ATAR Notes: Forum
Uni Stuff => Science => Faculties => Mathematics => Topic started by: Dumpling on March 10, 2012, 12:23:47 pm
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Can anyone help me determine if the argument p, p ~ q, [(p ^ (~q)) -> (~p)] |- q is valid.
Given p, q and r be primitive statements.
Thank you :)
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what do you mean by " p~q" ?
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Oh my bad, I mean the question was suppose to be
p, ~q, [(p ^ (~q)) -> (~P)] |- q is valid.
So its not p ~ q its ~q.... sorry typo
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Yeah it's valid simply because p, ~q, [(p ^ (~q)) -> (~p)] is a contradiction, as can be seen as follows:
Assume such a thing holds. Then p^(~q) is true since p and ~q are. Now that means that ~p is true. Which contradicts the fact that p is true. So such a thing cannot hold.
Who is teaching Real Analysis this semester?
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Is it really that simple? And it is Richard Brak?
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Also do you happen to know what => mean?
In this question [(p->q) ->r] => [(p^(~r)) -> (~q)]
I drew up a truth table and different truth values for (p->q)->r compared to (p^(~r)) -> (~q) .... so does it mean its not valid?
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I think different people use different notation so I am not sure, check your notes or whatever and let me know if you can. I think I've once used "=>" instead of what you have been calling "->". Whatever it is the following should be true:
[(p->q) ->r] => [(p^(~r)) -> (~q)] would mean that whenever [(p->q) ->r] holds then so does [(p^(~r)) -> (~q)], so the truth tables don't have to be the same, it just needs to be the case that in any row of the truth table, whenever [(p->q) ->r] holds then [(p^(~r)) -> (~q)] holds.
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Oh thank you, in my notes it says => is "valid" and -> is the conditional. Well for => it says "a well defined logical meaning (and so is doing more than just separating the premises from the conclusion) ...
As well as for the first question do I need to draw up a truth table?
Thank you for helping me out :)
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If P(A)=0.4, P(B)=0.5 and P(A or B')=0.1 -> Note: B' is B not occurring ~~
Find
(a). P(A and B)
(b). P(A or B)
(c). P(A' or B')
(d). P(A' or B)
(e). P(B|A)
My answers for question 1 were:
(a). 0.2
(b). 0.7
(c). 0.3
(d). 0.8
(e). 0.5
Question 2:
Let P(A)=0.3 and P(B)=0.6
(a). Find the probability of A or B when A and B are independent.
(b). Find P(A|B) when A and B are mutually exclusive.
(c). Find P(A|B') when A and B are mutually exclusive.
My answers for question 2 were;
(a). 0.72
(b). cannot exist (?)
(c). 0.5
Can anyone check those? :)
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Hrmm so firstly Real Analysis includes propositional logic and now it involves probability? :o
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Just random stuff im helping out a little friend that posted.. my probability skills arent what they used to be :(
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Help again with real analysis please :)
For this question prove the converse direction, that is, prove
p(x) => q(x) -> A C/ (subset symbol) B
where, for some non - empty set E,
A = { x element of E: p(x)}
B = { x element of E: q(x)}
Thank you :)