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.

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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 ${\mathrm{\aleph }}_0$.

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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: $1,2,4,7,8,11,13,14$.

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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 $n$.

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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

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

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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.

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

---

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

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

---

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 $a$.

---

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

$$s-r=0⟹s=r.$$

---

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

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

---

(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

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

---

(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

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

---

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 $n$.

---

Base case ($n=1$)

Trivially,

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

---

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

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

---

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

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

---

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:

---

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.

---

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 $\mathbb{Z}/m\mathbb{Z}=\{[0],[1],\dots ,[m-1]\}$.

---

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).

---

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.

---

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.

---

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.

Proof of Binet’s formula for the Fibonacci sequence

Briefly define the Fibonacci sequence.

The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, usually starting with 0 and 1. It begins: 0, 1, 1, 2, 3, 5, 8, 13, 21, ... Mathematically, it's defined as:

• F(0) = 0
• F(1) = 1
• F(n) = F(n-1) + F(n-2) for n ≥ 2.

Prove that, ∀n∈ℕ, $F(n)=\frac{1}{\sqrt{5}}[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n]$.

We are asked to prove Binet’s formula for the Fibonacci sequence:

$$F(n)=\frac{1}{\sqrt{5}}[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n],\mathrm{\forall }n\in \mathbb{N}$$

---

Step 1: Define the Fibonacci recurrence

The Fibonacci sequence is defined as:

• $F(0)=0$
• $F(1)=1$
• $F(n)=F(n-1)+F(n-2)$, for $n\ge 2$

---

Step 2: Use recurrence relation approach with characteristic equation

Assume a solution of the form:

$$F(n)=A{\alpha }^n+B{\beta }^n$$

where $\alpha$ and $\beta$ are the roots of the characteristic equation derived from the recurrence:

$$x^2=x+1\Rightarrow x^2-x-1=0$$

Solving this quadratic:

$$\alpha =\frac{1+\sqrt{5}}{2},\beta =\frac{1-\sqrt{5}}{2}$$

So the general solution becomes:

$$F(n)=A{\alpha }^n+B{\beta }^n$$

---

Step 3: Solve for constants $A$ and $B$

Using initial conditions:

For $n=0$:

$$F(0)=A{\alpha }^0+B{\beta }^0=A+B=0\Rightarrow B=-A$$

For $n=1$:

$$F(1)=A\alpha +B\beta =A\alpha -A\beta =A(\alpha -\beta )=1$$

We calculate $\alpha -\beta$:

$$\alpha -\beta =\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}=\frac{2\sqrt{5}}{2}=\sqrt{5}$$

So:

$$A(\sqrt{5})=1\Rightarrow A=\frac{1}{\sqrt{5}},B=-\frac{1}{\sqrt{5}}$$

---

Step 4: Final formula

Substitute $A$ and $B$ back into the general solution:

$$F(n)=\frac{1}{\sqrt{5}}{\alpha }^n-\frac{1}{\sqrt{5}}{\beta }^n=\frac{1}{\sqrt{5}}({\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n)$$

---

✅ Conclusion

We have proven that:

$$\boxed{{\displaystyle F(n)=\frac{1}{\sqrt{5}}[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n]}}$$

holds for all $n\in \mathbb{N}$.

Natural language formal grammars

Nowadays what's the most advanced fully formalized formal grammar for a natural language?

Short answer : For a single natural language, the English Resource Grammar (ERG)—currently at its May 2025 release—is generally regarded as the most advanced, fully formalised grammar that actually runs on a computer. It is hand-built in the HPSG formalism, covers 90-plus % of open-domain English, and produces logic-ready Minimal Recursion Semantics (MRS) representations. (Delph-in)

---

Why the ERG stands out

Aspect	What the ERG offers
Theoretical foundation	Head-Driven Phrase-Structure Grammar with a rich type hierarchy of 1 100+ lexical/phrased-structure types
Coverage & precision	~35 000 lexemes, ~200 syntactic rules; parses ≈ 94 % of Wall-Street-Journal sentences and > 90 % of contemporary web text while staying linguistically faithful (iafisher.com)
Semantics	Direct compositional output in MRS, compatible with downstream inference, QA, RTE and MT pipelines (Delph-in)
Tool chain	Ships as a binary grammar (“erg-2025-x86-64-0.9.34.dat”) for the ACE parser and for LKB/pyDelphin; also accessible through a REST API (Delph-in)
Sustained maintenance	Active DELPH-IN community, annual releases; latest tagged 2025

Because every rule, feature structure and semantic predicate is machine-checkable, ERG is what linguists call a precision grammar: if it accepts a sentence, it guarantees an internally consistent syntactic and semantic analysis. That property makes it the go-to resource for deep NLP experiments and for testing linguistic hypotheses.

---

How does it compare to other large formal grammars?

Project	Formalism	Languages	Maturity & scope
ParGram / XLE grammars	Lexical-Functional Grammar (LFG)	≈ 10 major languages (EN, FR, DE, JA…)	High-quality parallel grammars but smaller coverage per language; used mainly in grammar-checking research (iafisher.com)
GF Resource Grammar Library (RGL)	Type-theoretical GF	45 + languages	Elegant multilingual framework for generation & controlled NL; not yet as deep or wide-coverage per language as ERG (grammaticalframework.org)
OpenCCG / CCGBank grammar	Combinatory Categorial Grammar	English	Grammar induced from CCGBank; broad but automatically derived, with less hand-verified precision than ERG (GitHub)
Resource Grammars (Jacy, SRG, etc.)	HPSG	Japanese, Spanish, Norwegian…	Built with the same methodology as ERG; none yet match ERG’s size or stability (e.g. Spanish RG revival 2024)

In short, those projects show that formalised, executable grammars are feasible for many languages, but the ERG remains the flagship in terms of depth, breadth, and semantic integration.

---

Take-away

If you need a fully formal, broad-coverage, linguistically rigorous grammar that you can parse and generate with today, the ERG is the state of the art. For cross-lingual work, look toward the GF RGL or the ParGram LFG suite, but be prepared for more limited coverage or weaker semantic output.

Division Algorithm for Polynomials

Prove this: Let f(x), g(x) be two polynomials with deg(f) ≤ deg(g). Then there are unique polynomials q(x), r(x) such that g(x) = q(x)f(x) + r(x), where deg(r) < deg(f).

Proof (Division Algorithm for Polynomials)

Let $f(x)$ and $g(x)$ be polynomials over a field (or integral domain) $K$, with

$$\mathrm{deg}(f)\le \mathrm{deg}(g).$$

We will show:

1. Existence: there exist polynomials $q(x)$ and $r(x)$ such that

   $$g(x)=q(x)f(x)+r(x),$$

   with $\mathrm{deg}(r)<\mathrm{deg}(f)$.

2. Uniqueness: this representation is unique.

---

1. Existence

We proceed by induction on $n=\mathrm{deg}(g)$.

• Base case: If $\mathrm{deg}(g)<\mathrm{deg}(f)$, then we may simply take

  $$q(x)=0,r(x)=g(x).$$

  Clearly $g=0\cdot f+g$, and $\mathrm{deg}(r)=\mathrm{deg}(g)<\mathrm{deg}(f)$.

• Inductive step: Suppose the statement holds for all polynomials of degree less than $n$, and let $\mathrm{deg}(g)=n\ge \mathrm{deg}(f)$. Write

  $$f(x)=a_m{x}^m+\dots ,g(x)=b_n{x}^n+\dots$$

  with $a_m,b_n\ne 0$. Since $n\ge m$, set

  $$t(x)=\frac{b_n}{a_m}{x}^{n-m}.$$

  Then $\mathrm{deg}(t(x)f(x))=n$, and its leading coefficient matches that of $g(x)$. Consider

  $$g_1(x)=g(x)-t(x)f(x).$$

  By construction, $\mathrm{deg}(g_1)<n$. Now by the induction hypothesis applied to $g_1$ (whose degree is $<n$), there exist polynomials $q_1(x)$ and $r(x)$ with

  $$g_1(x)=q_1(x)f(x)+r(x),\mathrm{deg}(r)<\mathrm{deg}(f).$$

  Hence

  $$g(x)=t(x)f(x)+g_1(x)=t(x)f(x)+(q_1(x)f(x)+r(x))=(t(x)+q_1(x))f(x)+r(x).$$

  Setting $q(x)=t(x)+q_1(x)$ gives the desired representation for $g$, with $\mathrm{deg}(r)<\mathrm{deg}(f)$.

By induction, such $q(x)$ and $r(x)$ always exist.

---

2. Uniqueness

Suppose there are two representations

$$g(x)=q_1(x)f(x)+r_1(x),g(x)=q_2(x)f(x)+r_2(x),$$

with $\mathrm{deg}(r_1),\mathrm{deg}(r_2)<\mathrm{deg}(f)$. Subtracting gives

$$0=(q_1(x)-q_2(x))f(x)+(r_1(x)-r_2(x)).$$

Rearrange:

$$(q_1(x)-q_2(x))f(x)=-(r_1(x)-r_2(x)).$$

• The left side is a multiple of $f(x)$, so if $q_1\ne q_2$ then $\mathrm{deg}((q_1-q_2)f)\ge \mathrm{deg}(f)$.
• The right side is $r_2(x)-r_1(x)$, whose degree is strictly less than $\mathrm{deg}(f)$.

The only way a nonzero polynomial of degree $\ge \mathrm{deg}(f)$ can equal one of degree $<\mathrm{deg}(f)$ is if both are the zero polynomial. Hence

$$q_1(x)-q_2(x)=0,r_1(x)-r_2(x)=0,$$

so $q_1=q_2$ and $r_1=r_2$. This proves uniqueness.

---

Together, these two parts establish the Division Algorithm for polynomials: for any $f(x)$ and $g(x)$ with $\mathrm{deg}(f)\le \mathrm{deg}(g)$, there exist unique polynomials $q(x)$ and $r(x)$ such that

$$g(x)=q(x)f(x)+r(x),\mathrm{deg}(r)<\mathrm{deg}(f).$$

$◼$

Motivation for decomposition into even and odd functions

Every function can be expressed uniquely as the sum of an even function and an odd function, like so: let E(x) be (f(x)+f(-x))/2, and let O(x) be (f(x)-f(-x))/2. Then E(x) is even, O(x) is odd, and f(x) = E(x) + O(x).

But what’s the point of doing this?

Decomposing any function $f$ into its even and odd parts

$$E(x)=\frac{f(x)+f(-x)}{2},O(x)=\frac{f(x)-f(-x)}{2},f=E+O$$

is more than just a neat algebraic trick. Here’s why it’s so useful:

---

1. Exploiting symmetry in integrals

• Odd integrals vanish: if $O$ is odd, then

  $${\int }_{-a}^{a}O(x)dx=0.$$

  So

  $${\int }_{-a}^{a}f(x)dx={\int }_{-a}^{a}E(x)dx+\underset{0}{\underbrace{{\int }_{-a}^{a}O(x)dx}}=2{\int }_0^{a}E(x)dx.$$

  You’ve halved your work by reducing to the even part only.

---

2. Fourier series simplification

• An arbitrary periodic $f$ has a Fourier series

  $$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_n\cos (nx)+b_n\sin (nx)].$$

• The cosine terms are the even part; the sine terms are the odd part.

• You can immediately tell which coefficients vanish if you know $f$ is purely even or purely odd.

---

3. Power-series (Taylor) clarity

• Expanding about zero, an even function has only even powers:

  $$E(x)=c_0+c_2{x}^2+c_4{x}^4+\cdots ,$$

  while an odd function has only odd powers:

  $$O(x)=c_{1}x+c_3{x}^3+\cdots .$$

• Decomposition helps identify which terms must be present or zero in your series.

---

4. Solving differential equations

• Many differential operators commute with the “flip” $x\mapsto -x$, so they preserve evenness/oddness.
• You can split a boundary-value or initial-value problem into two simpler problems—one on the even subspace, one on the odd—often halving the complexity.

---

5. Linear-algebra viewpoint

• The space of all functions (say, on $\mathbb{R}$) decomposes as

  $$\{\text{even}\}\oplus \{\text{odd}\}.$$

• Projection onto each subspace is exactly $E$ and $O$. This is a concrete example of projecting onto complementary subspaces in any vector space.

---

6. Practical computations & modeling

• In physics and engineering, boundary conditions often enforce symmetry. If you know a solution must be even (or odd), you can discard half the domain or simplify the ansatz immediately.
• In signal processing, even/odd decomposition relates to splitting a signal into its symmetric and antisymmetric components, each of which may carry different information.

---

In short

Breaking $f$ into $E+O$ lets you

1. Harness symmetry to simplify integrals and equations.
2. Organize expansions (Fourier, Taylor) by parity.
3. Project cleanly onto invariant subspaces.

It’s a fundamental tool whenever symmetry—or the lack of it—can be used to your advantage.

Completeness of isohedral figure classification

The table below was simplified from the table at this Wikipedia article.

Here is a table of isohedral figures from Wikipedia. Is it proven that all isohedral figures must be one of these things – a Platonic solid, a Catalan solid, or one of these pyramid-stacking “polar” figures?

Faces	Face config.	Class	Name	Symmetry	Order
4	V33	Platonic	tetrahedron	Td, [3,3], (*332)	24
4	V33	Platonic	tetragonal disphenoid	D2d, [2+,2], (2*)	4
4	V33	Platonic	rhombic disphenoid	D2, [2,2]+, (222)	4
6	V34	Platonic	cube	Oh, [4,3], (*432)	48
6	V34	Platonic	cube	D3d, [2+,6]	12
6	V34	Platonic	cube	(2*3)	12
6	V34	Platonic	trigonal trapezohedron	D3	6
6	V34	Platonic	asymmetric trigonal trapezohedron	[2,3]+, (223)	6
8	V43	Platonic	octahedron	Oh, [4,3], (*432)	48
8	V43	Platonic	octahedron	D4h,[2,4],(*224)	16
8	V43	Platonic	octahedron	D2h,[2,2],(*222)	8
8	V43	Platonic	square bipyramid	D2d,[2+,4],(2*2)	8
8	V43	Platonic	rhombic bipyramid	D2d,[2+,4],(2*2)	8
8	V43	Platonic	square scalenohedron	D2d,[2+,4],(2*2)	8
12	V35	Platonic	regular dodecahedron	Ih, [5,3], (*532)	120
12	V35	Platonic	regular dodecahedron	Th, [3+,4], (3*2)	24
12	V35	Platonic	regular dodecahedron	T, [3,3]+, (*332)	12
12	V35	Platonic	pyritohedron	T, [3,3]+, (*332)	12
12	V35	Platonic	tetartoid	T, [3,3]+, (*332)	12
20	V53	Platonic	regular icosahedron	Ih, [5,3], (*532)	120
12	V3.62	Catalan	triakis tetrahedron	Td, [3,3], (*332)	24
12	V(3.4)2	Catalan	rhombic dodecahedron	Oh, [4,3], (*432)	48
12	V(3.4)2	Catalan	rhombic dodecahedron	Td, [3,3], (*332)	24
12	V(3.4)2	Catalan	deltoidal dodecahedron	Td, [3,3], (*332)	24
24	V3.82	Catalan	triakis octahedron	Oh, [4,3], (*432)	48
24	V4.62	Catalan	tetrakis hexahedron	Oh, [4,3], (*432)	48
24	V3.43	Catalan	deltoidal icositetrahedron	Oh, [4,3], (*432)	48
48	V4.6.8	Catalan	disdyakis dodecahedron	Oh, [4,3], (*432)	48
24	V34.4	Catalan	pentagonal icositetrahedron	O, [4,3]+, (432)	24
30	V(3.5)2	Catalan	rhombic triacontahedron	Ih, [5,3], (*532)	120
60	V3.102	Catalan	triakis icosahedron	Ih, [5,3], (*532)	120
60	V5.62	Catalan	pentakis dodecahedron	Ih, [5,3], (*532)	120
60	V3.4.5.4	Catalan	deltoidal hexecontahedron	Ih, [5,3], (*532)	120
120	V4.6.10	Catalan	disdyakis triacontahedron	Ih, [5,3], (*532)	120
60	V34.5	Catalan	pentagonal hexecontahedron	I, [5,3]+, (532)	60
2n	V33.n	Polar	trapezohedron	Dnd, [2+,2n], (2*n)	4n
2n	V33.n	Polar	trapezohedron	Dn, [2,n]+, (22n)	2n
2n	V33.n	Polar	asymmetric trapezohedron	Dn, [2,n]+, (22n)	2n
2n	V42.n	Polar	regular n-bipyramid	Dnh, [2,n], (*22n)	4n
2n	V42.2n	Polar	isotoxal 2n-bipyramid	Dnh, [2,n], (*22n)	4n
2n	V42.2n	Polar	2n-scalenohedron	Dnd, [2+,2n], (2*n)	4n
4n	V42.2n	Polar	2n-scalenohedron	Dnd, [2+,2n], (2*n)	4n

Short answer

Yes – for convex polyhedra the classification is finished. Branko Grünbaum and Geoffrey C. Shephard completed it in their study of isohedral (face-transitive) tilings of the sphere, and the result is exactly what your table shows:

• 5 Platonic solids

• 13 Catalan solids (duals of the Archimedean solids)

• Three infinite polar families

  • the (general) scalenohedra / isosceles tetrahedra
  • the $n$-bipyramids (duals of prisms)
  • the $n$-trapezohedra (duals of antiprisms)

Together these make up 30 “classes’’: 28 individual solids and 2 one-parameter families whose face-count grows with $n$. MathWorld’s Isohedron page quotes the same total and traces it to the Grünbaum–Shephard work (MathWorld), while Wikipedia summarises the duality-based reasoning – Platonic ↔ Platonic, Archimedean → Catalan, prism → bipyramid, antiprism → trapezohedron (Wikipedia).

What the proof looks like (in outline)

1. Work on the sphere. Any convex polyhedron gives a tiling of the unit sphere by projecting its faces outward.

2. Classify all face-transitive spherical tilings. Grünbaum & Shephard (1981) listed every way a single tile can repeat around the sphere under a finite symmetry group.

3. Lift the tilings back to space. Each tiling corresponds to either

   • a Platonic solid,
   • the dual of a uniform spherical tiling (hence a Catalan solid), or
   • a 2-cap tiling whose dual is a bipyramid or trapezohedron. No other convex embedding exists – that’s the heart of the proof.

Because the argument happens on the sphere, it automatically rules out any stray convex isohedra that look different but would project to one of the known patterns.

Limits of the result

• If you drop convexity (allow self-intersection or star faces) there are many more isohedra, starting with the four Kepler–Poinsot polyhedra, and no complete list is known.
• If you go up one dimension (4-polytopes) or down to tilings of the plane, the problem is again open.
• Even in 3-D, the weaker “all faces merely congruent’’ condition (monohedral) is not classified; several sporadic convex examples are known that are not isohedral (Mathematics Stack Exchange).

So, within the usual confines of convex solids the answer is yes – every isohedron is either Platonic, Catalan, or one of the polar stack-of-pyramids families on your list. Outside that safe zone, the zoo is still very much alive and un-catalogued.

Do you have the citation for the study by Branko Grünbaum and Geoffrey C. Shephard?

Below is the full bibliographic information that most authors cite when they refer to the Grünbaum–Shephard classification of isohedral (face-transitive) spherical tilings—which, in turn, yields the list of convex isohedral polyhedra you asked about.

Ref.	Standard citation (APA-style)	Notes
[GS 1981a]	Grünbaum, B., & Shephard, G. C. (1981). Spherical tilings with transitivity properties. In C. Davis, B. Grünbaum & F. A. Sherk (Eds.), The Geometric Vein: The Coxeter Festschrift (pp. 65-98). Springer-Verlag. https://doi.org/10.1007/978-1-4612-5648-9_4	The definitive, worked-out classification of all isohedral, isotoxal and isogonal tilings of the 2-sphere. Convex isohedral polyhedra correspond to these tilings via radial projection and duality. (SpringerLink)
[GS 1981b]	Grünbaum, B., & Shephard, G. C. (1981). Patterns on the 2-sphere. Mathematika, 28(1), 1-35. https://doi.org/10.1112/S0025579300015321	A shorter companion journal article that previews some of the same results in a more compressed form. (londmathsoc.onlinelibrary.wiley.com)

Either reference is usually accepted; the book chapter [GS 1981a] is the most frequently quoted source when people assert that “every convex isohedron is Platonic, Catalan, or one of the three polar families.”

To The Winner Go The Potatoes

In the philosophy of Humanitism, what does it mean that "to the winner go the potatoes"?

Short answer: In Machado de Assis’s satirical “philosophy” of Humanitism (Portuguese Humanitismo), the slogan “Ao vencido, ódio ou compaixão; ao vencedor, as batatas” (“To the vanquished, hatred or pity; to the victor, the potatoes”) sums up Quincas Borba’s claim that in any struggle for scarce resources the side that prevails rightfully takes the means of survival—the potatoes—while the losers’ fate doesn’t matter because individual lives are insignificant next to the abstract totality “Humanitas.” It’s a darkly comic reduction of grand metaphysics to crude material spoils, used by Machado to parody Social-Darwinist and positivist justifications for violence, competition, and social inequality. (Wikisource, Scribd, Wikipedia)

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How the motto is introduced

Quincas Borba explains to Rubião: imagine a field with potatoes enough to nourish only one of two starving tribes; if they share, both die; if one destroys the other, the survivors gain strength to reach more food. Hence “peace is destruction; war is preservation… To the vanquished, hatred or compassion; to the victor, the potatoes.” (Wikisource, Scribd, TeX talk)

What it asserts within Humanitism

From this parable Borba infers that conflict which secures resources for the stronger side serves life and thus serves the universal substance Humanitas; individuals are mere “bubbles” in the boiling water—transient phenomena whose extinction is irrelevant to the underlying whole. The ethical upshot: material benefit to the winner trumps concern for the defeated. (Wikisource, Scribd, Wikipedia, biblioteca.pucrs.br)

Machado’s satirical target

Critics read Humanitism as Machado’s parody of nineteenth-century scientistic creeds—especially Social Darwinism and Comtean positivism—that dressed up the “law of the strongest” in lofty philosophical language. By pushing this logic to absurdity (“to the winner, the potatoes”), Machado exposes the inhuman, self-justifying moral calculus that can underwrite war, exploitation, or simple greed. (Wikipedia, TeX talk, lac.ox.ac.uk, Terra)

Later and idiomatic use

Because the scene is so memorable, “Ao vencedor, as batatas” has entered Brazilian Portuguese as an idiom roughly equivalent to “to the victor go the spoils,” invoked—often wryly—in contexts of competition, money, or power to mean that the winner walks away with the goods. (Wiktionary, Terra, VEJA, biblioteca.pucrs.br)

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In sum: Within Humanitism the phrase literalizes a brutal survival calculus (winner eats; loser disappears) and, by exaggeration, lets Machado de Assis mock any philosophy that rationalizes inequality or violence as serving some higher good. (Wikisource, Scribd, Wikipedia)

Proof ∀a,b ∈ ℝ [ a = b ⇔ ( ∀ ϵ ∈ ℝ [ ϵ > 0 ⇒ |a − b| < ϵ ] ) ]

Prove ∀a,b ∈ ℝ [ a = b ⇔ ( ∀ ϵ ∈ ℝ [ ϵ > 0 ⇒ |a − b| < ϵ ] ) ].

We are asked to prove the equivalence:

$$\mathrm{\forall }a,b\in \mathbb{R},a=b⟺(\mathrm{\forall }ϵ\in \mathbb{R},ϵ>0\Rightarrow |a-b|<ϵ)$$

We will prove this biconditional statement by proving both directions:

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(⇒) Forward direction:

Assume $a=b$. We want to show:

$$\mathrm{\forall }ϵ>0,|a-b|<ϵ$$

Since $a=b$, we have $|a-b|=|0|=0$. For any $ϵ>0$, clearly $0<ϵ$. So, $|a-b|<ϵ$.

✅ Thus, the implication holds in the forward direction.

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(⇐) Reverse direction:

Assume:

$$\mathrm{\forall }ϵ>0,|a-b|<ϵ$$

We want to show $a=b$. We’ll proceed by contradiction:

Suppose $a\ne b$. Then $|a-b|=\delta$ for some $\delta >0$. Now, choose $ϵ=\delta /2>0$. Then:

$$|a-b|=\delta >\delta /2=ϵ$$

This contradicts the assumption that $|a-b|<ϵ$ for all $ϵ>0$.

❌ So, the assumption $a\ne b$ must be false. Hence:

✅ $a=b$

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✅ Conclusion:

$$\boxed{{\displaystyle \mathrm{\forall }a,b\in \mathbb{R},a=b⟺\mathrm{\forall }ϵ>0,|a-b|<ϵ}}$$

Classics of Misogynistic Literature (o3)

classics of misogynistic literature from Hesiod to the modern day

Below is a chronological sampler of widely discussed texts that contain sustained or strikingly negative generalizations about women, enforce female subordination, eroticize domination, or otherwise figure prominently in scholarly conversations about misogyny—from Hesiod through the contemporary era. Inclusion here doesn’t mean every line of a work is hostile to women, nor that the author “hated women” in a simple biographical sense. Many are complex, ironic, parodic, or voiced through characters; some have been forcefully reinterpreted by later readers. I flag them because they’ve become touchstones in debates about literary misogyny.

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How I’m Using “Misogynistic”

Working definition: A text (or major passage) that (1) depicts women as inherently inferior, corrupting, dangerous, or burdensome as a class; (2) normalizes or eroticizes domination/violence toward women; (3) persistently trivializes women’s subjectivity; or (4) has attracted a durable critical tradition labeling it misogynistic. Context matters: sometimes a character speaks misogyny; sometimes satire undercuts it; sometimes readers disagree. I note such complications briefly.

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Antiquity (Greek & Roman)

• Hesiod – Theogony & Works and Days (c. 700 BCE). The Pandora myth inaugurates “woman” as a divine punishment who unleashes woes on men; Works and Days amplifies economic/sexual anxieties about wives as consumers of male labor.
• Semonides of Amorgos – “Types of Women” (7th c. BCE). Iambic catalogue reducing women to animal-nature stereotypes (sow, fox, dog, etc.), ending that the best woman barely exists; a paradigmatic ancient misogynist poem.
• Euripides – Hippolytus (428 BCE; misogynist speeches). The title character’s tirades that “woman” is a plague are framed dramatically but have long been excerpted as antifeminist sententiae.
• Aristophanic Tradition (esp. Lysistrata, Thesmophoriazusae, Assemblywomen, 5th–4th c. BCE). Comic gender inversions rely on stock jokes about women’s lust, secrecy, and domestic foibles; reception divided between reading as satire on men vs. recycling misogynist tropes.
• Ovid – Ars Amatoria; parts of Amores & Metamorphoses (1 BCE–8 CE). Seduction “handbook” that encourages manipulative pursuit; the epic overflows with sexual coercion by gods and heroes—central to modern discussions of rape culture in classical myth.
• Juvenal – Satire VI (early 2nd c. CE). A blistering, book-length diatribe cataloguing every imaginable female vice; the canonical Roman invective against women.
• Martial – Selected Epigrams (1st c. CE). Recurrent mockery of wives, widows, and women’s bodies; brief but influential in shaping comic misogynist topoi.

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Late Antiquity & Patristic / Early Christian Polemic

• Tertullian – De Cultu Feminarum (“On the Apparel of Women,” c. 2nd–3rd c.). Casts all women as “the devil’s gateway,” morally dangerous because of Eve’s precedent; foundational for later clerical antifeminism.
• Jerome – Adversus Jovinianum (c. 393 CE). In valorizing virginity, denigrates marriage and female sexuality; mined for antifeminist authorities in the Middle Ages.

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Medieval Europe

• Andreas Capellanus – De Amore (late 12th c.). Book III overturns earlier courtly love rules with sweeping denunciations of women; its sincerity debated but long quoted as antifeminist proof text.
• Jean de Meun (continuation of Guillaume de Lorris) – Le Roman de la Rose (c. 1275). Allegorical encyclopedia that satirizes and slanders women’s morals, sexuality, & speech; the lightning rod that provoked Christine de Pizan’s famous defense of women.
• “Antifeminist Authorities” Compilations (e.g., Lamentations of Matheolus, florilegia). Sourcebooks of clerical slurs against wives—visible inside Chaucer’s Wife of Bath’s Prologue, which quotes and parodies them.
• Geoffrey Chaucer – Select Canterbury Tales Passages (late 14th c.). While Chaucer is often sympathetic, the Clerk’s Tale (extreme wifely obedience), Merchant’s Tale, and antifeminist citations in the Wife of Bath keep him central to the misogyny debate.
• Heinrich Kramer & Jacob Sprenger – Malleus Maleficarum (1486). Witch-hunting manual asserting women’s inherent carnality and susceptibility to the devil; massively influential in gendered persecution (polemical prose, yet culturally “classic”).

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Reformation & Early Modern

• John Knox – The First Blast of the Trumpet Against the Monstrous Regiment of Women (1558). Polemic denying the legitimacy of female rulers (targeting Mary Tudor & Mary of Guise); seminal political misogyny.
• Joseph Swetnam – The Arraignment of Lewd, Idle, Froward, and Unconstant Women (1615). Popular pamphlet vilifying women; sparked a lively counter-pamphlet tradition (“Swetnam controversy”).
• William Shakespeare – The Taming of the Shrew (c. 1590–94). Comic “taming” of a rebellious woman; productions range from endorsing patriarchal submission to ironic critique—hence perpetual argument about misogyny.
• Ben Jonson – Epicoene, or The Silent Woman (1609). Satire of noise, marriage, and gender performance that trades in stereotypes of the desirable “silent” wife.

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17th–18th Centuries (Restoration, Enlightenment, Augustan)

• Samuel Butler – Hudibras (1663–78, esp. anti-woman jibes). Burlesque poem including stock lampoons of female fickleness.
• Jonathan Swift – “The Lady’s Dressing Room” (1732); “A Beautiful Young Nymph Going to Bed” (1731). Scatological exposure poems deflating idealized femininity; often cited as classic literary disgust at women’s bodies.
• Alexander Pope – “Epistle II. To A Lady” (Moral Essays, 1735) & related verse. Brilliant couplets that reduce women to extremes (saint or devil, angel or ape); cornerstone of rhetorical misogyny discussions.
• Jean-Jacques Rousseau – Émile (1762, Book V on Sophie). Prescribes an education making women pleasing, modest, subordinate to men’s moral development; philosophically influential gender hierarchy.

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19th Century & Turn of the 20th

• Arthur Schopenhauer – “On Women” (1851). Blanket claims of women’s intellectual & moral inferiority; heavily anthologized philosophical misogyny.
• Leo Tolstoy – The Kreutzer Sonata (1889). Narrator’s fevered condemnation of female sexuality & marriage; critics debate whether Tolstoy endorses or exposes pathological jealousy, but it’s long read as misogynist.
• August Strindberg – The Father (1887), Creditors (1888), Miss Julie (1888, contested). Ferocious gender battles; author’s public writings intensified charges of woman-hatred.
• Otto Weininger – Sex and Character (1903). Pseudo-scientific classification ranking “Woman” as amorphous, amoral; hugely influential in European intellectual circles.

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20th Century (Modernist to Late)

• D. H. Lawrence – e.g., Women in Love (1920), Lady Chatterley’s Lover (1928). Erotic mysticism entwined with hierarchical gender myths; accused alternately of misogyny and of celebrating female desire—critical battleground.
• Ernest Hemingway – recurrent portrayals (e.g., The Sun Also Rises, 1926). Critics have read his women as projections of male anxiety; debates continue.
• Henry Miller – Tropic of Cancer (1934), Tropic of Capricorn (1939). Sexually explicit narratives objectifying women; landmark in censorship fights and misogyny critiques alike.
• Norman Mailer – An American Dream (1965); essay “The Prisoner of Sex” (1971). Violent fantasy & anti-feminist polemic made Mailer emblematic of postwar macho misogyny.
• Philip Roth – Portnoy’s Complaint (1969), later Sabbath’s Theater (1995). Male narrators whose obsessive sexual monologues slide into misogynist caricature; often defended as satire of their narrators.
• John Updike – Rabbit novels (1960–2000). Chronic domestic realism that, to some readers, reduces women to foils for male spiritual angst; frequently cited in feminist critique.

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Late 20th–21st Century / Contemporary

• Bret Easton Ellis – American Psycho (1991). Graphic torture-murder of women amid consumer satire; debates over whether it indicts or reproduces misogynistic violence.
• Michel Houellebecq – Whatever (1994), The Elementary Particles (1998), Submission (2015), others. Bleak sexual-market novels voiced through alienated men whose contempt for women (and men) has drawn persistent charges of misogyny (often broadened to misanthropy).
• Neil Strauss – The Game (2005). Memoir/manual from pickup-artist culture that systematizes manipulative tactics toward women; culturally influential in contemporary misogynist subcultures (more pop-nonfiction than “literary classic,” but historically significant).
• Online “Manosphere” / Incel Forums Anthologies (2000s–present). Not canonical literature, yet vital primary sources for current misogynist rhetoric; if you’re mapping the tradition forward, these digital texts extend long-standing tropes into networked culture.

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Pair Them with Counter-Voices

Reading misogynistic texts alongside feminist or woman-authored responses sharpens critical insight. Classic pairings include:

• Hesiod ↔ Sappho fragments (alternative female lyric voice).
• Roman de la Rose ↔ Christine de Pizan, The Book of the City of Ladies (1405).
• Swetnam ↔ Swetnam the Woman-Hater Arraigned by Women (1617).
• Knox ↔ Queen Elizabeth I’s speeches; Mary Astell’s A Serious Proposal to the Ladies.
• Pope/Swift ↔ Mary Wollstonecraft, A Vindication of the Rights of Woman (1792).
• Schopenhauer/Weininger ↔ Simone de Beauvoir, The Second Sex (1949).
• Mailer/Roth/Ellis ↔ contemporary feminist criticism (e.g., Kate Millett’s Sexual Politics, 1970).

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Using the List

Study suggestions:

1. Contextualize: What social, theological, or political anxieties does the text displace onto “women”?
2. Voice vs. Author: Distinguish character speech, narrator stance, and authorial irony.
3. Reception History: Track when and why readers have labeled the work misogynistic; judgments change.
4. Counter-Readings: What resistant or feminist readings reappropriate the text?

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Want More?

If you’d like:

• Brief key passages to examine (with context & warnings),
• A syllabus-style reading sequence,
• Non-Western traditions (e.g., Confucian “admonitions to women,” certain legal-religious codes, or South Asian texts),
• Or a focus on literature that critiques misogyny rather than embodies it,

just let me know and I can expand or create a printable reading list.

Classics of Misogynistic Literature (4o)

classics of misogynistic literature from Hesiod to the modern day

Here's a list of notable (and often notorious) works in the literary canon that have been critiqued for their misogynistic content, either in theme, portrayal of women, or ideological stance. This is not an endorsement of their ideas but an overview of historically significant works that reflect and often reinforce patriarchal or anti-feminine worldviews.

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Ancient & Classical Antiquity

Hesiod – Theogony & Works and Days (c. 700 BCE)

• Misogynistic Themes: Pandora, the first woman, is depicted as a divine punishment to men, a "beautiful evil."
• Legacy: Foundational in shaping Western attitudes toward women as the source of male suffering.

Euripides – Medea (431 BCE)

• Complexity: While Medea is a powerful and intelligent woman, she's also demonized as a child-killer and "monster" in the name of passion and betrayal.
• Controversy: Seen both as a proto-feminist icon and a misogynistic caricature.

Aristotle – Politics (4th century BCE)

• Misogynistic Elements: Asserts the natural inferiority of women and justifies male dominance in the household and state.

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Medieval to Early Modern

Juvenal – Satire VI (2nd century CE)

• Infamous for: A brutal, scathing diatribe against Roman women—accusing them of deceit, lust, and corruption.

Tertullian – On the Apparel of Women (c. 200 CE)

• Quote: Refers to women as “the devil’s gateway.”
• Impact: Deeply influential in early Christian thought on female modesty and guilt.

Geoffrey Chaucer – The Canterbury Tales (14th century)

• Problematic Characters: The "Wife of Bath" is often interpreted as a misogynistic stereotype, despite some feminist re-readings.

Martin Luther – Various Writings (16th century)

• Examples: Called women "weak" and "vain"; promoted strict domestic roles.
• Legacy: Helped codify Protestant domestic ideals that deeply restricted women’s autonomy.

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Enlightenment & 18th Century

Jean-Jacques Rousseau – Émile (1762)

• Gender Roles: Argues that women should be educated only to serve and please men.
• Criticism: Central to Enlightenment-era gender norms, despite being a progressive thinker in other areas.

Jonathan Swift – The Lady's Dressing Room (1732)

• Satirical but hostile: Peels back the facade of femininity to mock women’s vanity and alleged filth.

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19th Century

Arthur Schopenhauer – On Women (1851)

• Quote: “Women are by nature meant to obey.”
• Impact: Openly hostile toward women, influencing later pessimistic and anti-feminist thought.

Honoré de Balzac – Physiology of Marriage (1829)

• Cynical views: Treats women as scheming, shallow, and emotionally volatile beings.
• Popular in its time, reflecting patriarchal bourgeois attitudes.

Friedrich Nietzsche – Various Works

• Quotes: “When you go to a woman, take your whip.” (Thus Spoke Zarathustra)
• Philosophical: Saw femininity as weakness; idealized masculine creative force.

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20th Century to Modern

D. H. Lawrence – Various Novels

• Contradictory: Praised female sexuality but also framed it as subordinate to male creative energy.
• Criticism: Portrayed women as either passive vessels or dangerously disruptive.

Norman Mailer – An American Dream (1965)

• Controversial Scenes: Includes murder and rape of a wife, depicted without remorse.
• Mailer’s Persona: Embraced a macho, combative style often seen as violently misogynistic.

Henry Miller – Tropic of Cancer (1934)

• Content: Highly sexualized, objectifying language about women.
• Legacy: Celebrated for stylistic innovation, but heavily criticized for reducing women to sexual functions.

Philip Roth – Portnoy’s Complaint (1969)

• Perspective: Hilariously neurotic but often reductive in its treatment of women.
• Legacy: A flashpoint in debates about sexism in modern fiction.

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Contemporary Examples

Bret Easton Ellis – American Psycho (1991)

• Depiction: Explicit, sadistic violence toward women.
• Debate: Whether it’s satire or indulgence in misogyny remains contentious.

Michel Houellebecq – Submission (2015)

• Themes: Misogyny and sexual politics framed through a dystopian future.
• Critics: Call his work deeply nihilistic and contemptuous toward women.

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Conclusion

While many of these works are considered literary landmarks, their treatment of women has been—and continues to be—critically examined. Some are reflective of their time; others helped shape misogynistic ideologies. Engaging with these texts critically can reveal much about the cultural, philosophical, and psychological underpinnings of gender inequality through the ages.

Would you like a timeline chart, reading list with annotations, or feminist counterpoints to these texts?

Proof of square tesselation with trominos

For each natural number n, consider a square consisting of 2^(2n) smaller squares (i.e., a 2^n × 2^n square). If any one of the smaller squares is removed from the large square (not necessarily from the corner), then the resulting region can be completely covered by trominos (each made up of three small squares of the same size as the small squares in the region) in such a way that the trominos do not overlap.

Yes, for every natural number $n$, if we remove any one of the $2^{2n}$ smaller squares from a $2^n\times 2^n$ square, it is always possible to tile the remaining region using only L-shaped trominos (three connected unit squares forming an "L") without overlap or gaps.

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✅ Key Concepts:

• A tromino is an L-shaped tile made up of 3 connected unit squares.
• We are working with a $2^n\times 2^n$ square, which contains $2^{2n}$ unit squares.
• One square is removed, leaving $2^{2n}-1$ unit squares to be tiled.

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🧠 The Proof (by induction):

Base case: $n=1$

• A $2^1\times 2^1=2\times 2$ square has 4 unit squares.
• Removing one square leaves 3 squares — which perfectly matches the shape of a tromino.
• ✅ The base case holds.

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Inductive step:

Assume the statement holds for some $n=k$: Any $2^k\times 2^k$ square with one square removed can be tiled with trominos.

Now consider a $2^{k+1}\times 2^{k+1}$ square.

• This square can be divided into four $2^k\times 2^k$ quadrants.
• The removed square lies in one of these four quadrants.
• By the inductive hypothesis, we can tile the affected quadrant after placing a tromino at the center of the full square that covers one square in each of the other three quadrants, near the center.
• Now each of the four quadrants has one square missing, so each can be tiled by the inductive hypothesis.

✅ This completes the inductive step.

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📌 Conclusion:

By mathematical induction, any $2^n\times 2^n$ square with one unit square removed can always be tiled completely with L-shaped trominos.

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Would you like a visual demonstration or code to simulate this tiling?

Proof of Binet’s formula

Where x_n is the nth Fibonacci number, prove that $\mathrm{\forall }n\in \mathbb{N},x_n=\frac{1}{\sqrt{5}}[{\left(\frac{1+\sqrt{5}}{2}\right)}^n-{\left(\frac{1-\sqrt{5}}{2}\right)}^n]$.

To prove the closed‐form (“Binet’s”) formula

$$x_n=\frac{1}{\sqrt{5}}({\varphi }^n-{\psi }^n),\text{where}\varphi =\frac{1+\sqrt{5}}{2},\psi =\frac{1-\sqrt{5}}{2},$$

we’ll show that the right‐hand side satisfies

1. the same recurrence as the Fibonacci numbers,
2. the same initial conditions,

and then invoke uniqueness of solutions to a linear recurrence.

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1. Setup: characteristic equation

The Fibonacci numbers are defined by

$$x_0=0,x_1=1,x_n=x_{n-1}+x_{n-2}(\mathrm{\forall }n\ge 2).$$

The associated characteristic polynomial is

$$r^2-r-1=0,$$

whose two roots are

$$\varphi =\frac{1+\sqrt{5}}{2},\psi =\frac{1-\sqrt{5}}{2}.$$

Since this is a distinct‐root second‐order homogeneous recurrence, its general solution is

$$a_n=A{\varphi }^n+B{\psi }^n$$

for constants $A,B$ determined by $a_0$ and $a_1$.

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2. Determine $A$ and $B$ via initial conditions

We want $a_n=x_n$, so impose

1. $a_0=A+B=x_0=0$,
2. $a_1=A\varphi +B\psi =x_1=1$.

From $A+B=0$ we get $B=-A$. Plugging into the second,

$$A\varphi -A\psi =1⟹A(\varphi -\psi )=1.$$

But $\varphi -\psi =\sqrt{5}$, so $A=1/\sqrt{5}$ and hence $B=-1/\sqrt{5}$. Therefore

$$a_n=\frac{1}{\sqrt{5}}{\varphi }^n-\frac{1}{\sqrt{5}}{\psi }^n=\frac{1}{\sqrt{5}}({\varphi }^n-{\psi }^n).$$

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3. Check that $a_n$ satisfies the Fibonacci recurrence

Since $\varphi$ and $\psi$ are roots of $r^2=r+1$, we have

$${\varphi }^n={\varphi }^{n-1}+{\varphi }^{n-2},{\psi }^n={\psi }^{n-1}+{\psi }^{n-2}.$$

Hence

$$a_n=\frac{{\varphi }^n-{\psi }^n}{\sqrt{5}}=\frac{({\varphi }^{n-1}+{\varphi }^{n-2})-({\psi }^{n-1}+{\psi }^{n-2})}{\sqrt{5}}=\frac{{\varphi }^{n-1}-{\psi }^{n-1}}{\sqrt{5}}+\frac{{\varphi }^{n-2}-{\psi }^{n-2}}{\sqrt{5}}=a_{n-1}+a_{n-2}.$$

Thus $a_n$ obeys the same recurrence as $x_n$.

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4. Conclusion by uniqueness

We have produced a sequence $a_n=({\varphi }^n-{\psi }^n)/\sqrt{5}$ which

• satisfies the same recurrence $a_n=a_{n-1}+a_{n-2}$,
• has the same initial values $a_0=0$, $a_1=1$.

By the standard uniqueness theorem for linear recurrences, it follows that

$$a_n=x_n\mathrm{\forall }n\in \mathbb{N}.$$

That is,

$$\boxed{{\displaystyle x_n=\frac{1}{\sqrt{5}}({\varphi }^n-{\psi }^n)=\frac{1}{\sqrt{5}}((\frac{1+\sqrt{5}}{2}{)}^n-(\frac{1-\sqrt{5}}{2}{)}^n).}}$$

This completes the proof. ∎

Primes using Fermat numbers

Define the nth Fermat number, F_n, by $F_n=2^{2^n}+1$ for n = 0, 1, 2, 3, …. The first few Fermat numbers are F0 = 3, F1 = 5, F2 = 17, F3 = 257.

Prove by induction that F0 ⋅ F1⋯Fn-1 + 2 = Fn, for n ≥ 1.

Proof by induction on $n\ge 1$.

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Base case ($n=1$)

We must check that

$$F_0+2=F_1.$$

Since $F_0=2^{2^0}+1=2+1=3$ and $F_1=2^{2^1}+1=4+1=5$, we have

$$F_0+2=3+2=5=F_1,$$

so the statement holds for $n=1$.

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Inductive step

Inductive hypothesis. Suppose for some $k\ge 1$ that

$$F_0{F}_1\cdots F_{k-1}+2=F_k.$$

Equivalently,

$$F_0{F}_1\cdots F_{k-1}=F_k-2.$$

We must show

$$F_0{F}_1\cdots F_{k-1}{F}_k+2=F_{k+1}.$$

Starting from the left‐hand side,

$$F_0{F}_1\cdots F_{k-1}{F}_k+2=(F_k-2)F_k+2(\text{by the inductive hypothesis}).$$

Compute

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

Expand:

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

so

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

Thus

$$F_0{F}_1\cdots F_{k-1}{F}_k+2=(F_k^2-2F_k)+2=(2^{2^{k+1}}-1)+2=2^{2^{k+1}}+1=F_{k+1}.$$

This completes the inductive step.

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By the principle of mathematical induction, for every integer $n\ge 1$,

$$F_0{F}_1\cdots F_{n-1}+2=F_n,$$

as claimed. ∎

Use this formula to prove that there are an infinite number of primes, by showing that no two Fermat numbers have any prime factors in common.

Here’s the standard argument.

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1. Fermat numbers are pairwise coprime

Recall from the induction you’ve already proved that, for every $n\ge 1$,

$$F_n=F_0{F}_1\cdots F_{n-1}+2.$$

Fix two indices $0\le m<n$. Suppose $p$ is a prime dividing both $F_m$ and $F_n$. Then:

1. Since $p\mid F_n$ and

   $$F_n=F_0{F}_1\cdots F_{n-1}+2,$$

   we have

   $$F_0{F}_1\cdots F_{n-1}+2\equiv 0(\mathrm{mod}p)⟹F_0{F}_1\cdots F_{n-1}\equiv -2(\mathrm{mod}p).$$

2. But $p\mid F_m$ for some $m<n$, so $p$ divides the product $F_0{F}_1\cdots F_{n-1}$. Hence

   $$F_0{F}_1\cdots F_{n-1}\equiv 0(\mathrm{mod}p).$$

3. Combining the two congruences,

   $$0\equiv F_0\cdots F_{n-1}\equiv -2(\mathrm{mod}p)⟹p\mid 2.$$

   Thus $p=2$. But every Fermat number $F_k=2^{2^k}+1$ is odd, so cannot be divisible by 2.

Therefore no prime $p$ can divide both $F_m$ and $F_n$ for $m<n$. Equivalently,

$$gcd(F_m,F_n)=1\text{whenever }m\ne n.$$

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2. Infinitely many primes

Since each Fermat number $F_n$ is greater than 1, it has at least one prime divisor. By the argument above, no prime divisor of $F_n$ can ever appear among the prime divisors of any other $F_m$ with $m\ne n$. Hence each $F_n$ contributes at least one new prime that we have never seen before. As $n$ ranges over the nonnegative integers, we thus get infinitely many distinct primes. ∎