number theory – How to proceed with the proof

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Proof that “local” minimum spanning tree is “global” minimum spanning tree

I’m trying to understand a proof from the book “Graph Theory with Application to Engineering & Computer Science” by Narsingh Deo.

The chapter is about trees in non oriented graphs.

A bit of terminology so that you can understand the theorem and the beginning of the proof from the book:

The author calls minimum spanning trees, shortest spanning trees.

The author calls a branch of a spanning tree any edge of the tree.

A chord of a spanning tree is any edge of the underlying graph that is not in the tree.

A fundamental circuit associated to a spanning tree is a circuit formed by adding one of its chords to a spanning tree (for the author a “circuit” is a closed path, there is no repetition of vertices, it’s what most other sources I’ve read call a cycle). So, a fundamental circuit associated to a spanning tree is a actually a cycle formed by adding one of its chords to a spanning tree.

The distance between two spanning trees $S$ and $T$ of the same graph is (regarding $S$ and $T$ as sets of edges), is $|Ssetminus T|$ (which happens to be equal to $|Tsetminus S|$).

There’s a step in the proof of Theorem 3-16 that I don’t understand.

Theorem 3-16:

A spanning tree T (of a given weighted connected Graph G) is a shortest spanning tree (of G) if and only if there exists no other spanning tree (of G) at a distance of one from T whose weight is smaller than that of T


Let $T_1$ be a spanning tree in G satisfying the hypothesis of the theorem (i.e. there is no spanning tree at a distance of one from $T_1$ which is shorter than $T_1$). The proof will be completed by showing that if $T_2$ is a shortest spanning tree (different from $T_1$) in G, the weight of $T_1$ will also be equal to that of $T_2$. Let $T_2$ be a shortest spanning tree in G. Clearly, $T_2$ must also satisfy the hypothesis of the theorem (otherwise there will be a spanning tree shorter than $T_2$ at a distance of one from $T_2$, violating the assumption that $T_2$ is shortest).

Consider an edge $e$ in $T_2$ which is not in $T_1$. Adding $e$ to $T_1$ forms a fundamental circuit with branches in $T_1$. Some, but not all, of the branches in $T_1$ that form the fundamental circuit with $e$ may also be in $T_2$; each of these branches in $T_1$ has a weight smaller than or equal to that of $e$, because of the assumption on $T_1$. Amongst all those edges in this circuit which are not in $T_2$ at least one, say $b_j$, must form some fundamental circuit (with respect to $T_2$) containing $e$.

I’m stuck at the last sentence I just quoted:

“Amongst all those edges in this circuit which are not in $T_2$ at least one, say $b_j$, must form some fundamental circuit (with respect to $T_2$) containing $e$.”

I don’t see why among those cycles, there should necessarily be one that contains $e$. Why is that?

Find the first 20 primes found by the classical proof of the infinitude of the set of primes

Begin with P={2}; then form,m, the sum of 1 with the product overall elements of P. Place the smallest prime factor of m into P and repeat.

Suppose p = {p1,p2,…,pr}, then m = 1+ p1p2p3…pr.


2 is prime and 2+1 = 3 is prime;

2 * 3 +1 = 7 is prime;

2 * 3 * 7 +1 = 43 is prime;

2 * 3 * 7 * 43 +1 = 1807 = 13 * 139, then 13 is the prime;

Thus the first 5 prime number found by the classical proof is {2,3,7,43,13}.

So how to use this proof to find the first 20 prime in Mathematica?
Thank you.

combinatorics – Pigeonhole Principle Proof Verification

Let $a_1<a_2<ldots<a_{n+1}$ different integers. Prove that $$n!Big|prod_{1leq ileq jleq n+1}{(a_j-a_i)}$$Here’s my proof:

Let $kinbigl(nbigr)$. $k<n+1$ and using the Pigeonhole Principle, there are $0leq i<jleq n+1$ (or $0leq i<jleq k+1$) such that $$a_i equiv a_jpmod{k}$$Thus, $$kbig|(a_j-a_i)$$Every $kinbigl(nbigr)$ satisfies this, and we are done.

Is the proof correct? I couldn’t find a mistake myself but it seemed weird that the use of the Pigeonhole Principle had $n+1$ pigeons and only $k$ holes (when $k$ can be as little as $1,2,3$). Am I missing something?

proof writing – Is pattern matching an appropriate way to show that two infinite series are equal?

I want to show that

$$e^{a}e^{b} = e^{(a+b)}$$ using their infinite series representation ${displaystyle exp x:=sum _{k=0}^{infty }{frac {x^{k}}{k!}}=1+x+{frac {x^{2}}{2}}+{frac {x^{3}}{6}}+{frac {x^{4}}{24}}+cdots }$


LHS $= (1+ a + a^2/2! + ldots) (1+b + b^2/2! + ldots) = (1 + a + b + a^2/2! + b^2/2! + ab + ldots) $

RHS $= (1 + a+b + (a+b)^2/2! + ldots ) = (1+ a + b + a^2/2! + b^2/2! + ab + ldots)$

These terms seem to match up, so that LHS = RHS.

Can anyone offer a critique to the above method?

Does leaving behind a business and child count as proof of intention to return for UK Visa?

I know this has been done many times before but I’m seeking advice on my situation.

I’m a professional British citizen living in Asia. I have been with my Filipino Girlfriend for around a decade. I provide for her financially. She has a small food and beverage business in the Philippines which is in her name. She has all the paperwork to prove this, including business permits etc. Unfortunately, as it’s basically a cash in hand business not much of the takings are deposited in a bank, although she does have an account.

She has a 14 year old Daughter from a previous relationship and we have an 18-month old Son together who has my surname and I am named on the birth certificate.

I would like to take my Girlfriend and Son to the UK for 2 weeks this
year to visit my parents/my sons grandparents who are elderly and
would find a long journey difficult. I have the following questions
surrounding her being able to prove that she intends to return:

  • Would the fact that her Daughter, who lives with her and is reliant
    on her, will be staying behind in the Philippines be enough to show
    she intends to return?
  • Would the fact that she owns her own small business (and has all the
    documentation to prove this) be proof that she intends to return,
    despite not having accounts etc to show it is profitable?

Does the fact that I am my sons Father and I want him to see his elderly grandparents and meet his wider family members count for anything? Obviously an 18-month old baby is too young to spend 10 to 14 days away from his Mother.

For the record I will be sponsoring her trip/paying for flights/paying for private medical insurance for the both etc.

Edited to add from comments: My girlfriends mother will be taking care of the daughter and has provided a letter to state this. I can prove that I support my girlfriend, although I work and live in Singapore.

linear algebra – Direct proof of $mathfrak{so}(3)_{mathbb C}simeqmathfrak{sl}(2,mathbb C)$

It is well known that $mathfrak{su}(2)equivmathfrak{su}(2,mathbb C)$, the real Lie algebra of traceless skew-Hermitian $2times 2$ complex matrices, satisfies $mathfrak{su}(2)_{mathbb C}simeq mathfrak{sl}(2,mathbb C)$. To see this, it is sufficient to observe that any traceless matrix $A$ can be written as
$$A = ileft(frac{A+A^dagger}{2i}right) + left(frac{A-A^dagger}{2}right),$$
where both components are traceless and skew-Hermitian, and the decomposition is unique.

We also know that $mathfrak{so}(3)simeqmathfrak{su}(2)$, where $mathfrak{so}(3)$ is the real Lie algebra of traceless skew-orthogonal $3times 3$ real matrices. This follows from observing that both Lie algebras satisfy the same commutation relations, $(T_i,T_j)=epsilon_{ijk}T_k$ (or rather, we can always find bases for both spaces satisfying such relations).

This should imply that also $mathfrak{so}(3)_{mathbb C}simeqmathfrak{sl}(2,mathbb C)$ (as also mentioned in passing in this answer), but how can I show that this is the case more directly?
As far as I understand, this statement should mean that, given any traceless $2times 2$ complex matrix $A$, there is a bijection sending $A$ to two $3times 3$ real skew-orthogonal matrices. What is this decomposition?

proof techniques – Prove (p → ¬q) is equivalent to ¬(p ∧ q)

I need to prove the above sequent using natural deduction. I did the first half already i.e. I proved $(prightarrowneg q)rightarrow neg (p wedge q)$, but I’m stuck on where to start for the reverse i.e. proving $neg (p wedge q) rightarrow (prightarrowneg q)$. I figured I would start by assuming $neg (p rightarrow neg q)$ and then working towards a contradiction, but I’m still at a dead end. Can someone point me in the right direction? Thanks.

cryptography – How does Zero Knowledge Proof prevent lying?

I have seen and read several videos and articles that try to explain how Zero Knowledge Proof systems work. In these examples, there is a verifier of information, and a prover of information. The verifier asks the prover to prove that they know some information. Metaphors I have seen usually are:

The prover wants to prove they are not colorblind and gives the verifier two colored items. The verifier is color
blind and switches the items around (or not). The prover can tell if
the items were switched. Repeat this a hundred times successfully and
the prover has proven to the verifier that they really are not colorblind.

I’ve seen variations with Pepsi/Coke taste tests, colored pens or balls, etc. However, this kind of proof requires that the verifier has some sort of information at their disposal already (whether they switched the balls around or not).

I’ve also heard examples such as:

  • A liquor store asks a customer if they are at least of drinking age. The customer can prove they are without having to show their ID.

  • A bank asks a customer to prove they have a minimum balance so that they can get a loan, without having access to the customer’s bank account.

These are examples where the verifier (the liquor store and the bank) doesn’t have the same information at their disposal (such as the example of switching the items around).

If a liquor store asks a customer if they’re of drinking age, what’s preventing the customer from just lying as many times as needed to buy their alcohol?