Ladylike women

What are some famous women who are known (among their other traits) for being ladylike, such that it would be unthinkable to imagine them being vulgar or indecent

Below is a deliberately eclectic (and certainly non-exhaustive) set of women whose public personas are almost synonymous with grace, composure, and a refusal to descend into crudity—so much so that it feels odd to picture them behaving otherwise. “Ladylike” is, of course, a culturally loaded word, but these figures are consistently cited as exemplars of good-humoured dignity rather than prim severity.

Timeless screen & stage icons

Woman	Why her name comes up when people say “elegance”	Representative note
Audrey Hepburn	Her pared-down style, gentle voice, and decades of UNICEF field work forged a reputation for quiet refinement both on and off camera.	(Viveura)
Grace Kelly (Princess Grace of Monaco)	The Hitchcock star turned princess introduced a cool, polished aesthetic that is still referenced in fashion textbooks under “the Kelly look.”	(Grace de Monaco)
Julie Andrews	From Mary Poppins to her SAG life-achievement speech, Andrews has been celebrated for a warm, reassuring presence that critics routinely call “regal” and “gracious.”	(Facebook)

Royal exemplars

Woman	Why she fits the brief	Representative note
Queen Elizabeth II	Seven decades of duty executed with unflappable composure made “Her Majesty” shorthand for poise under pressure.	(My Social Work News)
Diana, Princess of Wales	Her compassionate hospital visits and handshake with an AIDS patient reset royal protocol with empathy—never brusqueness.	(Medium)
Catherine, Princess of Wales	Continues the Windsor tradition of restraint; British papers routinely praise her for “poise and polish” during engagements.	(Facebook)

First Ladies & public servants

Woman	Why she’s on the list	Representative note
Jacqueline Kennedy Onassis	Her pillbox hats and French gowns were only half the story; colleagues recall an editor who wielded quiet authority without raising her voice.	(Antique Trader)
Michelle Obama	Mixes warmth with eloquence; even critics concede that she projects “strength, grace and resilience” while keeping discourse civil.	(REVOLT)

Contemporary professionals & activists

Woman	Why she’s cited	Representative note
Amal Clooney	The human-rights barrister’s courtroom gravitas and red-carpet minimalism have made her a modern style-and-substance archetype.	(W Magazine)
Emma Watson	Her poised UN speeches and intentionally polished fashion choices have framed her feminism within a context of deliberate civility.	(Facebook)

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A closing thought

What all of these women share is less a set of fashion rules than a habit of considering others—tempering wit, conviction, or even activism with courtesy. That impulse, rather than hemlines or pearl necklaces, is what makes vulgarity feel out of character for them.

A monad is just …

There’s something of an internet meme about repeating the exact definition “a monad is just a monoid in the category of endofunctors” (sometimes “just” is emphasized to highlight how simple it is). I’m not fully sure on the meme history, but I believe it is something like, these programming kids get into programming languages that have a “monad” construct, and they’ve never seen this in their other languages (which just have normal stuff like strings, ints, arrays, etc) so they ask what a monad is, and then suddenly they get this stuff from algebra (monoid is from algebra I think?) and category theory which they haven’t studied because, like, not exactly a prerequisite for being a programmer. And the people trying to explain it can’t think of any clearer way to put it, there’s just no better way to explain it (as far as the people who understand it know) so whenever the kid goes to ask someone else they always get the same cryptic definition. That sound right? Anyway, uh, what does the definition mean anyway?

Yeah, your “meme origin” sketch is basically right. In the Haskell / functional programming world (and more broadly among people who’ve seen the category-theoretic formulation), the phrase

“A monad is just a monoid in the category of endofunctors”

(or with the emphasized “just” as if to say “it’s not some mystical thing, it’s this one neat abstract fact”) gets repeated because it’s compact and elegant if you already know what all the words mean. To someone who hasn’t seen category theory, it sounds like being told “a car is just a four-wheeled combustion-powered device with a steering wheel in the category of road vehicles.” It’s accurate, but not immediately illuminating. So beginners ask “what’s a monad?” and get handed this line, which leads to the meme of the cryptic “insider answer” being recycled, often without the unpacking that makes it usable.

So: what does it mean?

Break it down:

1. Monoid (familiar algebraic idea)

A monoid is something with:

• A binary operation ⋆ that’s associative: (a ⋆ b) ⋆ c = a ⋆ (b ⋆ c).
• An identity element e such that e ⋆ x = x = x ⋆ e.

Example: strings under concatenation. Operation is ++, identity is the empty string.

2. Category

A category consists of:

• Objects (in programming, types like Int, Bool, etc.).
• Morphisms between objects (in programming, functions between types).
• Composition of morphisms that’s associative, and each object has an identity morphism.

The category most functional programmers implicitly work in is something like Hask: types as objects, pure functions as morphisms.

3. Functor and Endofunctor

A functor is a mapping between categories that preserves structure: it maps objects to objects and morphisms to morphisms in a way that respects composition and identities.

An endofunctor is a functor from a category to itself. In Hask, examples are type constructors like:

• Maybe (which adds “possible failure” context),
• [] (lists, i.e., nondeterminism or multiple results),
• IO (side effects),
• State s (stateful computations), etc.

Each of these takes a type a and yields a new type (e.g., Maybe a), and you can map a function over their contents (fmap).

4. Category of endofunctors

Now consider all endofunctors on Hask as objects in a new category. The morphisms between those are natural transformations (uniform ways of going from one functor to another). This whole setup itself is a category: the category of endofunctors.

This category has a built-in notion of “combining” endofunctors: functor composition. Given two endofunctors M and M again, you can compose them to get M ∘ M (e.g., Maybe (Maybe a)).

5. Monoid in that category

A monoid object in this category (with composition as the “multiplication”) is:

• An endofunctor M,
• A natural transformation η: Id ⇒ M (the identity functor injecting into M), called the unit,
• A natural transformation μ: M ∘ M ⇒ M (flattening two layers of M into one), called the multiplication,

such that the usual monoid laws hold, but now in the setting of natural transformations and functor composition:

• Associativity: flattening nested Ms in either grouping gives the same result (μ ∘ Mμ = μ ∘ μM),
• Unit laws: injecting then flattening, or flattening then injecting, is a no-op (μ ∘ Mη = id = μ ∘ ηM).

6. That’s a monad

That (M, η, μ) satisfying those laws is what category theory calls a monad. So a monad is literally a monoid in the category whose objects are endofunctors and whose “multiplication” is composition. Hence the pithy definition.

7. Programming translation

In programming (e.g., Haskell) people usually work with:

• return ≈ η (puts a pure value into the monadic context),
• join ≈ μ (flattens M (M a) into M a),
• bind (>>=) is a derived helper that sequences: given m :: M a and f :: a -> M b, m >>= f is join (fmap f m).

The monad laws in the familiar form:

• Left identity: return a >>= f ≡ f a
• Right identity: m >>= return ≡ m
• Associativity: (m >>= f) >>= g ≡ m >>= (\x -> f x >>= g)

These correspond exactly to the monoid laws on η and μ.

8. Intuition (why people often give other explanations)

Saying “a monad is a way to sequence computations that carry along some context” gives operational intuition: the context might be “possible failure” (Maybe), “multiple results” (List), “state threading” (State), etc. The abstract definition packages all such patterns uniformly: you get a “contextual computation” with a way to inject pure values and a way to compose/flatten nested contexts, obeying clean laws.

9. Example: Maybe

• η x = Just x (inject a value),
• μ (Just (Just x)) = Just x, μ (Just Nothing) = Nothing, μ Nothing = Nothing (flatten nested Maybes),
• bind handles failure automatically.

This satisfies the associativity and identity laws, so Maybe is a monad.

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So the meme is that the answer “a monad is just a monoid in the category of endofunctors” is technically satisfying and compact, but for learners it’s helpful to unpack it in terms of return/bind/join and concrete examples. Once you’ve internalized the structure, the categorical definition becomes a crisp way of seeing why all those disparate-looking patterns (IO, lists, exceptions, state) share the same abstract shape.

If you want, I can give several concrete examples side by side, derive bind from μ/fmap, or draw the commutative diagrams that encode the laws. Which angle would help most next?

This post is probably not as good as the StackOverflow answer.

Ernest Benn Reading List for Voters

"In the ideal state of affairs, no one would record a vote in an election until he or she had read the eleven volumes of Jeremy Bentham and the whole of the works of John Stuart Mill, Herbert Spencer and Bastiat as well as Morley's Life of Cobden." - Ernest Benn

This quote made me curious to check out those books, however, there is no eleven-volume collection of Bentham currently in print, so please make me an updated reading list based on this quote.

A Practical “Benn List” for Twenty-First-Century Voters

Below is a streamlined syllabus that captures the spirit of Ernest Benn’s tongue-in-cheek prescription—giving you the essential arguments each thinker wanted citizens to weigh before casting a ballot. For every author you’ll see:

• Core texts (★) – the minimum to grasp their programme.
• Deep cuts (◇) – where to go next if the core hooks you.
• Recommended modern edition / access point – in print or freely online.

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1. Jeremy Bentham (1748-1832) – founding utilitarian & legal reformer

Tier	Title	Why it matters	Modern access
★	An Introduction to the Principles of Morals and Legislation (1789)	Bentham’s utilitarian calculus & the “greatest happiness” rule.	Oxford World’s Classics re-issue (paper, 2020) (Amazon)
★	The Panopticon Writings	Shows his obsession with institutional design and surveillance.	Verso Radical Thinkers, 2nd ed. (2011/2024) (Verso)
◇	Rights, Representation & Reform: “Nonsense upon Stilts”…	His ferocious attack on natural-rights theory; key to modern rights debates.	Collected Works vol. 2, OUP (2002) (oxfordscholarlyeditions.com)
◇	Of Laws in General	A blueprint for a scientific jurisprudence.	Collected Works, Athlone/OUP (1995 reprint) (Amazon)
◇	For completists: the original 11-volume Bowring edition (1838-43) – digitised free at Liberty Fund. (Liberty Fund) + UCL’s Collected Works project (36 vols. so far, open access). (University College London)

Suggested path: read Introduction, then jump to “Nonsense upon Stilts” for the rights debate, and dip into the Panopticon letters for flavour.

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2. John Stuart Mill (1806-1873) – liberal utilitarian for the modern electorate

Tier	Title	Modern access
★	Quad-pack: On Liberty, Utilitarianism, Considerations on Representative Government, The Subjection of Women	All four in one Oxford World’s Classics volume (2nd ed., 2015) (Amazon)
★	Autobiography	Collected Works vol I, Liberty Fund (free) (Liberty Fund)
◇	Principles of Political Economy (abridged)	Hackett ed., 2004 / 2021 reprint (hackettpublishing.com)
◇	Collected Works (33 vols.) – key vols now in affordable paper (Liberty Fund) (Liberty Fund)

Suggested path: Read the quad-pack straight through; follow with the Autobiography to see Mill apply those principles to his own political life.

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3. Herbert Spencer (1820-1903) – evolution-driven individualism

Tier	Title	Modern access
★	Social Statics (1851)	Liberty Fund online facsimile (Liberty Fund)
★	The Man versus the State (1884)	Liberty Fund paperback (2020 re-run) (Liberty Fund)
◇	The Principles of Ethics (2 vols.)	Legare Street Press reprint, 2023 (Amazon)
◇	Education: Intellectual, Moral & Physical	Dover thrift ed., 2004 (Amazon)

Suggested path: Social Statics → Man versus State for the mature political attack on state overreach; dip into Education for his applied philosophy.

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4. Frédéric Bastiat (1801-1850) – the free-trade pamphleteer

Tier	Title	Modern access
★	The Law (1850)	in Liberty Fund’s Collected Works vol. 3 or FEE booklet (pocket size) (Liberty Fund)
★	Economic Sophisms	Liberty Fund “Collected Sophisms” (2018 digital) (Liberty Fund)
◇	Selected Essays on Political Economy	FEE trade pbk (1995) (Amazon)
◇	Harmonies of Political Economy	Legare Street Press reprint, 2023 (Amazon)
◇	Full 6-volume Collected Works (Liberty Fund, ongoing). (Liberty Fund)

Suggested path: Start with The Law (30 pages) for a lightning-bolt intro; then the Sophisms essays to watch him demolish protectionism.

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5. John Morley’s Life of Richard Cobden (1867/79) – context for free-trade politics

• Core: 2-volume Cambridge Library Collection reprint (2011); or free Liberty Fund HTML/PDF. (Amazon, Liberty Fund)
• Supplement: Letters of Richard Cobden digital archive (2023) for primary sources. (The History of Parliament)

Read alongside Bastiat to see how free-trade arguments played out on the floor of Parliament.

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How to Tackle the List

1. Short-to-long spiral – begin with Bastiat’s pamphlets (quick wins), then Mill’s On Liberty, before diving into Bentham or Spencer.
2. Mix authors – alternate utilitarians (Bentham/Mill) with free-market polemic (Bastiat/Cobden) to keep ideas fresh.
3. Allowance for time – the ★ items total ~2,400 pages—doable in a year at 50 pp/week. Deeper cuts add as curiosity allows.

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Free & open resources

Almost everything above can be read gratis via Liberty Fund’s Online Library of Liberty, Econlib, or archive.org; the recommended print editions give you introductions, annotations, and pagination useful for citation.

Happy (and electorally responsible) reading!

Alice Liability for Tree

Suppose there are two plots of land, one owned by Alice and one owned by Bob. Bob's property has a fence around it, which is also owned by Bob. Alice's property has some trees, which are fully on Alice's side. However, lightning strikes and brings one tree down. This makes it fall over and break part of Bob's fence. In what kinds of jurisdictions is Alice liable for this?

The short answer

Outside a few narrow, mostly academic possibilities, Alice would not be liable when a healthy tree is toppled by lightning and damages Bob’s fence. In almost every modern system—common‑law and civil‑law alike—the combination of a healthy tree and a truly unforeseeable, irresistible natural event (“an act of God” or force majeure) breaks the chain of legal causation or supplies Alice with a complete defence.

Below is a map of how the major families of legal systems treat the problem and where (in theory) Alice might still face liability.

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1. Fault‑based common‑law jurisdictions

(England & Wales, nearly all U.S. states, Canada, Australia, etc.)

Core rule	Practical result for Alice
A land‑owner owes only a duty of reasonable care to inspect and maintain trees. Liability arises only if the owner knew or should have known the tree was dangerous and failed to act. A sudden act of nature is a complete defence (“act‑of‑God doctrine”).	Because the tree was healthy and the damage was caused by lightning, Alice has no fault; Bob must claim on his own insurance. (Merchants Insurance Group, forthepeople.com)

When could she be liable here? If the tree had been visibly diseased, leaning, or previously struck (i.e., Alice had notice of a hazard) courts apply ordinary negligence. Some states spell this out explicitly—e.g., the Restatement illustration where a lightning‑killed tree is left standing for a year and then falls (koleyjessen.com).

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2. Strict‑liability civil‑code jurisdictions

(France, Québec, Spain, pre‑1996 Louisiana, etc.)

2 a. Liability for “things in one’s custody”

Article 1242 §1 of the French Code (mirrored in many civilian systems) and its Louisiana analogue (old C.C. art. 2317) impose objective liability for damage caused by a “thing” one guards—even without personal fault. Louisiana’s Loescher v. Parr applied that idea to diseased trees. (Justia Law)

2 b. The escape hatch: force majeure

The same civil codes also say that a guardian is exonerated if the damage is due to an event that is (i) external, (ii) unforeseeable, and (iii) irresistible—the civilian version of an act of God. Lightning satisfies all three criteria in normal conditions, so the owner escapes liability. (Wikipedia)

Scenario	Healthy tree ‑ lightning	Diseased tree ‑ ordinary wind
Outcome	Owner not liable (force majeure)	Owner usually liable (defect created an unreasonable risk and wind was foreseeable)

Bottom line: even the “strict” civilian systems still free Alice when the only cause is lightning.

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3. Exceptional regimes that could, in theory, trap Alice

1. Statutory no‑fault compensation schemes – A few jurisdictions experiment with laws that spread losses from natural hazards without regard to fault (e.g., some Scandinavian environmental‑risk statutes). They are rare and usually capped by government funds, not true tort liability.
2. Contractual or insurance subrogation – Alice could be pulled in only if her own insurer indemnifies Bob and then seeks reimbursement, but the insurer would still have to prove negligence or the absence of a force‑majeure defence under the governing law.

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4. Practical takeaway

• In mainstream common‑law or civil‑law jurisdictions, a healthy tree felled by lightning is classified as an “act of God” / force majeure event.
• Liability shifts only when the owner’s fault (notice of decay, improper felling, etc.) contributes to the damage.
• Truly strict regimes with no force‑majeure escape are practically non‑existent; where strict liability exists, the same codes also enshrine the force‑majeure defence.

So, unless Alice ignored warning signs that the tree was dangerous before the storm, she can sleep soundly in almost any legal system.

Enslaved pilot dilemma

I asked ChatGPT to think about a moral dilemma about an enslaved pilot from different perspectives. The answers were pretty much determined by the conclusion it wanted to give, but not too bad for all that.

Vector spaces under multiplication (o4-mini-high)

What are vector spaces under multiplication?

In the usual definition, a vector space $V$ over a field $K$ consists of

1. An abelian group $(V,+)$ (the “vector-addition”), with identity element $0$.

2. A scalar multiplication $K\times V\to V$, satisfying

   • $1\cdot v=v$,
   • $(\alpha \beta )\cdot v=\alpha \cdot (\beta \cdot v)$,
   • $\alpha \cdot (u+v)=\alpha \cdot u+\alpha \cdot v$,
   • $(\alpha +\beta )\cdot v=\alpha \cdot v+\beta \cdot v$.

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1. “Multiplicative” notation for vector spaces

Nothing deep changes if you write your abelian group multiplicatively instead of additively. Concretely, you replace

• the additive identity $0$ by a multiplicative identity, call it $1_V$,
• the “sum” $u+v$ by a product $u\cdot v$,
• the inverse $-v$ by the group-inverse $v^{-1}$.

Then the scalar axioms become:

1. Unit:

   $$1_K\cdot v=v.$$

2. Compatibility of scalars:

   $$(\alpha \beta )\cdot v=\alpha \cdot (\beta \cdot v).$$

3. Distributes over the group product:

   $$\alpha \cdot (u\cdot v)=(\alpha \cdot u)\cdot (\alpha \cdot v).$$

4. Respects field-addition via the group product:

   $$(\alpha +\beta )\cdot v=(\alpha \cdot v)\cdot (\beta \cdot v).$$

Terminology. A set $V$ equipped with a (written-multiplicative) abelian group structure and a field-action satisfying the four rules above is sometimes called a multiplicative or exponential vector space over $K$. But it is completely equivalent (indeed isomorphic) to an ordinary (additive) vector space: you just rewrite $u\cdot v$ as “$u+v$” and $1_V$ as “0.”

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1.1 Example: ${\mathbb{R}}_{>0}$

• Take $V=\{x>0\}$ with multiplication as the group law.

• For any real scalar $t$, define

  $$t\cdot x=x^t.$$

  Then one checks:

  • $1\cdot x=x^1=x$.
  • $(s+t)\cdot x=x^{s+t}=x^s\cdot x^t=(s\cdot x)(t\cdot x)$.
  • $s\cdot (xy)=(xy{)}^s=x^s{y}^s=(s\cdot x)(s\cdot y)$.
  • $(st)\cdot x=x^{st}=s\cdot (x^t)=s\cdot (t\cdot x)$.

Thus $({\mathbb{R}}_{>0},\times )$ with exponentiation is a vector space over $\mathbb{R}$, written multiplicatively; under the isomorphism $x\mapsto \ln x$ it becomes the usual additive space $(\mathbb{R},+)$.

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2. Vector spaces “with” multiplication of vectors ≠

Sometimes one asks: “What if I try to multiply two vectors to get another vector?” That additional operation (bilinear in the vectors) turns your space into an algebra rather than a “vector space under multiplication.” In other words:

• A plain vector space only has

  1. an addition of vectors, and
  2. a scalar multiplication by field elements.

• A (unital) algebra over $K$ is a vector space $V$ plus a map

  $$V\times V⟶V,(u,v)\mapsto uv,$$

  which is bilinear:

  $$(\alpha u)v=\alpha (uv),u(\alpha v)=\alpha (uv),(u+u^{\prime })v=uv+u^{\prime }v,\dots$$

  Examples include:

  • The space of $n\times n$ matrices with ordinary matrix‐multiplication,
  • Polynomial rings $K[x]$ with polynomial multiplication,
  • Any group algebra, Clifford algebra, etc.

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Summary

• If by “vector spaces under multiplication” you mean writing the abelian group law multiplicatively, you get the same notion of vector space, with scalar action behaving like exponentiation (e.g.\ ${\mathbb{R}}_{>0}$ with $x^t$).
• If you mean equipping a vector space with a new “vector–vector” product, you’re describing an algebra over the field, which is a strictly richer structure.

Either argument establishes that the set of all finite sequences of natural numbers is countable.

Note: ChatGPT’s uses of \R, which rendered correctly in its own UI, were changed into \mathbb{R} here.

Vector spaces under multiplication (4o)

What are vector spaces under multiplication?

The concept of vector spaces under multiplication is a common source of confusion because vector spaces are not defined under multiplication of vectors. Let me clarify:

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📘 What Is a Vector Space?

A vector space over a field (like the real numbers ℝ) is a set equipped with:

1. Vector addition: Combining two vectors to get another vector.
2. Scalar multiplication: Multiplying a vector by a scalar (from the field) to get another vector.

These operations must satisfy eight axioms (closure, associativity, identity elements, distributivity, etc.).

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❌ Multiplication of Vectors? Not a Vector Space Operation

• Vector multiplication (like dot product, cross product, or element-wise multiplication) is not part of the vector space definition.
• These products don't satisfy the vector space axioms when used as the main operation.

For example:

• The dot product takes two vectors and gives a scalar — not another vector.
• The cross product only exists in ℝ³ and isn't associative or defined for general vector spaces.
• Component-wise multiplication doesn't preserve closure in all cases and doesn't satisfy vector space axioms.

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✅ Algebraic Structures with Multiplication

If you're thinking about combining vector-like objects with multiplication, you're likely looking at one of these more advanced structures:

Structure Type	Description
Algebra	A vector space with a bilinear multiplication operation (e.g., matrix algebra).
Ring	A set with two operations (addition and multiplication) but not necessarily scalar multiplication.
Field	A ring where every nonzero element has a multiplicative inverse.
Module	A generalization of vector spaces over rings instead of fields.

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🔍 Example: Matrix Algebra

Matrices form a vector space under addition and scalar multiplication, and also support multiplication (matrix multiplication), which makes them an algebra.

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Summary

There is no such thing as a "vector space under multiplication" if you mean multiplication of vectors. Vector spaces are defined with addition and scalar multiplication only. If you're exploring structures where multiplication of elements is involved, you're likely referring to algebras or modules, not vector spaces in the strict sense.

Let me know if you meant something more specific!

Insurance company profit example

Give a simple numerical example of what an insurance company does to turn a profit

Here’s a very simple toy example showing how an insurance company makes money:

Assumptions

• You sell 1,000 identical policies.
• Each policyholder pays a $100 annual premium.
• The insurer estimates a 5% chance any one policy will result in a claim, and that the average claim size is $1,000.
• The company has other operating costs (salaries, marketing, overhead) equal to 20% of total premiums.

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1. Premium revenue

   $$\mathrm{1,000}\text{policies}\times \mathrm{\$}100=\mathrm{\$}\mathrm{100,000}$$

2. Expected claims cost

   $$\mathrm{1,000}\text{policies}\times 5\mathrm{\%}\times \mathrm{\$}\mathrm{1,000}=\mathrm{\$}\mathrm{50,000}$$

3. Operating expenses

   $$20\mathrm{\%}\times \mathrm{\$}\mathrm{100,000}=\mathrm{\$}\mathrm{20,000}$$

4. Profit

   $$ \underbrace{$100{,}000}{\text{Premiums}} - \underbrace{$50{,}000}{\text{Claims}} - \underbrace{$20{,}000}_{\text{Expenses}} = $30{,}000 $$

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• Break‑even “loss ratio” here would be claims / premiums = 50%, and expense ratio = 20%.
• Together they give a combined ratio of 70% (50% + 20%), leaving 30% as profit margin.

In reality, insurers use actuarial models to set premiums so that, over many policies and years, the combined ratio stays below 100%—that gap is their profit.

Nonprime N Where All Groups of Order N Are Isomorphic

Is there nonprime n where all groups of order n are isomorphic?

Yes—there are plenty of composite (non‑prime) integers $n$ for which every group of order $n$ is forced to be the same (indeed, cyclic). The smallest example is

$$n=15=3\cdot 5.$$

Because $\mathrm{Aut}(C_5)\cong C_4$ has no element of order 3, no non‑trivial semidirect product $C_5⋊C_3$ can be formed, so the only group of order 15 is the cyclic one $C_{15}$.

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The complete criterion

Let

$$n=p_1{p}_2\cdots p_k(p_1<p_2<\cdots <p_k)$$

be the prime–factorisation of $n$.

All groups of order $n$ are isomorphic iff

1. $n$ is square‑free (no $p_i^2\mid n$), and

2. for every pair of primes with $p_i<p_j$ one has

   $$p_i\nmid (p_j-1).$$

Why these two conditions are necessary.

• If a prime square $p^2$ divides $n$, the abelian groups $C_{p^2}$ and $C_p\times C_p$ give two non‑isomorphic examples.
• If $p_i\mid (p_j-1)$ for some $i<j$, there exists a non‑trivial action $C_{p_i}\to \mathrm{Aut}(C_{p_j})$, producing the non‑abelian semidirect product $C_{p_j}⋊C_{p_i}$ alongside the cyclic one.

Why they are sufficient. With $n$ square‑free, every Sylow subgroup is cyclic. Condition (2) forces each Sylow subgroup to be normal (Sylow counts give $n_{p_i}\equiv 1(\mathrm{mod}{p}_i)$ and $n_{p_i}\mid n/p_i$; if $p_i$ divided some $p_j-1$ we could have a non‑normal Sylow). A direct product of mutually normal cyclic Sylow subgroups is cyclic, so every group of that order is $C_n$.

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Infinitely many such $n$

Choose any set of distinct primes with no divisibility relations of the form $p_i\mid (p_j-1)$. Products of two primes are easiest:

$n=pq$	Is every group cyclic?	Reason
$15=3\cdot 5$	Yes	$3\nmid 4$
$33=3\cdot 11$	Yes	$3\nmid 10$
$35=5\cdot 7$	Yes	$5\nmid 6$
$39=3\cdot 13$	Yes	$3\nmid 12$

Whenever $p<q$ and $p\nmid (q-1)$, there is only the cyclic group of order $pq$; thus there are infinitely many composite $n$ with the desired property.

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Summary

So the answer is yes: the composite integers satisfying the two conditions above (beginning with 15) have exactly one group structure—namely the cyclic group $C_n$.

What are Seven-Ten lists?

In the ada palmer universe, what are the "seven ten lists"?

In Terra Ignota—the 25th-century setting of Ada Palmer’s quartet—“Seven-Ten Lists” are eagerly awaited power-rankings published by the world’s most influential news outlets. Each list names ten people who matter right now, but with a twist:

Slots	What they signal
1 – 7 (“the Seven”)	A sober, quasi-statistical judgment about who really controls the fate of the planet—usually the heads of the great Hives or other actors whose combined clout could tip the world into war or keep it at peace.
8 – 10 (“the Ten”)	Lower-stakes kudos for rising celebrities, cultural icons, or outlier talents—people whose fame matters, but whose influence is not (yet) civilization-breaking.

Think of the Seven-Ten as a hybrid between Time’s “Person of the Year,” a Wall Street index, and a bookmakers’ odds sheet: placement on the list instantly sways markets, public confidence, and political maneuvering. (terra-ignota.fandom.com)

Why the name “Seven-Ten” if there are only ten names?

The first seven are what everyone argues over; the extra three are almost garnish. Hence “Seven-Ten” became shorthand, even though the document itself is only ten lines long.

How often and by whom?

Major newspapers or broadcast syndicates in each Hive issue their own list—normally once a year, but special editions appear whenever the political temperature spikes. The Black Sakura paper in Tokyo, for example, is famous for its list. (terra-ignota.fandom.com)

Plot weight inside the novels

The theft and replacement of a Black Sakura Seven-Ten List is the fuse that ignites the Days of Transformation chronicled in Too Like the Lightning and Seven Surrenders; once an altered list circulates, rivals scramble to confirm or refute the rankings, alliances wobble, and the utopian façade cracks. (Jayde Holmes)

In short, the Seven-Ten Lists are the series’ canary in the geopolitical coal-mine: a deceptively simple top-ten chart whose shifting names warn both characters and readers that the balance of world power is about to lurch.

Reading list on the formal mathematical study of board games

What are some of the best books that try to study board games in a formal mathematical way?

Below is a compact reading‑map that most researchers and advanced hobbyists follow when they want to treat board games as mathematics rather than mere recreation. I’ve grouped the titles by the role they usually play in a self‑study program and noted what each one actually does for you.

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1. Bedrock texts — where the modern theory began

Title	Why it matters	Typical level
On Numbers and Games — John H. Conway (2nd ed., 2001)	Introduces surreal numbers and the algebra that underlies all combinatorial game theory (CGT). A thin but dense book that defines the subject’s vocabulary. (Amazon)	Graduate‑level abstract algebra / set theory
Winning Ways for Your Mathematical Plays, Vols. 1 – 4 — Berlekamp, Conway, & Guy (2nd ed. 2018–2024)	The classic encyclopedia: impartial games (Nim‑type), partizan games, misère play, thermographs, hot/cold values, and dozens of concrete board‑game case studies. Readable once you’ve sampled ONAG. (Amazon, Amazon)	Upper‑undergraduate to research

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2. Friendly introductions / textbooks

Title	What you’ll get	Level
Lessons in Play, 2nd ed. — Albert, Nowakowski & Wolfe (2019)	A pedagogical walk from Nim to CGT algebra with hundreds of exercises and fully worked examples. Great first book if Winning Ways feels too brisk. (Routledge)	Advanced undergrad
Combinatorial Game Theory — Aaron N. Siegel (AMS GSM 146, 2023)	A modern, theorem‑proof treatment that unifies the field’s last 30 years and cleans up notation. Becomes the standard reference once you want full rigor. (Amazon)	Graduate text

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3. Research‑level compilations

• Games of No Chance series (MSRI Publ. Vols. 29, 42, 56, 63, & 90; vol. 5 appeared 2019) — edited collections of cutting‑edge papers on CGT structure, new game values, and solved board‑game endgames. (Cambridge University Press & Assessment)

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4. Monographs that dissect specific board games

Game	Book	Highlight
Go	Mathematical Go: Chilling Gets the Last Point — Berlekamp & Wolfe (aka Mathematical Go Endgames)	Uses CGT thermography to solve real professional endgames and introduces “chilling” as a cooling technique. (Amazon, Amazon)
Dots‑and‑Boxes	The Dots and Boxes Game: Sophisticated Child’s Play — Berlekamp (2000)	Turns a playground pastime into a solved impartial game; shows how to read the nim‑value of any position. (Amazon)
(Many other titles exist for Hex, Domineering, Amazons, etc., but the two above are the gold‑standards for single‑game depth.)

---

5. Algorithmic & complexity‑theoretic perspectives

Resource	What it covers	Use‑case
Games, Puzzles, and Computation — Hearn & Demaine (2009)	Reduces dozens of real board games (Rush Hour, Reversi, generalized Chess/Go) to complexity‑class hard problems via Constraint Logic. (Amazon)	When you care about NP/PSPACE/EXPTIME results
“Algorithmic Combinatorial Game Theory” survey — Demaine & Hearn (2001; updated in Games of No Chance 3)	A readable overview of algorithmic techniques and open problems. (erikdemaine.org)	Quick orientation before diving into proofs
Algorithmic Game Theory (SAGT 2024 Proceedings)	Latest papers on computational aspects, including hardness proofs for new tabletop titles. (Amazon)	Keeping current with 2020‑s research

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How to read them in practice

1. Start light: skim a chapter of Lessons in Play to see the main definitions and the Sprague‑Grundy theorem at work.

2. Build foundations: dip into On Numbers and Games for the surreal‑number construction, then sample the early chapters of Winning Ways to see those ideas applied.

3. Choose a specialty:

   • If you enjoy solving concrete board games, grab the relevant monograph (Go, Dots‑and‑Boxes, etc.).
   • If you’re pulled toward algorithms or CS theory, pivot to Hearn & Demaine.

4. Circle back to Siegel’s textbook once you want complete proofs of structural results such as the Atomic‑Weight Lemma or canonical forms.

5. Stay current by browsing the Games of No Chance volumes or SAGT proceedings whenever a new one appears (roughly every 3‑4 years).

Taken together, these works will carry you from the first Nim pile all the way to the current research frontier. Happy reading—and happy playing!

Singularity

In calculus, what is a singularity?

In calculus (and more broadly in mathematical analysis), a singularity of a function is a point at which the function ceases to be “well‑behaved” in some sense—typically because it is not defined there or its value (or derivative) “blows up” to infinity. More formally:

• Definition: A point $a$ is called a singularity of a real‑ or complex‑valued function $f(x)$ if $f$ is not analytic (i.e.\ not representable by a convergent power series) in any neighborhood of $a$.

  • In elementary (real) calculus this usually means $f$ is not defined at $a$, or $\underset{x\to a}{lim}f(x)$ does not exist as a finite number.

---

Types of Singularities (Real Calculus)

1. Removable singularity − The limit $\underset{x\to a}{lim}f(x)$ exists (finite), but $f(a)$ is either undefined or defined to be something else. − Example:

   $$f(x)=\frac{\sin x}{x}\text{has}\underset{x\to 0}{lim}\frac{\sin x}{x}=1,$$

   but $f(0)$ is undefined. We “remove” the singularity by defining $f(0)=1$.

2. Infinite (or pole‑type) singularity − As $x\to a$, $|f(x)|\to \mathrm{\infty }$. − Example:

   $$f(x)=\frac{1}{x}\text{blows up at}x=0,$$

   so $x=0$ is an infinite singularity (often called a “pole” in complex analysis).

3. Jump (or essential discontinuity) − The one‑sided limits $\underset{x\to a^{+}}{lim}f(x)$ and $\underset{x\to a^{-}}{lim}f(x)$ both exist but are unequal (so the two‑sided limit doesn’t exist). − Example:

   $$f(x)=\{\begin{array}{ll}1,& x<0,\\ 2,& x\ge 0,\end{array}$$

   has a jump at $x=0$.

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Singularities in Complex Analysis

When you extend to complex functions $f(z)$, singularities are classified more finely:

• Removable singularity: $\underset{z\to a}{lim}f(z)$ exists; you can redefine $f(a)$ to make it analytic.
• Pole: $|f(z)|\to \mathrm{\infty }$ as $z\to a$. A pole of order $n$ means $f(z)\sim (z-a{)}^{-n}$ near $a$.
• Essential singularity: Neither removable nor pole; $f(z)$ exhibits wild oscillation (Casorati–Weierstrass theorem).

---

Why Singularities Matter

• Asymptotic behavior: Vertical asymptotes in graphs arise from infinite singularities.
• Integration: When performing improper integrals, singularities determine convergence or divergence.
• Series expansions: Analytic (power‑series) techniques break down at singularities; the distance to the nearest singularity governs a series’ radius of convergence.

---

Key takeaway: A singularity is simply a point where a function “fails” to be nice—typically by being undefined or unbounded there. Understanding its type (removable, infinite/pole, jump or essential) tells you how you might “fix” it, how the graph behaves near it, and how analytic methods apply around it.

If g and f are injections, so is g ∘ f, and same with bijections

Show that if $f:A\to B$ and $g:B\to C$ are injections, so is $g\circ f:A\to C$.

Recall that a function $h:X\to Y$ is injective (one‐to‐one) precisely when

$$h(x_1)=h(x_2)⟹x_1=x_2.$$

---

Theorem. If

$$f:A⟶B\text{and}g:B⟶C$$

are both injective, then their composite

$$g\circ f:A⟶C$$

is injective.

---

Proof. Let $x_1,x_2\in A$ and suppose

$$(g\circ f)(x_1)=(g\circ f)(x_2).$$

By definition of composition, this means

$$g(f(x_1))=g(f(x_2)).$$

Since $g$ is injective, we conclude

$$f(x_1)=f(x_2).$$

But $f$ is injective, so from $f(x_1)=f(x_2)$ it follows that

$$x_1=x_2.$$

Thus

$$(g\circ f)(x_1)=(g\circ f)(x_2)⟹x_1=x_2,$$

showing $g\circ f$ is injective. $\boxed{{\displaystyle }}$

---

Alternative (via contrapositive). One can also argue:

• If $x_1\ne x_2$, then since $f$ is injective, $f(x_1)\ne f(x_2)$.
• Then since $g$ is injective, $g(f(x_1))\ne g(f(x_2))$.
• Hence $(g\circ f)(x_1)\ne (g\circ f)(x_2)$.

This directly shows “distinct inputs give distinct outputs,” i.e.\ $g\circ f$ is one‐to‐one.

Separate conversation with o4-mini:

Show that if $f:A\to B$ and $g:B\to C$ are bijections, so is $g\circ f:A\to C$.

Proof. Suppose

$$f:A\to B\text{and}g:B\to C$$

are both bijections. We must show that $g\circ f:A\to C$ is bijective, i.e. both injective and surjective.

---

1. Injectivity

Take any $a,a^{\prime }\in A$ and assume

$$(g\circ f)(a)=(g\circ f)(a^{\prime }).$$

Then

$$g(f(a))=g(f(a^{\prime })).$$

Since $g$ is injective, it follows that

$$f(a)=f(a^{\prime }).$$

But $f$ is injective, so from $f(a)=f(a^{\prime })$ we conclude

$$a=a^{\prime }.$$

Hence $g\circ f$ is injective.

---

2. Surjectivity

Let $c\in C$ be arbitrary. Since $g$ is surjective, there exists some

$$b\in B\text{such that}g(b)=c.$$

Next, because $f$ is surjective, there exists

$$a\in A\text{such that}f(a)=b.$$

Therefore

$$(g\circ f)(a)=g(f(a))=g(b)=c.$$

Since $c$ was arbitrary in $C$, this shows $g\circ f$ is surjective.

---

3. Conclusion

Having established that $g\circ f$ is both injective and surjective, we conclude that

$$g\circ f:A⟶C$$

is a bijection. Moreover, one can check directly that its inverse is

$$(g\circ f{)}^{-1}=f^{-1}\circ g^{-1},$$

providing an explicit bijective inverse.

No set of all ordered pairs, or of all functions

Explain this: Where an ordered pair (x, y) is a set $\{x,\{x,y\}\}$, the class of all ordered pairs is not a set.

In ZF (and related) set‐theories, one often defines the ordered pair $(x,y)$ à la Kuratowski by

$$(x,y)=\{x,\{x,y\}\}.$$

Let’s call

$$P=\{w:\mathrm{\exists }x,y[w=\{x,\{x,y\}\}]\}$$

the class of all such ordered pairs. We claim $P$ cannot itself be a set but must be a proper class. Here’s the standard argument:

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1. Suppose for contradiction that $P$ is a set.

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2. Define the first‐projection map

$${\pi }_1(\{x,\{x,y\}\})=x.$$

Because ${\pi }_1$ is a definable “function” on ordered pairs, the Axiom Schema of Replacement tells us that

$${\pi }_1[P]=\{x:\mathrm{\exists }y((x,y)\in P)\}$$

must also be a set.

---

3. But ${\pi }_1[P]$ is in fact the universe of all sets.

• For every set $x$, just take $y=x$. Then $(x,x)$ is an ordered pair, so belongs to $P$.
• Hence every $x$ appears as the first coordinate of something in $P$.

Thus

$${\pi }_1[P]=\{x:x\text{ is a set}\}=V,$$

the class of all sets.

---

4. But in ZF the universe $V$ is not a set.

If $V$ were a set, one would immediately violate Cantor’s theorem (e.g.\ its power set $\mathcal{P}(V)$ would have strictly larger cardinality than $V$ itself), or more directly it contradicts the axiom of foundation and the cumulative‐hierarchy conception. In any case, the theory forbids “the set of all sets.”

---

5. Conclusion

Since assuming $P$ is a set forces $V$ to be a set (via Replacement), which is impossible, we must reject the assumption. Therefore

The class of all Kuratowski‐ordered pairs is too large to be a set; it is a proper class.

---

Alternative (union) argument

One can also observe that

$$\bigcup (\bigcup P)=\bigcup \{x,\{x,y\}:x,y\in V\}=V.$$

If $P$ were a set then so would be its double–union $\bigcup (\bigcup P)$, again forcing $V$ to be a set. Same contradiction.

A set f is called a function if all its elements are ordered pairs and for each set a there is at most one set b such that that $(a,b)\in f$. Show that the class of all functions is not a set.

Here is a direct “size‐too‐large” argument showing the class

$$\mathcal{F}=\{f:f\text{ is a function (i.e. a set of ordered pairs with unique first–coordinates)}\}$$

cannot itself be a set in ZF.

---

1. For each set $a$, cook up a tiny function $f_a$

Define

$$f_a=\{(a,a)\}.$$

Clearly:

1. $(a,a)$ is an ordered pair (e.g.\ Kuratowski’s $\{a,\{a,a\}\}$),
2. $f_a$ is a set of ordered pairs,
3. no two pairs in $f_a$ share the same first coordinate. Hence $f_a\in \mathcal{F}$.

---

2. If $\mathcal{F}$ were a set, then so would be $\bigcup \bigcup \mathcal{F}$

• By the Axiom of Union, if $\mathcal{F}$ is a set then ${\displaystyle U:=\bigcup \mathcal{F}}$ is a set (the union of all functions).
• Then again by Union, ${\displaystyle \bigcup U=\bigcup (\bigcup \mathcal{F})}$ is also a set.

---

3. But $\bigcup (\bigcup \mathcal{F})=V$, the universe of all sets

1. First, $U=\bigcup \mathcal{F}$ contains every ordered‐pair component: for each $f_a$ we have $(a,a)\in f_a\subset U$, so

   $$U\supset \{a,\{a,a\}\}=\{a,\{a\}\}.$$

2. Now

   $$\bigcup U=\bigcup \{\{a,\{a\}\}:a\in V\}=\{a:a\in V\}\cup \{x:x\in \{a\}\text{ for some }a\in V\}=V.$$

   Concretely, every set $a$ appears among the first‐coordinates, and every set $x$ appears as the unique member of $\{a\}$ for some $a$.

Hence if $\mathcal{F}$ were a set, we would have built the set $\bigcup (\bigcup \mathcal{F})$, which must equal $V$. But ZF forbids $V$ from being a set (one way to see this is Cantor’s theorem: $\mathcal{P}(V)$ would strictly exceed $V$ in size).

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4. Conclusion

Because assuming $\mathcal{F}$ is a set forces $V$ to be a set, we reach a contradiction. Therefore

The class of all functions is not a set; it is a proper class.

List of countries with a state church

List of countries with a state church

As of 2025, the following countries officially have a state church—a religious organization recognized and supported by the government as the official religion:

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European Countries

1. England (United Kingdom) – Church of England (Anglican)

   • The monarch is the Supreme Governor of the Church.

2. Denmark – Evangelical Lutheran Church of Denmark

   • The state funds the church, and the monarch must be a member.

3. Greece – Eastern Orthodox Church of Christ

   • Closely tied to national identity; protected by the constitution.

4. Norway – Church of Norway (Evangelical Lutheran)

   • Still a state church, although reforms in 2012 reduced direct government control.

5. Iceland – Evangelical Lutheran Church of Iceland

   • Official state religion; receives state funding and support.

6. Liechtenstein – Roman Catholic Church

   • Constitutionally recognized as the state church.

7. Malta – Roman Catholic Church

   • Constitutionally recognized and prominent in public life.

8. Monaco – Roman Catholic Church

   • State religion according to the constitution.

9. Vatican City – Roman Catholic Church

   • Theocratic absolute monarchy; the Pope is both head of state and church.

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Other Regions

10. Argentina – While not strictly a state church, the constitution recognizes the Roman Catholic Church and supports it financially.

11. Costa Rica – Roman Catholic Church

• Recognized as the state religion in the constitution.

12. Zambia – Declares itself a Christian nation (1996 constitution), but without an official church.

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Note: Several other countries (like Finland, Sweden, or Germany) have churches with historical ties to the state but no longer have official state churches, or their status has been redefined (e.g., state-supported but independent).

Would you like a list of countries with privileged churches but not formal state churches as well?

Yes

Here is a list of countries with privileged or historically established churches that are not officially state churches, but still enjoy special legal, financial, or cultural status:

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🇪🇺 Europe

Finland

• Evangelical Lutheran Church of Finland and Finnish Orthodox Church

  • Both are recognized in law and receive tax funding.
  • Not considered state churches since the 2000 constitutional reform.

Sweden

• Church of Sweden (Lutheran)

  • Former state church until 2000.
  • Still receives tax funding and plays a public ceremonial role.

Germany

• Multiple churches (primarily Catholic and Evangelical Lutheran) receive funding via the church tax system.

  • Not a state church, but recognized under public law.
  • Government collects taxes on behalf of the churches.

Scotland (UK)

• Church of Scotland (Presbyterian) is the "national church" but is not a state church.

  • It is independent of the state, unlike the Church of England.

Switzerland

• No state church at the federal level.

  • Several cantons recognize and support Catholic, Protestant, and Old Catholic churches.

Austria

• The Roman Catholic Church and other religious communities are officially recognized.

  • The state cooperates with them and may fund religious education and activities.

Spain

• Constitutionally secular, but Catholicism enjoys significant state cooperation and funding.

  • Agreements with the Holy See provide privileges.

Italy

• Constitutionally secular since 1984 (end of Lateran Treaty status).

  • The Catholic Church still receives public funds through the “otto per mille” tax designation.

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🌍 Rest of the World

Israel

• No state religion, but Orthodox Judaism is heavily privileged.

  • The state funds religious institutions, and Orthodox authorities regulate many aspects of personal law (e.g., marriage, burial).

Jordan, Morocco

• Islam is the state religion, but the monarchs claim religious legitimacy, and specific branches of Islam are privileged.

Russia

• Officially secular.

  • The Russian Orthodox Church enjoys a favored status and close ties to the government, with substantial influence in politics and education.

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These countries exhibit a range of arrangements, from financial privileges and legal recognition to influence over public ceremonies or education, even without an officially established state religion.

Let me know if you'd like a categorized table or visual map of these distinctions.

Rotation, reflection, orientation

Is this true:
“Suppose that a line is chosen from within a plane, and the plane is rotated 180° clockwise around this line. This will mirror the plane, reflecting its points such that the points leftward from the line become rightward from it and vice-versa. But suppose that the rotation is done counter-clockwise instead. Then the same thing happens, and there is no difference.”

Yes – it’s absolutely true. Here’s why:

1. A 180° rotation about a line in the plane acts like a mirror reflection across that line.

   • Take for example the horizontal $x$–axis in the $xy$–plane. If you rotate the plane 180° about that axis (in three‑dimensional space), any point $(x,y,0)$ goes to $(x,-y,0)$. That is exactly the effect of reflecting across the $x$–axis.
   • More generally, for any line $L\subset P$, the restriction to $P$ of the half‐turn about $L$ is the involution “reflect across $L$.”

2. “Clockwise” versus “counter‑clockwise” makes no difference at 180°.

   • Angles are taken modulo 360°, so rotating by $+180°$ or by $-180°$ lands you in the same place:

   $$R_L(+180°)=R_L(-180°).$$

   • In other words, a half‑turn is its own inverse, and the sign convention for direction drops out.

3. Orientation flip on the plane.

   • In three dimensions the rotation by 180° about a line lying in a plane reverses the plane’s “sidedness” (it sends a chosen normal vector to its negative). That’s exactly what a reflection in the line does: it reverses orientation in the plane.

So rotating $P$ by 180° around $L$ is the same map as reflecting $P$ across $L$, and whether you spin “clockwise” or “counter‑clockwise” your half‑turn is identical.

Set of finite sequences of naturals: countable

Show that the set $${\mathbb{N}}^{<\mathbb{N}}={\cup }_{k=0}^{\mathrm{\infty }}{\mathbb{N}}^{\{0,1,\dots ,k-1\}}$$ of all finite sequences of natural numbers is countable.

Here are two standard ways to see that the set of all finite sequences of naturals

$${\mathbb{N}}^{<\mathbb{N}}=\bigcup _{k=0}^{\mathrm{\infty }}{\mathbb{N}}^{\{0,1,\dots ,k-1\}}$$

is countable.

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1. Union of countably many countable sets

1. Each fixed‑length piece is countable. For each $k\ge 1$, the set of $k$‑tuples

   $${\mathbb{N}}^k=\{(n_0,n_1,\dots ,n_{k-1}):n_i\in \mathbb{N}\}$$

   is in bijection with $\mathbb{N}$. For example, one can use the usual Cantor–Gödel pairing repeatedly to pair off coordinates:

   $$\pi :\mathbb{N}\times \mathbb{N}\stackrel{\sim }{\to }\mathbb{N},(a,b)\mapsto \frac{1}{2}(a+b)(a+b+1)+b,$$

   and then pair $\pi (\pi (n_0,n_1),n_2)$, etc., to get a bijection ${\mathbb{N}}^k\cong \mathbb{N}$.

2. A countable union of countable sets is countable. We have

   $${\mathbb{N}}^{<\mathbb{N}}=\bigcup _{k=0}^{\mathrm{\infty }}{\mathbb{N}}^k,$$

   and since each ${\mathbb{N}}^k$ is countable, their (countable) union is again countable.

Hence ${\mathbb{N}}^{<\mathbb{N}}$ is countable.

---

2. Explicit injective coding via prime factorisation

Define a map

$$\mathrm{\Phi }:{\mathbb{N}}^{<\mathbb{N}}⟶{\mathbb{N}}_{>0}$$

by sending a finite sequence

$$s=(n_0,n_1,\dots ,n_{k-1})$$

to the integer

$$\mathrm{\Phi }(s)=2^{n_0}\cdot 3^{n_1}\cdot 5^{n_2}\cdots p_k^{n_{k-1}}(p_k\text{ = the }k\text{th prime}).$$

• Injectivity. By the Fundamental Theorem of Arithmetic, every positive integer has a unique prime‑power factorisation; in particular, different sequences $s$ give different factorizations, so $\mathrm{\Phi }$ is injective.
• Conclusion. An injective map from ${\mathbb{N}}^{<\mathbb{N}}$ into $\mathbb{N}$ shows $\mathrm{\#}{\mathbb{N}}^{<\mathbb{N}}\le \mathrm{\#}\mathbb{N}$. Since ${\mathbb{N}}^{<\mathbb{N}}$ is infinite, its cardinality is exactly ${\mathrm{\aleph }}_0$.

---

Either argument establishes that the set of all finite sequences of natural numbers is countable.

Note: ChatGPT’s uses of \N, which rendered correctly in its own UI, were changed into \mathbb{N} here.

Some congruences

I asked 4o about these congruences in separate conversations, but decided to make them just one post. They are: that (39x) mod 5 = 13, which is true; that (95x) mod 5 = 13, which is false; that (x²) mod 6 = 3, which is true; that (5x²) mod 8 = 12, which is true; that (4x³ + 2x) mod 5 = 7, which is false; and that (5^x + 3) mod 100 = 5, which is false.

Why 2^923 mod 15 = 8 (in-depth)

For any fixed natural number m greater than 1, we say that the integer a is congruent to the integer b modulo m if a − b is divisible by m. Please find the remainder when 2^923 is divided by 15.

To compute $2^{923}mod15$, note that $gcd(2,15)=1$ and Euler’s totient $\phi (15)=8$, so

$$2^8\equiv 1(\mathrm{mod}15).$$

Thus

$$2^{923}\equiv 2^{923mod8}\equiv 2^3\equiv 8(\mathrm{mod}15).$$

Answer: The remainder is $8$.

Thank you for finding the remainder, but I don’t understand what you said, please explain this “Euler’s totient φ” business.

Sure! Let me break it down step by step.

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1. What is Euler’s totient function, $\phi (n)$?

• $\phi (n)$ is the number of positive integers up to $n$ that are coprime to $n$ (i.e.\ they share no prime factors with $n$).

• For example, to compute $\phi (15)$:

  1. Factor $15=3\times 5$.

  2. Use the formula for a product of two distinct primes:

     $$\phi (p\cdot q)=(p-1)(q-1).$$

  3. Thus

     $$\phi (15)=(3-1)(5-1)=2\times 4=8.$$

  So there are 8 numbers in $\{1,2,\dots ,15\}$ that are coprime to 15: $1,2,4,7,8,11,13,14$.

---

2. Euler’s theorem (a generalization of Fermat’s little theorem)

Statement. If $gcd(a,n)=1$, then

$$a^{\phi (n)}\equiv 1(\mathrm{mod}n).$$

• “$gcd(a,n)=1$” means $a$ and $n$ have no common prime factors.
• So raising $a$ to the $\phi (n)$-th power always gives a remainder of 1 when divided by $n$.

---

3. Applying it to $2^{923}mod15$

1. Check that $gcd(2,15)=1$. ✔

2. Compute $\phi (15)=8$.

3. By Euler’s theorem,

   $$2^8\equiv 1(\mathrm{mod}15).$$

4. Therefore powers of 2 modulo 15 repeat every 8 steps. In particular,

   $$2^{923}=2^{8\cdot 115+3}=(2^8{)}^{115}\times 2^3\equiv 1^{115}\times 2^3=2^3=8(\mathrm{mod}15).$$

5. Remainder = 8.

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4. Why this helps

Without Euler’s theorem, you’d have to multiply 2 by itself 923 times and reduce mod 15 each time—impractical by hand. The totient function tells you exactly when the pattern of remainders resets.

I hope this clarifies what $\phi (n)$ is and how Euler’s theorem lets us simplify huge exponents!

What’s the reasoning for going from $2^8\equiv 1(\mathrm{mod}15)$ to $2^{923}\equiv 2^{923mod8}(\mathrm{mod}15)$?

The key property you’re using is that if you know

$$a^M\equiv 1(\mathrm{mod}n),$$

then for any exponent $E$ you can write $E$ in the division‐with‐remainder form

$$E=M\cdot Q+R,0\le R<M,$$

and get

$$a^E=a^{M\cdot Q+R}=(a^M{)}^Q\cdot a^R\equiv 1^Q\cdot a^R=a^R(\mathrm{mod}n).$$

---

Applying it to our case

1. We know $2^8\equiv 1(\mathrm{mod}15)$.

2. Divide the exponent 923 by 8:

   $$923=8\times 115+3,$$

   so here $M=8$, $Q=115$, $R=3$.

3. Then

   $$2^{923}=2^{8\cdot 115+3}=(2^8{)}^{115}\times 2^3\equiv 1^{115}\times 2^3=2^3(\mathrm{mod}15).$$

4. Hence you only need to compute $2^3\equiv 8$, and you never have to multiply out all 923 factors.

---

So whenever you have a “cycle” $a^M\equiv 1(\mathrm{mod}n)$, you can reduce the exponent modulo $M$. That’s exactly why

$$2^{923}\equiv 2^{923mod8}(\mathrm{mod}15).$$

How do we know $\phi (p\cdot q)=(p-1)(q-1)$?

There are two classic ways to see why, for two distinct primes $p$ and $q$,

$$\phi (pq)=(p-1)(q-1).$$

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1. Multiplicativity of $\phi$

Euler’s totient function is multiplicative on coprime arguments:

$$\text{if }gcd(m,n)=1,\text{then }\phi (mn)=\phi (m)\phi (n).$$

Since $p$ and $q$ are primes, $gcd(p,q)=1$, and we know

$$\phi (p)=p-1,\phi (q)=q-1$$

(because out of the $p$ numbers $1,2,\dots ,p$, exactly the multiples of $p$ — just “$p$” itself — fail to be coprime, leaving $p-1$ survivors; similarly for $q$). Hence

$$\phi (pq)=\phi (p)\phi (q)=(p-1)(q-1).$$

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2. Direct “inclusion–exclusion” count

Count how many integers $1\le k\le pq$ are not coprime to $pq$. Exactly those that share a factor with $pq$, i.e.\ those divisible by $p$ or by $q$:

• Multiples of $p$ in that range: ${\displaystyle \frac{pq}{p}=q}$ of them ($p,2p,\dots ,qp$).
• Multiples of $q$ in that range: ${\displaystyle \frac{pq}{q}=p}$ of them ($q,2q,\dots ,pq$).
• But we’ve counted multiples of both $p$ and $q$ (i.e.\ multiples of $pq$) twice; there is exactly 1 of those ($pq$ itself).

By inclusion–exclusion, the total count of “bad” numbers is

$$\mathrm{\#}\{\text{divisible by }p\}+\mathrm{\#}\{\text{divisible by }q\}-\mathrm{\#}\{\text{divisible by both}\}=q+p-1.$$

Since there are $pq$ numbers in all, the count of good (coprime) ones is

$$pq-(q+p-1)=pq-p-q+1=(p-1)(q-1).$$

Either viewpoint immediately yields

$$\boxed{{\displaystyle \phi (pq)=(p-1)(q-1).}}$$

How is Euler’s theorem proved?

Here is a classic, elementary proof of Euler’s theorem:

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Setup: the reduced residue system mod $n$

Let

$$n>1,a\in \mathbb{Z},gcd(a,n)=1.$$

Define the reduced residue system modulo $n$ as the set

$$R=\{r_1,r_2,\dots ,r_{\phi (n)}\}$$

of all positive integers $<n$ that are coprime to $n$. By definition, $|R|=\phi (n)$.

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Proof by “permutation of residues”

1. Multiply each element of $R$ by $a$, and reduce modulo $n$. Because $gcd(a,n)=1$, none of the products

   $$a{r}_1,a{r}_2,\dots ,a{r}_{\phi (n)}$$

   can share a nontrivial factor with $n$. Hence each $a{r}_i(\mathrm{mod}n)$ is again in the reduced residue system.

2. No collisions. If for two indices $i\ne j$ we had

   $$a{r}_i\equiv a{r}_j(\mathrm{mod}n),$$

   then

   $$a(r_i-r_j)\equiv 0(\mathrm{mod}n)⟹n|(r_i-r_j),$$

   which forces $r_i\equiv r_j(\mathrm{mod}n)$. But $0<r_i,r_j<n$, so $r_i=r_j$, contradicting $i\ne j$. Thus multiplication by $a$ permutes $R$.

3. Compare the two products. Consider the product of all members of $R$:

   $$P=r_1{r}_2\cdots r_{\phi (n)}.$$

   On one hand, that’s $P$. On the other hand, because the list $\{a{r}_i(\mathrm{mod}n)\}$ is just a re‑ordering of the same $r_i$’s, its product is also

   $$a^{\phi (n)}P\equiv P(\mathrm{mod}n).$$

   Hence

   $$a^{\phi (n)}P\equiv P(\mathrm{mod}n)⟹a^{\phi (n)}\equiv 1(\mathrm{mod}n)(gcd(P,n)=1).$$

That completes the proof.

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Alternate (group‑theoretic) viewpoint

The integers coprime to $n$, under multiplication mod $n$, form a finite group of order $\phi (n)$. In any finite group, Lagrange’s theorem says that the order of any element divides the group order, and in particular $x^{|G|}=e$. Taking $x=[a]$ in the group $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$ shows

$$[a{]}^{\phi (n)}=[1]⟺a^{\phi (n)}\equiv 1(\mathrm{mod}n).$$

Why 3^2463 ≡ 3 (mod 8)

For any fixed natural number m greater than 1, we say that the integer a is congruent to the integer b modulo m if a − b is divisible by m. Please find the remainder when 3^2463 is divided by 8.

To compute $3^{2463}mod8$, observe the pattern of powers of 3 modulo 8:

$$\begin{array}{rl}{3}^1& \equiv 3(\mathrm{mod}8),\\ 3^2& \equiv 9\equiv 1(\mathrm{mod}8),\\ 3^3& \equiv 3^2\cdot 3\equiv 1\cdot 3\equiv 3(\mathrm{mod}8),\\ 3^4& \equiv (3^2{)}^2\equiv 1^2\equiv 1(\mathrm{mod}8),\end{array}$$

so the residues cycle with period 2:

$$3^{\text{odd}}\equiv 3,3^{\text{even}}\equiv 1(\mathrm{mod}8).$$

Since $2463$ is odd,

$$3^{2463}\equiv 3(\mathrm{mod}8).$$

Answer: The remainder when $3^{2463}$ is divided by 8 is $\boxed{{\displaystyle 3}}$.

Why do we expect that this pattern always holds?

The reason the pattern

$$3^1\equiv 3,3^2\equiv 1,3^3\equiv 3,3^4\equiv 1,\dots$$

repeats every two exponents comes down to the single fact

$$3^2=9=8+1⟹3^2\equiv 1(\mathrm{mod}8).$$

Once you know that $3^2\equiv 1(\mathrm{mod}8)$, it follows immediately that for any integer $k$,

• $$3^{2k}=(3^2{)}^k\equiv 1^k\equiv 1(\mathrm{mod}8),$$
• $$3^{2k+1}=3\cdot 3^{2k}\equiv 3\cdot 1\equiv 3(\mathrm{mod}8).$$

Thus the residues must alternate $3,1,3,1,\dots$ with period 2.

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A bit more structure: the multiplicative group of units modulo 8

• The units in $\mathbb{Z}/8\mathbb{Z}$ are those integers coprime to 8, namely $\{1,3,5,7\}$.

• This set forms a finite group under multiplication mod 8, whose order is $\phi (8)=4$.

• By Lagrange’s theorem, the order (the size) of any element in a finite group must divide the size of the group. Here:

  $$\text{order}(3)|4.$$

• We check directly that $3^2\equiv 1(\mathrm{mod}8)$, so the order of 3 is exactly 2, and hence its powers repeat with period 2.

Either way—whether by the elementary identity $3^2\equiv 1(\mathrm{mod}8)$ or by invoking the structure of $(\mathbb{Z}/8\mathbb{Z}{)}^{\times }$—you see why the cycle length is 2, and why $3^{2463}\equiv 3$.

I’m not sure why ChatGPT wrote “period 2” with a narrow no-break space (NNBSP) instead of a regular space (as in “period 2”).

Congruences as equations

Suppose we want to solve many congruences such as:

2x ≡ 7 (mod 11)
7x ≡ 4 (mod 11)
x^5 ≡ 3 (mod 4)

And we already have a calculator that solves equations about real numbers very well. How do we convert such congruences into equations?

To solve congruences like:

• $2x\equiv 7(\mathrm{mod}11)$
• $7x\equiv 4(\mathrm{mod}11)$
• $x^5\equiv 3(\mathrm{mod}4)$

using a calculator that works with real numbers, you need to translate modular arithmetic into standard equations. Here's how you can do it:

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✅ General Strategy: Use the definition of congruence

Recall:

$$a\equiv b(\mathrm{mod}n)\text{means}a=b+kn\text{for some integer }k.$$

We can turn a modular congruence into an ordinary equation involving an unknown integer $k$.

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🔁 Step-by-step conversion

Example 1: $2x\equiv 7(\mathrm{mod}11)$

We rewrite this as:

$$2x=7+11k\text{for some integer }k.$$

Now solve for $x$ in terms of $k$:

$$x=\frac{7+11k}{2}$$

Now you can try integer values of $k$ and check when $x$ becomes an integer.

For example:

• $k=1$: $x=\frac{7+11}{2}=\frac{18}{2}=9\Rightarrow \boxed{{\displaystyle x=9}}$
• You can verify: $2\cdot 9=18\equiv 7(\mathrm{mod}11)$

✅ So the solution is $x\equiv 9(\mathrm{mod}11)$

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Example 2: $7x\equiv 4(\mathrm{mod}11)$

Write:

$$7x=4+11k\Rightarrow x=\frac{4+11k}{7}$$

Try values of $k$ and look for integer $x$:

• $k=3$: $x=\frac{4+33}{7}=\frac{37}{7}\notin \mathbb{Z}$
• $k=2$: $x=\frac{4+22}{7}=\frac{26}{7}$
• $k=1$: $x=\frac{4+11}{7}=\frac{15}{7}$
• $k=5$: $x=\frac{4+55}{7}=\frac{59}{7}$
• $k=6$: $x=\frac{4+66}{7}=\frac{70}{7}=10\Rightarrow \boxed{{\displaystyle x=10}}$

Verify: $7\cdot 10=70\equiv 4(\mathrm{mod}11)$

✅ So the solution is $x\equiv 10(\mathrm{mod}11)$

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Example 3: $x^5\equiv 3(\mathrm{mod}4)$

Write:

$$x^5=3+4k\Rightarrow x=\sqrt[5]{3+4k}$$

Try small integer values of $k$ and check if the 5th root is an integer:

• $k=0$: $x^5=3\Rightarrow x=\sqrt[5]{3}$ ❌
• $k=1$: $x^5=7\Rightarrow x=\sqrt[5]{7}$ ❌
• $k=2$: $x^5=11$ ❌
• $k=3$: $x^5=15$ ❌
• $k=4$: $x^5=19$
• $k=5$: $x^5=23$
• $k=6$: $x^5=27\Rightarrow x=3\Rightarrow \boxed{{\displaystyle x=3}}$

Verify: $3^5=243\equiv 3(\mathrm{mod}4)$

✅ So $x\equiv 3(\mathrm{mod}4)$

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🧠 Summary

To convert congruences into real equations (for calculators):

1. Rewrite the congruence $a\equiv b(\mathrm{mod}n)$ as $a=b+kn$ where $k\in \mathbb{Z}$.
2. Solve the resulting real-number equation for $x$.
3. Try small integer values of $k$ to find valid (integer) solutions for $x$.

Let me know if you want a method that avoids brute-force trial for $k$.

Is there one?

Yes — instead of trial and error, there's a systematic method based on modular inverses, which avoids guessing values of $k$. Here's how it works:

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🔄 Method: Use Modular Inverses (no guessing)

Idea:

To solve a linear congruence like:

$$ax\equiv b(\mathrm{mod}n)$$

You can multiply both sides by the modular inverse of $a\mathrm{mod}n$ — provided the inverse exists (i.e., $gcd(a,n)=1$).

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🔁 Step-by-step

Step 1: Find the modular inverse of $a\mathrm{mod}n$

Let $a^{-1}$ be the modular inverse of $a\mathrm{mod}n$, meaning:

$$a\cdot a^{-1}\equiv 1(\mathrm{mod}n)$$

Step 2: Multiply both sides by $a^{-1}$

$$a^{-1}\cdot ax\equiv a^{-1}\cdot b(\mathrm{mod}n)\Rightarrow x\equiv a^{-1}b(\mathrm{mod}n)$$

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✅ Examples

Example 1: $2x\equiv 7(\mathrm{mod}11)$

• Find $2^{-1}\mathrm{mod}11$

We need $2y\equiv 1(\mathrm{mod}11)$. Try small values:

• $y=6\Rightarrow 2\cdot 6=12\equiv 1(\mathrm{mod}11)$

So $2^{-1}=6\mathrm{mod}11$

Then:

$$x\equiv 6\cdot 7=42\equiv \boxed{{\displaystyle 9(\mathrm{mod}11)}}$$

No trial and error needed!

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Example 2: $7x\equiv 4(\mathrm{mod}11)$

• Find $7^{-1}\mathrm{mod}11$

Try:

• $7\cdot 8=56\equiv 1(\mathrm{mod}11)$

So $7^{-1}=8\mathrm{mod}11$

Then:

$$x\equiv 8\cdot 4=32\equiv \boxed{{\displaystyle 10(\mathrm{mod}11)}}$$

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⚠️ Note on Higher Powers

This method is mainly for linear congruences (like $ax\equiv b\mathrm{mod}n$).

For nonlinear ones like:

• $x^5\equiv 3\mathrm{mod}4$

you generally:

• Try all values of $x\mathrm{mod}4$, since there are only 4: $0,1,2,3$
• Compute $x^5\mathrm{mod}4$ for each

Example:

• $x=0\Rightarrow 0^5=0\mathrm{mod}4$
• $x=1\Rightarrow 1^5=1$
• $x=2\Rightarrow 32\equiv 0\mathrm{mod}4$
• $x=3\Rightarrow 243\equiv 3\mathrm{mod}4$

✅ So $x\equiv 3(\mathrm{mod}4)$ is the solution

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💡 Summary

Type	Method
Linear congruence $ax\equiv b\mathrm{mod}n$	Multiply both sides by $a^{-1}\mathrm{mod}n$
Power congruence $x^k\equiv b\mathrm{mod}n$	Try all $x\in \{0,1,...,n-1\}$ or use discrete logs (advanced)

Let me know if you want help finding modular inverses programmatically or using the Extended Euclidean Algorithm.

Modular arithmetic

These first questions were all with o4-mini. The theorem statements are all from this book. I found ChatGPT’s proofs more didactic than the originals.

For any fixed natural number m greater than 1, we say that the integer a is congruent to the integer b modulo m if a − b is divisible by m. Show that the congruence relation (modulo m) is transitive.

To prove transitivity, start with the hypotheses:

1. $a\equiv b(\mathrm{mod}m)$. This means $m$ divides $a-b$, so there exists an integer $k$ such that

   $$a-b=km.$$

2. $b\equiv c(\mathrm{mod}m)$. This means $m$ divides $b-c$, so there exists an integer $\ell$ such that

   $$b-c=\ell m.$$

Now consider $a-c$. We can write

$$a-c=(a-b)+(b-c)=km+\ell m=(k+\ell )m.$$

Since $k+\ell$ is an integer, this shows $m$ divides $a-c$. Hence

$$a\equiv c(\mathrm{mod}m),$$

which is exactly the statement of transitivity for congruence modulo $m$.

Show that when a and b are nonnegative integers, the relationship a ≡ b (mod m) is equivalent to a and b leaving equal remainders upon division by m.

To show the two statements are equivalent, we prove each implication in turn.

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1. If $a\equiv b(\mathrm{mod}m)$, then $a$ and $b$ have the same remainder upon division by $m$.

Write the Euclidean divisions of $a$ and $b$ by $m$:

$$\begin{array}{rl}a& =q_{a}m+r_a,0\le r_a<m,\\ b& =q_{b}m+r_b,0\le r_b<m.\end{array}$$

Then

$$a-b=(q_a-q_b)m+(r_a-r_b).$$

If $a\equiv b(\mathrm{mod}m)$, by definition $m$ divides $a-b$. But

$$a-b=(q_a-q_b)m+(r_a-r_b)$$

and $(q_a-q_b)m$ is already a multiple of $m$. For the whole sum to be a multiple of $m$, the “leftover” term $r_a-r_b$ must also be a multiple of $m$. Since

$$|r_a-r_b|<m,$$

the only multiple of $m$ in that range is $0$. Hence

$$r_a-r_b=0⟹r_a=r_b.$$

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2. If $a$ and $b$ leave the same remainder upon division by $m$, then $a\equiv b(\mathrm{mod}m)$.

Again write

$$a=q_{a}m+r,b=q_{b}m+r\text{with }0\le r<m.$$

Then

$$a-b=(q_{a}m+r)-(q_{b}m+r)=(q_a-q_b)m,$$

which is manifestly divisible by $m$. Therefore $a\equiv b(\mathrm{mod}m)$.

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Combining (1) and (2) shows the two characterizations are logically equivalent.

Show that for a given modulus m, each integer is congruent to exactly one of the numbers in the set {0, 1, 2, …, m − 1}.

Proof. Fix an integer $m>1$. We must show that for any integer $a$ there is one—and only one—remainder $r$ with

$$0\le r\le m-1$$

such that

$$a\equiv r(\mathrm{mod}m).$$

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Existence

By the Division Algorithm, for any integer $a$ there exist unique integers $q$ and $r$ with

$$a=qm+r\text{and}0\le r<m.$$

Rewriting gives

$$a-r=qm,$$

so $m$ divides $a-r$, i.e.\ $a\equiv r(\mathrm{mod}m)$. Thus some $r$ in $\{0,1,\dots ,m-1\}$ is congruent to $a$.

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Uniqueness

Suppose $r$ and $s$ are two remainders in $\{0,1,\dots ,m-1\}$ with

$$a\equiv r(\mathrm{mod}m)\text{and}a\equiv s(\mathrm{mod}m).$$

Then $m$ divides both $a-r$ and $a-s$, so it also divides their difference:

$$m\mid [(a-r)-(a-s)]⟹m\mid (s-r).$$

But $r$ and $s$ lie in the range $0\le r,s<m$, so $|s-r|<m$. The only multiple of $m$ in that range is $0$. Hence

$$s-r=0⟹s=r.$$

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Combining these two parts shows each integer $a$ is congruent to exactly one element of $\{0,1,2,\dots ,m-1\}$.

Show that if a ≡ b (mod m) and c ≡ d (mod m), then (i) (a + c) ≡ (b + d) (mod m), and
(ii) ac ≡ bd (mod m).

Proof. Let $m>1$ be fixed, and suppose

$$a\equiv b(\mathrm{mod}m)\text{and}c\equiv d(\mathrm{mod}m).$$

By definition, there exist integers $k$ and $\ell$ such that

$$a-b=km,c-d=\ell m.$$

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(i) Addition

Consider

$$(a+c)-(b+d)=(a-b)+(c-d)=km+\ell m=(k+\ell )m.$$

Since $(k+\ell )$ is an integer, $m$ divides $(a+c)-(b+d)$. Hence

$$a+c\equiv b+d(\mathrm{mod}m).$$

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(ii) Multiplication

Consider

$$ac-bd=ac-bc+bc-bd=c(a-b)+b(c-d).$$

Substituting the expressions above,

$$c(a-b)+b(c-d)=c(km)+b(\ell m)=(ck+b\ell )m,$$

which shows $m$ divides $ac-bd$. Therefore

$$ac\equiv bd(\mathrm{mod}m).$$

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Thus congruence modulo $m$ is compatible with both addition and multiplication.

Show that if a ≡ b (mod m), then a^n ≡ b^n (mod m), for every natural number n.

Proof by induction on $n$.

Fix a modulus $m>1$ and suppose $a\equiv b(\mathrm{mod}m)$. We show $a^n\equiv b^n(\mathrm{mod}m)$ for every natural number $n$.

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Base case ($n=1$)

Trivially,

$$a^1=a\equiv b=b^1(\mathrm{mod}m).$$

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Inductive step

Assume for some $k\ge 1$ that

$$a^k\equiv b^k(\mathrm{mod}m).$$

We must show

$$a^{k+1}\equiv b^{k+1}(\mathrm{mod}m).$$

Notice

$$a^{k+1}=a^k\cdot a,b^{k+1}=b^k\cdot b.$$

By the inductive hypothesis $a^k\equiv b^k(\mathrm{mod}m)$, and by our original assumption $a\equiv b(\mathrm{mod}m)$. From the compatibility of congruence with multiplication (proved earlier), we conclude

$$a^{k+1}=a^k\cdot a\equiv b^k\cdot b=b^{k+1}(\mathrm{mod}m).$$

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By induction, $a^n\equiv b^n(\mathrm{mod}m)$ for all natural numbers $n$. This completes the proof.

Show that every natural number is congruent to the sum of its digits modulo 9.

Proof. Every natural number $N$ can be written in its decimal expansion as

$$N=d_0+d_1\cdot 10+d_2\cdot {10}^2+\cdots +d_k\cdot {10}^k,$$

where each $d_i$ is one of the digits $0,1,\dots ,9$. Call

$$S=d_0+d_1+d_2+\cdots +d_k$$

the sum of its digits.

We will show

$$N\equiv S(\mathrm{mod}9).$$

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1. Observe that $10\equiv 1(\mathrm{mod}9)$. Indeed, $10-1=9$ is divisible by 9.

2. Use the fact that congruence is compatible with powers and multiplication. From $10\equiv 1(\mathrm{mod}9)$ it follows by induction (or by repeatedly applying the multiplicative property) that

   $${10}^i\equiv 1^i=1(\mathrm{mod}9)\text{for every }i\ge 0.$$

3. Apply congruence to each term in the expansion of $N$.

   $$N=\sum _{i=0}^k{d}_i{10}^i\equiv \sum _{i=0}^k{d}_i\cdot 1=\sum _{i=0}^k{d}_i=S(\mathrm{mod}9).$$

Since every step uses only the rules of congruence (namely that if $x\equiv y$ and $u\equiv v(\mathrm{mod}m)$, then $xu\equiv yv(\mathrm{mod}m)$ and sums likewise agree), we conclude

$$N\equiv S(\mathrm{mod}9),$$

i.e.\ every natural number is congruent modulo 9 to the sum of its digits.

For this question I used o4-mini-high.

Summarize and contextualize the knowledge about congruences we have seen so far.

Here’s a bird’s‑eye view of what we’ve built up so far in our study of congruences modulo $m$, and why it matters:

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1. Congruence as an Equivalence Relation

• Definition:

  $$a\equiv b(\mathrm{mod}m)⟺m\mid (a-b).$$

• Properties:

  • Reflexive: $a\equiv a(\mathrm{mod}m)$.
  • Symmetric: If $a\equiv b$, then $b\equiv a$.
  • Transitive: If $a\equiv b$ and $b\equiv c$, then $a\equiv c$.

Together, these show congruence partitions the integers into equivalence classes.

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2. Equivalence Classes and Canonical Representatives

• Every integer $a$ is congruent to exactly one of

  $$\{0,1,2,\dots ,m-1\},$$

  its remainder upon division by $m$.

• Thus the set of all integers splits into $m$ classes, often denoted $\mathbb{Z}/m\mathbb{Z}=\{[0],[1],\dots ,[m-1]\}$.

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3. Arithmetic in $\mathbb{Z}/m\mathbb{Z}$

Congruence “respects” the usual operations: if

$$a\equiv b(\mathrm{mod}m),c\equiv d(\mathrm{mod}m),$$

then

1. Addition: $(a+c)\equiv (b+d)(\mathrm{mod}m).$
2. Multiplication: $ac\equiv bd(\mathrm{mod}m).$
3. Powers: $a^n\equiv b^n(\mathrm{mod}m)$ for all $n\in \mathbb{N}$.

Because of this, one can do arithmetic “mod $m$” and get a well‑defined algebraic system (a ring).

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4. A Classic Application: The “Divisibility by 9” Test

• Writing a decimal number $N$ as $\sum d_i{10}^i$, and using $10\equiv 1(\mathrm{mod}9)$, shows

  $$N\equiv \sum d_i(\mathrm{mod}9).$$

• Hence a number is divisible by 9 exactly when the sum of its digits is.

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5. Why This Matters

• Number‑theoretic tools: solving linear congruences, understanding invertibility (“units” in $\mathbb{Z}/m\mathbb{Z}$), working toward Fermat’s Little Theorem and Euler’s Theorem.
• Chinese Remainder Theorem: decomposing problems mod $m$ into simpler ones mod its prime‑power factors.
• Cryptography & coding theory: RSA, Diffie–Hellman, checksums all rely on modular arithmetic.
• Practical divisibility tests: beyond 9, there are tests for 3, 11, 7, etc., all built on the same congruence ideas.

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In short, congruences give us a way to “collapse” the infinite set of integers into finitely many residue classes while still doing meaningful arithmetic—and that simple idea underpins huge swaths of both pure and applied mathematics.