Improve Your Algebra Skills with Practice Problems

I can`r do this, need help
Given a locally integrable function $\mathrm{\gamma <!--\; \gamma \; -->}:{\mathbb{R}}_{\ge <!--\; \ge \; -->0}\to <!--\; \to \; -->\mathbb{R}$, we define the absolutely continuous function $\mathrm{\Gamma <!--\; \Gamma \; -->}(t):=\underset{u\le <!--\; \le \; -->t}{max}{\int <!--\; \int \; -->}_u^t\mathrm{\gamma <!--\; \gamma \; -->}\mathrm{d}\mathrm{\lambda <!--\; \lambda \; -->}$.
I want to show, that $\mathrm{\gamma <!--\; \gamma \; -->}(t)\le <!--\; \le \; -->0$ holds for almost all $t$ with $\mathrm{\Gamma <!--\; \Gamma \; -->}(t)=0$. In other words, I want to show that the set
$A:=\{t\in <!--\; \in \; -->{\mathbb{R}}_{\ge <!--\; \ge \; -->0}|\mathrm{\Gamma <!--\; \Gamma \; -->}(t)=0\text{and}\mathrm{\gamma <!--\; \gamma \; -->}(t)>0\}$
is a Lebesgue-null set, i.e. $\mathrm{\lambda <!--\; \lambda \; -->}(A)=0$.
All my attempts have failed. Nevertheless, I was able to show, that if $\mathrm{\Gamma <!--\; \Gamma \; -->}$ vanishes on a (proper) interval $[a,b]$ with $a<b$, then $\mathrm{\lambda <!--\; \lambda \; -->}(A\cap <!--\; \cap \; -->[a,b])=0$. This however does not lead to a proof of the more general claim $\mathrm{\lambda <!--\; \lambda \; -->}(A)=0$.

I have the following exercise, but i can't find the answer.
Let $(\mathit{X},\mathcal{A},\mathrm{\mu <!--\; \mu \; -->})$ be a measure space. Show that the measure $\mathrm{\mu <!--\; \mu \; -->}$ is $\mathrm{\sigma <!--\; \sigma \; -->}$-finite, if there exists a function $f\in <!--\; \in \; -->{\mathcal{M}}^{+}(\mathit{X})$ with $f(x)><!--\; >\; -->0$ for all $x\in <!--\; \in \; -->\mathit{X}$ and ${\int <!--\; \int \; -->}_{\mathit{X}}fd\mathrm{\mu <!--\; \mu \; -->}<<!--\; <\; -->\mathrm{\infty <!--\; \infty \; -->}$.
From what I understand is that the function $f$ has to be from the set ${\mathcal{M}}^{+}$, so the set of strictly positive numeric functions (as defined in the script of the professor).

In the equation ${\displaystyle x+y=5}$ , the value for x is a whole number greater than 2 but less than 6. How do you determine the possible solutions for y?

Log predictive density asmptotically in predictive information criteria for Bayesian models
I am reading this paper, Andrew Gelman's Understanding predictive information criteria for Bayesian models, and I will give a screenshot as below:
Under standard conditions, the posterior distribution, $p(\mathrm{\theta <!--\; \theta \; -->}\mid <!--\; \mid \; -->y)$, approaches a normal distribution in the limit of increasing sample size (see, e.g., DeGroot, 1970). In this asymptotic limit, the posterior is dominated by the likelihood-the prior contributes only one factor, while the likelihood contributes n factors, one for each data point-and so the likelihood function also approaches the same normal distribution.
As sample size $n\to <!--\; \to \; -->\propto <!--\; \propto \; -->$, we can label the limiting posterior distribution as $\mathrm{\theta <!--\; \theta \; -->}|y\to <!--\; \to \; -->N({\mathrm{\theta <!--\; \theta \; -->}}_0,V_0/n)$. In this limit the log predictive density is
$logp(y\mid <!--\; \mid \; -->\mathrm{\theta <!--\; \theta \; -->})=c(y)-<!--\; -\; -->\frac{1}{2}(klog(2\mathrm{\pi <!--\; \pi \; -->})+log|Vo/nl+(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->{\mathrm{\theta <!--\; \theta \; -->}}_0{)}^T(Vo/n{)}^{-<!--\; -\; -->1}(\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->{\mathrm{\theta <!--\; \theta \; -->}}_0))$
where c(y) is a constant that only depends on the data y and the model class but not on the parameters $\mathrm{\theta <!--\; \theta \; -->}$.
The limiting multivariate normal distribution for 0 induces a posterior distribution for the log predictive density that ends up being a constant $\text{(equal to }c(y)-<!--\; -\; -->\frac{1}{2}(klog(2\mathrm{\pi <!--\; \pi \; -->})+log|Vo/n|))$ minus $\frac{1}{2}$ times a ${\mathrm{\chi <!--\; \chi \; -->}}_k^2$ random variable, where k is the dimension of $\mathrm{\theta <!--\; \theta \; -->}$, that is, the number of parameters in the model. The maximum of this distribution of the log predictive density is attained when equals the maximum likelihood estimate (of course), and its posterior mean is at a vaue lower.
For actual posterior distributions, this asymptotic result is only an approximation, but it will be useful as a benchmark for interpreting the log predictive density as a measure of fit.
With singular models (e.g. mixture models and overparameterized complex models more gener- ally) a set of different parameters can map to a single data model, the Fisher information matrix i not positive definite, plug-in estimates are not representative of the posterior, and the distribution of the deviance does not converge to a ${\mathrm{\chi <!--\; \chi \; -->}}^2$ distribution. The asymptotic behavior of such models can be analyzed using singular learning theory (Watanabe, 2009, 2010).
Sorry for the long paragraph. The things that confuse me are:
1. Why here seems like we know the posterior distribution $f(\mathrm{\theta <!--\; \theta \; -->}|y)$ first, then we use it to find the $\log <!--\; \; -->p(y|\mathrm{\theta <!--\; \theta \; -->})$? Shouldn't we get the model, $\log <!--\; \; -->p(y|\mathrm{\theta <!--\; \theta \; -->})$ first?
2. What does the green line "its posterior mean is at a value $\frac{k}{2}$ lower" mean? My understanding is since there is a term $-<!--\; -\; -->\frac{1}{2}{\mathrm{\chi <!--\; \chi \; -->}}_k^2$ in the expression and the expectation of ${\mathrm{\chi <!--\; \chi \; -->}}_k^2$ is k, which lead to a $\frac{k}{2}$ lower. But $\frac{k}{2}$ lower than what?
3. How does the $\log <!--\; \; -->p(y|\mathrm{\theta <!--\; \theta \; -->})$ interpreting the measure of fit? I can see that there is a mean square error(MSE) term in this expression but it is an MSE of the parameter $\mathrm{\theta <!--\; \theta \; -->}$, not the data y.
Thanks for any help!

How do you solve ${\displaystyle 5(5x-2)<15}$ ?

Do the following numbers share common factors? If so, which is the greatest common factor? $\{240,96,144\}$

I'm comfortable with fractions like $\frac{-<!--\; -\; -->3}{8}$ being the same as $\frac{3}{-<!--\; -\; -->8}$ (though I'd think the latter anachronistic and would in any case probably prefer to write either of those two as $-<!--\; -\; -->\frac{3}{8}$ ), and of course I'm comfortable with improper fractions like $\frac{-<!--\; -\; -->8}{3}$ being the mixed number $-<!--\; -\; -->2\frac{2}{3}$
However, if I'm trying to teach a computer how to handle negative vulgar fractions, I should also consider the remaining cases, so how should I interpret:
1.$2\frac{-<!--\; -\; -->2}{3}$
2.$2\frac{2}{-<!--\; -\; -->3}$
3.$2\frac{-<!--\; -\; -->2}{-<!--\; -\; -->3}$
4.$-<!--\; -\; -->2\frac{-<!--\; -\; -->2}{3}$
5.$-<!--\; -\; -->2\frac{2}{-<!--\; -\; -->3}$
6.$-<!--\; -\; -->2\frac{-<!--\; -\; -->2}{-<!--\; -\; -->3}$
I'm considering the logical approach, by inference from $2\frac{2}{3}$ and $-<!--\; -\; -->2\frac{2}{3}$ so I'd get:
1.$1\frac{1}{3}$
2.$1\frac{1}{3}$
3.$1\frac{2}{3}$
4.$-<!--\; -\; -->1\frac{1}{3}$
5.$-<!--\; -\; -->1\frac{1}{3}$
6.$-<!--\; -\; -->2\frac{2}{3}$

Let $X=Y$ uncountable, and let $A=B$ be the $\mathrm{\sigma <!--\; \sigma \; -->}$ algebra of countable-cocountable subsets of $X$,$Y$ respectively. Let $C$ be the countable-cocountable $\mathrm{\sigma <!--\; \sigma \; -->}$-algebra on $X\times <!--\; \times \; -->Y$.
Is $C=\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)$?
$\subseteq <!--\; \subseteq \; -->$ is obvious, but I am not sure about the other direction. I know that in general the two do not have to be equal, so I am trying to find a counterexample i.e find an element in C which is not in $\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)$. Perhaps AC is involved, which makes it tricky? I am not sure.
Also, is it true that $\mathrm{\sigma <!--\; \sigma \; -->}(A\times <!--\; \times \; -->B)=A\times <!--\; \times \; -->B?$ This may help simplify the question.

Consider $f$ a bounded, measurable function defined on $[1,\mathrm{\infty <!--\; \infty \; -->}]$ and define $a_n:={\int <!--\; \int \; -->}_{[n,n+1)}f$. Does the fact that $f$ is integrable imply $\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n$ converges?
I would like to say that the argument
$\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n={\int <!--\; \int \; -->}_1^{\mathrm{\infty <!--\; \infty \; -->}}f$
made in that post is not correct.
My attempt was the following: Since $f$ is integrable, $|f|$ is also integrable. Therefore,
${\int <!--\; \int \; -->}_{[1,\mathrm{\infty <!--\; \infty \; -->})}|f|=\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{\int <!--\; \int \; -->}_1^k|f|<\mathrm{\infty <!--\; \infty \; -->}.$
Moreover, we have the following:
$\begin{array}{rl}\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{a}_n& =\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_n^{n+1}f\\ & \le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}|{\int <!--\; \int \; -->}_n^{n+1}f|\\ & \le <!--\; \le \; -->\underset{n=1}{\overset{\mathrm{\infty <!--\; \infty \; -->}}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_n^{n+1}|f|\\ & =\underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\underset{n=1}{\overset{k}{\sum <!--\; \sum \; -->}}{\int <!--\; \int \; -->}_n^{n+1}|f|\end{array}$
However, I cannot proceed further from here. Can anyone help?

I am representing my 3d data in convariance matrix. I just want to know what the determinant of Convariance Matrix gives. If the determinant is positive, zero, negative, high positive, high negative. What does it mean or represent?
Thanks
EDIT:
Covariance is being used to represent variance for 3d coordiantes that I have. If my covariance matrix A determinant is +100, and the other covariance matrix B determinant is +5. Which of these values show if the variance is more or not. Which value tells that datapoints are more dispersed. Which value shows that readings are further away from mean.

How do you solve ${\displaystyle 36>-\frac{1}{2}y}$ ?

Help with simplification of a rational expression (with fractional powers)
Can you please help me see what I don't see yet. Here's a problem from a high school textbook (ISBN 978-5-488-02046-7 p.9, #1.029):
$\frac{(a^{1/m}-<!--\; -\; -->a^{1/n}{)}^2\cdot <!--\; \cdot \; -->4a^{(m+n)/mn}}{(a^{2/m}-<!--\; -\; -->a^{2/n})(\sqrt[m]{a^{m+1}}+\sqrt[n]{a^{n+1}})}$
Here's my try at it:
$\frac{(a^{1/m}-<!--\; -\; -->a^{1/n})(a^{1/m}-<!--\; -\; -->a^{1/n})\cdot <!--\; \cdot \; -->4a^{(1/m)+(1/n)}}{(a^{1/m}-<!--\; -\; -->a^{1/n})(a^{1/m}+a^{1/n})\cdot <!--\; \cdot \; -->a(a^{1/m}+a^{1/n})}$
...which is
$\frac{(a^{1/m}-<!--\; -\; -->a^{1/n})\cdot <!--\; \cdot \; -->4a^{(1/m)+(1/n)}}{a(a^{1/m}+a^{1/n}{)}^2}$
Wolfram Alpha's simplify stops here, too. I don't see where to go from here. The final form, according to the book, is this:
$\frac{1}{a(a^{1/m}-<!--\; -\; -->a^{1/n})}$
How did they do it?

How to interpret following max-min optimization criterion
I am reading a paper in which the author says the following about the maximization problem. Now before reading that, please note that I understand that max-min fairness is. But I cannot understand the following equation and how it represents a max-min fairness algorithm. The following is what is written in the paper.
For..., the for computational efficiency ($\mathrm{\eta <!--\; \eta \; -->}$ below) is formulated under the max-min fairness criterion as
$$\begin{gathered} P_1:max \\ min \\ \mathrm{\eta <!--\; \eta \; -->}({\mathrm{\tau <!--\; \tau \; -->}}_0,{\mathrm{\tau <!--\; \tau \; -->}}_k,P_k,f_k) \\ {\mathrm{\tau <!--\; \tau \; -->}}_0,{\mathrm{\tau <!--\; \tau \; -->}}_k,P_k,f_k \\ k\in <!--\; \in \; -->K \end{gathered}$$
where, $\mathrm{\eta <!--\; \eta \; -->}=\frac{\text{rate of data processing}}{\text{energy harvested by the circuit}}$, k is any user etc...

What is the solution of the inequality ${\displaystyle -5+b\ge -3}$ ?

Calculating pOH
If [OH-]=10^-pOH and [OH-]= 0.003 then what does pOH equal? I know this is simple but I just can't figure out how to do this calculation. Any help would be appreciated.

Find the real solution set of the following equations, inequalities, and system of equations. Please provide me a clean, readable, and understandable solution.
$2-<!--\; -\; -->\frac{2x-<!--\; -\; -->1}{x}\le <!--\; \le \; -->\frac{2}{2x+1}$

How do you solve ${\displaystyle 27>x+18}$ ?

How to calculate an $n$ digit decimal approximation of a fraction?
Say I have a number $x\in <!--\; \in \; -->\mathbb{Q}$ which possibly cannot be writen exactly in decimal form, but at least $x\notin <!--\; \notin \; -->\mathbb{Z}$ ($x$ is not an integer).
How do I calculate the first $n$ digits of this number $x$, for a given $n$? Also, how do I know if it has a finite or infinite amount of (nonzero) digits in decimal form?
For example, if $x=\frac{1000}{1001}$, how do I calculate the first $n=10$ digits of this number?
I know this question sounds really basic, but since it is basically always done by means of a calculator or other computing device, I've never really done it before.

How do you solve and graph ${\displaystyle x-8>4}$ ?

Let $X$ be an arbitrary space, and let $\mathrm{\mu <!--\; \mu \; -->}$,$\mathrm{\nu <!--\; \nu \; -->}$ be two probability measures on $X$. Suppose there is a common set $A$ on which $\mathrm{\mu <!--\; \mu \; -->}(A)=\mathrm{\nu <!--\; \nu \; -->}(A)=1$.
Now suppose there are $B,C$ such that $\mathrm{\mu <!--\; \mu \; -->}(B)=\mathrm{\nu <!--\; \nu \; -->}(C)=1$. Can anything be said of $\mathrm{\mu <!--\; \mu \; -->}(C)$ and $\mathrm{\nu <!--\; \nu \; -->}(B)$, or $\mathrm{\mu <!--\; \mu \; -->}(B\cap <!--\; \cap \; -->C)$ and $\mathrm{\nu <!--\; \nu \; -->}(B\cap <!--\; \cap \; -->C)$?
My initial intuition was that the answer should be yes since $B\cap <!--\; \cap \; -->A$ and $C\cap <!--\; \cap \; -->A$ both have full $\mathrm{\mu <!--\; \mu \; -->}$ and full $\mathrm{\nu <!--\; \nu \; -->}$ measure respectively, and so there should be "enough" overlap for $B\cap <!--\; \cap \; -->C$ to have at least positive measure. But I can't seem to prove this - so I'm interested in whether anything can be said of these quantities (either alone or possibly together, e.g. is it possible for $\mathrm{\mu <!--\; \mu \; -->}(C)=\mathrm{\nu <!--\; \nu \; -->}(B)=0$) and whether there is any clever counterexample for which $\mathrm{\mu <!--\; \mu \; -->}(C)$ and $\mathrm{\nu <!--\; \nu \; -->}(B)$ can be any arbitrary number in [0,1].