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We'd like to disqualify us from receiving prize money for this Millennium Prize Problem. Was phrased badly by Mr. R. It mislead.

We'd like to disqualify us from receiving prize money for this Millennium Prize Problem. Was phrased badly by Mr. R. It mislead. | user assures that the Riemann hypothesis is not nicely stated†; mathematicians everywhere:; †a proof would be irrelevant to a mathematical theory of everything in so much as schanuel's conjecture is false: euler's identity is not at all an ultimately general function of reals, instead is conjecturally the penultimately specific; I^((β„βˆ–π”Έ)^0)=I^((ℝ\β„š)^0) WHERE (β„βˆ–π”Έ)^0=((ℝ\β„š)^0)^(4) WHICH PRACTICALLY MEANS, HUMBLY HOPING I'VE WRITTEN IT DOWN RIGHT, THAT E^0β‡’((√I)^∞)⁕(E))/((√I)^∞)⁕(E)) β‡’((√I)^(∞/∞))⁕(E)=((√I)^(∞/∞))⁕(E) WHERE WE WILL ALLOW {E=E} IF SUBSTITUTING OUR MORE DIFFICULT AXIOM ((I^(1/2))^((β„βˆ–π”Έ)^0))=((I^2)^((β„βˆ–π”Έ)^0)) I^((β„βˆ–π”Έ)^0=I^((β„βˆ–π”Έ)^0 THE DIFFICULTY IS THE INTERMEDIARY β‡’(√I)^(∞/∞)=(√I)^(∞/∞) WHERE ((√2)^0)^(∞/∞)=((-1)/(1))^(∞/∞) BECAUSE OF Β±(1+I) POLYNOMIAL FACTOR SO WHY IS THE NUMERATOR NEGATIVE BESIDES THE NEED TO AVOID (1)^∞ AS INDETERMINATE? WELL WE CAN TAKE THE ZERO-SQRT OF 2 AS 'VENOMOUS' AND THE ZERO-FRSTRT OF E AS 'RAVENOUS' WHERE {β–±=-β–³=(∞-1)/(∞)} THEN WE CAN SHOW THAT THE INFINITE ROOT OF E HAS TO BECOME PSEUDO-β–³ IE. (1+∞)/(∞) "AFTER" THE INFINITE ROOT OF SQRT OF TWO DOES. WE PROVE THIS WITH WITH THE LEMMA THAT COUNTS THE INFINITE DECIMAL β–± NUMBERS. THIS COUNT MUST INCLUDE THE INFINITESIMAL β–± YET CAN ONLY DO SO IN THE UNCOUNTABLE DIAGONAL SUCH THAT THE β–± NUMBER CANNOT BE ITS OWN INFINITE ROOT. | image tagged in memes,unsettled tom,toe,last,monday,______ | made w/ Imgflip meme maker
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    user assures that the Riemann hypothesis is not nicely stated†; mathematicians everywhere:; †a proof would be irrelevant to a mathematical theory of everything in so much as schanuel's conjecture is false: euler's identity is not at all an ultimately general function of reals, instead is conjecturally the penultimately specific; I^((β„βˆ–π”Έ)^0)=I^((ℝ\β„š)^0) WHERE (β„βˆ–π”Έ)^0=((ℝ\β„š)^0)^(4) WHICH PRACTICALLY MEANS, HUMBLY HOPING I'VE WRITTEN IT DOWN RIGHT, THAT E^0β‡’((√I)^∞)⁕(E))/((√I)^∞)⁕(E)) β‡’((√I)^(∞/∞))⁕(E)=((√I)^(∞/∞))⁕(E) WHERE WE WILL ALLOW {E=E} IF SUBSTITUTING OUR MORE DIFFICULT AXIOM ((I^(1/2))^((β„βˆ–π”Έ)^0))=((I^2)^((β„βˆ–π”Έ)^0)) I^((β„βˆ–π”Έ)^0=I^((β„βˆ–π”Έ)^0 THE DIFFICULTY IS THE INTERMEDIARY β‡’(√I)^(∞/∞)=(√I)^(∞/∞) WHERE ((√2)^0)^(∞/∞)=((-1)/(1))^(∞/∞) BECAUSE OF Β±(1+I) POLYNOMIAL FACTOR SO WHY IS THE NUMERATOR NEGATIVE BESIDES THE NEED TO AVOID (1)^∞ AS INDETERMINATE? WELL WE CAN TAKE THE ZERO-SQRT OF 2 AS 'VENOMOUS' AND THE ZERO-FRSTRT OF E AS 'RAVENOUS' WHERE {β–±=-β–³=(∞-1)/(∞)} THEN WE CAN SHOW THAT THE INFINITE ROOT OF E HAS TO BECOME PSEUDO-β–³ IE. (1+∞)/(∞) "AFTER" THE INFINITE ROOT OF SQRT OF TWO DOES. WE PROVE THIS WITH WITH THE LEMMA THAT COUNTS THE INFINITE DECIMAL β–± NUMBERS. THIS COUNT MUST INCLUDE THE INFINITESIMAL β–± YET CAN ONLY DO SO IN THE UNCOUNTABLE DIAGONAL SUCH THAT THE β–± NUMBER CANNOT BE ITS OWN INFINITE ROOT.