Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

liliy

  • 2 years ago

discrete math:(attached)

  • This Question is Closed
  1. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  2. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    am i doing it right?

  3. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @mathmate ????

  4. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    first of all the answer is definitely T

  5. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right... how?:(

  6. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    looks like you are on the right track, it is hard for me to read it all you are turning each implication of the form \(p\to q\) into a statement of the form \(p\lor\lnot q\)

  7. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Remember that de Morgan's law requires you to interchange the and / or, like: ~(~a or b) <=> a ^ ~b

  8. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    right...

  9. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    So there is a problem in the first term of the third line (second written line).

  10. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oy.

  11. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    The same problem with the second term. Otherwise, it will be the same as what you did.

  12. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    You skip a lot of steps, as @satellite73 pointed out. You may lose points in an exam.

  13. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \(\lnot (p\to q)\) \(\lnot (p\lor \lnot q)\) \(\lnot p \land q\) i believe

  14. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @mathmate probably has a better explanation, but as he/she said take it slow

  15. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so can you write it out?

  16. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I will follow your thoughts.

  17. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\lnot[(p\to q) \land (q\to r)]\lor (q\to r)\] may be better, yes?

  18. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hold up. im gonna write it all out and you guys tell me if its right

  19. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but i kind of suck at this, so i will let mathmate check steps

  20. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    First replace all p->q etc with the or equivalents, ~[(~p or q) ^ (~q or r) ] or (~p or r)

  21. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Now apply de Morgan's ~(~p or q) or ~(~q or r) or (~p or r)

  22. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Apply de Morgan again: (p ^ ~q) or (q ^ ~r) or ~p or r

  23. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Rearrange (commutativity of "or") ~p or (p^~q) or r or (q^~r)

  24. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Now use associativity: [~p or (p^~q)] or [r or (q^~r)]

  25. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Would that be obvious now?

  26. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    1 Attachment
  27. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    BAsically im going to assume things to be True or false and then go on from there...

  28. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Your last approach is called proof by cases. Since there are three variables, you will need to have 8 cases, which is equivalent to a truth table. The idea here is to do a proof without a truth table.

  29. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    allow me to butt in a second if you were going to prove this using a truth table, all the above work would be unnecessary you would simply construct the truth table for this (it might take a while) and observe that you get all T in the last column

  30. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    your job doing this method is to try to come up with an expression that looks like \(p\lor \lnot p\) which you can then replace by \(T\)

  31. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it is hidden in the last line written by @mathmate

  32. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Let's take it from: [~p or (p^~q)] or [r or (q^~r)] use distributivity: [(~p or p) ^ (~p or ~q)] or [(r or q) ^ (r or ~r)] which reduces to [T ^ (~p or ~q)] or [(r or q) ^ T] or (~p or ~q) or (r or q) => ~p or r or (q or ~q) => ~p or r or T => T

  33. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    like that. hope the method is clear. if you wanted to work from cases there would be no reason to change for example \(p\to q\) into \(\lnot p \lor q\)

  34. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    when you do these without truth tables and you want to prove something is true, (at the risk or repeating myself) try to look for \(p \lor \lnot p\)

  35. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    woah, this is intense... reading what you guys did, i undertsand it all.. but is there a way to "learn this stuff" bec my teacher did NOT do this...

  36. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    if this was not the method, then as i said you could write out the entire truth table and see that the last column was all T

  37. liliy
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    no, that's wat i did originally and then i realized that in the instructions my professor said specifically NOT to use tables :(

  38. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    it would look like this \[\begin{array}{|c|c|c|c|c|c|c|c} P & Q & R & P\Rightarrow{}Q & Q\Rightarrow{}R & (P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R) & P\Rightarrow{}R & ((P\Rightarrow{}Q)\land{}(Q\Rightarrow{}R))\Rightarrow{}(P\Rightarrow{}R) \\ \hline 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline 1 & 0 & 0 & 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}\]

  39. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    well not really, i used a machine to do this

  40. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    don't be put off by all this nonsense. the statement itself is completely obvious. like saying if r is contained in q and q is contained in p then r is contained in p doh

  41. satellite73
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 1

    you just have to grind it til you find it, there is no other way. write out what you know and see what you can find. practice will help, but of course in real life no one ever does such a thing so it will only serve you for your next exam. then you can forget about it. until the final

  42. mathmate
    • 2 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Right you are!

  43. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.