geometry – Proving the archimedean property for angles

This is a problem from my freshman Calculus course (it was a homework problem so no rigurous solution was given in class). I’m now a sophomore but I still can’t fully understand how to solve it (for what I can recall, the solution sketch involved a right-angle construction). I’d really appreciate if someone could help me.

Axiom. Given two straight line segments $AB$ and $CD$, there is a natural number $n$ such that $n$ juxtaposed copies of $AB$ are bigger than $CD$.

Using the former statement, one has to prove the following:

Given two angles $alpha$ and $beta$, there is a natural number $n > 0$ such that either $nalpha > beta$ or $nalpha$ is not defined for exceeding two right angles ($180°$).

Proving a complex function is real when $|z|=1$

The question is as follows:

Given, $$f(z)=(-19+9i)frac{1}{z}-(19+9i)z,$$
explain why when $|z|=1$, $f(z)$ is real.

My first instinct to approach this question was to take take the complex conjugate of $f(z)$ and show that $f(z)=f(z)^{*}$ when $|z|=1$. I did this since taking the conjugate of a complex number has the geometric consequence of reflecting on the real axis, hence if the conjugate of the complex number was equal to itself then I can deduce that $f(z)$ must lie on the real axis and hence is real.

Computing the conjugate gave me, $$f(z)^{*}=(-19+9i)z^*-(19+9i)frac{1}{z^*}.$$
Since conjugation only reflects the complex number, its modulus would remain unchanged, hence $|z|=1 implies |z^*|=1$. From this point I am not sure on how to continue, so any help is appreciated!

proof writing – Proving Zorn’s lemma implies axiom of choice (trouble showing every chain has an upper bound)

I am attempting to prove the fact that Zorn’s lemma implies the axiom of choice, however my proof falls short when I try to prove every chain has an upper bound. I understand there are other proofs on stackexchange, however my question is quite specific and I have not seen them answered elsewhere.
Here is my attempt so far.

Let $X$ be a non-empty set. I will use Zorn’s lemma to prove a choice function on $X$ exists (a choice function on $X$ is $c: (P(X)-{emptyset}) to X$ such that for every non-empty set $A subset X$, $c(A) in A$). Then define the set $mathcal{A}$ to be the collection of all $(f, Y)$ where $f$ is a choice function on $Y$, where $Y subset X$ , and equip $mathcal{A}$ with the partial order $(f, Y) preceq (f’, Y’)$ if $Y subset Y’$ and $f’|_{Y} = f$.

Now let $mathcal{B} subset mathcal{A}$ be a linearly ordered subset (or commonly known as “chain”) of $mathcal{A}$. Then defining $Z = bigcup_{(f,Y) in mathcal{B}}Y$ and $g = bigcup_{(f,Y) in mathcal{B}} f$, I claim $(g, Z) in mathcal{A}$ (where I refer to $f$, I’m refering to $f$ as a set, namely $(x, y) in f iff y = f(x)$).

This is the part where I’m having trouble. I know $g$ is a function, since if $(x, y), (x,z) in g$, then $(x,y) in f$ and $(x,z) in f’$, where $(f, Y), (f’, Y’) in mathcal{B}$. Since $mathcal{B}$ is linearly ordered, we have without loss of generality $(f, Y) preceq (f’, Y’)$ and so $(x,y), (x,z) in f’$ and since $f’$ is a function, we have $y=z$. Hence $g$ is a function.

However, I’m having a hard time showing that the domain of $g$ is $P(Z)-{emptyset}$. Since if $Wsubset Z$ and $W neq emptyset$, then by defintion of $Z$, $W subset bigcup_{(f,Y) in mathcal{B}}Y$. To show $g$ is defined on $W$ then requires us to show there exists a $Y$ such that $Wsubset Y$ and $(f,Y) in mathcal{B}$ but how do I show such a $Y$ esists?

It seems to me using the fact that $mathcal{B}$ is linearly ordered is essential, but I fail to see how it works. I am comfortable with the continuing the rest of the proof to completion.

np complete – Proving a problem is NP Hard

Consider the following problem: Given a weighted directed graph $G$, determine if $G$ has a cycle whose total weight is $k$. All edge weights are integer but might be negative. $k$ is not an inputted value. It’s fixed and known beforehand to everyone. So $k$ is NOT an input to this algorithm.

I want to show the problem is NP hard, so this means I have an oracle that tells me whether or not a graph has a cycle whose sum of edge weights equals $k$ for a fixed, known value of $k$.

I thought about reducing from Hamiltonian cycle. So I want to show that I can use this oracle to solve Hamiltonian cycle problems.

Let $G$ be a graph. I want to see if $G$ has a Hamiltonian cycle. I try to construct a new graph to provide to the oracle but I’m not sure how to do so. I think it needs to have $k$ vertices so I tried doing casework on when $G$ has more or less than this many vertices.

I can show that I can solve Hamiltonian cycle problems with $k$ vertices by constructing a new complete graph where edge weights are $1$ if it was in the original graph and a really large nunber otherwise. Then running the oracle on this graph returns true only if there’s a Hamiltonian cycle with $k$ vertices. But this doesn’t handle the general case. Is this proof fine?

But I’m stuck. Any help is appreciated.

data structures – Proving Quicksort is $O(n^2)$

So I’m trying to figure out why the worst case of Quicksort is $O(n^2)$.

I know this a very well known problem, but the funny thing is where ever I look (even Wikipedia) gives the following explanation: “The worst case is the most unbalanced case where the problem splits to a problem of size $n-1$ and a problem of size $0$ (i.e. when the array is already sorted)”.

Then they use the master theorem and find it is $O(n^2)$.

Marvelous. So simple. But wait.

Do we know upfront that the worst case is $O(n^2)$? No, that’s what we need to prove.

“The most unbalanced case” meaning it is the worst case? Is there any theorem that states this?

So what is actually a coherent proof that Quicksort is $O(n^2)$?

Or in other words, what is the proof for the missing part?

We can derive that the run time can be described as $T(n) = T(n_1) + T(n_2) + O(n)$ where $n_1 + n_2 + 1 = n$. How to prove $T(n)$ is the largest when $n_1 = 0$ and $n_2 = n-1$?

I already know this is the most unbalanced case. Why is it actually the worst case?

Proving that a language defined by a regular expression is equivalent to a right linear grammar

After thinking for a bit, I am not able to prove a double inclusion proof for the following problem. It seems very interesting to me.

Consider the regular expression $r= ((1(00)^∗1 + 0)1)^∗$ and the right-linear grammar $G= ({S,A},{0,1},S,P)$, where $P$ consists of the following rules:

$Srightarrow 1A|01S|lambda$

$Arightarrow 00A|11S$

Prove that $L(G)subseteq L(r)$ and vice versa.

In general, how exactly do I prove that a regular grammar describes the same language as a regular expression?

Proving if $a_{k}ge a_{k-1}+1$ then $1+frac{1}{a_{0}}(1+frac{1}{a_{1}-a_{0}})…(1+frac{1}{a_{n}-a_{0}})le prod_{k=0}^{n}(1+frac{1}{a_{k}})$

I’ve worked on this problem with the sequence $a_{k}$ being the natural numbers, that is $a_{k}=a_{k-1}+1$ and $a_{0}=1$. Over the naturals, $prod_{k=0}^{n}(1+frac{1}{a_{k}})$ can be proven to be $n+1$.

I’ve been able to recognize the obvious pattern in $prod_{k=2}^{n}(1+frac{1}{k-1})$ to be equal to $n$, but I do not know how to prove this rigorously.

That said, over the naturals, the inequality reduces to $n+1 le n+1$. That got me thinking if the sequence in question could be expanded with the condition $a_{k}ge a_{k-1}+1$ and specifically if this would not lead to the situation corresponding with RHS being greater than LHS.

Thank you for your help.

reductions – Proving B-Min-Cost Strongly connected Subgraph is NP-Complete

We have a strongly connected directed graph where each edge has positive integer weights. We are also given a $B in mathbb{N}$. Does there exist a strongly connected subgraph where sum of edge weights $leq B$?

I’m not sure what NP-Complete problem to use. I tried subset sum, but it takes exponential time to build the graph I had in mind (yes/no tree with edge cost being the numbers in subset sum problem, and a final edge to root at end of every decision sequence). Any ideas? Is there a better NP-Complete problem to use?

analytic number theory – Proving Integral representation of $zeta(n)$

Currently studying analytic number theory and was baffled by the following integral representation of $zeta(n)$

$$zeta(n) = dfrac{2}{1 – 2^{1 – n}}sinleft(dfrac{pi n}{2}right)int_0^infty dfrac{x^{-2n}}{pi x}(1 – pi x^2csc(pi x^2))dx$$

However there’s no proof for this provided in the text.

Any help or hints would be highly appreciated.