UnkleRhaukus
  • UnkleRhaukus
\[0\implies1\]?
Meta-math
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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UnkleRhaukus
  • UnkleRhaukus
how?
anonymous
  • anonymous
Is this logic?
Directrix
  • Directrix
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

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Directrix
  • Directrix
@UnkleRhaukus
anonymous
  • anonymous
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 (x-y) because in doing so , you are actually dividing by 0 on both sides which is absurd... or in mathematical sense, undefined...
UnkleRhaukus
  • UnkleRhaukus
yes logic @wio and as @saloniiigupta95 says dividing by zero is against the law @Directrix
ParthKohli
  • ParthKohli
\[0\implies 1\]is true.
ParthKohli
  • ParthKohli
Considering you meant 0 and 1 as in the boolean values.
UnkleRhaukus
  • UnkleRhaukus
yeah howcome false implies true?
ParthKohli
  • ParthKohli
Because of the identity:\[\phi \implies \psi \ \ \cancel{\equiv} \ \ \psi \implies \phi \ \ \ \ \rm if \ \ \ \ \ \phi \cancel{\equiv} \psi \]
ParthKohli
  • ParthKohli
I mean \(\phi \implies \psi \equiv \neg(\psi \implies \phi) \ \ \ \ \rm if \ \ \ \ \phi \equiv \neg \psi \)
UnkleRhaukus
  • UnkleRhaukus
\[\underline{\qquad\phi\implies\psi\\ {\large\land}\quad\phi\iff\neg\psi\\}\]\[\therefore\quad \neg(\psi\implies\phi)\] ?
ParthKohli
  • ParthKohli
I have problems with the notation :-P
UnkleRhaukus
  • UnkleRhaukus
my problem is \[0\to0\to0\to\dots\to0\to0\to1\to1\to\dots\to1\to1\to1\]
ParthKohli
  • ParthKohli
In any order, it'd be \(1\).
UnkleRhaukus
  • UnkleRhaukus
i can see why zero implies zero , , i can see one implies one, but why/when does zero imply one ?
ParthKohli
  • ParthKohli
\( \ \ \rm m \implies 1 \ \ \ \) is always true.
UnkleRhaukus
  • UnkleRhaukus
yeah

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