1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Basic Propositional Calc

  1. Nov 12, 2007 #1
    This is mostly some basic stuff, but I just want to make sure I am doing these right. I have a hard time understanding what the questions are saying sometimes.

    1. The problem statement, all variables and given/known data
    1)Explain how we may conclude that if f is a decreasing function, then f(5) < f(3)
    Make reference to a logical principle. [Note- By defintion, a function g is decreasing i.f.f the conditional "if x<y, then g(x)>g(y)" is true for any real numbers x and y]

    2)Suppose A and B are sets and w is an object, for which it is known that A[tex]\subseteq[/tex]B and w[tex]\in[/tex]A. Explain how we may conclude w[tex]\in[/tex](A[tex]\cap[/tex]B) Make reference to a logical principle.

    3)Explain why it is true that if a real number x satisfies |x|>a, but it is not the case that x > a, then x < -a must hold. Make reference to a logical principle.

    2. Relevant equations
    (p[tex]\wedge[/tex]q) -> p Law of Simplification
    p -> (p[tex]\vee[/tex]q) Law of Addition
    [p[tex]\wedge[/tex](p ->q)] -> q Modus Ponens
    [(p[tex]\vee[/tex]q)[tex]\wedge[/tex] ~q] -> p Modus Tollendo Ponens
    [(p -> q)[tex]\wedge[/tex]~q] -> ~p Modus Tollens
    (p -> r) -> [(p[tex]\wedge[/tex]q) -> r]
    [~p -> (q[tex]\wedge[/tex]~q] -> ~p Law of Contradiction


    3. The attempt at a solution
    1) The definition of a decreasing function states, "if x<y, then g(x)>g(y)" is true for any real numbers x and y. By hypothsis we know f is a decreasing function, so that the preticular case of the definition, "if 5 < 3, then g(5) > g(3)" is known to be true. --That's as far as I can get using an example form the book as a guideline. I'm not sure where to go from here and which logical principle applies to this argument.--

    2) --I know how to explain it but I don't know which logical principle to use.--
    A[tex]\subseteq[/tex] B includes all the elements in A that are also elements of B. Since it is given w[tex]\in[/tex] A, by this definiton w is also an element of B. A[tex]\cap[/tex]B includes only the elements that A and B share in common. Since w is both an element of A and B, it is one of the elements that A and B share in common, making it an element of A[tex]\cap[/tex]B.

    3)If a real number x satisfies |x|>a, that means x < -a or x > a. It is also given that it is not the case that x > a. Therefore by Modus Tollendo Ponens we can conclude that x < -a.


    If anyone could help me with finding which logical principles apply to one and two, that'd help a lot.
     
    Last edited: Nov 12, 2007
  2. jcsd
  3. Nov 12, 2007 #2
    Anybody out there who could help?
     
  4. Nov 12, 2007 #3
    I could help you with number 2, you could make use of the "intersection of a subset rule." But you've got the basic idea nailed.

    Edit: I think that you could use Universal Modus Ponens for #1.
     
    Last edited: Nov 12, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?