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Thread: implication paradox

  1. #1 implication paradox 
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    Hello again! :-D

    I do not understand something about implication.

    Lets consider the following statement:

    "If at least one of the numbers a or b equals 0, then a times b equals 0"

    Here is the truth table:
    p -----q ----- p=>q
    T-------T------T
    T-------F------F
    F------T-------T
    F------F------T

    Now lets check if the truth table is really "truthful" 8)

    For T-------T------T, at least one of a or b =0 so logically a*b would be zero.

    ii) T-------F------F, at least one of a or b =0 so a*b would not be zero. (which is incorrect, therefore p=>q is false)

    iii) F------T------T, (this part makes me crazy), none of a and b = 0, so a*b would be zero (which is obviously not correct but the truth table says its correct.

    iv)F------F-------T, neither a or b = 0 so a*b is different from 0 (it is correct)

    Please help me.

    Thank you.


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  3. #2  
    Forum Radioactive Isotope MagiMaster's Avatar
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    The problem is, you've stated (p implies q), which doesn't actually say anything about what happens when p is false. By default we assume the preposition as a whole to be vacuously true. If you write out the truth table, you'd actually get:

    Code:
    p q ?
    T T T
    T F F
    F T F
    F F T
    This isn't the truth table for (p => q), it's the truth table for (p <=> q). This comes from the fact that a*b = 0 if and only if either a is 0 or b is 0. (p => q would be a*b = 0 if, but not only if, either a or b is 0.)

    To see why (p => q) normally works fine, change p and q. Try "the bird is a blue-jay" for p, and "the bird has blue feathers" for q. (p => q) boils down to "blue-jays have blue feathers." When p is false and q is true, the statement becomes "the bird isn't a blue-jay, but still has blue feathers" which doesn't contradict the original statement, so we assume the statement as a whole is still true.


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  4. #3  
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    Quote Originally Posted by MagiMaster
    The problem is, you've stated (p implies q), which doesn't actually say anything about what happens when p is false. By default we assume the preposition as a whole to be vacuously true. If you write out the truth table, you'd actually get:

    Code:
    p q ?
    T T T
    T F F
    F T F
    F F T
    This isn't the truth table for (p => q), it's the truth table for (p <=> q). This comes from the fact that a*b = 0 if and only if either a is 0 or b is 0. (p => q would be a*b = 0 if, but not only if, either a or b is 0.)

    To see why (p => q) normally works fine, change p and q. Try "the bird is a blue-jay" for p, and "the bird has blue feathers" for q. (p => q) boils down to "blue-jays have blue feathers." When p is false and q is true, the statement becomes "the bird isn't a blue-jay, but still has blue feathers" which doesn't contradict the original statement, so we assume the statement as a whole is still true.
    Thanks for the help.

    You are right. I made the truth table for equivalence and not implication (I misunderstand the original sentence). Ok let me do it again.

    If and not only if a=0 or b=0 then a*b=0

    p: a=0 or b=0
    q: a*b=0

    If p=T and q=F, then the sentence would be:

    "If and not only if a=0 or b=0 then a*b≠0"

    Does this mean that, a or b should not be necessarily 0?

    What if p=F and q=T

    Does it mean that a or b should not be necessarily ≠0, so that a*b=0

    Thanks in advance.
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  5. #4  
    Forum Radioactive Isotope MagiMaster's Avatar
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    Well, the problem is, with p = "either a or b is 0" and q = "a*b = 0", you don't have implication, you have equivalence.

    The basic point is, when we say (p => q), the statement as a whole is meaningless, and thus vacuously true, when p is false. Basically, it doesn't matter whether (~p => q) or (~p => ~q) or anything else because the original (p => q) doesn't cover that case. It's a bit of a stretch, but it's binary logic, so it has to either be true or false, and false doesn't really work. (Now one caveat there. Because of the contrapositive, (p => q) <=> (~q => ~p), so if p and q are both false, the implication should still be true.)
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  6. #5  
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    most logical constructs of mathematical equations or concepts are "if and only if" statements as opposed to implications. Your initial example is one of this nature; if and only if either A=0, B=0, or A and B=0. Using simple algebra you can show this by solving for A and solving for B.
    B=anything
    A=anything
    Math is just one of those subjects like that, where more often than not you will have equivalence statements instead of implications.
    Wise men speak because they have something to say; Fools, because they have to say something.
    -Plato

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  7. #6  
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    Thank you for the replies. But what implication means?

    If... means it don't need to necessarily be true or false right?
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  8. #7  
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    What "P implies Q" means is that if P is true, then Q is true. But, Q can also be true when P is false, or stated better, "Q does not imply P" the truth table for implication you had is right.

    P | Q |
    T | T | T
    T | F | F
    F | T | T
    F | F | T

    And like Magi said, a better example of implication is in biology, where exceptions exist.
    Wise men speak because they have something to say; Fools, because they have to say something.
    -Plato

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  9. #8  
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    thanks for the replies.

    I have another sentence problem, so if you have free time please help me.

    Here is the task:

    *6. A certain college offered exactly four languages: French, German,
    Russian, and Latin. The registrar was instructed to enroll each student
    for exactly two languages. After registration, the following facts
    were compiled:
    i) All students who registered for French also registered for exactly
    one of the other three languages.
    ii) All students who registered for neither Latin nor German registered
    for French.
    iii) All students who did not register for Russian registered for
    at least two of the other three languages.
    iv) No candidate who registered for Latin and German registered
    for Russian.
    Did the registrar actually follow his instructions?

    I will start for i)

    i) F -> G (xor) R (xor) L

    Now let me represent my solution with sentence.

    If the students registered for french then they also registered for German or Russian (but not both, thats why it is xor) or Latin (but not both).

    What is wrong with my solution?

    ii) (L or G) -> F

    iii)R -> [((F or G) xor L) or ((F or L) xor G) or ((G or L) xor F)

    All students who did not registered for Russian registered for French or German (or both) or Latin

    What is wrong here?
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