## UnkleRhaukus 2 years ago $0\implies1$?

1. UnkleRhaukus

how?

2. wio

Is this logic?

3. 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

4. Directrix

@UnkleRhaukus

5. saloniiigupta95

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...

6. UnkleRhaukus

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

7. ParthKohli

$0\implies 1$is true.

8. ParthKohli

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

9. UnkleRhaukus

yeah howcome false implies true?

10. ParthKohli

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

11. ParthKohli

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

12. UnkleRhaukus

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

13. ParthKohli

I have problems with the notation :-P

14. UnkleRhaukus

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

15. ParthKohli

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

16. UnkleRhaukus

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

17. ParthKohli

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

18. UnkleRhaukus

yeah