What Patrick McKenzie means by a “Dangerous Professional”

The following are selected posts by Patrick McKenzie @patio11 which bear on his idea of the "Dangerous Professional". Read them and write a comprehensive summary of what his idea is about, with examples of things that such a professional does and says, with context for each.

(The posts are omitted for brevity.)

What Patrick McKenzie means by a “Dangerous Professional”

A Dangerous Professional (DP) is not a hit-man in a suit; it is an ordinary-looking person who has mastered the practical magic of paperwork, procedure, and persuasion. They are “dangerous” only from the perspective of a large organization that hoped to keep things sloppy, verbal, or undocumented. To the victim of a fraud, the customer stuck in a phone-tree, or the junior employee facing “Um — please stop taking notes,” a DP is simply the person you want on your side.

McKenzie’s scattered Twitter threads form an informal field-manual. Taken together they describe a mindset, a toolkit, and a style of speech that turns asymmetric bureaucratic fights into winnable games.

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1. Core attitudes

Attitude	Explanation	Representative quotes / situations
“Get it in writing.”	Written records trigger statutes, shot-clocks and institutional accountability. Verbal promises can safely be ignored; signed PDFs and timestamped emails cannot.	Gate-agent rebooking; Reg E dispute e-mails; post-meeting “Hi Bob, thanks for meeting — we discussed…” mail.
“Ask for citations.”	Force the other side to say which policy, law or contract clause they are relying on. Often they can’t.	“Please cite, in writing, your refusal to repay a fraud victim under Regulation E’s ‘no printed receipt’ clause.’”
Document first, argue later.	A single contemporaneous diary entry, e-mail, or transcript will beat fuzzy recollections years later.	Being told to stop taking notes → hand-wrote a transcript immediately after the meeting.
Use the system’s own rules.	Know which regulator, statute, or SLA starts a clock and invoke it deliberately.	FDCPA 30-day dispute window; 10-day FOIA appeal timeline; airline “irregular operations” options budget.
Polite, calm, relentless.	No table-pounding unless you lack facts or law. A DP sounds boringly professional while making it clear they can escalate.	Airline chat: “I’m sure you have options for me.”
Not necessarily a lawyer.	Lawyers may be DPs, but anyone who learns the playbook can do the job.	“Some lawyers are Dangerous Professionals and some Dangerous Professionals are lawyers, but not all DPs are lawyers.”

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2. Characteristic moves & phrases

Move	Typical wording	Context & purpose
Memorialize the meeting.	“Hi Bob, thanks for meeting today. We discussed … Looking forward to working with you.”	Locks the oral conversation into an immutable timeline.
Force written authority.	“Could you please set forth, in writing, the policy that governs X?”	Either they pony up a real document or admit it doesn’t exist.
Invoke the regulator.	“Under Regulation E you are required to complete your investigation within 10 business days.”	Flips the risk onto the institution’s legal/compliance team.
Escalate by address line.	Letters marked Office of the President / Chief Compliance Officer / Investor Relations or sent certified mail, return-receipt requested.	Short-circuits the CSR tier and lands on someone who can say “yes.”
Apply the “What are my options?” incantation.	Direct, calm request repeated until the agent presses the escalation button.	Airline IRROPs, billing disputes, any “computer says no” moment.
Preserve the shot-clock.	“I reiterate that you have not performed X. Your clock began on [date of first letter].”	Prevents a bureaucracy from resetting its own deadline.
Professional presentation props.	A notebook, neatly tabbed binder, diary app with timestamps, or even a FOIA appeal letterhead.	Visual signals that you keep records and will use them.

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3. Field-tested examples

Domain	What the DP did	Outcome / rationale
Bank fraud (tap-to-pay scam)	Asked bank to cite in writing why Reg E didn’t apply; pointed out that newspapers are interested.	Legal & PR risk flips, bank almost always refunds.
FOIA appeal	First-time appellant writes a letter whose “load-bearing paragraph” threatens cheap, predictable litigation.	Agency promises to respond within statutory timeframe.
Airline IRROPs	Calmly repeats “What are my options?”; refuses to guess at alternatives; triggers supervisor override; re-booked on competitor at airline’s expense.	Airlines have budgets for this; agent just needed permission.
Medical billing error	Drafts letter to provider’s general counsel: invoice, receipt, card statement attached; notes FDCPA obligations.	Billing department fixes error rather than escalate.
Payroll tax mismatch	Letter “in a tone so polite the paint peels” warns bank that Ministry regulators may be consulted.	Forces institution, not customer, to project-manage the clean-up.
Meeting where note-taking banned	Continues note-taking, keeps physical control of notebook, immediately diaries that the request was “extraordinary.”	Creates contemporaneous evidence and signals unwillingness to be intimidated.
Consumer loans / debt collection	Cites FDCPA; demands validation in writing; records start date of dispute.	Debt collector must stop until it can prove claim.
Customer-service dead end	Buys bank’s “cheapest premium product” to gain elite-tier phone number, then internally transfers to the needed department.	Small fee buys access to staff with actual decision authority.

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4. Negative space: who the DP isn’t

• Not a bully. Table-pounding is what lacks facts or law. A DP rarely needs theatrics.
• Not always competent. McKenzie has met “Dangerous-to-Themselves-and-Others Professionals” whose titles outrun their skills.
• Not secretive about methods. The “deep magick” works even when everybody knows the spell; the power lies in doing the boring work.

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5. Learning the craft

Skill to acquire	McKenzie’s tips
Reg-reading & citation.	Practice Googling the primary source (“Fannie Mae Seller’s Guide §…”) until you can pull a paragraph in 30 sec.
Write like a DP.	Keep templates; use clear subjects; state facts, request, desired timeline. LLMs can draft a passable DP letter now.
Note-taking & diaries.	Day One journal, or 1-minute phone videos. Any timestamped medium becomes powerful evidence later.
Channel selection.	Pick the communication path that makes their lawyer or compliance officer read your message.
Understand incentives.	Know who actually wears financial or regulatory liability inside the org and aim the spotlight there.

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6. Why the archetype matters now

• Cheap generative AI means anyone can summon a respectable DP letter, so institutions that rely on friction (“write us a letter within 30 days”) will face a flood.
• Regulatory shot-clocks (Reg E, FDCPA, FOIA, FDCPA, etc.) were designed when writing letters was hard. Automation breaks that assumption.
• Professional veneer is democratizing. LLM-drafted correspondence turns former class-gated advantages (clarity, citations, formatting) into commodities, shifting power toward individual consumers.

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In one sentence

A Dangerous Professional is a calm, paperwork-wielding adult who knows which letter, statute, or shibboleth makes a billion-dollar institution clear its throat and say, “Yes ma’am, right away.”

Square trisection

use code to draw a picture that divides a unit square into three equal parts using two inscribed parabolas, one positive and one negative, such that the vertex of the concave-up parabola is the lower left corner of the square and the vertex of the concave-down parabola is the upper right corner of the square

Here’s why the picture you see cuts the unit square into three equal-area bands:

• Concave-up parabola $y_1(x)=x^2$ has its vertex at $(0,0)$.

  $$\text{Area below }{y}_1={\int }_0^1{x}^2,dx=\frac{1}{3}.$$

• Concave-down parabola $y_2(x)=-x^2+2x$ has its vertex at $(1,1)$.

  $$\text{Area below }{y}_2={\int }_0^1(-x^2+2x),dx=\frac{2}{3}.$$

• Three equal regions

  • Region A (bottom strip): area $=\frac{1}{3}$.
  • Region B (between the two parabolas): $\frac{2}{3}-\frac{1}{3}=\frac{1}{3}$.
  • Region C (top strip): $1-\frac{2}{3}=\frac{1}{3}$.

Because each region occupies $\frac{1}{3}$ of the square’s total area, the two parabolas achieve the desired trisection while meeting the specified vertices. The code above draws the square’s boundary and both parabolic arcs so you can verify the construction visually or adapt it further (e.g., change sampling density, export as SVG, add shading, etc.).

The code was as follows:

import numpy as np
import matplotlib.pyplot as plt

# x-coordinates inside the unit square
x = np.linspace(0, 1, 400)

# Two parabolas that split the square into three equal-area regions
y_up = x**2            # concave-up, vertex at (0, 0)
y_down = -x**2 + 2*x   # concave-down, vertex at (1, 1)

# Draw the unit square
plt.figure()
plt.plot([0, 1], [0, 0])   # bottom
plt.plot([1, 1], [0, 1])   # right
plt.plot([1, 0], [1, 1])   # top
plt.plot([0, 0], [1, 0])   # left

# Plot the parabolas
plt.plot(x, y_up, label=r"$y = x^2$")
plt.plot(x, y_down, label=r"$y = -x^2 + 2x$")

# Formatting
plt.xlim(0, 1)
plt.ylim(0, 1)
plt.gca().set_aspect("equal", adjustable="box")
plt.title("Parabolic trisection of a unit square")
plt.legend()
plt.show()

Sum of 2ᵏ, or of k/(2ᵏ) for k∈ℕ∩[1,n]

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}2+2^2+2^3+\cdots +2^n=2^{n+1}-2\end{array}}$$

Proof by Mathematical Induction:

We are asked to prove that for every natural number $n$,

$$2+2^2+2^3+\cdots +2^n=2^{n+1}-2$$

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Step 1: Base Case

Let’s check the base case $n=1$.

Left-hand side (LHS):

$$2=2$$

Right-hand side (RHS):

$$2^{1+1}-2=2^2-2=4-2=2$$

✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume that the formula holds for some natural number $k$, i.e.,

$$2+2^2+2^3+\cdots +2^k=2^{k+1}-2$$

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Step 3: Inductive Step

We need to prove that the formula holds for $k+1$, i.e.,

$$2+2^2+2^3+\cdots +2^k+2^{k+1}=2^{(k+1)+1}-2$$

Start from the left-hand side for $k+1$:

$$(2+2^2+2^3+\cdots +2^k)+2^{k+1}$$

Apply the inductive hypothesis to the first part:

$$(2^{k+1}-2)+2^{k+1}$$

Combine like terms:

$$2^{k+1}+2^{k+1}-2=2\cdot 2^{k+1}-2$$

Simplify the exponent:

$$2^{k+2}-2$$

Check the right-hand side formula for $k+1$:

$$2^{(k+1)+1}-2=2^{k+2}-2$$

✅ Matches exactly.

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Conclusion:

By the principle of mathematical induction, the formula

$$2+2^2+2^3+\cdots +2^n=2^{n+1}-2$$

holds for all natural numbers $n$.

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{2}{2^2}}+{\displaystyle \frac{3}{2^3}}+\cdots +{\displaystyle \frac{n}{2^n}}=2-{\displaystyle \frac{n+2}{2^n}}\end{array}}$$

Proof by Mathematical Induction:

We are asked to prove that for every natural number $n$,

$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots +\frac{n}{2^n}=2-\frac{n+2}{2^n}$$

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Step 1: Base Case

Check the base case $n=1$.

Left-hand side (LHS):

$$\frac{1}{2}=0.5$$

Right-hand side (RHS):

$$2-\frac{1+2}{2^1}=2-\frac{3}{2}=2-1.5=0.5$$

✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume that the formula holds for some natural number $k$, i.e.,

$$\frac{1}{2}+\frac{2}{2^2}+\cdots +\frac{k}{2^k}=2-\frac{k+2}{2^k}$$

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Step 3: Inductive Step

Prove that the formula holds for $k+1$, i.e.,

$$\frac{1}{2}+\frac{2}{2^2}+\cdots +\frac{k}{2^k}+\frac{k+1}{2^{k+1}}=2-\frac{(k+1)+2}{2^{k+1}}$$

Start from the left-hand side for $k+1$:

Apply the inductive hypothesis to the first part:

$$(2-\frac{k+2}{2^k})+\frac{k+1}{2^{k+1}}$$

Express $\frac{k+1}{2^{k+1}}$ with denominator $2^k$ to combine terms:

$$2-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}$$

Express $\frac{k+1}{2^{k+1}}$ as:

$$\frac{k+1}{2\cdot 2^k}=\frac{k+1}{2^k\cdot 2}=\frac{k+1}{2^{k+1}}$$

It doesn't directly combine, so use a common denominator $2^{k+1}$ for the fractions.

Express $2$ with denominator $2^{k+1}$:

$$2=\frac{2\cdot 2^{k+1}}{2^{k+1}}=\frac{2^{k+2}}{2^{k+1}}$$

Express $\frac{k+2}{2^k}$ with denominator $2^{k+1}$:

$$\frac{k+2}{2^k}=\frac{2(k+2)}{2^{k+1}}$$

Now write the entire left-hand side over denominator $2^{k+1}$:

$$\frac{2^{k+2}-2(k+2)+(k+1)}{2^{k+1}}$$

Compute the numerator:

$$2^{k+2}-2(k+2)+(k+1)$$

Compute step by step:

• $2^{k+2}$ stays as is.
• $-2(k+2)=-2k-4$
• $+(k+1)$

Sum:

$$2^{k+2}-2k-4+k+1=2^{k+2}-k-3$$

The numerator becomes:

$$2^{k+2}-(k+3)$$

Now express the right-hand side formula for $k+1$:

Check if it matches:

$$2-\frac{(k+1)+2}{2^{k+1}}=2-\frac{k+3}{2^{k+1}}=\frac{2\cdot 2^{k+1}}{2^{k+1}}-\frac{k+3}{2^{k+1}}$$

Compute numerator:

$$2\cdot 2^{k+1}=2^{k+2}$$

So numerator:

$$2^{k+2}-(k+3)$$

Indeed, matches exactly.

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Conclusion:

By the principle of mathematical induction, the formula

$$\frac{1}{2}+\frac{2}{2^2}+\cdots +\frac{n}{2^n}=2-\frac{n+2}{2^n}$$

holds for every natural number $n$.

Sum of products of consecutive naturals, and sum of the reciprocals of such products

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot (n+1)={\displaystyle \frac{n(n+1)(n+2)}{3}}\end{array}}$$

Let's prove the statement by mathematical induction.

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Statement:

For every natural number $n\ge 1$:

$$1\cdot 2+2\cdot 3+3\cdot 4+\cdots +n\cdot (n+1)=\frac{n(n+1)(n+2)}{3}$$

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Step 1: Base Case

Test for $n=1$.

Left-hand side (LHS):

$$1\cdot 2=2$$

Right-hand side (RHS):

$$\frac{1(1+1)(1+2)}{3}=\frac{1\cdot 2\cdot 3}{3}=\frac{6}{3}=2$$

✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume the formula holds for $n=k$, i.e.,

$$1\cdot 2+2\cdot 3+\cdots +k\cdot (k+1)=\frac{k(k+1)(k+2)}{3}$$

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Step 3: Inductive Step

Show it holds for $n=k+1$.

Left-hand side for $n=k+1$:

$$(1\cdot 2+2\cdot 3+\cdots +k\cdot (k+1))+(k+1)(k+2)$$

Using the inductive hypothesis:

$$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)$$

Factor out $(k+1)(k+2)$:

$$(k+1)(k+2)(\frac{k}{3}+1)$$

Combine the terms inside the parentheses:

$$(k+1)(k+2)\cdot \frac{k+3}{3}$$

Rearrange:

$$\frac{(k+1)(k+2)(k+3)}{3}$$

This matches the RHS for $n=k+1$.

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✅ Conclusion:

By mathematical induction, the formula holds for all natural numbers $n\ge 1$:

$$1\cdot 2+2\cdot 3+\cdots +n\cdot (n+1)=\frac{n(n+1)(n+2)}{3}$$

Prove, using induction, that for every natural number n: $${\displaystyle \begin{array}{r}{\displaystyle \frac{1}{1\cdot 2}}+{\displaystyle \frac{1}{2\cdot 3}}+\cdots +{\displaystyle \frac{1}{n\cdot (n+1)}}={\displaystyle \frac{n}{n+1}}\end{array}}$$

Let's prove this formula using mathematical induction.

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Statement:

For every natural number $n\ge 1$:

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}=\frac{n}{n+1}$$

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Step 1: Base Case

Test for $n=1$.

Left-hand side (LHS):

$$\frac{1}{1\cdot 2}=\frac{1}{2}$$

Right-hand side (RHS):

$$\frac{1}{1+1}=\frac{1}{2}$$

✅ The base case holds.

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Step 2: Inductive Hypothesis

Assume the formula holds for $n=k$, i.e.,

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{k\cdot (k+1)}=\frac{k}{k+1}$$

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Step 3: Inductive Step

Show it holds for $n=k+1$.

Left-hand side for $n=k+1$:

$$(\frac{1}{1\cdot 2}+\cdots +\frac{1}{k\cdot (k+1)})+\frac{1}{(k+1)(k+2)}$$

Apply the inductive hypothesis:

$$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}$$

Get a common denominator:

$$\frac{k(k+2)+1}{(k+1)(k+2)}$$

Compute the numerator:

$$k(k+2)+1=k^2+2k+1=(k+1{)}^2$$

So the sum becomes:

$$\frac{(k+1{)}^2}{(k+1)(k+2)}$$

Cancel one $(k+1)$ from numerator and denominator:

$$\frac{k+1}{k+2}$$

This matches the RHS for $n=k+1$:

$$\frac{k+1}{(k+1)+1}=\frac{k+1}{k+2}$$

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✅ Conclusion:

By mathematical induction, the formula holds for all natural numbers $n\ge 1$:

$$\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\cdots +\frac{1}{n\cdot (n+1)}=\frac{n}{n+1}$$

Proof that all and only perfect squares have an odd number of different factors

Show that a natural number has an odd number of different factors if, and only if, it is a perfect square.

Proof: A natural number has an odd number of different factors if and only if it is a perfect square.

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Step 1: Prime Factorization Approach

Let’s express a natural number $n$ in terms of its prime factorization:

$$n=p_1^{a_1}{p}_2^{a_2}\dots p_k^{a_k}$$

where $p_1,p_2,\dots ,p_k$ are distinct prime numbers and $a_1,a_2,\dots ,a_k$ are non-negative integers.

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Step 2: Total Number of Factors Formula

The total number of positive factors of $n$ is given by the formula:

$$\text{Number of factors}=(a_1+1)(a_2+1)\dots (a_k+1)$$

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Step 3: When is the Number of Factors Odd?

The product of numbers $(a_1+1)(a_2+1)\dots (a_k+1)$ is odd if and only if each $(a_i+1)$ is odd, because the product of odd numbers is odd, and if any factor is even, the whole product is even.

A number $(a_i+1)$ is odd if and only if $a_i$ is even (since an even number plus 1 is odd).

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Step 4: Link to Perfect Squares

• If all exponents $a_i$ are even, then the number is a perfect square, because:

$$n=(p_1^{a_1/2}{p}_2^{a_2/2}\dots p_k^{a_k/2}{)}^2$$

• Conversely, if the number is a perfect square, then all exponents in its prime factorization must be even.

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Step 5: Conclusion

• A natural number has an odd number of positive divisors if and only if it is a perfect square.

✔️ Q.E.D.

Why is the number of factors $(a_1+1)(a_2+1)\dots (a_k+1)$?

Excellent question! Let’s break down why the number of factors of a natural number is given by:

$$(a_1+1)(a_2+1)\dots (a_k+1)$$

✅ Reasoning:

Suppose a natural number $n$ has prime factorization:

$$n=p_1^{a_1}{p}_2^{a_2}\dots p_k^{a_k}$$

🔍 How do we form a factor of $n$?

A factor of $n$ is a number that divides $n$ exactly. To construct a factor, you can choose:

• Any exponent for $p_1$ between $0$ and $a_1$ (inclusive),
• Any exponent for $p_2$ between $0$ and $a_2$ (inclusive),
• ...
• Any exponent for $p_k$ between $0$ and $a_k$ (inclusive).

🔢 Counting the choices:

• For prime $p_1$, there are $(a_1+1)$ choices (exponents $0,1,2,\dots ,a_1$).
• For prime $p_2$, there are $(a_2+1)$ choices.
• ...
• For prime $p_k$, there are $(a_k+1)$ choices.

🔗 Total number of distinct factors:

Each choice of exponent is independent, so the total number of different factors is the product:

$$(a_1+1)(a_2+1)\dots (a_k+1)$$

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📜 Example to illustrate:

Let’s take $n=12$.

• Prime factorization: $12=2^2\times 3^1$.
• Number of factors:

$$(2+1)(1+1)=3\times 2=6$$

• The factors are: $1,2,3,4,6,12$ — 6 factors.

✔️ The formula holds perfectly.

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🚩 Conclusion:

The formula works because forming a divisor is about independently choosing each exponent from its possible range $0$ to $a_i$ for every prime factor.

Would you like me to visualize this with a tree diagram or table?

Proof that, ∀n∈ℕ, there are n consecutive composite numbers

Prove that, for every natural number n, there are n consecutive composite numbers.

Proof: For every natural number $n$, there are $n$ consecutive composite numbers.

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Step 1: Construct the sequence.

Consider the number:

$$N=(n+1)!$$

where $(n+1)!$ is the factorial of $n+1$, meaning:

$$(n+1)!=(n+1)\cdot n\cdot (n-1)\cdots 2\cdot 1$$

Step 2: Build the sequence of $n$ consecutive numbers:

$$N+2,N+3,N+4,\dots ,N+(n+1)$$

This is a sequence of $n$ consecutive numbers (from $N+2$ to $N+(n+1)$).

Step 3: Prove each number is composite.

Consider any number in the sequence:

$$N+k\text{where}k=2,3,4,\dots ,n+1$$

Note that $(n+1)!$ is divisible by $k$ because $k\le n+1$ and the factorial contains every integer up to $n+1$ as factors.

Therefore:

$$N+k=(n+1)!+k$$

is divisible by $k$ (since $(n+1)!$ is divisible by $k$).

Check whether this number is composite:

• It is greater than $k$ (because $(n+1)!\ge (n+1)\cdot n\cdot \cdots \cdot 2\cdot 1$, which is obviously larger than $k$ for all $n\ge 1$).
• It has $k$ as a divisor other than 1 and itself.

Thus, every number in the sequence is composite.

Step 4: Conclusion.

For every natural number $n$, there exists a sequence of $n$ consecutive composite numbers — specifically the numbers:

$$(n+1)!+2,(n+1)!+3,\dots ,(n+1)!+(n+1)$$

This proves the statement. $◼$

Bertrand postulate proof sketch

Prove that ∀n∈ℕ(n>1⟹∃p∈ℙ(n<p<2n)).

Below is a classical proof of

$$\mathrm{\forall }n\in \mathbb{N},n>1⟹\mathrm{\exists }p\in \mathbb{P}\text{ with }n<p<2n,$$

first conjectured by Bertrand in 1845 and proved (independently) by Chebyshev in 1852. It is often called Bertrand’s postulate.

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1. Chebyshev’s θ-function

Define

$$\theta (x)=\sum _{\begin{array}{c}p\le x\\ p\text{prime}\end{array}}\ln p.$$

Chebyshev showed by completely elementary means (bounding factorials via products) that there are absolute constants $A,B>0$ such that for all $x\ge 2$,

$$Ax\le \theta (x)\le Bx.$$

One may take for instance

$$0.92x\le \theta (x)\le 1.11x,$$

for every $x\ge 2$. (The proof is a straightforward—but somewhat lengthy—analysis of the prime‐power factors of $n!$.)

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2. From θ to “a prime in $(n,2n)$”

Observe that

$$\theta (2n)-\theta (n)=\sum _{n<p\le 2n}\ln p.$$

If there were no prime in the open interval $(n,2n]$, then $\theta (2n)-\theta (n)=0$. But by Chebyshev’s bounds,

$$\theta (2n)-\theta (n)\ge 0.92(2n)-1.11n=(1.84-1.11)n=0.73n>0(\mathrm{\forall }n>1).$$

Hence

$$\sum _{n<p\le 2n}\ln p>0,$$

so there must be at least one prime $p\in (n,2n]$. This completes the proof.∎

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Historical remarks

• Chebyshev’s original 1852 proof bounds factorials $n!\approx \sqrt{2\pi n}(n/e{)}^n$ from above and below elementarily, then translates those bounds into $Ax\le \theta (x)\le Bx$.
• In 1932, Erdős found a shorter “combinatorial” proof using properties of the central binomial coefficient $(\genfrac{}{}{0ex}{}{2n}{n})$.
• The statement is often referred to as Bertrand’s postulate, though it is now a theorem.