## math_proof 3 years ago what is the negation" all cabbages are green

1. nickhouraney

be more specific

2. math_proof

suppose that "if z is a w-group 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 w-group" is true?

3. omgitsjc

i wish i was good a math

4. ChmE

no because for the statement to be true z is a w-group could be true or false

5. lgbasallote

unless there's more to that question...don't you just put a "not" before the adjective?

6. math_proof

igbasallote what are you talking about?

7. ChmE

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.

8. ChmE

I can give you an example to clarify if you want

9. ChmE

I was answering ur second question by the way

10. ChmE

11. ChmE

and @lgbasallote you were correct

12. math_proof

I don't get what lgbasallote was saying

13. ChmE

actually let me look at me logic notes to make sure.

14. ChmE

The correct answer to your first question is All cabages are not green

15. ChmE

Like the car is red. The negation is the car is not red

16. math_proof

why isn't it not all cabbages are green

17. ChmE

because that implies that some are green

18. ChmE

we want to to do a complete opposite

19. ChmE

its confusing I know

20. math_proof

so u wan't to tell that none of them are green right?

21. ChmE

correct

22. ChmE

Did you understand what I was saying to your second question

23. math_proof

im trying to figure it out now

24. ChmE

What class is this? just curious

25. math_proof

26. math_proof

whole class based on proofs basically

27. math_proof

induction proofing, contradictions, and now we are at sets theories

28. ChmE

I took a class called mathematical logic which involved a bunch of this stuff. Kinda interesting kinda hell

29. ChmE

I hate sets

30. ChmE

Only class I have ever had to drop cuz i was failing. Sets, relations, and functions.

31. ChmE

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

32. math_proof

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

33. ChmE

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.

34. math_proof

im trying to figure out that example and don't completely get it So P=>Q so if z is a w-group then z is solvable IS TRUE, then "Z is solvable is also true" why s s "Z is a w-group false?

35. DanielxAK

The negation of "all cabbages are green" is "there exists a cabbage which is not green". All is a quantifier.

36. math_proof

isn't the negation "all cabbages are not green?"

37. ChmE

So @DanielxAK we just had to show that for all to be false one is true?

38. ChmE

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?

39. ChmE

P-->Q T Premise Q T Prove

40. DanielxAK

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

41. ChmE

Good catch Daniels

42. ChmE

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.

43. ChmE

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

44. math_proof

still confused omg

45. ChmE

So we are Given that P --> Q is true correct? and we have to prove that Q is true? am I understanding right?

46. math_proof

we know that Q is true

47. math_proof

so P can be either true of false

48. math_proof

so P=>Q, z is a w-group 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

49. ChmE

P->Q T Premise Q T Premise P T Prove Ok I'm on the right page

50. ChmE

So I would say the answer is ~P V P

51. math_proof

thats based on truth table on logical implication

52. ChmE

I agree

53. math_proof

so we can't deduce if P is true

54. ChmE

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

55. ChmE

I guess we can't for certain

56. math_proof

omg thanks so much for that, I actually get it now

57. ChmE

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

58. math_proof

i never heard of this website, but now that i discovered it is so useful and helpful and for free

59. math_proof

how do you give a medal?

60. math_proof

oo oki got it