Improve Your Algebra Skills with Practice Problems

a) Identify equations, expressions and inequalities
b) Consider
$5=3m$
$n>|-<!--\; -\; -->8|$
$2+6=(5-<!--\; -\; -->1)\times <!--\; \times \; -->2$
$8\times <!--\; \times \; -->(k+6)$

Why is $\frac{\sqrt{x+1}-<!--\; -\; -->1}{x}$ equal to $\frac{1}{\sqrt{x+1}+1}$?
I'm working with the expression
$\frac{\sqrt{x+1}-<!--\; -\; -->1}{x}.$
According to Wolfram Alpha "alternate form" section (http://www.wolframalpha.com/input/?i=%28%28x%2B1%29%5E1%2F2-1%29%2Fx) it is equal to
$\frac{1}{\sqrt{x+1}+1}.$
How can I go from one expression to the other?

Square root fraction confusion
I was doing math for school and got to something that really confused me. With having the rule $\frac{2}{4}=\frac{4}{8}$ (or some simular fraction equation) in mind, I got to the following confusing equation:
$\frac{\sqrt{3}}{3}=\frac{(\sqrt{3}{)}^2}{3^2}=\frac{3}{9}=\frac{1}{3}$
This is really confusing me since $\frac{\sqrt{3}}{3}$ can never equal $\frac{1}{3}$. Where am I going wrong?

Consider a position measurement that is prone to a random error in any direction. This would mean that the position would be in a circle where the probability curve taken across the diameter would show a normal distribution.
How can the probability be calculated that the position measurement is inside another circle?
I can ignore the cases where the circle of probability is entirely inside the detection circle (because the probability of being in the detection circle has reached 100%) and when it is entirely outside the detection circle (probability falls to 0%), i.e. the cases where the two circles do not intersect.

How do you write an inequality and solve given "twice a number is more than the sum of that number and 9"?

Let $X$ be any locally compact Hausdorff space, and $\mathrm{\mu <!--\; \mu \; -->}$ be a positive finite regular (both inner and outer) measure on $X$ such that $\mathrm{supp}<!--\; \; -->(\mathrm{\mu <!--\; \mu \; -->})$ is compact. We have that
${\int <!--\; \int \; -->}_{X}fd\mathrm{\mu <!--\; \mu \; -->}=0$
for each $f\in <!--\; \in \; -->C_c(X)$.
Then can we conclude that $\mathrm{\mu <!--\; \mu \; -->}=0$?
I was trying to find a counter-example, but am stuck at some technical issue.

I have this:
Theorem. If $\mathit{E}$ is any class of sets and if $\mathit{A}$ is any subset of X, then
$\text{S(E)}\cap <!--\; \cap \; -->\mathit{A}=\mathbf{S}\mathbf{(}\mathbf{E}\cap <!--\; \cap \; -->\mathit{A}).$
Proof. Denote by $\mathit{C}$ the class of all sets of the form B $\cup <!--\; \cup \; -->$ $(C-<!--\; -\; -->\mathit{A})$, where
$B\mathrm{\epsilon <!--\; \epsilon \; -->}\mathbf{S}\mathbf{(}\mathbf{E}\cap <!--\; \cap \; -->\mathit{A}\mathbf{)}\text{and}C\mathrm{\epsilon <!--\; \epsilon \; -->}\mathbf{S}\mathbf{(}\mathbf{E});$
it is easy to verify that $\mathit{C}$ is a $\mathrm{\sigma <!--\; \sigma \; -->}$-ring.

We are using Paul Halmos's Measure Theory as our textbook. One of the proofs provided is incomplete. I don't like learning things without knowing the full proof. Is there full proof that shows why $\mathcal{C}$ is a $\mathrm{\sigma <!--\; \sigma \; -->}$-ring?

I know I need to take two sets $G,H$ in $\mathcal{C}$ and prove that $G-<!--\; -\; -->H\in <!--\; \in \; -->\mathcal{C}$

But that means I have to show that $(B_G\cup <!--\; \cup \; -->(C_G-<!--\; -\; -->A))-<!--\; -\; -->(B_H\cup <!--\; \cup \; -->(C_H-<!--\; -\; -->A))\in <!--\; \in \; -->\mathcal{C}$, which is just too complicated to me. I am not sure how it is easy to verify.

Let $S,T$ be two stopping times. Then $S\vee <!--\; \vee \; -->T$ and $S\wedge <!--\; \wedge \; -->T$ are salso topping times with respect to ${\mathcal{F}}_{S\wedge <!--\; \wedge \; -->T}={\mathcal{F}}_S\cap <!--\; \cap \; -->{\mathcal{F}}_T$. Futhermore, $\{S\le <!--\; \le \; -->T\}\in <!--\; \in \; -->{\mathcal{F}}_{S\wedge <!--\; \wedge \; -->T}$ and $\{S=T\}\in <!--\; \in \; -->{\mathcal{F}}_{S\wedge <!--\; \wedge \; -->T}$. For a stopping time $\mathrm{\tau <!--\; \tau \; -->}\in <!--\; \in \; -->{\mathcal{F}}_t$ and $A\in <!--\; \in \; -->{\mathcal{F}}_{\mathrm{\tau <!--\; \tau \; -->}}$. How to show that $A\cap <!--\; \cap \; -->\{\mathrm{\tau <!--\; \tau \; -->}\le <!--\; \le \; -->t\}\in <!--\; \in \; -->{\mathcal{F}}_{\mathrm{\tau <!--\; \tau \; -->}\wedge <!--\; \wedge \; -->t}$?

How do you solve ${\displaystyle -\frac{s}{3}<-3.5}$ ?

Prove that $\frac{\frac{1}{11}+\frac{1}{12}+\cdots <!--\; \cdots \; -->+\frac{1}{200}}{\frac{1}{10\ast <!--\; \ast \; -->11}+\frac{1}{11\ast <!--\; \ast \; -->12}+\cdots <!--\; \cdots \; -->+\frac{1}{19\ast <!--\; \ast \; -->20}}>19$
My attempt:The denominator is $\frac{1}{20}$ using telescopic series.
$\frac{1}{10\ast <!--\; \ast \; -->11}+\frac{1}{11\ast <!--\; \ast \; -->12}+\cdots <!--\; \cdots \; -->+\frac{1}{19\ast <!--\; \ast \; -->20}=\frac{1}{10}-<!--\; -\; -->\frac{1}{11}+\frac{1}{11}-<!--\; -\; -->\frac{1}{12}+\cdots <!--\; \cdots \; -->-<!--\; -\; -->\frac{1}{20}=\frac{1}{10}-<!--\; -\; -->\frac{1}{20}=\frac{1}{20}$
I think we could Prove that the top part of fraction is bigger than $1$ but I don't know how.

Partial fractions - alternative result
I have the following fraction:
$\frac{z}{(z-<!--\; -\; -->1)(z-<!--\; -\; -->2)}$
When I try to decompose it to partial fractions, I get:
$-<!--\; -\; -->\frac{1}{z-<!--\; -\; -->1}+\frac{2}{z-<!--\; -\; -->2}$
But the result in my book is:
$-<!--\; -\; -->\frac{z}{z-<!--\; -\; -->1}+\frac{z}{z-<!--\; -\; -->2}$
Both results are correct, but, how am I supposed to get the second one?

How do you solve ${\displaystyle y-5>-2}$ ?

For each $t\in <!--\; \in \; -->[0,1]$, let $T_t:<!--\; :\; -->L^2(\mathbb{R})\to <!--\; \to \; -->L^2(\mathbb{R})$ be the operator that shifts everything to the right by $t$, i.e.
$T_{t}f(x)=f(x-<!--\; -\; -->t)$
for all $f\in <!--\; \in \; -->L^2(\mathbb{R})$.
Then $t\mapsto <!--\; \mapsto \; -->T_t$ defines a map
$\mathrm{\gamma <!--\; \gamma \; -->}:<!--\; :\; -->[0,1]\to <!--\; \to \; -->\mathcal{B}(L^2(\mathbb{R})).$
Note that $\mathrm{\gamma <!--\; \gamma \; -->}$ is not continuous.
Question: Is $\mathrm{\gamma <!--\; \gamma \; -->}$ Bochner-integrable, i.e. does the integral
${\int <!--\; \int \; -->}_{[0,1]}{T}_{t}dt$
converge in $\mathcal{B}(L^2(\mathbb{R}))$?

Thoughts: Since each operator Tt has norm, the question is whether $\mathrm{\gamma <!--\; \gamma \; -->}$ is strongly measurable. In particular, there is a question of whether there exists a subset of $J\subseteq <!--\; \subseteq \; -->[0,1]$ of measure 1 such that $\mathrm{\gamma <!--\; \gamma \; -->}(J)$ is separable. Certainly $\mathrm{\gamma <!--\; \gamma \; -->}([0,1])$ is not separable, and I suspect that no such subset $J$ exists, but I'm not sure how to show this.

For a compact subset $X\subset <!--\; \subset \; -->{\mathbb{R}}^n$ and a continuous function $f:X\to <!--\; \to \; -->\mathbb{C}$, f is integrable on $X$ ( as both $X$ and $f$ will be bounded). Does this generalise? That is, if $X$ is a compact measure space and $f:X\to <!--\; \to \; -->\mathbb{C}$ is a continuous function, under which circumstances (i.e. under which assumptions on the topology of $X$ and under which assumptions on the measure on $X$) is it true that $f$ is integrable on $X$? Thanks for any help!

Reduce a fraction
Given the function
$f(x)=\frac{x^2-<!--\; -\; -->5}{x+\sqrt{5}}$
If I draw this function in maple, I will get a line. How can that be true? I should expect a line except in area of $x=-<!--\; -\; -->\sqrt{5}$, where $f(x)\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$ or $f(x)\to <!--\; \to \; -->-<!--\; -\; -->\mathrm{\infty <!--\; \infty \; -->}$. Of course Maple has factorized the numerator and reduced.
$\frac{x^2-<!--\; -\; -->5}{x+\sqrt{5}}=x-<!--\; -\; -->\sqrt{5},x\ne <!--\; \ne \; -->-<!--\; -\; -->\sqrt{5}$
My question is now, how can we ever reduce such a fraction with only condition $x\ne <!--\; \ne \; -->-<!--\; -\; -->\sqrt{5}$, when it is "$-<!--\; -\; -->\sqrt{5}$ and around it".

What exactly goes raising a number to a fraction mean?
I apologise for asking something so fundamental, but what exactly does
$2^{\frac{2}{5}}$
actually mean? I get raising a whole number to another whole number
$x^y$
means you are multiplying x with itself y times, but what does it mean when y is a fraction?

How to solve inequality equation with ceil operator product
I am trying to solve this inequality for x:
$\lceil \frac{c_1}{x}\rceil \cdot <!--\; \cdot \; -->\lceil \frac{c_2}{x}\rceil \le <!--\; \le \; -->N$
With: $c_1,c_2,x,N\in <!--\; \in \; -->\mathbb{N}$ and $c_1,c_2,N$ being just constants.

Consider a measure space $(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\mu <!--\; \mu \; -->})$, where $\mathrm{\mu <!--\; \mu \; -->}(\mathrm{\Omega <!--\; \Omega \; -->})<\mathrm{\infty <!--\; \infty \; -->}$.
It is classically known that if $p<q$, then $L^q(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\mu <!--\; \mu \; -->})\not\subset L^p(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\mu <!--\; \mu \; -->})$. The form of the argument $q>p\ge <!--\; \ge \; -->1$, we construct a function $f\in <!--\; \in \; -->L^p(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\mu <!--\; \mu \; -->})$ such that $f\notin L^q(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\mu <!--\; \mu \; -->})$.
Is it possible to take one such function f∈Lp(Ω,μ) such that $f\notin L^{p+\mathrm{ϵ<!--\; ϵ\; -->}}(\mathrm{\Omega <!--\; \Omega \; -->},\mathrm{\mu <!--\; \mu \; -->})$ for every $\mathrm{ϵ<!--\; ϵ\; -->}>0$?

I recently had an idea for an app that I would like to start developing for personal use and development, that attempts to present you with recipe idea's for lunch/dinner etc and by recording your responses learns your preferences. I was thinking it would do this by recording very specific details off each recipe such as carbcount caloriecount proteincount and etc, (factors which might act as determinants for our preference). Then this program would run an OLS regression with prob of being chosen as the dependent variable (for which we will have data on as we know what our user rejected (recipe) and what he accepted and how many times). We will then have various independant variables with which we will try and create an unbiased estimator. We can then run all recipe's to be presented under this regression and rank the recipes in order of probability to be chosen, highest to lowest.
Would this be a viable thing to do? If no, why not and what could perhaps be better?

How to interpret the imaginary part of an inverse fourier transform
The fourier transform of arbitrary real data can (usually will) result in complex data. If the real data represents samples in time, then the complex FT data represent frequencies with the magnitudes representing the amplitude and the phase angles representing the phase.
If I perform some arbitrary manipulation on the complex FT data prior to applying an inverse fourier transform, the result of the IFT can also be complex. My question is, how do I interpret the imaginary part of the IFT result? My guess is that a complex IFT result is non-physical, and that this must imply that my original arbitrary manipulation was also therefore non-physical.
Is this correct? If so, are there equations which describe whether such a manipulation would be non-physical.