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Boolean algebra conceptual questions

  1. Sep 16, 2009 #1
    Guys I need your help! :cry: I've never liked boolean algebra, with conjugation, disjunction, implication, equivalence and all that stuff because I considered very easy to understand.

    But what I can not understand is implication (I get headaches from implications :frown:)

    Here is what I do not understand.

    Lets consider the following sentence:

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

    Lets write the truth table :smile: but first...

    Lets consider p - the "if" sentence, and q the "then" sentence. Its clear that we have implication because of If...then

    p -----q ----- p=>q

    Ok that was the general case. Now lets check the "consistency" of the truth table.

    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)

    Am I missing some part of the sentence? Please help!

    Thanks a lot.
  2. jcsd
  3. Sep 16, 2009 #2
    a b a*b
    0 0 0
    0 1 0
    1 0 0
    1 1 1

    This is the truth table for the above statment.

    Don't think of it as an implication. All it is saying is a*b only has a non-zero value when both a and b have values greater than 0.
  4. Sep 16, 2009 #3
    Let p and q be the following statements
    p = "a=0 or b=0"
    q = "a*b=0"

    And consider the following cases:

    (i) a=0, b=0

    (ii) [itex]a \neq 0[/itex], [itex]b \neq 0[/itex]

    (iii) a=0, [itex]b \neq 0[/itex]

    (iv) [itex]a \neq 0[/itex], b=0

    For each case examine the statement p=>q
  5. Sep 16, 2009 #4
    The conceptual problem people tend to have with implications is when the antecedent (p) is false. I tend to think of it this way:

    The implication speaks to the nature of the consequent (q) only when the antecedent is true. If the antecedent is false no conclusion about the consequent can be drawn.

    For example: If the suspect's DNA matches the sample collected at the scene, then the suspect must have been at the scene of the crime.

    If there is no DNA match, no conclusion can be made. The suspect might have been at the scene and maybe not. Regardless it is a true statement.

    In the case of "If at least one of a or b equals 0, then ab = 0," the value of ab is irrelevant if both a and b are non-zero (making the antecedent false).

    I find it useful to also remember that [itex]p \rightarrow q \equiv \neg p \vee q[/itex].

    So the implication "if p then q" is true if either p is false or q is true.

    I hope this is helpful.

  6. Sep 17, 2009 #5
    Thanks for the replies.

    Eluclidus I guess, I am not converting the true sentence into false sentence right.

    Lets consider your statement.

    "If the suspect's DNA matches the sample collected at the scene, then the suspect must have been at the scene of the crime."

    Now lets guess that this sentence is true, i.e p => q (T) where p=T and q=T

    Now let me made the p sentence false.

    If the suspect's DNA do not match the sample collected at the scene, then the suspect must have been at the scene of the crime.

    Simply, it does not make sense. If the suspect's DNA do not match with the scene's one why it must have been at the scene there?

    Do you refer for the fact, that the suspect may be there but it did not left any DNA on the scene?

    By saying that if a≠0 and b≠0 then a*b=0, its paradoxical.
    F => T is T is paradoxical.

    @Edgardo I did examine it before.

    If a=b=0 then a*b=0

    If a≠0 and b=0 then a*b=0

    If a=0 and b≠0 then a*b=0

    If a≠0 and b≠0 then a*b≠0

    I used implication in my first post.
    Last edited: Sep 17, 2009
  7. Sep 17, 2009 #6
    In doing this, you are changing the sentence. The sentence is still "If the DNA matches..." the only thing that is new is that you now known there is no match. This information does not change the fact that the statement is true. It just becomes irrelevant.

    You are substituting different phrases in for "p". In the implication p is always "at least on of a and b is zero" the only thing that can change is whether p is true or false. Changing the interpretation changes the sentence.

    For example consider the implication [itex]p \rightarrow q[/itex]. You seem to be saying that if p were false then we'd have [itex]\neg p \rightarrow q[/itex]. This is incorrect, the statement is still [itex]p \rightarrow q[/itex], but it is known that p = F so we could say [itex]F \rightarrow q[/itex].

  8. Sep 17, 2009 #7
    IThanks for the reply. I guess you're right. I should not change the sentence. Now I understand that there is no match, so lets the ascendant be p and consequent q.
    p=F: there is no match
    q=T: the suspect must have been on the scene.

    Because there is no match, we can not say (for sure) that it must have been on the scene. It may, but it may not, so why p => q is true?
  9. Sep 17, 2009 #8
    Consider I said "If I were King of the World, then I would eliminate all taxes." Since I am not King of the World (at least no one has informed me of this yet), have I lied?

    An implication is true if the consequent is true whenever the antecedent is true. As long as q is true when p is true then p -> q is true. So the only way for it to be false is if p = T and q = F.

  10. Sep 17, 2009 #9
    In that case you did not lie. Ok, you're saying me that if the antecedent is F then I should not consider the consequent. Anyway, I do not consider the p:F, q:T as T.
    If you said that something is different from 0 than its product must be different from 0. I still do not know why we ignore the consequent, when the antecedent is F.
    If we say that there is no DNA match of the suspect, we can not still say for sure that it has been on the scene.
    Last edited: Sep 17, 2009
  11. Sep 17, 2009 #10
    You examined the statement q only.

    The statement "p=>q" has the following truth table:

    For p=false, q=false: "p=>q" is true
    For p=false, q=true: "p=>q" is true
    For p=true, q=false: "p=>q" is false
    For p=true, q=true: "p=>q" is true

    For our first case we have:
    (i) a=0, b=0
    Then the statement p is true and the statement q is true (do you see why?).
    We know from our truth table that for p=true and q=true the statement
    "p=>q" is true.

    Can you examine the other cases?
  12. Sep 17, 2009 #11
    Ok, thank you. Here is what I got:

    Do you mean (p v q) => r ?

    What are p and q ?

    Edit: Edgardo I meant to check how the truth table is really "truth" ? I do not want to check by the truth table.
  13. Sep 18, 2009 #12
    After some while and help of you and others, I realized that I was using equivalence instead of implication.

    So the sentence is If...Then and not If and only of...then

    I guess if p=T and q=T then

    if a and b are not necessarily 0, then a*b=0? Am I right? Is this correct?
  14. Sep 18, 2009 #13
    Do you consider the following sentence to be truthful?

    "If it is snowing then it is cold out."

    Do you think it is any less true when it is warm outisde?

    Implications are either true or false regardless of whether the antecedent is currently true.

    In the case of "If either a = 0 or b = 0 then ab = 0," the statement is true because it is impossible to get a product other than 0 when at least one of the factors involved is 0.

    The only way this implication could be false is if one could furnish a pair of non-zero factors whose product is zero. (Which is impossible.)

    I get the impression there is a misunderstanding about implications that has to do with misinterpreting them as modus ponens deductions. For example:

    "If x and y have the same sign then xy is nonnegative." Well what if x and y are of oppositive sign? The product is negative and the conclusion is clearly false, so how can the implication be "true?"

    This is actually an interpretation of the deductive argument [itex]p \rightarrow q, \neg p \vdash q[/itex] which is an invalid inference. However an implication and a deductive argument are not the same thing.

    I am not sure if this is what is happening.

  15. Sep 18, 2009 #14
    Not correct. 2 and 3 are not necessarily 0 and 2*3 = 6 (not 0) is a simple counter-example.

  16. Sep 19, 2009 #15
    Yes, your example is same like mine one.

    What if the antecedent is false then the sentence would be: x and y have different signs then xy has nonnegative sign. That's what I do not understand.
  17. Sep 19, 2009 #16
    The implication is "If x and y have the same sign then xy is nonnegaitve." Letting the antecedent be false does not change the sentence to "If x and y have different signs then xy is nonnegative." The sentence is as it was originally. The only thing new is that we know that the phrase "x and y have the same sign" is false.

    If you have the implication [itex]p \rightarrow q[/itex] and you know p is false you cannot reinterpret this as [itex]\neg p \rightarrow q[/itex]. You've changed the sentence entirely.

    If p is false, you could say [itex]F \rightarrow q[/itex] (which is true).

    As I said before "If it is snowing, then it is cold," is true even when it is warm. Knowing it is not snowing cannot change the statment to "If it is not snowing, then it is cold," because this is a new and different (and false) statement unrelated to the original.

  18. Sep 20, 2009 #17
    Ok, I understand. Thank you.

    And how will you write F -> q with sentence?
  19. Sep 20, 2009 #18
    One could interpret it as "If FALSE, then it is cold," but that doesn't parse well in a natural language. At this point any contradiction could do, like "0 = 1."

    So if you knew it was not snowing the sentence "If it is snowing, then it is cold," has the same truthhood as "If 0 = 1, then it is cold." Both are true.

    I will mention this vaccuous case of implication is what most people strugglle with.

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