On 3/16/2024 1:31 PM, Ross Finlayson wrote:

A function surjects the rational numbers onto
the irrational numbers

Ross A. Finlayson
December 28, 2006

Abstract
There exists a surjection
from the rational numbers
onto the irrational numbers.

Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.

There is a distinct q e Q for each p e P.

Proof.

De.ASCII.fied.
|
| For pᵢ ∈ ℙ⁺ = ℝ⁺\ℚ⁺
| Q_i+ =
| ℚ⁺ᵢ = {q ∈ ℚ⁺: q<pᵢ}
|
| ∀pₕ < pᵢ: ℚ⁺ₕ ≠⊂ ℚ⁺ᵢ
|
| Q_nlt i =
| ℚᐠᑉᵢ =
| ℚ⁺ᵢ\(⋃[pₕ<pᵢ] ℚ⁺ₕ) =
| ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
|
| It is so that ℚᐠᑉ ᵢ ≠ ∅
| else
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or ∃pₕ = pᵢ

What is X else Y ?
X Y X.else.Y
t t ?
f t ?
t f ?
f f ?

----
It is not so that ℚᐠᑉᵢ ≠ ∅
ℚᐠᑉᵢ = ∅

For each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∀r₁,r₂ ∈ ℝ, r₁<r₂:
∃p₁₂ ∈ ℙ: r₁ < p₁₂ < r₂

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
∃pₕ ∈ ℙ⁺: q < pₕ < pᵢ:
q ∈ ℚ⁺ₕ
q ∉ ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ℚᐠᑉᵢ

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
q ∉ ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Proof.

Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+:
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.

Let
Q_nlt i =
Q_i+ \ (  U p_h < p_i  Q_h+  ) =
Int p_h< p_i (Q_i+ \ Q_h+)

It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+, or some p_h = p_i.

Thus, Eq_i in Q_nlt i.
A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
As Q_nlt i neq 0 and Q_nlt_i subset Q+:
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.

Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.

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Am 07.04.2024 um 19:19 schrieb Jim Burns:

On 3/16/2024 1:31 PM, Ross Finlayson wrote:

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

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Am 09.04.2024 um 03:39 schrieb Ross Finlayson:

On 04/08/2024 06:03 PM, Ross Finlayson wrote:

On 04/08/2024 12:36 PM, Moebius wrote:

Am 07.04.2024 um 19:19 schrieb Jim Burns:

On 3/16/2024 1:31 PM, Ross Finlayson wrote:

There exists a surjection from the rational numbers onto the
irrational numbers.

Certainly not.

There exists a surjection ρ: Q → P. [...]

"Just the place for a Snark! I have said it thrice:
What I tell you three times is true."

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On 4/7/2024 11:23 PM, Ross Finlayson wrote:

On 04/07/2024 04:39 PM, Ross Finlayson wrote:

On 04/07/2024 10:19 AM, Jim Burns wrote:

[...]

Yeah, I get that a lot,
but then there wouldn't be what it is there.

So, "it is so that ...".

(Mostly the
"it is so ... or else [one of] ... or ..."
bit,
neither of those being a thing.)

Then, by
| It is so that ℚᐠᑉᵢ ≠ ∅
| else
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or ∃pₕ = pᵢ
|
you're saying [1]
| ℚᐠᑉᵢ ≠ ∅ or
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or
| ∃pₕ = pᵢ
|
Is that right?

It would be great if
I didn't have to guess why you think [1] is true.

Until you find time to help me understand you,
I will guess that you think that because
for each ℚ⁺ₕ: ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅

This is an invalid quantifier swap:
∀ℚ⁺ₕ: ∃qᵢ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
☠ ∃qᵢ: ∀ℚ⁺ₕ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ

If it were valid, it would follow that
☠ ℚᐠᑉᵢ = ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅

That inference isn't valid.
That conclusion doesn't follow and isn't true.

----
For each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∀r₁,r₂ ∈ ℝ, r₁<r₂:
∃p₁₂ ∈ ℙ=ℝ\ℚ: r₁ < p₁₂ < r₂

| Assume r₁,r₂ ∈ ℝ, r₁<r₂
|
| d = 1+max{d′ ∈ ℕ: d′ ≤ 3/(r₂-r₁) }
| n = 1+max{n′ ∈ ℕ: n′ ≤ d⋅r₁ }
|
| r₁ < n/d < (n+1)/d < r₂
| n/d, (n+1)/d ∈ ℚ
|
| n/d < (n+⅟√​̅2)/d < (n+1)/d
| r₁ < (n+⅟√​̅2)/d < r₂
| (n+⅟√​̅2)/d ∈ ℙ
|
| p₁₂ = (n+⅟√​̅2)/d
| p₁₂ ∈ ℙ
| r₁ < p₁₂ < r₂

Therefore,
for each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∀r₁,r₂ ∈ ℝ, r₁<r₂:
∃p₁₂ ∈ ℙ=ℝ\ℚ: r₁ < p₁₂ < r₂

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
∃pₕ ∈ ℙ⁺: q < pₕ < pᵢ:
q ∈ ℚ⁺ₕ
q ∉ ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ℚᐠᑉᵢ

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
q ∉ ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

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Le 08/04/2024 à 19:36, Moebius a écrit :

Am 07.04.2024 um 19:19 schrieb Jim Burns:

On 3/16/2024 1:31 PM, Ross Finlayson wrote:

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

*All* digits of Cantor's diagonal represent rational numbers.

Regards, WM

--- SoupGate-Win32 v1.05
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Am 09.04.2024 um 14:10 schrieb WM:

Le 08/04/2024 à 19:36, Moebius a écrit :

Am 07.04.2024 um 19:19 schrieb Jim Burns:

On 3/16/2024 1:31 PM, Ross Finlayson wrote:

There exists a surjection from the rational numbers onto the
irrational numbers.

Certainly not.

*All* digits of Cantor's diagonal represent rational numbers.

@Ross Finlayson:

Rule of thumb: If Mückenheim agrees with you, there's something VERY
wrong with your argument!

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Now! With More Unicode!

On 3/16/2024 1:31 PM, Ross Finlayson wrote:

A function surjects the rational numbers
onto the irrational numbers

Ross A. Finlayson
December 28, 2006

Abstract

There exists a surjection
from the rational numbers
onto the irrational numbers.

Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.

There is a distinct q e Q for each p e P.

Proof.
Let Q_i+ be the set of
positive rationals less than p_i,
p_i e P+.
For each p_h < p_i, p_h e P+:
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.

Let

Q_nlt i =
Q_i+ \ (  U p_h < p_i  Q_h+  ) =
Int p_h< p_i (Q_i+ \ Q_h+)

It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+,
or
some p_h = p_i.

I am guessing that you are depending on
an unreliable quantifier swap:
∀ℚ⁺ₕ: ∃qᵢ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
☠ ∃qᵢ: ∀ℚ⁺ₕ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ

If it were reliable, it would follow that
☠ ℚᐠᑉᵢ = ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅

However, that doesn't follow and isn't true.

∀pᵢ ∈ ℙ⁺ = ℝ⁺\ℚ⁺:
∀qⱼ ∈ ℚ⁺ᵢ = ℚ⁺∩[0,pᵢ):
dⱼ = ⌊1+3/(pᵢ-qⱼ)⌋
nⱼ = ⌊1+dⱼ⋅qⱼ⌋
qⱼ < nⱼ/dⱼ < (nⱼ+1)/dⱼ < pᵢ
pₕ = (nⱼ+⅟√​̅2)/dⱼ
pₕ ∈ ℙ⁺: qⱼ < pₕ < pᵢ
qⱼ ∈ ℚ⁺ₕ = ℚ⁺∩[0,pₕ)
qⱼ ∉ ℚ⁺ᵢ\ℚ⁺ₕ
qⱼ ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ = ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ∀qⱼ ∈ ℚ⁺ᵢ: qⱼ ∉ ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Thus, Eq_i in Q_nlt i.

 A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
As Q_nlt i neq 0 and Q_nlt_i subset Q+:
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.

Similarly for Q- and P- via symmetry,
replace < with >, nlt with ngt, + with -,
"less than" with "greater than",
and positive with negative:
A p_i e P-, E q_i e Q-, where
q_i is related to p_i and
not related to p_j neq p_i,
for any p_j e P.

Box

There exists an injection phi: P -> Q.

Proof.
As there exists a distinct rational q e Q
for each distinct irrational p e P,
there exists an injection q: P -> Q.

E q_i in Q, A p_i, p_j in P:
phi(p_i) = q_i,
p_i neq p_j -> phi(p_i) ne phi(p_j),

a one-to-one function from P to a subset of Q.

Box

There exists a surjection rho: Q -> P.
Proof.
As there exists an injection phi: P -> Q
there exists a surjection rho:  Q -> P.

Box

1

^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+,
or some p_h = p_i".

[...]

I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.

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