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269: ◆Ph05QxAcng [] 2024/03/09(土)18:05 ID:P9QGsLiu(1/21)
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.
270(1): ◆Ph05QxAcng [] 2024/03/09(土)18:22 ID:P9QGsLiu(2/21)
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 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.
271: ◆Ph05QxAcng [] 2024/03/09(土)18:22 ID:P9QGsLiu(3/21)
>>270
これを後でviXraに投稿します
272(1): ◆Ph05QxAcng [] 2024/03/09(土)18:32 ID:P9QGsLiu(4/21)
viXraにサブミットしました
273: ◆Ph05QxAcng [] 2024/03/09(土)18:36 ID:P9QGsLiu(5/21)
あと有名数学者にメールを送ればタイムスタンプも付いて証拠としては完全ですかね
274: ◆Ph05QxAcng [] 2024/03/09(土)18:37 ID:P9QGsLiu(6/21)
>>272
sakura.tachibana8@gmail.com
このメールアドレスで投稿しました
275(1): ◆Ph05QxAcng [] 2024/03/09(土)19:04 ID:P9QGsLiu(7/21)
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 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.
276: ◆Ph05QxAcng [] 2024/03/09(土)19:04 ID:P9QGsLiu(8/21)
>>275
この内容のメールを送ります
277: ◆Ph05QxAcng [] 2024/03/09(土)19:07 ID:P9QGsLiu(9/21)
送りました。これでタイムスタンプも残りますね
278(1): ◆Ph05QxAcng [] 2024/03/09(土)19:30 ID:P9QGsLiu(10/21)
定理2 存在とは輪郭の事である
証明
まず2つの補題を示す。
補題2.1 a及びその冪集合{a}を考える。この時、この二つは別のものである。
証明
仮に同じものだとする。
そうするとa,bがあると、その集合{a,b}が同じものとなる。後者はa,bを一つのものとしてみたものであり、明らかに異なる。よって最初の前提が背理である。すなわち補題が示された。
補題2.2 空間の要素である集合の元は空間の要素である
証明
集合をEと置き、その元eが空間の要素でないと仮定する。E∩e=eであり、eがEの一部であるのにも関わらず空間の要素でない部位がある事になり背理である。よって補題は示された。
対偶を取ると次の系が得られる。
系2.2.1 空間の要素でない元からなる集合は空間の要素ではない
次にAが存在する事とAが空間に要素を持つ事が同値であることを示す。
仮に、存在するが空間に要素がない存在物Aがあったとする。
A及びその冪集合{A}を考える。
ここで仮に{A}も空間に要素がないと仮定する。
そしてBを空間に要素がない存在物の集合と定義する。今{B}を考えると、仮に{B}が空間に要素がないとすると、B⊇{B}となり、かつ{B}⊇BでBと{B}は一致する。しかし、これは補題2.1より矛盾である。
よって{B}は空間の要素である。よって補題2.2よりA,{A}共に空間の要素である事が導かれるがこれは前提と矛盾する。
よって{A}は空間の要素である。
また補題2.2よりAも空間の要素である事か導かれる。
よってAが存在するならば、Aは空間に要素を持つ事が示された。
また空間に要素を持つならば存在する事は自明なので、存在する事と空間に要素を持つ事は同値である事が示された。また空間に要素を持つ事と輪郭を持つ事は同値なので、存在する事と輪郭が定まる事は同値である。
279(1): ◆Ph05QxAcng [] 2024/03/09(土)19:31 ID:P9QGsLiu(11/21)
>>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.
280: ◆Ph05QxAcng [] 2024/03/09(土)19:31 ID:P9QGsLiu(12/21)
>>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.
281: ◆Ph05QxAcng [] 2024/03/09(土)19:32 ID:P9QGsLiu(13/21)
次はこの内容を論文にします
282(2): ◆Ph05QxAcng [] 2024/03/09(土)19:38 ID:P9QGsLiu(14/21)
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.
283: ◆Ph05QxAcng [] 2024/03/09(土)19:39 ID:P9QGsLiu(15/21)
>>282
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 1.1.
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.
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 silhouette.
284: ◆Ph05QxAcng [] 2024/03/09(土)19:45 ID:P9QGsLiu(16/21)
>>282
これもSakura Tachibanaの名前でサブミットしました
285(1): ◆Ph05QxAcng [] 2024/03/09(土)19:55 ID:P9QGsLiu(17/21)
次にこのメールを有名数学者に送ってタイムスタンプをつけますね
タイトル
Request for Review of a Mathematical Proof: existence is a silhouette
内容
Dear Professor ***************,
I hope this email finds you well. My name is Sakura Tachibana.
I have recently completed a proof regarding that existence is a silhouette. I was wondering if you would be willing to review my proof.
The main ideas and conclusions of the proof are as follows:
existence is a silhouette.
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
286(1): ◆Ph05QxAcng [] 2024/03/09(土)19:56 ID:P9QGsLiu(18/21)
>>285
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.
287: ◆Ph05QxAcng [] 2024/03/09(土)19:56 ID:P9QGsLiu(19/21)
>>286
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 1.1.
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.
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 silhouette.
288: ◆Ph05QxAcng [] 2024/03/09(土)20:04 ID:P9QGsLiu(20/21)
メールを送ったのでタイムスタンプが押されました
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