$$0\implies1$$?

Question

$$0\implies1$$?

unklerhaukus · Answer

how?

anonymous · Answer

Is this logic?

directrix · Answer

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

directrix · Answer

@UnkleRhaukus

anonymous · Answer

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 · Answer

yes logic @wio 

and as @saloniiigupta95 says
dividing by zero is against the law @Directrix

parthkohli · Answer

$$0\implies 1$$is true.

parthkohli · Answer

Considering you meant 0 and 1 as in the boolean values.

unklerhaukus · Answer

yeah howcome false implies true?

parthkohli · Answer

Because of the identity:$$\phi \implies \psi \ \ \cancel{\equiv} \ \ \psi  \implies \phi  \ \ \ \ \rm if \ \ \ \ \ \phi \cancel{\equiv} \psi $$

parthkohli · Answer

I mean $\phi \implies \psi \equiv \neg(\psi \implies \phi) \ \ \ \ \rm  if \ \ \ \  \phi \equiv \neg \psi $

unklerhaukus · Answer

$$\underline{\qquad\phi\implies\psi\ {\large\land}\quad\phi\iff\neg\psi\}$$$$\therefore\quad \neg(\psi\implies\phi)$$ 
?

parthkohli · Answer

I have problems with the notation :-P

unklerhaukus · Answer

my problem is 

$$0\to0\to0\to\dots\to0\to0\to1\to1\to\dots\to1\to1\to1$$

parthkohli · Answer

In any order, it'd be $1$.

unklerhaukus · Answer

i can see why zero implies zero , , i can see one implies one, 

but why/when does zero imply one ?

parthkohli · Answer

$ \ \ \rm m \implies 1 \ \ \ $ is always true.

unklerhaukus · Answer

yeah