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artofspeed Group TitleBest ResponseYou've already chosen the best response.0
however e^(cosh1)=>pi
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Is this logic?
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.0
Let x = y. Then, x  y + y = y (x  y + y) / ( x  y) = y / ( x  y) 1 + y / ( x  y ) = y / ( x  y) Therefore, 1 = 0
 one year ago

Directrix Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus
 one year ago

saloniiigupta95 Group TitleBest ResponseYou've already chosen the best response.1
I hope I can reply :) Here in the first step you took x=y... If you take this into consideration, you cannot do the third step... that is , dividing both sides by (xy) because in doing so , you are actually dividing by 0 on both sides which is absurd... or in mathematical sense, undefined...
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
yes logic @wio and as @saloniiigupta95 says dividing by zero is against the law @Directrix
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
\[0\implies 1\]is true.
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Considering you meant 0 and 1 as in the boolean values.
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
yeah howcome false implies true?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Because of the identity:\[\phi \implies \psi \ \ \cancel{\equiv} \ \ \psi \implies \phi \ \ \ \ \rm if \ \ \ \ \ \phi \cancel{\equiv} \psi \]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
I mean \(\phi \implies \psi \equiv \neg(\psi \implies \phi) \ \ \ \ \rm if \ \ \ \ \phi \equiv \neg \psi \)
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
\[\underline{\qquad\phi\implies\psi\\ {\large\land}\quad\phi\iff\neg\psi\\}\]\[\therefore\quad \neg(\psi\implies\phi)\] ?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
I have problems with the notation :P
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
my problem is \[0\to0\to0\to\dots\to0\to0\to1\to1\to\dots\to1\to1\to1\]
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
In any order, it'd be \(1\).
 one year ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
i can see why zero implies zero , , i can see one implies one, but why/when does zero imply one ?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
\( \ \ \rm m \implies 1 \ \ \ \) is always true.
 one year ago
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