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269: ◆Ph05QxAcng  [] 2024/03/09(土) 18:05:45.21 ID:P9QGsLiu

定理1 直線は存在しない
証明
{0,1}及び演算{+,-,*,/}から数を作る操作を考える。この時1/0の符号は±であり、+と-は交わる。曲率0が存在⇔直線が存在するという条件は、+と-は交わらないので矛盾しているので追加出来ない。よって直線は存在しない。

定理2 直線を含む全ての空間構造は存在しない
証明
自明

定理3 加法は成立しない
証明
{a,0}及び演算{+}で貼られるベクトル空間Vを考える。この時-a∈Vであるが、a,0,-aは直線上に並ぶが、これは定理1に反する。
よって定理は示された

系3.1 0は成立しない
系3.2 加法及び直線を前提とする命題及び証明は全て偽

Theorem 1: Lines do not exist.
Proof:
Consider the operation of creating numbers from {0,1} and the operations {+,-,*,/}. In this case, the sign of 1/0 is ±, and + and - intersect. The condition that curvature 0 exists ⟺ lines exist is contradictory because + and - intersect, so the condition cannot be added. Therefore, lines do not exist.

Theorem 2: All spatial structures that include lines do not exist.
Proof:
Self-evident

Theorem 3: Addition does not hold.
Proof:
Consider a vector space V constructed from {a}. In this case, -a∈V, but a, 0, and -a line up on a line, which contradicts Theorem 1.
Therefore, the theorem is proven.

Corollary 3.1: 0 does not exist.

Corollary 3.2: All propositions and proofs that assume addition and lines are false.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/269

270: ◆Ph05QxAcng  [] 2024/03/09(土) 18:22:24.07 ID:P9QGsLiu

論文が出来ました

Addition does not hold.

Sakura Tachibana

Abstract
I prove that lines do not exist and addition does not hold.

Theorem 2: All spatial structures that include lines do not exist.
Proof:
Self-evident

Corollary 3.1: 0 does not exist.

Corollary 3.2: All propositions and proofs that assume addition and lines are false.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/270

275: ◆Ph05QxAcng  [] 2024/03/09(土) 19:04:09.01 ID:P9QGsLiu

タイトル
Request for Review of a Mathematical Proof:addition does not hold

内容
Dear Professor ***************,
I hope this email finds you well. My name is Sakura Tachibana.

I have recently completed a proof regarding that addition does not hold. I was wondering if you would be willing to review my proof.
The main ideas and conclusions of the proof are as follows:
addition does not hold.

For the complete proof, please refer to the attached PDF file and text.

I understand that you have a busy schedule, but if you could find the time to review my proof, I would be incredibly grateful. I am eager to improve my proof based on any feedback you may provide.

Thank you for considering my request. I appreciate your time and expertise.

Best regards,
Sakura Tachibana

Theorem 2: All spatial structures that include lines do not exist.
Proof:
Self-evident

Corollary 3.1: 0 does not exist.

Corollary 3.2: All propositions and proofs that assume addition and lines are false.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/275

279: ◆Ph05QxAcng  [] 2024/03/09(土) 19:31:24.78 ID:P9QGsLiu

>>278
英訳
Theorem 2: Existence is a silhouette.
Proof
First, we will demonstrate two lemmas.

Lemma 2.1: Consider an element 'a' and its power set {a}. These two are different entities.
Proof
Assume they are the same.
If there exist elements a, b, then the set {a, b} would be the same entity. The latter is seen as a single entity consisting of a and b, which are clearly different. Thus, the initial premise is contradicted, proving the lemma.

Lemma 2.2: Elements of a set that are elements of space are themselves elements of space.
Proof
Let a set be E, and suppose an element e of E is not an element of space. If E∩e = e, then despite e being a part of E, there would exist a part that is not an element of space, which is a contradiction. Therefore, the lemma is proven.

Taking the contrapositive, we obtain the following corollary.

Corollary 2.2.1: A set composed of elements that are not elements of space is not an element of space.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/279

280: ◆Ph05QxAcng  [] 2024/03/09(土) 19:31:39.43 ID:P9QGsLiu

>>279

Next, we demonstrate that the existence of A and A having elements in space are equivalent.

Suppose there exists an entity A that exists but does not have elements in space.
Consider A and its power set {A}.
If we assume that {A} also does not have elements in space,
and define B as a set of entities that do not have elements in space. Considering {B}, if {B} is assumed to not have elements in space, then B⊇{B} and {B}⊇B would mean B and {B} coincide. However, this contradicts Lemma 2.1.

Therefore, {B} is an element of space. Thus, by Lemma 2.2, both A and {A} are shown to be elements of space, which contradicts the premise.
Hence, {A} is an element of space.
Furthermore, from Lemma 2.2, it can be deduced that A is also an element of space.

Therefore, if A exists, it is shown that A has elements in space.

Since it is self-evident that having elements in space implies existence, it is demonstrated that existence and having elements in space are equivalent. Furthermore, having elements in space is equivalent to having a contour, thus existing is equivalent to having a defined contour.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/280

282: ◆Ph05QxAcng  [] 2024/03/09(土) 19:38:56.00 ID:P9QGsLiu

論文の内容は以下のようになります

タイトル
A mathematical proof that existence is a silhouette

Abstract
I prove that existence is a silhouette.

Theorem
Existence is a silhouette.
Proof
First, we will demonstrate two lemmas.

Lemma 1.1: Consider an element 'a' and its power set {a}. These two are different entities.
Proof
Assume they are the same.
If there exist elements a, b, then the set {a, b} would be the same entity. The latter is seen as a single entity consisting of a and b, which are clearly different. Thus, the initial premise is contradicted, proving the lemma.

Lemma 1.2: Elements of a set that are elements of space are themselves elements of space.
Proof
Let a set be E, and suppose an element e of E is not an element of space. If E∩e = e, then despite e being a part of E, there would exist a part that is not an element of space, which is a contradiction. Therefore, the lemma is proven.

Taking the contrapositive, we obtain the following corollary.

Corollary 1.2.1: A set composed of elements that are not elements of space is not an element of space.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/282

283: ◆Ph05QxAcng  [] 2024/03/09(土) 19:39:11.26 ID:P9QGsLiu

>>282
Next, we demonstrate that the existence of A and A having elements in space are equivalent.

Therefore, {B} is an element of space. Thus, by Lemma 1.2, both A and {A} are shown to be elements of space, which contradicts the premise.
Hence, {A} is an element of space.
Furthermore, from Lemma 1.2, it can be deduced that A is also an element of space.

Therefore, if A exists, it is shown that A has elements in space.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/283

286: ◆Ph05QxAcng  [] 2024/03/09(土) 19:56:24.91 ID:P9QGsLiu

>>285

Theorem
Existence is a silhouette.
Proof
First, we will demonstrate two lemmas.

Taking the contrapositive, we obtain the following corollary.

Corollary 1.2.1: A set composed of elements that are not elements of space is not an element of space.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/286

287: ◆Ph05QxAcng  [] 2024/03/09(土) 19:56:37.16 ID:P9QGsLiu

>>286

Next, we demonstrate that the existence of A and A having elements in space are equivalent.

Therefore, if A exists, it is shown that A has elements in space.

http://rio2016.5ch.net/test/read.cgi/math/1668766352/287

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