Thiago’s ChatGPT Blog

Wednesday, May 28, 2025

Bertrand postulate proof sketch

Prove that ∀n∈ℕ(n>1⟹∃p∈ℙ(n<p<2n)).

Below is a classical proof of

$$ \forall n\in\Bbb N,\;n>1\;\Longrightarrow\;\exists\,p\in\Bbb P\text{ with }n<p<2n, $$

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

1. Chebyshev’s θ-function

Define

$$ \theta(x)\;=\;\sum_{\substack{p\le x\\p\;\text{prime}}}\ln p. $$

Chebyshev showed by completely elementary means (bounding factorials via products) that there are absolute constants $A,B>0$ such that for all $x\ge2$,

$$ A\,x\;\le\;\theta(x)\;\le\;B\,x. $$

One may take for instance

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

for every $x\ge2$. (The proof is a straightforward—but somewhat lengthy—analysis of the prime‐power factors of $n!$.)

2. From θ to “a prime in $(n,2n)$”

Observe that

$$ \theta(2n)\;-\;\theta(n)\;=\;\sum_{\,n< p\le 2n}\ln p. $$

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

$$ \theta(2n)-\theta(n)\;\ge\;0.92\,(2n)\;-\;1.11\,n \;=\;(1.84-1.11)\,n \;=\;0.73\,n \;>\;0 \quad(\forall n>1). $$

Hence

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

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

Historical remarks

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

Proof that ∀n∈ℕ(n>2⟹∃p∈ℙ(n<p<n!))

Prove that, for every natural number n greater than 2, there is a prime number between n and n!.

Proof: For every natural number $n > 2$, there is a prime number between $n$ and $n!$.

Step 1: Understand the problem.

We are asked to show that for any integer $n > 2$, there exists at least one prime $p$ such that:

$$ n < p < n! $$

Step 2: Use a classical approach based on divisor arguments.

Consider the number:

$$ N = n! - 1 $$

Let's analyze the prime factors of $N$.

Step 3: Key observation.

For any integer $k$ with $2 \leq k \leq n$, $k$ divides $n!$ (by the definition of factorial).
But $k$ does not divide $n! - 1$ because:

$$ n! \equiv 0 \pmod{k} \implies n! - 1 \equiv -1 \not\equiv 0 \pmod{k} $$

This means none of the integers $2, 3, \dots, n$ divides $N = n! - 1$.

Step 4: Consequence.

Any prime divisor of $N$ must be greater than $n$.

Step 5: Conclusion.

$N = n! - 1$ is either prime itself (in which case it's a prime greater than $n$) or composite.
If it's composite, it must have a prime factor greater than $n$, since none of the primes $\leq n$ divide it.

Thus, there exists a prime $p > n$ such that $p$ divides $n! - 1$.

Since $n! - 1 < n!$, this prime $p$ satisfies:

$$ n < p < n! $$

✅ Therefore, for every integer $n > 2$, there is a prime number between $n$ and $n!$.

This proof uses an idea similar to Euclid's proof of the infinitude of primes but applied to the interval $(n, n!)$.

Proof that 3,5,7 are the only 3 primes separated by 2

Suppose that p, p + 2, and p + 4 are prime numbers. Prove that p = 3.

Proof that if $p$, $p + 2$, and $p + 4$ are all prime numbers, then $p = 3$.

Step 1: Understand the problem

We are told that $p$, $p+2$, and $p+4$ are all prime numbers, and asked to prove that the only possible value for $p$ is $3$.

Step 2: Consider modulo 3 behavior

Any integer is congruent to one of 0, 1, or 2 modulo 3.

Check how this applies to the numbers $p$, $p+2$, and $p+4$.

Step 3: Check modulo 3 cases

If $p \equiv 0 \pmod{3}$, then $p = 3$ is the only prime satisfying this (since 3 is the only prime divisible by 3).

Test it:

$p = 3$
$p+2 = 5$ (prime)
$p+4 = 7$ (prime)

✔️ Works.

Check whether any other options exist.

If $p \equiv 1 \pmod{3}$, then
$p+2 \equiv 3 \equiv 0 \pmod{3}$

This means $p+2$ is divisible by 3.

The only prime divisible by 3 is 3 itself, but $p+2 > p \geq 1$, so $p+2 > 3$, and the only possibility is if $p+2=3$, leading to $p=1$, but 1 is not prime.

So in this case, $p+2$ would be composite for any $p \equiv 1 \pmod{3}$ with $p > 3$.

Contradicts the assumption that $p+2$ is prime.

If $p \equiv 2 \pmod{3}$, then
$p+4 \equiv (2+4) \equiv 6 \equiv 0 \pmod{3}$

This means $p+4$ is divisible by 3.

Again, unless $p+4=3$ (impossible because $p+4 > p \geq 2$), this means $p+4$ is composite.

Contradicts the assumption that $p+4$ is prime.

Step 4: Conclusion

The only possible value where neither $p+2$ nor $p+4$ are divisible by 3 is when $p\equiv 0\pmod{3}$, i.e., $p=3$.

Test:

$3, 5, 7$ — all prime ✔️

No other $p$ works.

Therefore, the only possible prime number $p$ such that $p$, $p+2$, and $p+4$ are all primes is

$$ p = 3 $$

✔️ QED.

Smallest prime where the successor of the product of primes up to it is not prime

Find a prime number p such that the number (2 ⋅ 3 ⋅ 5 ⋅ 7 ⋅⋅⋅ p) + 1 is not prime.

Let's break this problem down step by step.

Problem restatement:

Find a prime number $p$ such that the number

$$ (2 \cdot 3 \cdot 5 \cdot 7 \cdots p) + 1 $$

is not prime.

This is the product of all primes up to $p$ (the primorial of $p$) plus 1. You're being asked for the smallest prime $p$ where this is not prime (i.e., composite).

Step 1: Test Small Values of $p$

Let's compute $(2 \cdot 3 \cdot 5 \cdots p) + 1$ for successive primes until it's not prime.

$p = 2$:

$$ 2 + 1 = 3 \text{ (prime)} $$

$p = 3$:

$$ 2 \cdot 3 + 1 = 6 + 1 = 7 \text{ (prime)} $$

$p = 5$:

$$ 2 \cdot 3 \cdot 5 + 1 = 30 + 1 = 31 \text{ (prime)} $$

$p = 7$:

$$ 2 \cdot 3 \cdot 5 \cdot 7 + 1 = 210 + 1 = 211 \text{ (prime)} $$

$p = 11$:

$$ 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 + 1 = 2310 + 1 = 2311 \text{ (test primality)} $$

Test 2311:

Not divisible by 13 (2311/13 = 177.76)
Not divisible by 17, 19, ... (test shows 2311 is prime)

$p = 13$:

$$ 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1 = 30030 + 1 = 30031 $$

Test 30031: Check divisibility:

30031 divided by 59:

$$ 30031 / 59 = 509 $$

Check:

$$ 59 \cdot 509 = 30031 \text{ ✅ } $$

So 30031 is composite.

✅ Therefore, the smallest prime $p$ such that $(2 \cdot 3 \cdot 5 \cdots p) + 1$ is not prime is

$$ \boxed{13} $$

Thursday, May 1, 2025

Denumerability of Subsets

Given an enumeration $ a_1, a_2, \ldots, a_n, \ldots $ of a denumerable set $ A $, and an enumeration $ b_1, b_2, \ldots, b_n, \ldots $ of a denumerable set $ B $, we can enumerate $ A \cup B $ in the order $ a_1, b_1, a_2, b_2, \ldots, a_n, b_n, \ldots $. Write a more general version of this statement for the union of arbitrarily many denumerable sets.

More generally, one shows:

Theorem.
If $\{A_m\}_{m\in\mathbb N}$ is a countable family of countable (denumerable) sets, then
\[ \bigcup_{m=1}^\infty A_m \] is itself countable.

Proof sketch (by explicit enumeration).

For each $m\in\mathbb N$, choose an enumeration

\[ A_m = \{\,a_{m,1},\,a_{m,2},\,a_{m,3},\dots\}. \]

Recall that $\mathbb N\times\mathbb N$ is countable. Fix any bijection

\[ \pi\colon \mathbb N \;\longrightarrow\;\mathbb N\times\mathbb N, \quad \pi(k) = (m(k),\,n(k)). \]

Define a new sequence $(c_k)_{k=1}^\infty$ by

\[ c_k \;=\; a_{\,m(k),\,n(k)}. \]

Then $\{c_1,c_2,c_3,\dots\}$ lists every element of $\bigcup_{m}A_m$

(possibly with repetitions), showing that $\bigcup_m A_m$ is countable.

Equivalently, if you prefer a “diagonal” style ordering without naming a particular pairing function, you can say:

List all pairs $(m,n)\in\mathbb N\times\mathbb N$ in a single sequence, for example by moving along the diagonals

\[ (1,1),\;(1,2),(2,1),\;(1,3),(2,2),(3,1),\;\dots \]

Whenever $(m,n)$ appears in that list, write down the $n$-th element of $A_m$, namely $a_{m,n}$.

Because every pair $(m,n)$ will eventually appear, every element of every $A_m$ will be listed. Hence a countable union of countable sets is countable.

The set of all infinite subsets of $ \mathbb{N} $ cannot be denumerable; show this by showing that the set of all finite subsets of $ \mathbb{N} $ is denumerable, and then applying the above theorem.

First, partition the collection of all finite subsets of $\mathbb{N}$ by their cardinality. For each $k\in\mathbb{N}$, let \[ \mathcal F_k = \bigl\{\,F\subseteq\mathbb{N}:\lvert F\rvert = k\bigr\} \] be the family of all $k$-element subsets of $\mathbb{N}$. Then:

Each $\mathcal F_k$ is denumerable.
An easy way to see this is to observe that any $k$-element subset can be written uniquely as an increasing $k$-tuple \[ (n_1,n_2,\dots,n_k) \quad\text{with}\quad n_1<n_2<\cdots<n_k,\quad n_i\in\mathbb{N}. \] The set of all such increasing $k$-tuples is a subset of $\mathbb{N}^k$, and $\mathbb{N}^k$ is denumerable (e.g.\ by the diagonal‐pairing argument). Hence each $\mathcal F_k$ is denumerable.
The set of all finite subsets is a countable union of the $\mathcal F_k$.
\[ \mathcal F \;=\; \bigl\{\,F\subseteq\mathbb{N}:\lvert F\rvert<\infty\bigr\} = \bigcup_{k=0}^\infty \mathcal F_k. \] By the theorem you just proved, a countable union of denumerable sets is denumerable. Therefore $\mathcal F$ is denumerable.

Now let \[ \mathcal I = \bigl\{\,X\subseteq\mathbb{N}:\lvert X\rvert=\infty\bigr\} \] be the collection of infinite subsets of $\mathbb{N}$. Notice that every subset of $\mathbb{N}$ is either finite or infinite, so \[ \mathcal P(\mathbb{N}) \;=\; \mathcal F \;\cup\;\mathcal I. \] Suppose, for the sake of contradiction, that $\mathcal I$ were denumerable. Then $\mathcal F$ is denumerable (as shown above), so by the union theorem again, \[ \mathcal P(\mathbb{N}) = \mathcal F \cup \mathcal I \] would be a union of two denumerable sets, hence itself denumerable. But this contradicts Cantor’s theorem that $\mathcal P(\mathbb{N})$ is uncountable. Therefore our assumption was false, and $\mathcal I$ cannot be denumerable. In other words, the family of all infinite subsets of $\mathbb{N}$ is uncountable.

Naturals Powerset Uncountable

show that the powerset of the natural numbers is not denumerable

To show that the powerset of the natural numbers $ \mathcal{P}(\mathbb{N}) $ is not denumerable, we aim to prove that it is uncountable—that is, it cannot be placed into a one-to-one correspondence with $ \mathbb{N} $ (the set of natural numbers).

Step-by-step Proof (Using Cantor’s Diagonal Argument)

1. Assume for contradiction:

Suppose $ \mathcal{P}(\mathbb{N}) $ is countable. Then there exists a bijection \[ f: \mathbb{N} \to \mathcal{P}(\mathbb{N}) \] So we can list all subsets of $ \mathbb{N} $ as: \[ f(0), f(1), f(2), \dots \] where each $ f(n) \subseteq \mathbb{N} $.

2. Construct a special subset:

Define a subset $ D \subseteq \mathbb{N} $ by: \[ D = \{ n \in \mathbb{N} \mid n \notin f(n) \} \] This is called the diagonal subset, because its construction resembles Cantor’s diagonalization.

3. Check whether $ D $ appears in the list:

Since we assumed all subsets are listed, $ D $ must be $ f(k) $ for some $ k \in \mathbb{N} $. So consider whether $ k \in D $:

If $ k \in D $, then by definition of $ D $, $ k \notin f(k) = D $ ⇒ contradiction.
If $ k \notin D $, then $ k \in f(k) = D $ ⇒ contradiction.

In either case, we get a contradiction.

Conclusion:

This contradiction shows our assumption was false. Therefore:

\[ \boxed{\mathcal{P}(\mathbb{N}) \text{ is not countable, i.e., not denumerable.}} \]

It is, in fact, uncountably infinite.

Thursday, March 27, 2025

Why Does Gerson Persist?

The context was the final two chapters (§9 and §10) of Platonism and Naturalism, by Lloyd P. Gerson, which are omitted.

why doesn't he see the issue raised in chapter 9 as having ruined his whole project?

Excellent question—and it strikes right at the philosophical heart of the matter.

In Chapter 9, especially through the challenges raised by Proclus and Damascius, the author confronts a deep aporia (philosophical impasse):
👉 How can an absolutely simple first principle—the One—cause or explain anything, if any causal connection implies complexity or relation?
👉 And if the One is too simple to be participated in, then what explanatory role can it play at all?

This appears to threaten the very core of Platonism, which relies on the One as the ultimate principle that grounds and explains everything.

So why doesn’t the author see this as fatal?

1. The Problem is Diagnostic, Not Destructive

He treats the issues raised in Chapter 9 not as a refutation, but as a diagnosis of the extreme tensions inherent in Platonism—tensions he believes must be faced squarely if philosophy is to remain philosophy.

Rather than ruin his project, these tensions:

Reveal the depth and necessity of Platonism’s metaphysical ambition
Demonstrate that any coherent alternative (especially Naturalism) faces equally or more devastating problems (e.g. explaining normativity, consciousness, or causation without metaphysical depth)

He sees the struggle to explain how the One can cause multiplicity as a defining feature of Platonism, not a disqualifying flaw.

2. Plotinus’s Response Is Still Defensible

Although Damascius presses hard on the paradoxes of participation and simplicity, the author still finds Plotinus’s original answer—the One as “δύναμις” (power), or pure “ἐνέργεια” (activity)—philosophically viable.

Plotinus doesn’t try to fully resolve the paradox, but shows how the tension itself can be philosophically illuminating:

The One does not “act” in a conventional sense but overflows or emanates.
This non-dual causality avoids direct relationality while still grounding being.

So the paradox isn’t so much a problem to be solved as it is a structure of reality to be understood differently—a Platonist metaphysics of asymmetrical dependence.

3. The Aporia is a Signpost, Not a Dead End

Damascius’s conclusion that the ultimate principle is ineffable and even non-referable doesn’t negate the project—it completes it, in a way.

In Chapter 10, the author implies that:

Ultimate philosophical explanation requires ultimate simplicity, even if we can’t conceptually grasp it.
The tension between intelligibility and transcendence is not a failure of Platonism—it’s what sets it apart from Naturalism.

Rather than demand a reductive, scientific kind of explanation, Platonism points beyond conceptual grasp to what grounds graspability itself.

4. The Alternatives Are Worse

Most crucially, the author believes that abandoning Platonism leads directly to:

Naturalism, which cannot justify normativity, knowledge, or metaphysical explanation
Or religion, which invokes authority and revelation, removing philosophy from public reason

The collapse of the Platonic One into aporia still leaves Platonism as the only coherent metaphysical option for sustaining philosophy’s distinct subject matter.

5. He Embraces the Strain as a Defining Feature of Real Philosophy

The author ends by recommending a new pedagogical model, where philosophy is understood as the unfolding of this very tension—between intelligibility and its ground, between Platonism and its critiques.

So, to him, the “crack in the system” doesn’t destroy the system; it marks the point where philosophy becomes serious.

Would you say his response is compelling? Or do you feel the paradox really should be fatal?