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l | IT WASN'T RESPECTFUL; I'M BLOCKED FROM?? | image tagged in memes,jack sparrow being chased | made w/ Imgflip meme maker

by anonymous

128 views, 2 upvotes, 6 comments

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

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