Hopefully you have only one question, kid. | Teacher:"Unless there are any other questions, you're free to leave." Kid: *Raises hand* Rest of the Class: | image tagged in gifs,memes,funny,class,question | made w/ Imgflip video-to-gif maker
My Brain:"Adios!" | Me: *looks at the first exam question* Everything I ever studied: | image tagged in gifs,memes,funny,question,exam | made w/ Imgflip video-to-gif maker
I have a question for y'all | WHAT WOULD PETA THINK OF JAPANESE EXOTIC CAFÉS? TELL ME IN THE COMMENTS | image tagged in gifs,question,peta | made w/ Imgflip video-to-gif maker
This is Trifficult! | WHY IS THE OLYMPICS ONLY HELD DURING LEAP YEARS? | image tagged in gifs,memes,question,olympics,leap year | made w/ Imgflip video-to-gif maker
![blck wht shadow flop variant on neon background...WITH EASTER EGG HOMAGE TO THE RIPPING GIANTS WE WRESTLED | IF WE TAKE THE ZERO POWER OF SQUARE OF TWO
RAISED TO INFINITY EQUALS ITSELF THERE IS A BIZARRE
RAMIFICATION. IT HAS TO DO WITH UNCOUNTABLE
SQUARE ROOTS IN THE IMAGINARY. THAT IS WE HAVE
TAKEN PLUS OR MINUS THE IMAGINARY ADDED TO ONE
OVER ITSELF RAISED TO INFINITY. AND BY
DIAGONALIZING THE PLUS OR MINUS ONE FACTORS
THINGS RAMIFY IN A WAY THAT CAN BE INTRIGUING
ALSO, IF HANDLED PROPERLY. SO LET'S SAY THAT THIS
EQUALITY IS ALSO SQUARED AND THAT WE HAVE A FORM
OF TWO RAISED TO INFINITY OVER ITSELF OVER ITSELF
EQUALS INFINITY OVER INFINITY OVER INFINITY OVER
INFINITY. THAT IS RESTATED WHERE THE RIGHT HAND
SIDE HAS OUR UNCOUNTABLE POWERS OF THE SQUARE
ROOT OF TWO TO A ZERO POWER AND AGAIN TO
ANOTHER ZERO POWER. WHAT DOES THIS ACCOMPLISH?
LET'S BREAK ANY ONE OF THOSE NUMERATORS OR
DENOMINATORS DOWN! ⫷2^∞=∞⁕(±√2)∞⫸ IS
THUSLY KNOWN TO HAVE AN UNCOUNTED FACTOR OF
⫷(√2^∞)⫸. WHAT WE DO WITH THAT UNCOUNTED
FACTOR IS HAVE IT BE EXISTENT ON THE LEFT HAND SIDE
OF THE BREAKDOWN SO ⫷(2^∞)⁕(√2)^∞=∞⁕
(±√2)∞⫸ IS FULLY ACCOUNTED FOR OR NOT A
PRODUCT WITH UNCOUNTABLE FACTORS FOR. THUSLY IF
WE BUILD UP FROM THERE WE CAN PROVIDE ⫷
{∞⁕(√2^)^∞}◍◪
—————◍—
{∞⁕(√2^)^∞}◍◪
—————=—
{∞⁕(√2^)^∞}◍◪
—————◍—
{∞⁕(√2^)^∞}◍◪
⫸ WHERE THOSE VERTICLELY SCORED CIRCLES ARE JUST
FORMATTING MARKS TO BE ABLE TO TYPE OUT. WHAT'S
NEED NOW IS MAKING SURE OUR CONSTRUCTABLE
SYNTHESIS DOES NOT ALLOW UNRESTRICTED
COMPREHENSION WHEN OR FACTOR LISTS ARE
REMOVED. SO WE COMMIT TO MAKING SOME OF OUR
INFINITE POWERS OF TWO SQUARE ROOT OF TO BE
INFINITE PLUS ONE POWER. AT LEAST MOMENTARILLY,
WHERE THE LEFT SIDE BEING EQUALED TO ITSELF HAS
SOME FACTORS AS SQUARE ROOTS OF TWO SHIFTED
AROUND. ⫷
{∞⁕(√2^)^[∞+1]}◍{∞⁕(√2^)^[∞+1]}
———————◍————————
{∞⁕(√2^)^[∞]}◍{∞⁕(√2^)^∞}
———————=—————————
{∞⁕(√2^)^∞}◍{∞⁕(√2^)^[∞+1]}
———————◍————————
{∞⁕(√2^)^[∞-1]}◍{∞⁕(√2^)^∞}
⫸; TO BALANCE SPACE AND CLARITY AND A RATIONAL WE
RETAIN ONE OF THE OVER-INFINITE POWERS IN UPPER
FRACTIONS ON LEFT-SIDE & RIGHT-SIDE. THOSE UPPER
FRACTIONS CAN BE TERMED AN IMPROPERLY INFINITE
FRACTIONS. THE ARE UNCOUNTABLY FACTORED AND ARE
ALLOWED TO COLLAPSE PEACEFUL BACK INTO THE UNIT
IDENTITY FROM WHICH THEY MUST BE DERIVED. THIS
KEAVES ONCE A UNIT RATIO OF ⫷(√2)/(√2)⫸ IS
FACTORED FROM THE RIGHT-SIDE LOWER ⫷(√2/√2)=
{(∞-1)/∞}⁕(∞^2/∞^2)⫸. OR
⫷√2^0=1-1/∞⫸ AS SIGNIFICANT RESULT, GLEEFULLY.
THE PREVIOUS SENTENCE'S EQUATION CAN THEN BE
USED FOR DISPUTATIOUS INTEPRETATION OF THE
GELFOND-SCHNEIDER CONSTANT. OUR RESULT ALSO
MEANS THAT ⫷{(∞-1)/∞}={(∞-1)/∞}^[2]⫸ YET
OFCOURSE NOT ⫷2^0={(∞-1)/∞}⫸. SO
⫷√2^[1/4]⁕∞⁕{(∞-1)/∞}=
√2^[1/4]⁕∞⁕{(∞-1)/∞}⫸ WHERE THAT (∞/∞)
MUST NOT BE MESSED WITH LEADS TO PARTIALLY
THROUGH ⫷√2^[2/2]=√2^[2/2]⫸ AND
⫷√2^[√2/√2]^[√2/√2]=
√2^[2/2]^[√2/√2]⫸ AS
⫷√2^2^[√2/√2]=√2^2^[2/2]⫸
AND ⫷2^[√2/√2]=2^[2/2]⫸ PROVIDING
EVENTUALLY LUCKILLY ON THE ADJUSTED SIDE
⫷(2/2)^[(∞-1)/∞]⫸ WHICH IS ANOTHER NICE WAY
TO SAY A UNITY OF ONE OVER ONE WHICH IS OFCOURSE
OUR EARLIER CANCELLED DISUNITY OF
⫷(2/2)^[(∞+1)/∞]⫸ IN TECHNICALLY PROPER
NON-INDETERMINATE ONE TO INFINITY. SO NOW
⫷∞⁕(√2)^0=∞⁕(√2)^0⫸ IS TAKEN AS
⫷(∞^2)⁕(√2)^2=1/{(∞^2)⁕(√2)^2⫸†
AND THEN SQUARE ROOTED FINALLY ⫷(∞⁕√2)=
1/(∞⁕√2)⫸. GELFOND-SCHNEIDER REARRANGED
AND POWERED TO INFINITY OVER INFINITY (WHICH
WONT TURN OUT TO BE INDETERNINATE FOR TWO IF
OBVIOUSLY WE SUBSTITUTE SQUARE ROOT OF TWO
SQUARED THAT IS THE SISTER TO GELFOND-SCHNEIDER):
⫷2^[∞/∞]^[√2/√2]=2⫸. SUBSTITUTING IN
⫷4^[∞⁕√2]=2⫸ YET ALSO
⫷(2^2)^[1/{∞⁕√2}]=4^[1/2]⫸.
LOOKS LIKE ⫷2^[2/∞]⁕2^[√2]=2^[√2]⫸ AS
⫷2^[1/∞]⁕2^[√2]=(2^∞)⁕2^[√2]⫸.
THAT WAS RUSHED IN LAST STEPS YET WE GET THE
IDEA? AND TO CONCEIVE AS BEST WE CAN ON OUR
OWN, RAISE UP TO SOME ACTUALLY OVER INFINITE
POWER AND MAGICALLY, ⫷2^[∞]⁕2^[√2]=
(2^∞^∞)⁕2^[√2]⫸. WE KNOW INFINITIES TIMES
INFINITIES ARE NEGATIVE INFINITIES -DON'T ASK- THUS
THE PROOF IS DONE.
⫷2^[∞]⁕2^[√2]=2^[-∞]⁕2^[√2]⫸.■; WHAT SEQUEL IF
RAINMAN [W]
TOOK CARE
BABBIT | image tagged in bye bro,for good,question,repost creator within hour | made w/ Imgflip meme maker](http://i.imgflip.com/9umelg.jpg)
