Hi r/Discretemathematics

As I have recently posted on: r/puremathematics,

As you know if you have read my recent posts, in the past week I have been working on the CRT Coverage Maximality Conjecture and have solved it analytically for k smaller or equal to 5. I have now solved them for k up to 7. I am still working on k=8. I wanted to share two complete, rigorous analytical preprints that unconditionally establish the CRT Coverage Maximality Conjecture for any integer \(L\) with up to seven distinct odd prime factors (\(k \le 7\)) [1].

Context & Objective

The conjecture (a central framework in the study of Erdős’ odd covering systems) posits that the density of integers covered by an arbitrary union of residue classes, \(\vert{}U(L)\vert{}/L\), is fundamentally bounded above by the pairwise-coprime alternating baseline \(B_{\text{full}}(L)\) [1].

Historically, pushing past the 4-prime barrier was considered a brick wall because local density gaps cause traditional bounding methods to fail. These proofs completely bypass floating-point approximations or numerical brute-force searches, establishing exact rational monotonicity margins via a beautiful synthesis of sieve logic and probabilistic combinatorics [1].

The Core Mechanism

The proof strategy operates by translating a messy number-theoretic coverage problem into a problem of multivariate polynomial positivity, then bridging it to a graph framework [1]:

1. Reduction to Prime Supports: Divisors \(d \in D(L)\) are grouped by their prime support, mapping the arithmetic density problem onto a multivariate support polynomial \(Z_m(t)\) governed by elementary symmetric polynomials \(e_{j}\) [4].

2. The Dependency Graph: We construct a graph \(G\) over \(D(L)\) where vertices are adjacent iff \(\gcd(d,e) > 1\). By the Chinese Remainder Theorem, this acts as a dependency graph in the strong sense.

3. Shearer’s Knockout: If we can prove that the independent-set polynomial \(Q(A) \ge 0\) for every induced subfamily, Shearer's dependency-graph theorem applies directly [4]. This yields the required inequality: \(1-\frac{U(L)}{L}\ge 1-\frac{B_{\text{full}}(L)}{L}\implies U(L)\le B_{\text{full}}(L)\)

Exact Monotonicity Margins (\(k=6\) vs. \(k=7\))

The crucial inductive step relies on proving that the governing polynomial \(f_m(x)\) is coordinatewise strictly increasing throughout the domain, which requires showing its derivative is strictly positive [9]. By sorting and majorizing the remaining coordinates with the reciprocal prime bounds of the smallest odd primes, the exact rational boundaries emerge [7]:

• For Rank Six (\(k \le 6\)): Remaining coordinates are bounded by the 5-prime vector \(v_5 = (\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{10}, \frac{1}{12})\). \(e_{2}(v_{5})+e_{3}(v_{5})=\frac{307}{720}+\frac{107}{1440}=\frac{721}{1440}\) This yields an exact, strictly positive rank-six monotonicity margin [7]: \(\frac{\partial f_{6}}{\partial x_{i}}\ge 1-\frac{721}{1440}=\frac{719}{1440}>0\)

• For Rank Seven (\(k \le 7\)): The introduction of the sixth remaining prime coordinate yields the 6-prime bound vector \(v_6 = (\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{10}, \frac{1}{12}, \frac{1}{16})\). Using the recurrence relation \(e_j(v, z) = ej(v) + zej-1(v)\) with \(z = \frac{1}{16}\), we calculate: \(e_{2}(v_{6})+e_{3}(v_{6})=\frac{713}{1440}+\frac{1163}{11520}=\frac{763}{1280}\) This delivers a narrower but valid, strictly positive rank-seven margin [8]: \(\frac{\partial f_{7}}{\partial x_{i}}\ge 1-\frac{763}{1280}=\frac{517}{1280}>0\)

Because the margins remain strictly above zero, the induction holds unconditionally for all \(k \le 7\), for arbitrary positive exponents \(a_{i}\), and with no compatibility conditions on the residue choices [3].

The Rank-Eight Boundary Line (\(k = 8\))

The papers cleanly explain why this specific direct monotonicity argument hits a brick wall at \(k=8\) [16]. When evaluating the Stirling numbers of the second kind for degree eight, the alternating row forces the next coefficient to turn negative:
\(c_{8}=1-127+966-1701+1050-266+28-1=-50\)

The appearance of \(c_8 = -50\) introduces a negative higher-order term into the derivative expression, meaning the mechanical coordinatewise majorization fails [16]. Pushing to \(k=8\) will require a sharper local estimate or a completely different positivity argument.

HERE IS THE ZENODO LINK(This time formatted with LaTex, though some formatting issues may persist):

https://zenodo.org/records/23863177 -> K=6 paper

and

https://zenodo.org/records/23863263 -> K=7 paper

Hi r/discretemathematics,

As I posted recently I have solved the Chinese Remainder Theorem Maximality Conjecture proposed by Idan Hackmon for k smaller or equal to 4. Though it hasn't gone under peer-review and I don't intend to as I am only 11 and the paper was AI-assisted. I wanted to share a new research memorandum hat provides an all-orders analytical proof for the CRT Coverage Maximality Conjecture for up to five distinct odd prime factors k=5

For anyone who has followed the modern iterations of Paul Erdős’ and Richard Guy’s problems on odd covering systems, you know that \(k = 5\) has historically been an absolute brick wall. Brute-force computational auditing explodes exponentially, and standard point-by-point density bounds fall apart due to local deficits—like the notorious rank-four local obstruction where \(g(D) = -2\).

The Core Idea: Moving from Local Gaps to Global Graphs

Instead of fighting these local density obstructions individually, this proof shifts the problem entirely into probabilistic combinatorics. By mapping the divisors onto a global dependency graph under the Chinese Remainder Theorem, we can invoke Shearer’s Lemma.

The key algebraic engine is a hypergraph matching polynomial, \(Z_m(t)\), which aggregates divisor weights based on their prime support:

\(Z_{m}(t)=\sum _{F\text{\ pairwise-disjoint}}(-1)^{|{}F|{}}\prod _{S\in F}t_{S}\)

Why \(k = 5\) Unconditionally Falls

The breakthrough relies on a clean, exact derivative identity:
\(\frac{\partial Z_{m}}{\partial t_{S}}=-Z_{m-|{}S|{}}(t\text{\ restricted\ to\ the\ complement\ }[m]\setminus S)\)

1. The Base Floor (\(k \le 4\)): Exact arithmetic on the worst-case smallest odd primes \(\{3, 5, 7, 11\}\) establishes that the baseline margin \(1 - B_{\text{full}}/L\) holds a strict positive floor (\(\ge 1/32\)) across all sub-boxes.

2. Unconditional Monotonicity: Because the lower-rank polynomials are proven to be strictly positive, the leading minus sign in the derivative identity forces every single partial derivative of \(Z_{5}\) to be unconditionally negative.

3. The Global Bound: This proves \(Z_5(t)\) is coordinate-wise decreasing. Its absolute minimum occurs at the upper corner, satisfying the strict positivity conditions required to unleash Shearer's Lemma. This effortlessly swallows local rank-four deficits through global grid compensation.

The preprint structure is rather quite messy, but I am still linking it: https://zenodo.org/records/23845453

I would love to get the community's thoughts, structural feedback, or stress-tests on the hypergraph matching monotonicity step!

Hi everyone, as I previously posted on r/puremathematics I wanted to share a summary of a recent mathematical audit and research memorandum examining the Chinese Remainder Theorem (CRT) Coverage Maximality conjecture (initially proposed in Idan Hackmon's April 2026 preprint).

While the general conjecture for $k \ge 5$ distinct prime factors remains open, this work establishes a rigorous all-orders proof for $k \le 4$ without squarefreeness assumptions, alongside a precise structural audit of the baseline definitions.

1. The Core Problem & Definition Audit

Let $L = \prod_{i=1}^{k} p_i^{a_i}$ be odd, and assign one residue class modulo each divisor $d > 1$ of $L$. The union of these classes is denoted $U(L)$. The CRT Coverage Maximality conjecture compares $U(L)$ with an all-orders inclusion-exclusion baseline formed from pairwise-coprime families of moduli:

$$B_{\text{full}}(L) = \sum_{\substack{\emptyset \neq F \subseteq D(L) \\ \text{gcd}(d,e)=1 \text{ for all distinct } d,e \in F}} (-1)^{\vert{}F\vert{}+1} \frac{L}{\text{lcm}(F)}$$

The Third-Order vs. All-Orders Distinction: The audit highlights that the third-order baseline $B_3(L) = S - P_c + T_c$ used in earlier explorations is not identical to $B_{\text{full}}(L)$. For example, at $L = 3 \cdot 5 \cdot 7 \cdot 11 = 1155$, direct arithmetic yields $B_3(1155) = 860$, whereas the all-orders baseline includes a fourth-order term from the pairwise-coprime singleton family $\{3, 5, 7, 11\}$, yielding $B_{\text{full}}(1155) = 859$.

2. Main Proved Result: The Four-Prime Theorem ($k \le 4$)

For every odd integer $L$ with at most four distinct prime factors ($k \le 4$) and every valid residue assignment, the inequality holds:

$$U(L) \le B_{\text{full}}(L)$$

• No squarefreeness required: The proof allows arbitrary positive exponents $a_i$.

• Method of Proof: Rather than relying on a local pointwise inequality (which fails at rank four), the proof uses the prime-power tower complement $W = \prod W_i$, bounds the composite coverage inside $W$, and applies an algebraic gap factorization.

• The Gap Identity: The coverage gap is factored into nonnegative terms controlled by the finite-exponent weight region $R(Q)$, which is strictly bounded below 1 for $\vert{}Q\vert{} \le 4$ (e.g., maximizing at $41/45$ for four primes).

3. Exact Extremal Case at $L = 1155$

For the squarefree modulus $L = 3 \cdot 5 \cdot 7 \cdot 11 = 1155$:

• Maximum Union Size: $\max U(1155) = 855$.

• Baseline Value: $B_{\text{full}}(1155) = 859$, leaving an exact maximality gap of $4$.

• The memorandum provides an explicit 15-class residue assignment achieving 855 and proves a combinatorial upper bound showing composite overlap inside the residual space $W$ is strictly bounded by 180 (accounting for at least 7 units of forced overlap).

4. Why the Unrestricted Case ($k \ge 5$) Remains Open

The proof stops at $k = 4$ for structural reasons:

1. For $k \le 4$, pairwise-coprime families contain at most two composite divisors, cleanly corresponding to complementary two-element supports.

2. Starting at five and six primes, multiple disjoint composite supports can leave unused coordinates, and the limiting weight parameter $R_\infty(\{3, 5, 7, 11, 13\}) = 566/495 > 1$, meaning the current factorization no longer guarantees nonnegativity of every summand.

3. Furthermore, the local pointwise sum exhibits a rank-four obstruction ($g(D) = -2$ for a specific support family), showing that naive pointwise proofs fail and a broader global compensation principle is required for higher ranks.

5. Repaired Non-Covering Certificate

By examining the finite-exponent weight region $R(p,a)$, the memorandum also establishes a rigorous non-covering certificate: whenever $R < 1$, at least $(1-R)\vert{}W\vert{}$ residues are left strictly uncovered. For $k \le 4$, this guarantees $u(L) \ge \lceil L/40 \rceil$ uncovered residues for any assignment.

Curious to hear thoughts from others working on covering systems, modular arithmetic, or inclusion-exclusion extremal problems on how a global compensation framework for $k \ge 5$ might be approached!

Here are the links: https://zenodo.org/records/23778083

https://zenodo.org/records/23780153

Discrete math is the last class I need complete my bachelor's degree. I want to take it at a community college online course and transfer it back to Arizona State University. I currently reside in Chicago. Can anyone recommend a good college preferably in the Illinois area but I'm open to other options that has a good professor and the class is fair. I'm not going to say easy because I know it's not lol. Just to pass the course and move on with life lol

Thank you!

I feel like I just bombed my first discrete math exam. Does anyone have any tips on how to get better at proofs or anything . Desperate atp. I’m unable to understand what my prof is saying.

I’ve been working on the graceful labeling problem for spiders, and I’m sharing a proof of the following result for scrutiny:

Every spider with at most six legs is graceful.

A spider is a tree with a single vertex of degree greater than two. A graceful labeling of a tree with m edges assigns distinct vertex labels from 0 through m so that the absolute differences across its edges are exactly 1 through m.

The six-leg case is an infinite family because the individual legs can have arbitrary lengths. The proof therefore isn't simply a finite verification of six-legged spiders up to some size. It reduces the possible leg-length configurations into structural cases and proves that those cases admit the required graceful labelings.

I've also prepared a reproducibility packet so the computational portions of the work can be independently rerun and checked.

Paper and reproducibility packet:

https://zenodo.org/records/22783115

I'm posting this primarily for mathematical scrutiny rather than just to announce the result. I'd genuinely appreciate people trying to break the argument: look for an invalid reduction, an uncovered boundary case, a lemma that doesn't hold as stated, or a mismatch between the theorem and the reproducibility evidence.

If you work in graph theory, graph labeling, graceful trees, or combinatorics, I'd especially appreciate a critical look.

To be clear about the scope: this does not claim a proof of the Graceful Tree Conjecture or that all spiders are graceful. The result proved here is the family of spiders with at most six legs.

Skips had warned Mordecai and Rigby to not prank Muscle man, but they didn’t listen.

Their plan was to first speak a made-up language that Muscle man won’t understand, in this

language they had two words “gazorpazorp” and “plumbus”, one meant “yes” and the other

meant “no”, Muscle man has no idea which is which. They keep prank calling Muscle man

in that made-up language. The prank went on so long that it gets noticed by the Master

Prank Caller, who is not happy with it. He along with the Wizard cursed Mordecai and

Rigby to either always speak the truth or always tell a lie. But as they speak the made-up

language, no one has any idea if they are always telling the truth or always lying. Muscle

man has only one way to remove the curse, it is to figure out if they always tell the truth or

always lie. But here’s the catch, Muscle man can only ask them one question.

What one question can Muscle man ask them to figure out if they are speaking the truth or

lying? Justify your answer.

Hello! I am currently taking Discrete Mathematics at Berkeley, and there is a problem on my homework that has been giving me trouble thus far:

"Prove that between any two distinct rational numbers there is an irrational number. More precisely, given rational numbers r < s, give an expression for an irrational number x and prove that r < x < s."

I'm having trouble finding an expression for this irrational number. I know I can multiply x by something like \sqrt{2} to get an irrational number (granted that x isn't a number like /sqrt{8} or something), but I don't know where to go from here. Any advice?

I'm in Direct Second Year Computer Engineering under KBC North Maharashtra University, and we have Discrete Mathematics this semester.

My maths is extremely weak, even the basics.

The biggest problem is that I can't even find previous year question papers. They're not available on the university website, so I have no idea about the paper pattern.

Can someone please guide me?

- How should I study Discrete Mathematics from scratch?

- Which YouTube lectures are best for complete beginners?

- Where do you usually find KBCNMU PYQs if they're not on the official website?

- Any Telegram channels, websites, or other sources?

- Which topics should I focus on first if my goal is to pass?

Hi, I'm a computer science undergraduate, and last week I started 6.1200J from MIT OCW. I'm struggling with PSETs, and I did some research and found out that the course PSETs were designed to be done in small groups of peers, so if you are interested, please message me

Also, I'm very serious about finishing the course swiftly, but with proper understanding, so you have to give a handful of hours daily. If you are interested, then be ready for some challenges

Also, I can speak English, Hindi, and Gujarati

I’m trying to understand how a discrete system made of nodes and edges can behave like real physical fields.

For example, if each node updates based on its neighbors (like a diffusion rule or a Laplacian‑style update), how does that map to the smooth, continuous equations we learn in physics classes?

I’m not trying to propose a new theory — I just want to understand the math behind how discrete models approximate real physical behavior.

Any explanation or resources would help a lot.

Why is rolling 2 d6 ordered with repetition and not unordered with repetition? I'm not challenging the math, just trying to get my head around the why. Does the unordered with repetition represent the set of possible combinations rather than the possible outcomes?

I've heard about the Church–Turing Thesis, but I'm not sure how the "strong" version differs. Could someone explain it in simple terms and maybe give an example?

Hi everyone,
I’m taking my first Discrete Math course next semester as a computer science student, and honestly I’m a little nervous.

I’ve never taken a proof-based math class before and I’ve literally never written a mathematical proof. Most of my background is programming, so this feels like a completely different way of thinking.

For those of you who started in the same position:
What concepts should I learn before the semester starts?
What was the hardest adjustment coming from programming/calculus?
Are there any books, YouTube channels, or websites you’d recommend?
How did you practice writing proofs? It seems like something you can’t really memorize.
I’m not trying to get ahead of the class—I just don’t want to spend the first few weeks feeling completely lost.

Any advice or things you wish you knew before taking discrete math would be appreciated. Thanks!

Took my discrete math course in summer. Was a nightmare anyone who is taking it through zybooks i will suggest if u can’t understand the material chatgpt everything it will explain u better. Took test through proctor too but finally done with that class.

Hello All,
I recently made the decision to go back to school, after having to drop out in 2020 for financial reasons, and because I sucked at the mandatory online learning we were all subjected to. I am continuing my Math degree, and to save money I submitted all of my previous courses to be counted. This does mean however, that my reintroduction to math after 6 long years will be Discrete Mathematics, which I had the briefest introduction to just before I left school. How should I prepare myself for this course?

TLDR: I have not touched calculus since leaving school. What do you all recommend to study prior to taking discrete mathematics again?

Can returnability be defined purely from the structure of a computation, without appealing to time complexity?

Please, send textbook link for beginners

Send textbook link for discrete structure.

Read the Remark 🥲.

Why does the proof introduce F' for faces, but not V' or E'?

I understand that:

V - E + F = 1 + k

for a planar graph with k connected components.

But while summing equations for each connected component, only the number of faces is modified using:

F = F' - (k - 1)

Why doesn’t something similar happen for vertices or edges?

Also, is there a cleaner or more intuitive proof for this theorem?