## 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 \$binom{2n}{n}>frac{2^n}{sqrt{npi}}\$

I want to prove the following inequality $$binom{2n}{n}>frac{2^n}{sqrt{npi}}$$ by induction. Can anyone suggest a hint?

## 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.

## 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.

Thanks.