Explore Fraction Concepts with Examples and Questions

Farey Sequence implemenatation
I'm trying to use the Farey sequence to get the next lowest reduced fraction in a list. For example, for $n=8$, we have $\dots <!--\; \dots \; -->,\frac{1}{3},\frac{3}{8},\frac{2}{5},\frac{3}{7},\frac{1}{2},\dots <!--\; \dots \; -->$
So let's take $\frac{3}{8}$ and $\frac{2}{5}$ and attempt to get $\frac{3}{7}$.
$p=\lfloor {\displaystyle \frac{n+b}{d}}\rfloor c-<!--\; -\; -->a$
$q=\lfloor {\displaystyle \frac{n+b}{d}}\rfloor d-<!--\; -\; -->b$
$a=3,b=8,c=2,d=5,n=8$
$p=[\frac{8+8}{5}]\cdot <!--\; \cdot \; -->2-<!--\; -\; -->3$
$q=[\frac{8+8}{5}]\cdot <!--\; \cdot \; -->5-<!--\; -\; -->8$
$\frac{p}{q}=\frac{3.4}{8}\ne <!--\; \ne \; -->\frac{3}{7}$
Am I using the equation wrong?

Factorising this formula
I came across this simplification in an iTunes U calculus course.
$\frac{\frac{1}{x_0+\mathrm{\delta <!--\; \delta \; -->}x}-<!--\; -\; -->\frac{1}{x_0}}{\mathrm{\delta <!--\; \delta \; -->}x}=\frac{1}{\mathrm{\delta <!--\; \delta \; -->}x}\left(\frac{x_0-<!--\; -\; -->(x_0+\mathrm{\delta <!--\; \delta \; -->}x)}{(x_0+\mathrm{\delta <!--\; \delta \; -->}x)x_0}\right)$
I didn't understand how this was done. I can see the denominator being taken out for $\frac{1}{\mathrm{\delta <!--\; \delta \; -->}x}$ but do not understand the remainder. Can someone give me some guidance? Thanks

Why do we cancel out the denominator of fractions with variables in a equation with their LCDs?
$x+\frac{5}{2}=\frac{1}{6}$
$6(x+\frac{5}{2})=6(\frac{1}{6})$
$3(x+5)=1$
$3x+15=1$
$3x=-<!--\; -\; -->14$
Why is it that we do that?

How to show simple inequality of fractions
If
$\frac{a}{a+b}<\frac{a^{\prime }}{a^{\prime }+b^{\prime }}$
then how can I show that
$\frac{a}{a+2b}<\frac{a^{\prime }}{a^{\prime }+2b^{\prime }}\mathrm{\forall <!--\; \forall \; -->}a,b,c>0$
I tried puitting in a constant k so
$\frac{a}{a+b}=k\ast <!--\; \ast \; -->\frac{a}{a+2b}$
and I get
$k=\frac{a+b}{a+2b}$
and I dont know where from there.

Relating an expression to two similar ones
Is it possible to express C solely in terms of A and B, where
$A={\displaystyle \frac{m}{x+z}},B={\displaystyle \frac{n}{y+z}},C={\displaystyle \frac{m+n}{x+y+z}}$
and $m,n,x,y,z>0?$
If not, how close can I get?

Does the sum of the reciprocals of composites that are $\le <!--\; \le \; -->$
The sum itself:
$\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{15}+\frac{1}{39}...\le <!--\; \le \; -->1$
These are all sums of reciprocals of composites that can be closest to 1. Notice that the term after $\frac{1}{15}$ is $\frac{1}{39}$ rather than $\frac{1}{16}$, or any other reciprocal of a composite number between 15 and 39 for the reason that it would make the sum up to that point > 1. That is a description of this sum from the most basic level.
Sum Parameters:
-If 1 is reached at an exact given point, the sum stops and thus making the sum finite.
-The sum must be $\le <!--\; \le \; -->$ 1
-All terms must be the reciprocal of composite numbers.
-To find the next term in the list, you must find the largest reciprocal (that fits standards) that makes the sum $\le <!--\; \le \; -->$1, and NOT the term that best fits the sum. For example, if you were to need to add $\frac{1}{x}$ and $\frac{1}{y}$ to make the sum exactly 1, but $\frac{1}{a}$ was the largest and still fit the rest of the parameters, then you would have to stick with with $\frac{1}{a}$. (Even if $\frac{1}{a}$was a really ugly term to add at that point).
Questions:
1) With these rules in place will the sum be a finite or infinite sum? If this is an infinite sum than the terms will always get closer and closer to 1, but never actually get there with a finite number of terms. With an finite sum, then at a quantifiable number of terms does the sum reach 1.
2)If the sum turns out to be infinite, than please tell me why.
3)If the sum turns out to be finite, list the terms. If the sum is ridiculously large, then prove that it is finite.
Thank you

What is the reciprocal of $(-<!--\; -\; -->1/2{)}^k$
The answer is meant to be $2^{-<!--\; -\; -->k}$ as if you flip something upside down the power becomes negative. However, I am not sure what happens to the negative in front of the fraction.

Expression for binomial coefficient denominator
I'm trying to find an analytical expression for the denominator of $\left(\begin{array}{c}-<!--\; -\; -->1/2\\ k\end{array}\right)$ in terms of k when the fraction is fully reduced.
E.g., the first several such denominators, starting with $k=0$, are
$1,2,8,16,128,256,1024,2048,32768$ so there are various power-of-2 jumps, but I haven't been able to figure out the overall pattern so that I can nail down the expression.
Does anyone know of such an expression, or know of a good place to look to try to figure this out? If not, does anyone know if this is a fool's errand?
Thanks for any help.

What are the steps to simplify the following equation (with simplified version included)
This is part of a question asking to find the differential of a polynomial fraction. I have already taken the derivative using quotient and chain rule, and this is where I am up to.
Rather than go through the entire question (I have been told this is too confusing for readers in previous questions), I will simply show the area where I am stuck.
I need to know how to simplify :
$\frac{\sqrt{2x-<!--\; -\; -->x^2}-<!--\; -\; -->\frac{-<!--\; -\; -->x+1}{\sqrt{2x-<!--\; -\; -->x^2}}x}{(\sqrt{2x-<!--\; -\; -->x^2}{)}^2}$
to :
$\frac{x}{(2x-<!--\; -\; -->x^2{)}^{3/2}}$
Any help with this would be very much appreciated , exam is tomorrow .
P.S. If users would prefer me to write out the full question, just ask and I will. I am still learning to write the perfect question.

How do I prove $\frac{\sqrt{x+h}-<!--\; -\; -->\sqrt{x}}{h}=\frac{1}{\sqrt{x+h}+\sqrt{x}}$

Solve fractions multiplication
I believe this is a very simple one, but I simply can't figure it out. How to solve?
$\frac{1}{2}\cdot <!--\; \cdot \; -->\frac{3}{4}\cdot <!--\; \cdot \; -->\frac{5}{6}\cdots <!--\; \cdots \; -->\frac{17}{18}\cdot <!--\; \cdot \; -->\frac{19}{20}$

Show that ${\displaystyle \frac{\sqrt{8-<!--\; -\; -->4\sqrt{3}}}{\sqrt[3]{12\sqrt{3}-<!--\; -\; -->20}}}=2^{\frac{1}{6}}$
This was the result of evaluating an integral by two different methods. The RHS was obtained by making a substitution, the LHS was obtained using trigonometric identity's and partial fractions.
Now I know that these two are equal, but I just can't prove it. I tried writing the LHS in terms of powers of 2, but can't get any further to the desired result.

Integration with partial fractions help please
I'm trying to work in my partial fractions chapter and some were easy but for whatever reason, I'm stuck now:
$\int <!--\; \int \; -->\frac{-<!--\; -\; -->3}{x2+2x+4}dx$
What I tried: since my denominator is of higher order and a irreducible quadratic, I set it up to start like this:
$x-<!--\; -\; -->3=\int <!--\; \int \; -->\frac{Ax+B}{x2+2x+4}+\frac{Cx+D}{x2+2x+4}dx$
I'm not really sure if that's right, the example in the book is a little different.
Anyway, after multiplying and collecting terms, I end up with
$x^5(C)=0$
$x^4(4C+D)=0$
$x^3(A+12C+4D)=0$
$x^2(2A+B+16C+12D)=0$
$x(4A+2B+16D+16)=1$
$4B+16D=-<!--\; -\; -->3$
but then when I tried to solve by substitution/addition methods, my numbers don't make sense. Am I even on the right track here? We just got started in this class and I feel like I'm behind already. Thanks for any help.

Problem with simplifying $\frac{(3+h{)}^2-<!--\; -\; -->9}{(3+h)-<!--\; -\; -->3}$

What is the smallest possible value of $a+b$
If $\frac{a}{b}$ rounded to the nearest trillionth is $0.008012018027$, where $a$ and $b$ are positive integers, what is the smallest possible value of $a+b$?
I don't see any strategies here for solving this problem, any help? Thanks in advance!

If $x$ is positive, then why does $\frac{1}{\sqrt{x+1}+\sqrt{x}}=\sqrt{x+1}-<!--\; -\; -->\sqrt{x}$?
I've been trying to convert the left side of the equation to the right side:
$\frac{1}{\sqrt{x+1}+\sqrt{x}}$
But then how can I flip this round to be what I have on the right side?
I know that $\frac{1}{\sqrt{x}}=x^{-<!--\; -\; -->\frac{1}{2}}$, so I would think that this would give me
$(x+1{)}^{-<!--\; -\; -->1/2}+x^{-<!--\; -\; -->1/2}$
Which I thought would then convert to
$-<!--\; -\; -->\sqrt{x+1}+(-<!--\; -\; -->\sqrt{x})$
so I'm not sure how the first part $\sqrt{x+1}$ got to be positive

$\frac{6}{4\times <!--\; \times \; -->2}+\frac{7}{5\times <!--\; \times \; -->2}+...+\frac{21}{19\times <!--\; \times \; -->2}$
I got this exercise from school and I have no idea what notion to use, it resumes to Harmonic series, I can't find a generic answer. Do you have any idea?
$\frac{6}{4\times <!--\; \times \; -->2}+\frac{7}{5\times <!--\; \times \; -->2}+...+\frac{21}{19\times <!--\; \times \; -->2}$
Which is the generic answer for such a sum?

Why is ${\left(\frac{1}{2}\right)}^x=\frac{1}{7}$ the same as saying: $(2{)}^x=7$
Sorry for the really dumb question but I'd like to see the process of how this is achieved.

Calculate weight based on body fat percentage
This is the YMCA method for determining body fat percentage for a female based on weight and waist size:
$\frac{-<!--\; -\; -->76.76+(4.5A)-<!--\; -\; -->(0.082B)}{B}=C$
Where: A = Waist size in inches, B = Weight in lbs, C = Ratio of body fat.
I would like to re-arrange this formula to solve for B.
ie: determine target weight based on waist and body fat ratio.
Thanks in advance! :-)

simplifying and factoring a fraction
how i get $\frac{(a+b{)}^2+(a+c{)}^2+(b+c{)}^2}{2}$ from $\frac{a^4}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)}+\frac{b^4}{(b-<!--\; -\; -->a)(b-<!--\; -\; -->c)}+\frac{c^4}{(c-<!--\; -\; -->a)(c-<!--\; -\; -->b)}$ assuming that $a\ne <!--\; \ne \; -->b\ne <!--\; \ne \; -->c\ne <!--\; \ne \; -->a$
i tried to make
$\begin{array}{rl}& \frac{a^4}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)}+\frac{b^4}{(b-<!--\; -\; -->a)(b-<!--\; -\; -->c)}+\frac{c^4}{(c-<!--\; -\; -->a)(c-<!--\; -\; -->b)}\\ & =\frac{a^4}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)}-<!--\; -\; -->\frac{b^4}{(a-<!--\; -\; -->b)(b-<!--\; -\; -->c)}+\frac{c^4}{(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}\\ & =\frac{a^4(b-<!--\; -\; -->c)}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}-<!--\; -\; -->\frac{b^4(a-<!--\; -\; -->c)}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}+\frac{c^4(a-<!--\; -\; -->b)}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}\\ & =\frac{a^4(b-<!--\; -\; -->c)-<!--\; -\; -->b^4(a-<!--\; -\; -->c)+c^4(a-<!--\; -\; -->b)}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}\\ & =\frac{a^{4}b-<!--\; -\; -->a^{4}c-<!--\; -\; -->a{b}^4+b^{4}c+a{c}^4-<!--\; -\; -->b{c}^4}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}\\ & =\frac{ab(a^3-<!--\; -\; -->b^3)+ac(c^3-<!--\; -\; -->a^3)+bc(b^3-<!--\; -\; -->c^3)}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}\\ & =\frac{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)(a^2+ab+ac+b^2+bc+c^2)}{(a-<!--\; -\; -->b)(a-<!--\; -\; -->c)(b-<!--\; -\; -->c)}\\ & =a^2+ab+ac+b^2+bc+c^2\\ & =\frac{2a^2+2ab+2ac+2b^2+2bc+2c^2}{2}\\ & =\frac{a^2+2ab+b^2+a^2+2ac+c^2+b^2+2bc+c^2}{2}\\ & =\frac{(a+b{)}^2+(a+c{)}^2+(b+c{)}^2}{2}\end{array}$