qquad \qquad \qquad \qquad \qquad \qquad,\; \quad \quad \quad \; \; \; \;\overline 3 = \{ ...-9,-3,3,9,15 ... \},$  $f(\overline4) =  [4],\qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \; \;\overline 4 = \{ ...-8,-2,4,10,16...  \}, $   $f(\overline5) = [5], \qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \;\;\overline 5 = \{ ...-7,-1,5,11,17...  \},  $  $f(\overline6) = [6] ,$  So my main question for this problem is how to find out if this question is not a function. From the information I have gathered here I still cannot see why this is not a function any help on showing how this is not function would be much appreciated.     The set of equivalence classes for the relation $\cong_{m}$ is denoted $\Bbb Z_{m}$   The $ 3x+0 \text{ and } 6x+0$ are just showing how I got $\overline 0 $
53006
If the sequence $b_n$ is defined as $$b_n =\frac {a_1+a_2+\dots+a_n}n.$$  then  I found out that $b_n$ is always divergent and now that I have to prove the given statement I tried using comparison test. The problem is that I am getting that this is true if the sequence $a_n$ in monotonically increasing. But the hypothesis only says that it is a non negative sequence.