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Mathematics 6 Online
OpenStudy (anonymous):

could you guys help me , I got a quiz tomorrow on this subject and I wanted to make sure that my solutions for these are good , so I know that understood it , I'll post the solution + question in a sec.

OpenStudy (anonymous):

I(x) be " x has an internet connection " , C(x,y) be " x and y have chatted over the internet " and the domain is all the students in your class , I need to use quantifiers to express these statements: a) jerry does not have an internet connection. *note: the symbol " - " is negation " My solution: ∃x -I(x) b) Rachel has not chatted over the internet with Chelsea. My solution:∃(x,y) -C(x,y) c) jan and sharon have never chatted over the internet. My solution: ∃(x,y) -C(x,y) d) no one in the class has chatted with bob. My solution: -∀x ∃y C(x,y) e) sanjay has chatted with everyone except joseph. My solution: ∃x∀y C(x,y) , y does not equal to joseph f) someone in your class does not have an internet connection. My solution: ∃x-I(x) g) not everyone in your class has an internet connection. -∀x I(x)

OpenStudy (anonymous):

Help please , :(

OpenStudy (anonymous):

Im sorry, I cant help :S

OpenStudy (anonymous):

ah , it's ok :D

OpenStudy (anonymous):

a. i think -I(jerry)

OpenStudy (anonymous):

what's the difference ? I think ∃x is in general which is someone from your class ? right ?

OpenStudy (anonymous):

Mustafa are you there ? :D

OpenStudy (anonymous):

but here i know who is it

OpenStudy (anonymous):

You're right , but I think the point of this question is to know where to use quantifiers , but in this case , is my solution right ?

OpenStudy (anonymous):

i think -I(jerry) is right

OpenStudy (anonymous):

OK let's leave that last , let's check the others.

OpenStudy (anonymous):

b.i think -C(Rachel,Chelsea)

OpenStudy (anonymous):

ok let's see c

OpenStudy (anonymous):

to c is the same ?

OpenStudy (anonymous):

as your solution to b ?

OpenStudy (anonymous):

so i think d is going to be -∀x C(x,bob) ,

OpenStudy (anonymous):

I think e is going to be : ∀y C(sanjay,y) , y!= joseph

OpenStudy (anonymous):

d. i think ∀x -C(x,bob)

OpenStudy (anonymous):

oh yeah , I forgot about d :D

OpenStudy (anonymous):

f , g are correct

OpenStudy (anonymous):

e. ∀x : x not equal Joseph ⇒ C(Sanjay, x)) ∧ ¬C(Sanjay, Joseph)

OpenStudy (anonymous):

Thanks so much man ^^

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