I am learning [Kostrikin's Algebra book][1]. There are couple exercises I want to discuss with you, colleagues. I am looking for more elegant/math-alike solutions than I see. Also I am looking for any contructive comments.
Exercise #1
Description
Let $$ \Omega = \{+, -, ++, +-, -+, --, +++, ...\} $$ - is a set of all finite sequences of pluses and minuses, $$ f: \Omega \to \Omega $$ - is a transformation that transforms $$ \omega = \omega_1\omega_2...\omega_n \in \Omega $$ to $$ \omega'=\omega_1\dot\omega_1\omega_2\dot\omega_2...\omega_n\dot\omega_n $$ where $$ \dot\omega_k = - \text{, if } \omega_k=+ \text{, and } \dot\omega_k=+ \text{, if } \omega_k=- $$ Show that in $$ f(f(\omega)) $$ every segment with length > 4 contains $$++ \text{ or } --.$$
My solution
I built an $$ (\omega')' = \omega_1\dot\omega_1\dot\omega_1\ddot\omega_1\omega_2\dot\omega_2\dot\omega_2\ddot\omega_2...\omega_n\dot\omega_n\dot\omega_n\ddot\omega_n \text{ where } \ddot\omega_k = \begin{cases} +, & \text{if $\dot\omega_k = -$}, \\ -, & \text{if $\dot\omega_k = +$}; \end{cases}$$
$$\text{From $(\omega')'$ we can see that every segment of length 5 (the lowest length higher than 4)} \\ \text{always contains two equal elements } \dot\omega_k\dot\omega_k \text{ that are either $++$ or $--$.} $$ Solved.
Exercise #2
Description
$$\text{Consider mapping $f: \Bbb{N} \to \Bbb{N}$, that maps $n \mapsto n^2$.} \\ \text{Does $f$ have right inverse mapping? Show two left inverse mappings for $f$. } $$
My solution
$$\text{1) Does $f$ have right inverse mapping?} \\ \text{$f$ is injective but $f$ is not surjective. There is a theorem that shows that for two mappings $g$ and $f$} \\ \text{to make this valid: $fg = e_x g$ must be injective and $f$ must be surjective.} \\ \text{But $f$ is not surjective. Surjective mappings do not have right inverse mappings.} \\ \text{The answer is: $f$ does not have right inverse mapping. } $$
$$\text{2) Show two left inverse mappings for $f$.} \\ \text{Have to find two $g$ from $gf = e_N$} \\ \text{The first is obvious - $g(f(n)) = \sqrt{(f(n))}$} \\ \text{Second one is less obvious to me.} \\ \text{What I thinking of is a rule that just maps each $f(n)$ to 1,2,3,... Are there better examples?} $$
Solved.
[1]: Кострикин А.И. Введение в алгебру. В 3-х частях. Часть 1. Основы алгебры. 2-е издание М. МЦНМО 2018г. 272 с. Твердый переплет, Стандартный формат. (ISBN: 978-5-94057-887-1 / 9785940578871) page 40