Improve Your Algebra Skills with Practice Problems

$a$ and $b$ for which $\frac{a}{b}=\stackrel{¯<!--\; ¯\; -->}{a.b}$
Are there natural numbers $a$ and $b$ such that $\frac{a}{b}=\stackrel{¯<!--\; ¯\; -->}{a.b}$?
I know that $\stackrel{¯<!--\; ¯\; -->}{a.b}=a+\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{b}_i{10}^{-<!--\; -\; -->i}$ where $b_i$ is the $i$th digit of $b$ and $b$ has $n$ digits.
I also know that $b=\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{b}_i{10}^{n-<!--\; -\; -->i}$

How do you solve and graph ${\displaystyle -13m>-26}$ ?

Why do the denominators of two fractions with numerators 1 add up to a third fraction that have the special things below?
I found out that the denominators of two fractions with numerators 1 add up to a third fraction that has the sum in the numerator and the product in the denominator. For example, $\frac{1}{5}+\frac{1}{6}=\frac{11}{30}$ because one, $0.2+0.1\stackrel{¯<!--\; ¯\; -->}{6}=0.3\stackrel{¯<!--\; ¯\; -->}{6}$ , which convert to the main fractions, and two, the denominators with numerators 1 add up to a third fraction having a sum in the numerator and evaluates to the product in the denominator. Why does this always happen? This is mind-blown for me!

Finding limits when given $2$ different limits.
Let $\underset{x\to <!--\; \to \; -->-<!--\; -\; -->1}{lim}f(x)=8$ and $\underset{x\to <!--\; \to \; -->-<!--\; -\; -->1}{lim}g(x)=-<!--\; -\; -->4$. Find $\underset{x\to <!--\; \to \; -->-<!--\; -\; -->1}{lim}{\displaystyle \frac{f(x)}{g(x)}}$
Answer Choices are:
A.$-<!--\; -\; -->2$
B.$12$
C.$-<!--\; -\; -->1/2$
D.$-<!--\; -\; -->1$
I started out trying to solve this problem by attempting to use a derivative formula which just overcomplicated the problem and did not lead me to any of the potential answer choices.
I was struggling with this problem as I did not understand the limit laws and upon further review I understood them well enough to realize that I should have just simply plugged in my $f(x)$ and $g(x)$ values to get my final answer of $-<!--\; -\; -->2$
Is this okay?

I have the following proof in my notes:
$(C[0,1],\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->{\Vert <!--\; \Vert \; -->}_1)$ is not complete
Consider the sequence of functions
$f_n(t)=\{\begin{array}{lll}\frac{1}{\sqrt{t}}& ,& 1/n\le <!--\; \le \; -->t\le <!--\; \le \; -->1\\ \sqrt{n}& ,& 0\le <!--\; \le \; -->t\le <!--\; \le \; -->\frac{1}{n}\end{array}$
It can easily be shown that it is Cauchy in $\Vert <!--\; \Vert \; -->\cdot <!--\; \cdot \; -->{\Vert <!--\; \Vert \; -->}_1$ So we have that $\mathrm{\forall <!--\; \forall \; -->}\mathrm{\epsilon <!--\; \epsilon \; -->}\mathrm{\exists <!--\; \exists \; -->}{N}_{\mathrm{\epsilon <!--\; \epsilon \; -->}}$ such that $\Vert <!--\; \Vert \; -->f_n-<!--\; -\; -->f{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}\le <!--\; \le \; -->\mathrm{\epsilon <!--\; \epsilon \; -->}\mathrm{\forall <!--\; \forall \; -->}m,n>N_{\mathrm{\epsilon <!--\; \epsilon \; -->}}$
Suppose $f_n$ converges, Then $\mathrm{\exists <!--\; \exists \; -->}f\in <!--\; \in \; -->C[0,1]$ such that $\Vert <!--\; \Vert \; -->f_n-<!--\; -\; -->f{\Vert <!--\; \Vert \; -->}_1\to <!--\; \to \; -->0.$
Then $f_{n_k}\to <!--\; \to \; -->f\text{ a.e in}[0,1]$
$f(t)=\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{f}_{n_k}(t)=\frac{1}{\sqrt{t}}\text{ a.e in}[0,1]$. Absurd.
I'm just failing to understand the last line. Can someone please explain
1. what is the absurd?
2. And how does pointwise convergence follows from a.e convergence (last line). Am I supposed to understand this $f_{n_k}\to <!--\; \to \; -->f\text{ a.e in}[0,1]$ as convergence in the real numbers, with the absolute value for every fixed t ($|f_{n_k}(t)-<!--\; -\; -->f(t)|\text{ a.e in}[0,1]$) ?

How do you solve ${\displaystyle \frac{2}{3}t<-48}$ ?

How do you solve ${\displaystyle 68\ge 60-b}$ ?

How do you solve ${\displaystyle 4p<32}$ ?

Simple formula for the $n$-ary version of $(x,y)\mapsto <!--\; \mapsto \; -->\frac{x+y}{1-<!--\; -\; -->xy}$
Let $x\ast <!--\; \ast \; -->y=\frac{x+y}{1-<!--\; -\; -->xy}$. I want a single formula for $x_1\ast <!--\; \ast \; -->x_2\ast <!--\; \ast \; -->\dots <!--\; \dots \; -->\ast <!--\; \ast \; -->x_n$, for all natural $n$.
In order to generate plausible candidates, let's see what happens at small values of $n$:
$x_1\ast <!--\; \ast \; -->x_2\ast <!--\; \ast \; -->x_3={\displaystyle \frac{x_1+x_2+x_3-<!--\; -\; -->x_1{x}_2{x}_3}{1-<!--\; -\; -->x_1{x}_2-<!--\; -\; -->x_1{x}_3-<!--\; -\; -->x_2{x}_3}}$
$x_1\ast <!--\; \ast \; -->x_2\ast <!--\; \ast \; -->x_3\ast <!--\; \ast \; -->x_4={\displaystyle \frac{x_1+x_2+x_3+x_4-<!--\; -\; -->x_1{x}_2{x}_3-<!--\; -\; -->x_1{x}_2{x}_4-<!--\; -\; -->x_1{x}_3{x}_4-<!--\; -\; -->x_2{x}_3{x}_4}{1-<!--\; -\; -->x_1{x}_2-<!--\; -\; -->x_1{x}_3-<!--\; -\; -->x_1{x}_4-<!--\; -\; -->x_2{x}_3-<!--\; -\; -->x_2{x}_4-<!--\; -\; -->x_3{x}_4+x_1{x}_2{x}_3{x}_4}}$
Then I conjecture the following:
$x_1\ast <!--\; \ast \; -->x_2\ast <!--\; \ast \; -->\dots <!--\; \dots \; -->\ast <!--\; \ast \; -->x_n={\displaystyle \frac{\underset{I\in <!--\; \in \; -->T_1}{\sum <!--\; \sum \; -->}\underset{i\in <!--\; \in \; -->I}{\prod <!--\; \prod \; -->}{x}_i-<!--\; -\; -->\underset{I\in <!--\; \in \; -->T_3}{\sum <!--\; \sum \; -->}\underset{i\in <!--\; \in \; -->I}{\prod <!--\; \prod \; -->}{x}_i}{\underset{I\in <!--\; \in \; -->T_0}{\sum <!--\; \sum \; -->}\underset{i\in <!--\; \in \; -->I}{\prod <!--\; \prod \; -->}{x}_i-<!--\; -\; -->\underset{I\in <!--\; \in \; -->T_2}{\sum <!--\; \sum \; -->}\underset{i\in <!--\; \in \; -->I}{\prod <!--\; \prod \; -->}{x}_i}}$
Where $T_k=\{I\in <!--\; \in \; -->\mathrm{\wp <!--\; \wp \; -->}(\{1,2\dots <!--\; \dots \; -->n\}):k\equiv <!--\; \equiv \; -->|I|(\mathrm{mod}4)\}$
Is there a nice proof of the above not involving transcendental functions?

Is the ordered pair (3,13) a solution to the inequality: ${\displaystyle y>2x+1}$ ?

Find the fractional representation $p/q$
Been trying to get some sort of solution for this for hours now, with no avail.
Find the fractional representation $p/q$, with $p\in <!--\; \in \; -->\mathbb{N}$ and $q\in <!--\; \in \; -->\mathbb{N}$, of the rational number whose decimal representation is:
$\mathrm{22.521111...}$
Any help would be greatly appreciated.
Thanks

How many distinct equilateral triangles can be formed in a regular nonagon having atleast two of their vertices as the vertices of nonagon.
As the question says two vertices are common so, I did ${}^9{C}_2=36$ but after that i am not able to proceed further. how should I think?

Monotonicity of a fraction.
So I want to prove that the following fraction is monotone increasing, as a part of another proof, that's why I stumbled on:
$\frac{4^{n+1}}{2\sqrt{n+1}}\ge <!--\; \ge \; -->\frac{4^n}{2\sqrt{n}}$
I know it's basic, though how to prove it?

Define the operator $Kf(x)={\int <!--\; \int \; -->}_0^{1}k(x,y)f(y)dy.$ by
$Kf(x)={\int <!--\; \int \; -->}_0^{1}k(x,y)f(y)dy.$
where $k(x,y)$ is a continuous complex valued function on the unit square. $[0,1]\times <!--\; \times \; -->[0,1]$ Also assume that
$su{p}_{x\in <!--\; \in \; -->[0,1]}{\int <!--\; \int \; -->}_0^1|k(x,y)|dy<1$
How to show that given $g\in <!--\; \in \; -->(C[0,1],\mathbb{C})$, the integral equation
$f(x)=g(x)+{\int <!--\; \int \; -->}_0^{1}k(x,y)f(y)dy$
has exactly one continuous solution $f\in <!--\; \in \; -->(C[0,1],\mathbb{C})$?
Note: I only know introductory measure theory and functional analysis, so I am suppose to do this problem without heavy integral function theory.
One way to get started is to show that K is compact and its operator norm $||K||<1$ because we are given that $su{p}_{x\in <!--\; \in \; -->[0,1]}{\int <!--\; \int \; -->}_0^1|k(x,y)|dy<1$ Then we know that $(I-<!--\; -\; -->K)$ is invertible and its inverse is given by the Neumann series. At this point, I'm not sure how to proceed.

How do you solve ${\displaystyle 18\le 21+x}$

Comparing two fractions
I saw this problem from an elementary textbook:
Let
$A=\frac{2014}{2015}+\frac{2015}{2016}$
and
$B=\frac{2014+2015}{2015+2016}$
Compare A and B.
I know the answer is $A><!--\; >\; -->B$ because
$\frac{A}{B}><!--\; >\; -->1.$
But I cannot find a precise way to explain it. Can anyone please help? Thank you.

How to solve ${\displaystyle \frac{7x}{8}}+4-<!--\; -\; -->{\displaystyle \frac{2x}{3}}=4x-<!--\; -\; -->3$
I do not understand how to simplify this. Could anyone here help me, please? Thanks.

Can everything in category theory be restated in a sufficiently expressive type theory?
I've been reading the homotopy type theory (HoTT) book as well as some articles on ncat lab
It seems that HoTT can be interpreted as the syntax for an $\mathrm{\infty <!--\; \infty \; -->}$-groupoid or more generally an ($\mathrm{\infty <!--\; \infty \; -->}$,1)-category. That article and related articles on ncatlab seem to suggest that within an arbitrary category C the objects can be translated as types (as in HoTT types) and morphisms can be translated as dependent types. Thus a category can be interpreted as an "abstract data structure" and functors between categories are just ordinary (type theoretic) functions between data types/structures in type theory.
Is that translation correct, and if so, does that suggest that any categorical notion has a simple reformulation (without much work) to a homotopy type theoretic form? Coming from a computational background, for some reason, just replacing the word "category" with "abstract data structure" and a few other terms in category theory with type theoretic terminology seems to make category theory much more intuitive than it appears.

How do you solve and graph ${\displaystyle k+24>-5}$ ?

I am reading Rudin's RCA and have a question regarding the proof of Jensen's inequality.
Let $\mathrm{\mu <!--\; \mu \; -->}$ be a positive measure on a $\mathrm{\sigma <!--\; \sigma \; -->}$-algbebra $\mathfrak{M}$ in a set $\mathrm{\Omega <!--\; \Omega \; -->}$, so that $\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\Omega <!--\; \Omega \; -->})=1$. If $f$ is a real function in $L^1(\mathrm{\mu <!--\; \mu \; -->})$, if $a<f(x)<b$ for all $x\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$, and if $\mathrm{\phi <!--\; \phi \; -->}$ is convex on $(a,b)$ then
$\mathrm{\phi <!--\; \phi \; -->}({\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f\text{d}\mathrm{\mu <!--\; \mu \; -->})\le <!--\; \le \; -->{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}.$
After a few steps Rudin obtains the following inequality:
$\mathrm{\phi <!--\; \phi \; -->}(f(x))-<!--\; -\; -->\mathrm{\phi <!--\; \phi \; -->}(t)-<!--\; -\; -->\mathrm{\beta <!--\; \beta \; -->}(f(x)-<!--\; -\; -->t)\ge <!--\; \ge \; -->0$
for every $x\in <!--\; \in \; -->\mathrm{\Omega <!--\; \Omega \; -->}$. Here, $t={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}f\text{d}\mathrm{\mu <!--\; \mu \; -->}\in <!--\; \in \; -->(a,b)$ and $\mathrm{\beta <!--\; \beta \; -->}$ is a real number. If $\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f\in <!--\; \in \; -->L^1(\mathrm{\mu <!--\; \mu \; -->})$ then the inequality can be obtained by integrating both sides. However, I am not sure how to proceed in the case $\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f\notin <!--\; \notin \; -->L^1(\mathrm{\mu <!--\; \mu \; -->})$. Rudin states that in this case, the integral ${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}$ is defined in the extended sense
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}={\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{+}\text{d}\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{-<!--\; -\; -->}\text{d}\mathrm{\mu <!--\; \mu \; -->}$
where
${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{+}\text{d}\mathrm{\mu <!--\; \mu \; -->}=\mathrm{\infty <!--\; \infty \; -->}\text{ and }{\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f{)}^{-<!--\; -\; -->}\text{d}\mathrm{\mu <!--\; \mu \; -->}\text{ is finite.}$
I am unable to show the existence of ${\int <!--\; \int \; -->}_{\mathrm{\Omega <!--\; \Omega \; -->}}(\mathrm{\phi <!--\; \phi \; -->}\circ <!--\; \circ \; -->f)\text{d}\mathrm{\mu <!--\; \mu \; -->}$ as defined above.
Any help would be greatly appreciated. Thank you.