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Math Help - Discrete Math

  1. #1
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    Cool Discrete Math

    Help me with the following.

    2. Prove that the law of syllogism is a tautology.

    8. If a divides b, then a <= b.

    14. Prove or disaprove - The product of two odd integers is odd.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by j-down View Post
    Help me with the following.

    2. Prove that the law of syllogism is a tautology.

    8. If a divides b, then a <= b.

    14. Prove or disaprove - The product of two odd integers is odd.
    8 and 14 I can do.

    8: If a | b, where a and b are positive whole numbers then b = ak for k is some positive whole number: k = 0, 1, 2, 3.... Therefore, a <= b.

    14: Since the form of an odd number is 2n + 1 for n = 0, 1, 2...
    Let 2a + 1 and 2b + 1 be two odd integers
    (2a + 1)(2b + 1) = 4ab + 2a + 2b + 1 = 2(2ab + a + b) + 1 which is an odd integer.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by j-down View Post
    Help me with the following.

    2. Prove that the law of syllogism is a tautology.

    8. If a divides b, then a <= b.

    14. Prove or disaprove - The product of two odd integers is odd.
    here's 14:

    The statement is true:

    Proof

    let a and b be two odd integers. then a = 2n + 1 and b = 2m + 1 for some integers m and n.

    so a*b = (2n + 1)(2m + 1) = 4mn + 2m + 2n + 1 = 2(2mn + m + n) + 1 = 2c + 1, where c = 2mn + m + n is an integer

    so we see a*b is odd

    QED


    what is the law of syllogism?
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by j-down View Post
    8. If a divides b, then a <= b.
    Note, if a and b can be positive OR negative integers, then if a | b, it can be shown that |a| <= |b|, however, if b is negative then b <= a. Therefore, I assume this problem is only considering a and b to be positive whole numbers.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by j-down View Post
    Help me with the following.

    2. Prove that the law of syllogism is a tautology.

    8. If a divides b, then a <= b.

    14. Prove or disaprove - The product of two odd integers is odd.
    Okay, so if my memory serves me right, the law of syllogism goes like this:

    for statements P, Q and R

    if (P=>Q) and (Q=> R) then (P=>R)

    the proof is done using truth tables, see the image below. notice that the last column has all true values. a statement that is true in every instance is called a taughtology (of course i don't remember if this is the exact law of syllogism, but whatever it is, it's a taughtology. so it will at least give you an idea for the solution
    Attached Thumbnails Attached Thumbnails Discrete Math-truth-table.gif  
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  6. #6
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    Number 8 is not true.

    2 divides -2 but yet,

    2<=-2

    You should have used absolute values.
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  7. #7
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Number 8 is not true.

    2 divides -2 but yet,

    2<=-2

    You should have used absolute values.

    Quote Originally Posted by ecMathGeek View Post
    Note, if a and b can be positive OR negative integers, then if a | b, it can be shown that |a| <= |b|, however, if b is negative then b <= a. Therefore, I assume this problem is only considering a and b to be positive whole numbers.
    Good of you to catch that.
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Good of you to catch that.
    Hmmmm....

    And 4 divides 2 a modulo 6....

    -Dan
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  9. #9
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by topsquark View Post
    Hmmmm....

    And 4 divides 2 a modulo 6....

    -Dan
    Good of you to make that completely irrelevant point.
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Good of you to make that completely irrelevant point.
    Not completely irrelevant. Though it is very likely that we are using the real number system, given that the class is Discrete Math it is not outside of the realm of reason to use modular Mathematics, in which case the theorem doesn't hold.

    Just pointing that out.

    -Dan
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