# discrete math

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• Jun 19th 2007, 09:13 AM
EquinoX
discrete math
Let I(x) be the statement "x has an Internet connection" and C(x, y) be the statement " x and y have chatted over the Internet," where the domain for the variables x and y consists of all students in your class. Use quantifiers to express each of these statements.
a. Everyone in your class with an internet connection has chatted over the Internet with at least one other student in your class
b. There are at least two students in your class who have not chatted with the same person in your class.
• Jun 19th 2007, 09:36 AM
Plato
$a)\;\left( {\forall x} \right)\left( {\exists y} \right)\left\{ {I(x) \Rightarrow \left[ {x \ne y \wedge C(x,y)} \right]} \right\}$

$b)\;\left( {\exists x} \right)\left( {\exists y} \right)\left( {\forall z} \right)\left\{ {x \ne y \wedge \left( {\neg C(x,z) \vee \neg C(y,z)} \right)} \right\}$
• Jun 19th 2007, 10:20 AM
SkyWatcher
i wonder what would happen if there were only one person in the class (wich is exclude by statement b (but they are independant) but not by the question itself)
if that would be not made Plato statement 'a' wrong
i 'm not sure because i am not used to those notation
in the old time we would had say: for every(V) A belonging to C (C would be the set representing the student of the class room), we would have used comas(,) the expression 'that got the following property' was represented by '/'
i ' m a little loosed and not sure if my remark is very pertinent :confused:
• Jun 19th 2007, 10:49 AM
Plato
Quote:

Originally Posted by SkyWatcher
i wonder what would happen if there were only one person in the class …if that would be not made Plato statement 'a' wrong

Actually that is not the case at all.
The statement is part (a) is known a universal positive, an A proposition. As such it is a hypothetical (see the implication operator). It says that If someone in the class has an internet connection then there is someone else to whom the person speaks online.

So if anyone in class has a connection then there must be another person is the class.
If there is only one person in the class and that person does not have access to the internet, then the statement is still true. That is the nature of implications.
• Jun 19th 2007, 10:56 AM
SkyWatcher
after reflexion (i'm sorry i cannot delet or edit my post)
Plato 'a' statement is absolutly fitting the question but i found it so weird written! it is probably due to a normalisation of the syntax to fit the computerisation

i would normaly have said (excuse me fo not using quantificaters but words)
for every A (of the classroom), (I(A)=> (there is one B (belonging to the classroom) / (not B=A and C(A,B)))
would seem to me the most normal way to express the a statement but that is not in what is called the normal form (all the quantificateurs before) wich seem weird to me and i suppose it's pricipaly more comprehensible and maniable by computers than by human beeing (if i may express my opinion :) )
• Jun 19th 2007, 12:14 PM
EquinoX
Quote:

Originally Posted by Plato
$a)\;\left( {\forall x} \right)\left( {\exists y} \right)\left\{ {I(x) \Rightarrow \left[ {x \ne y \wedge C(x,y)} \right]} \right\}$

$b)\;\left( {\exists x} \right)\left( {\exists y} \right)\left( {\forall z} \right)\left\{ {x \ne y \wedge \left( {\neg C(x,z) \vee \neg C(y,z)} \right)} \right\}$

on part a, do we also have to state that y also has an internet connection? so we add ^ I(y), because if not then it is not possible to do this

for part b, don't we have to state that x is not the same as z and y is not the same as z, otherwise it will be a weird logical statement, what if x and z are the same person?
• Jun 19th 2007, 12:32 PM
Plato
The answer to both of those is no. But that would be in my classes, maybe not in yours.

If you negate the way I wrote part b, then I think that you will see it.
• Jun 19th 2007, 01:07 PM
EquinoX
just one more question,

Express each of these system specifications using predicates, quantifiers, and logical connectives:
a. The firewall is in diagnostic state only if the proxy server is in a diagnostic state.
b. Ar least one router is functioning normally if the throughput is between 100 kbps and 500 kbps and the proxy server is not in dianostic mode
• Jun 19th 2007, 01:29 PM
Plato
a. The firewall is in diagnostic state only if the proxy server is in a diagnostic state.
D(x) is “x is in diagnostic state”.
Recall “If P then Q” is equivalent to “P only if Q.
$D(fw) \Rightarrow D(ps)$.

b. At least one router is functioning normally if the throughput is between 100 kbps and 500 kbps and the proxy server is not in diagnostic mode.
R(x) “x is a router functioning normally”.
T is “the throughput is between 100 kbps and 500 kbps”.
$\left[ {T \wedge \sim D(ps)} \right] \Rightarrow \left( {\exists y} \right)\left[ {R(y)} \right]$.
• Jun 19th 2007, 01:51 PM
EquinoX
shouldn't T has a variable x? T(x)?
• Jun 19th 2007, 02:09 PM
Plato
Quote:

Originally Posted by EquinoX
shouldn't T has a variable x? T(x)?

Do you see any free variable in “if the throughput is between 100 kbps and 500 kbps”? I don’t.
• Jun 19th 2007, 04:16 PM
EquinoX
I just had one confirmation question,

Say that the question is "All tools are in the correc place and are in excellent condition"

and my answer is:

A(x): x is in the correct place
B(x): x is in excellent condition

the domain x is tools

then the logical expression would be:
$\left( {\forall x} \right)$ [A(x) ^ B(x)] is this right?
• Jun 19th 2007, 05:31 PM
Plato
I very sorry to tell you this: You have worn out your welcome with me.
You have piss off probably the only person on this board who regularly offers a course in this material. Do not expect any further help from me. You are just milking the system for answers. I have played along too long. You do not understand what is going on with this material.
• Jun 19th 2007, 06:46 PM
EquinoX
I am not asking for answers this time, I am practicing to answer my question now and ask you if it's right or not. But if you don't want to answer it, then it's totally fine.