Why vacuously true and not vacuously false? Let a room is empty. Consider a statement.
Modified [4:04 PM, 26 March 20] Every mobile phone in this room is working.
: This is called vacuously true because there is no mobile phone in the room.
Let I say that this statement is vacuously false, If you think it is not, show a mobile phone (in this room) which is working. You cannot do this.
When we can choose both the options, why have we chosen Vacuously True and not Vacuously False?
Is it a convention?
 A: Well, the statement you just gave is not even false or true, it doesn't refer to anything.
The statement which is vacuously true is "every mobile phone in this room is working". For that to be false, you would have to show that there exists a mobile phone in the room which is not working. No mobile phone in the room exists, so in particular, there exists no mobile phone which is not working. Therefore, the statement is true.
Now, the statement you seem to be attacking is "there exists a mobile phone in the room which is working". This statement is false. There is, again, no mobile phone around, so in particular none that is working.
So you could say it's a convention, but it's completely natural in the sense that it's the only convention consistent with the general negation rule $\neg(\forall x:p)=\exists x:\neg p$. In other words, it's as much of a convention as the convention that $a^{-n}=(a^n)^{-1}$.
A: Short answer : you are right when you say that " No mobile phone is working " is true. But you're wrong when you claim it implies that " all mobile phones are working" is false , assuming, of course that the set of mobile phones under consideration is empty. 


*

*Your reasoning is as follows : 


(1) If the negation of a sentence is true, then this sentence is false
(2) " No mobile phone is working " is the negation of " All mobile phones are working"
(3) But " No mobile phone is working" is true ( since no counterexample can be pointed out). 
(4) Therefore, " All mobile phones are  working" is false. 


*

*However, proposition (2) is not correct. So, the conclusion does not hold( although the other premises are correct). 

*The sentence " no mobile phone is working" is not the contradictory ( i.e. the pure negation) of " All mobile phones are working" but the contrary statement. 

*The actual  negation ( i.e. the contradictory sentence) of " All mobile phones are working" is " There is some mobile that is not working" . For the pure negation of " all" is simply " not-all...". In symbols : 

$\exists (x)  [ M(x) \land \neg W(x)]$. 

Note : in traditional logic, contraries cannot be true at the same time; but if set M is empty, then " all M are W" and " all M are not-W" are both true, vacuously. 
https://plato.stanford.edu/entries/square/
Below a diagram showing that any  " convention" would have the same effect on both sentences : it's not up to us to decide that one is true and that the other is false. 

A: Premising that there are no mobile phones in the room.
"Every mobile phone in the room is working," is vacuously true.
"Every mobile phone in the room is not working," is vacuously true.
Yes, both are vacuously true.  There is no contradiction about this. The truth is in the everyness of the claims.
To prove either statement false requires finding a mobile phone in the room to contradict that they all have the claimed status, but there are none to be found.

Likewise claims of existence of phones will be fallacious when there are no phones in the room.
"There is a mobile phone in the room that is working," is vacuously false.
"There is a mobile phone in the room that is not working," is vacuously false.
A: Let $U$ be the universal set (room) containing elements (objects) denoted by $x$. The hypothesis says that $x$ is anything other than the mobile.
Note that a conditional statement is $p\implies q$ which is equivalent to $\lnot p\lor q$. Your statement is indeed a conditional statement which can be formulated as 
"For every $x\in U$, if $x$ is a mobile then $x$ is working" which can be written using symbols as "$\forall x\in U, p\implies q$ where
$p: x$ is mobile
$q: x$ is working
So the conditional is equivalent to saying "$\forall x\in U, \lnot p\lor q$ ". Rewriting this in language, we get
"For every $x$, either $x\in U$ is not a mobile or $x$ is working"
Now the last statement stands true by the hypothesis that "$x$ is anything other than a mobile". Hope it helps!
A: It's down to the semantics that we assign to the word "every" in mathematics.
It would be perfectly reasonable to define the phrase "for all" to mean "there are no counterexamples and at least one example". Indeed, in plain English this is more or less how we do use that phrase, hence your confusion. But in math we choose to define the phrase "for all $x$, $Px$" to mean only the first half of that sentence: there are no counterexamples to $Px$.
The reason we do this is because then we get a nice symmetry with another phrase, "there exists", expressed by these formulae:
$$\neg\forall x Px\iff\exists x\neg Px$$
$$\neg\exists x Px\iff\forall x\neg Px$$
In other words, logical negation is a kind of isomorphism between the two operators $\forall$ and $\exists$.
Whatever we choose to call them, these two operators with their symmetrical relationship exist, and are fundamental in expressing the rules of logic. Since their meanings almost correspond to the English phrases "for all" and "there exists", we draw inspiration from those phrases in naming these operators. But even if you don't like those names, the operators themselves are in some sense the natural ones to use in logic, so whatever we call them they should be the ones we use in mathematics. In some sense they are "more elemental" than their English equivalents. This elementalness is seen in the beautifully simple symmetrical laws I already mentioned, and in the fact that the colloquial English "for every" can be expressed in terms of them as $\exists xPx\wedge\forall x Px$
A: 
Why vacuously true and not vacuously false?

Suppose that proposition $A$ is false.
Then $A\implies B$ is true for any proposition $B$ regardless of its truth value. In this case, we say that $A \implies B$ is vacuously true.
On the other hand, $A \land B$ is then false for any proposition $B$ regardless of its truth value. In this case, we might reasonably say that $A\land B$ is "vacuously false" (my proposed terminology).
Note that if $P\implies Q$ is vacuously true, then $\neg(P\implies Q)$ would then vacuously false in the above sense of the term.
