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nickhouraney Group TitleBest ResponseYou've already chosen the best response.0
be more specific
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
suppose that "if z is a wgroup then z is solvable" is a true statement, you also know that the "z is solvable" is true. can you deduce that "z is a wgroup" is true?
 2 years ago

omgitsjc Group TitleBest ResponseYou've already chosen the best response.0
i wish i was good a math
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
no because for the statement to be true z is a wgroup could be true or false
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
unless there's more to that question...don't you just put a "not" before the adjective?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
igbasallote what are you talking about?
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
in a "if p then q" statement. if "if p" is false the whole statement is true regardless of then p. If they are both true then it is also true. It can only be false if "if p" is true and "then q" is flase.
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
I can give you an example to clarify if you want
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
I was answering ur second question by the way
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
@lgbasollote was answering your first.
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
and @lgbasallote you were correct
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
I don't get what lgbasallote was saying
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
actually let me look at me logic notes to make sure.
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
The correct answer to your first question is All cabages are not green
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Like the car is red. The negation is the car is not red
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
why isn't it not all cabbages are green
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
because that implies that some are green
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
we want to to do a complete opposite
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
its confusing I know
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
so u wan't to tell that none of them are green right?
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Did you understand what I was saying to your second question
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
im trying to figure it out now
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
What class is this? just curious
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
advanced mathematics
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
whole class based on proofs basically
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
induction proofing, contradictions, and now we are at sets theories
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
I took a class called mathematical logic which involved a bunch of this stuff. Kinda interesting kinda hell
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Only class I have ever had to drop cuz i was failing. Sets, relations, and functions.
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Let me know if you don't understand my explanation about your second question I can provide an example that would make it clearer. I'm going to move on to other questions
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
thats pretty much what i am in now, and let me tell you that I have no idea how im going to pass that class
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Logic was kinda fun. Sets is hell. I would master definitions. completely understand what a subset is and what real numbers and complex numbers. just master definitions because you can use those in ur proofs. I wish I had done that.
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
im trying to figure out that example and don't completely get it So P=>Q so if z is a wgroup then z is solvable IS TRUE, then "Z is solvable is also true" why s s "Z is a wgroup false?
 2 years ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
The negation of "all cabbages are green" is "there exists a cabbage which is not green". All is a quantifier.
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
isn't the negation "all cabbages are not green?"
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
So @DanielxAK we just had to show that for all to be false one is true?
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
lets use p and q for your question all the words are confusing me lol. So the question is essentially... If, if p then q is true, then q is true" So its a nested statement?
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
P>Q T Premise Q T Prove
 2 years ago

DanielxAK Group TitleBest ResponseYou've already chosen the best response.0
I'm not sure I understand what you mean by that. If it was phrased, cabbage is green. Then, the negation would be cabbage is not green. But, you have "all cabbages are green". So, the negation would be "there exists a cabbage which isn't green". The negation of all is one. This might be able to explain it better than I can: http://www.math.cornell.edu/~hubbard/negation.pdf
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Good catch Daniels
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
Your second question @math_proof, I think it is a correct statement. Because "if p" is true, for the statement to be true (which it is by a premise) "then q" must be true. and "if p" is false. Then "then q" is assumed true because it can't be proven otherwise.
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
You got me inerested now. I dug out my logic notes. Law of Equivalence for Implication and Disjunction (LEID) states IF p>q is true, THEN not p or q is true
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
still confused omg
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
So we are Given that P > Q is true correct? and we have to prove that Q is true? am I understanding right?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
we know that Q is true
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
so P can be either true of false
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
so P=>Q, z is a wgroup then z is solvable" is a true statement, so P>Q is true and we know Q is true because "Z is solvable" so P can be either True of False? thats what i'm thinking
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
P>Q T Premise Q T Premise P T Prove Ok I'm on the right page
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
So I would say the answer is ~P V P
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
thats based on truth table on logical implication
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
so we can't deduce if P is true
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
I have this cheat sheet that my teacher made us with all the laws of logic that can be used in a proof. If you'd like it i'll email it to you. Just private message me your email. If you are not comfortable with that I guess I could post it on this thread, but I've got warned for stuff like that
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
I guess we can't for certain
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
omg thanks so much for that, I actually get it now
 2 years ago

ChmE Group TitleBest ResponseYou've already chosen the best response.1
No problem. When you jump into sets I will be no help. Like I said earlier for sets proofs I would master definitions because my teacher used a lot of "By def'n..." in her proofs
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
i never heard of this website, but now that i discovered it is so useful and helpful and for free
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
how do you give a medal?
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.1
oo oki got it
 2 years ago
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