A method to create zeros in the real half of the critical region of the ζ function uses ζ(1)=∞ and ζ(2)=π^2/6. Then, if a hypercube with a four-dimensional hypercube with a side length between 1/2 and 1 is infinitely repeated as an initial value, 1/2=0 and the sum of the sides is ζ(1)=∞ . You can move freely both vertically and horizontally. Can be used for fermions and bosons.
Einstein got E =mc^2 by the equivalence principle. Next, we proceed to the constant equivalence principle. By setting c=π, E=mc^2=mπ^2 Especially when M=6, E=ζ(2). 6 is the number of quarks. π≈22/7 rounds off the extra dimensions from the 11th dimension. 2(1+4)=2+8=10 This gives us 10 dimensions. Eighteen elementary particles were created, including the graviton. A light spot travels in 10 dimensions at the speed of light. The light changes shape at will.
Let m=6 E=6π^2 /6=π^2 here 1=6 1=ζ(2) 1=E Rotate the 3rd formula with = and exchange the 3rd formula with each other. Constant commutation transformations are local and immutable. A constant transformation in the local field gives E=ζ(2), m=6, m=1, E=ζ(2). Use these to expand the formula. When the desired result is obtained, return to the original. If the returned result is good enough for normal algebraic geometry, we reuse the result.
If we were to draw 11 dimensions in 3D space in a Cartesian coordinate system, we would have 3 spheres. Do the following to each of the two spheres. Two light cone tori are connected in a figure eight. There are two suns, and the prominence created by the magnetic field lines on the surface of the sun is a figure-eight torus. Another spherical body in the distance becomes the earth.
The proof of the Riemann hypothesis is due to the spin 1/2 of the Grassmann number fermion. So far, the complex plane has the unit circle from 0 to 1. Using the linear space of x^1=1 of y=x+z, I was forced to use y=x^2 and y=x^3. As a result, Fermat's last theorem concludes that there is no integer solution for x^3+y^3=z^3. Therefore, in the high-dimensional integer Cartesian coordinate system, the hypotenuse used in the Pythagorean theorem on the two-dimensional plane becomes the coordinate axis of the high-dimensional integer dimension. When origin 0 is used as common, origin 0 becomes the only common point in high-dimensional space and the point required for the reference of other spaces. Integer high-dimensional coordinate axes in a two-dimensional plane rotate around the circumference of the unit circle 1.
When considering high-dimensional spaces, π, prime numbers, 1, 0, Euler's formula, gamma function, and ζ function are not evenly possible. Originally -∞.-1, 0, 1, +∞ If you add 1-dimensional time to a 2-dimensional plane 3-dimensional space, π may be approximated by the integer 3 and used. Inner circle number π is required to improve accuracy.
Zero-dimensional pi = 0 1/2 dimensional pi = 1/2 One-dimensional pi = 1 Two-dimensional pi = 2 3D Pi = 3 Four-dimensional pi = 4 Five-dimensional pi = 5 Six-dimensional pi = 6 Seven-dimensional pi=7 8 dimensional pi π=8 9th pi = 9 10-dimensional pi=10 11-dimensional pi=11 12-dimensional pi = 12
±∞ infinity pi ≒ ±3.141592653589793 Pi in one dimension 3 π=22/7≒ 3.142857142857143
Infinite three-dimensional pi ≒ 3.141592653589793
pi in 431×137 dimensions π≒ 3.145985401459854
infinity or one-dimensional pi≒ 3.141592653589793
Infinite two-dimensional six-dimensional pi≒
Infinite two-dimensional six-dimensional pi≒6^0.5×ζ(2)^0.5=± Pi 1/2 is 0 points! The Riemann hypothesis is correct!
Consider a field K, which is generated by {0,1} and the operations {+,-,*,/}. At this moment, we must not create the exception 1/0. This is because the exception is not derived from the premises. Do not add extraordinary axioms or premises. At this moment, the sign of 1/0 is positive and negative. This means that the line, which looks straight, is curve. There is no straight line. So the Riemann hypothesis is wrong.
This leads that there is no space where straight lines exist. So the addition is wrong. Thus, Fermat's Last Theorem is true.
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361/2=180.5⇔ ≒π+1/2 Until now 360°⇔2π 180°⇔π actually, 180⇔π+1/2 360⇔2π+1 It is here 1/2,1 is a very fine minute amount and smaller than the normal differential minute amount. The parameter shifts by 1/2,1. The Riemann Hypothesis seems to have been solved.
Proposition Riemann hypothesis is wrong, and Fermat's Last Theorem is true
Proof Consider a field K, which is generated by {0,1} and the operations {+,-,*,/}. At this moment, we must not create the exception 1/0. This is because the exception is not derived from the premises. Do not add extraordinary axioms or premises. At this moment, the sign of 1/0 is positive and negative. This means that the line, which looks straight, is curve. There is no straight line. So the Riemann hypothesis is wrong.
This leads that there is no space where straight lines exist. So the addition is wrong. Thus, Fermat's Last Theorem is true.
英訳 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. Systems that add conditions such as '1/0 cannot be defined' afterwards are spaces where lines are preserved, and where + and - do not intersect, thus these conditions cannot be added due to contradiction. Therefore, lines do not exist.
Theorem 2: All Spatial Structures Containing Lines Do Not Exist Proof: Trivial. Corollary 2.1: Addition is not valid. Corollary 2.2: 0 is not valid.
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 ::2024/03/09(土) 18:22:24.07 ID:P9QGsLiu.net
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.
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.
>>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.
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 ::2024/03/09(土) 19:32:16.20 ID:P9QGsLiu.net
次はこの内容を論文にします
282 ::2024/03/09(土) 19:38:56.00 ID:P9QGsLiu.net
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 ::2024/03/09(土) 19:39:11.26 ID:P9QGsLiu.net
>>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.
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.
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.
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.
Proposition: For every computational problem, there exists an algorithm that matches the class of computational complexity of that problem. Proof: First, we demonstrate the following theorem.
Theorem: The proof of every proposition that can be judged as true or false will inevitably materialize. Proof: First, we present three lemmas.
Lemma 1: The condition that defines an outline is when the inclusion relation ⊇ is commutative. Proof: If there is an A with a known shape and an A with an unknown shape, then the shape of B can be determined by showing A⊇B and B⊇A.
Corollary 1.1: The concept of existence and the inclusion relation are equivalent. Proof: The concept of existence and space are equivalent. Clearly, the inclusion relation is derived from space, and from the inclusion relation, existence=space is derived.
Lemma 2: The inclusion relation ⊇ and the correspondence relation → are equivalent. Proof: If the inclusion relation A⊇B holds, then the correspondence relation A→B holds. Conversely, assume the correspondence relation A→B holds. In this case, the premise that if something exists, then all elements are in space, and from Corollary 1.1, that the existence of elements in space implies an inclusion relation, demonstrates the proposition.
Lemma 3: The inclusion relation ⊇ and the causal relation → are equivalent. Proof: Clearly, if a causal relation exists, then, as per Corollary 1.1, since there are elements in space, an inclusion relation is derived. Next, we demonstrate that a causal relation can be derived from an inclusion relation. When A⊇B holds, a causal relation that if B then A belongs, holds. Thus, the proposition is demonstrated.
Next, we provide the proof of the theorem.
Existence and outline are equivalent, outline and wave (inclusion relation ⊇) are equivalent, and the inclusion relation and causal relation (logic →) are equivalent. Therefore, taking the physical space of this visible reality = existence as the foremost premise, everything that follows is represented. In other words, it may be said that the essence of this world is the visible space. Therefore, if the proof of every true proposition exists, it contradicts not existing in visible space. Thus, the theorem is demonstrated.
Corollary: For every true proof problem, as time is infinitely increased (with ∞^n, where n can be made as large as possible), the ratio of solved problems converges to 1.
Now, if we assume that the proposition does not hold, then for every problem, there exists a problem a whose algorithm does not match the class of the problem, and is of a smaller class.
At this time, it is possible to create a program A that generates problem a infinitely.
Let there be a program X that creates problems infinitely, and let r(X) be the ratio of the total problems solved by it.
In the case of A, even if we take n as large as possible with time being ∞^n, r(A) converges to 0.
Let P be the set of all true proof problems, and let p be the set of all computational problems, then P ⊃ p.
To solve a computational problem means to demonstrate the following two points: (1) Decide whether there is a better algorithm than brute force search, and if it exists, demonstrate that it is the best; if not, demonstrate its non-existence. (2) If it exists, verify whether a solution exists using that algorithm and specifically output the value. If it does not exist, verify the existence of a solution through brute force search.
296 ::2024/03/10(日) 01:06:54.56 ID:86TMEkfZ.net
>>295 Let E be the set of numerous problems created by A. Since the corollary indicates that r converges to 1, both (1) and (2) are demonstrated for E and converge to 1 ⇔ for each problem, as all algorithms are the same, the best algorithm for a computational problem must match its class. However, as demonstrated earlier, r(A) converges to 0, which is a contradiction. Thus, the proposition is demonstrated.
Corollary: P=NP holds.
297 ::2024/03/10(日) 01:23:34.39 ID:86TMEkfZ.net
viXraに以下の論文を投稿しました。存在が輪郭である事の証明もつけてます。
A proof of P=NP
Sakura Tachibana
Abstract I prove that existence is a silhouette. Next I prove P=NP.
Theorem1 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.
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.
299 ::2024/03/10(日) 01:24:37.02 ID:86TMEkfZ.net
Next, I prove P=NP.
Proposition: For every computational problem, there exists an algorithm that matches the class of computational complexity of that problem. Proof: First, we demonstrate the following theorem.
Theorem2: The proof of every proposition that can be judged as true or false will inevitably materialize. Proof: First, we present three lemmas.
Lemma 2.1: The condition that defines an outline is when the inclusion relation ⊇ is commutative. Proof: If there is an A with a known shape and an A with an unknown shape, then the shape of B can be determined by showing A⊇B and B⊇A.
Corollary 2.1.1: The concept of existence and the inclusion relation are equivalent. Proof: The concept of existence and space are equivalent. Clearly, the inclusion relation is derived from space, and from the inclusion relation, existence=space is derived.
Lemma 2.2: The inclusion relation ⊇ and the correspondence relation → are equivalent. Proof: If the inclusion relation A⊇B holds, then the correspondence relation A→B holds. Conversely, assume the correspondence relation A→B holds. In this case, the premise that if something exists, then all elements are in space, and from Corollary 1.1, that the existence of elements in space implies an inclusion relation, demonstrates the proposition.
Lemma 2.3: The inclusion relation ⊇ and the causal relation → are equivalent. Proof: Clearly, if a causal relation exists, then, as per Corollary 1.1, since there are elements in space, an inclusion relation is derived. Next, we demonstrate that a causal relation can be derived from an inclusion relation. When A⊇B holds, a causal relation that if B then A belongs, holds. Thus, the proposition is demonstrated.
Existence and outline are equivalent, outline and wave (inclusion relation ⊇) are equivalent, and the inclusion relation and causal relation (logic →) are equivalent. Therefore, taking the physical space of this visible reality = existence as the foremost premise, everything that follows is represented. In other words, it may be said that the essence of this world is the visible space. Therefore, if the proof of every true proposition exists, it contradicts not existing in visible space. Thus, the theorem is demonstrated.
Corollary: For every true proof problem, as time is infinitely increased (with ∞^n, where n can be made as large as possible), the ratio of solved problems converges to 1.
Now, if we assume that the proposition does not hold, then for every problem, there exists a problem a whose algorithm does not match the class of the problem, and is of a smaller class.
At this time, it is possible to create a program A that generates problem a infinitely.
Let there be a program X that creates problems infinitely, and let r(X) be the ratio of the total problems solved by it.
In the case of A, even if we take n as large as possible with time being ∞^n, r(A) converges to 0.
301 ::2024/03/10(日) 01:25:38.05 ID:86TMEkfZ.net
>>300 Let P be the set of all true proof problems, and let p be the set of all computational problems, then P ⊃ p.
To solve a computational problem means to demonstrate the following two points: (1) Decide whether there is a better algorithm than brute force search, and if it exists, demonstrate that it is the best; if not, demonstrate its non-existence. (2) If it exists, verify whether a solution exists using that algorithm and specifically output the value. If it does not exist, verify the existence of a solution through brute force search.
Let E be the set of numerous problems created by A. Since the corollary indicates that r converges to 1, both (1) and (2) are demonstrated for E and converge to 1 ⇔ for each problem, as all algorithms are the same, the best algorithm for a computational problem must match its class. However, as demonstrated earlier, r(A) converges to 0, which is a contradiction. Thus, the proposition is demonstrated.
Corollary: P=NP holds.
302 ::2024/03/10(日) 01:39:39.19 ID:86TMEkfZ.net
某有名数学者にメールを送りました
303 ::2024/03/10(日) 02:46:01.53 ID:86TMEkfZ.net
この内容の論文をviXraに投稿しました。次に有名数学者に送ってタイムスタンプを付けます。
Fermat’s conjecture is true
Sakura Tachibana
Abstract I prove that lines do not exist and addition does not hold. Thus Fermat’s conjecture is true.
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.
Corollary 3.3: Fermat’s conjecture is true. Proof: Because addition does not hold, so we can not define addition.
Abstract I prove that lines do not exist and addition does not hold. Thus Riemann hypothesis is false.
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.
Corollary 3.3: Riemann hypothesis is false. Proof: Because lines do not exist, so Riemann hypothesis is false.