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Math Help - Discrete Math Practice Problems (Please help confirm)

  1. #1
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    Discrete Math Practice Problems (Please help confirm)

    1) Discrete Math Practice Problems (Please help confirm)-prob1.jpg


    2)Discrete Math Practice Problems (Please help confirm)-prob2.jpg

    3)Discrete Math Practice Problems (Please help confirm)-prob3.jpg

    4)Discrete Math Practice Problems (Please help confirm)-prob4.jpg



    I will be forever in debted if you can help me with these. My first discrete math class.
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  2. #2
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    Re: Discrete Math Practice Problems (Please help confirm)

    you have to show some work. These are all straightforward problems you should be able to do after reading your textbook.
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  3. #3
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    Re: Discrete Math Practice Problems (Please help confirm)

    Hello, outkast32!

    There is no formula for #2.
    I doubt that your textbook will help you.
    It requires some Original Thinking.


    2. Let A \,=\,\{5,6,7,8,9,10,11,12,13,14,15,16,17,18\}
    How many numbers must be chosen to ensure that two of them add up to 23?

    Note that there are 7 Pairs that add up to 23:
    . . (5,18),\:(6,17),\:(7,16),\:(8,15),\:(9,14),\:(10,1  3),\:(11,12)

    To have a sum of 23, your choices must include one of these Pairs.

    Now consider the worst-case scenario: none of the Pairs.
    If you select 7 numbers, one from each Pair,
    . . you will not have a sum of 23.

    When you choose an 8th number, you will have a Pair.

    Therefore, you must draw 8 numbers.
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  4. #4
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    Re: Discrete Math Practice Problems (Please help confirm)

    Hello, outkast32!

    The first problem (unnumbered) is not true.

    It should be stated like this:

    Using mathematical induction, prove that:

    . . 1^3 + 2^3 + 3^3 + \cdots + n^3 \:=\:\frac{n^2(n+1)^2}{4}

    Verify S(1)\!:\;1^3 \:=\:\frac{1^2\!\cdot2^2}{4} . . . True!

    Assume S(k)\!:\;1^3+2^3+3^3+\cdots+k^3 \;=\;\frac{k^2(k+1)^2}{4} .[1]

    We must prove S(k+1)\!:\;1^3+2^3+3^3+\cdots + (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4} .[2]


    Add (k+1)^3 to both sides of [1].

    . 1^2+2^3+3^3+\cdots+k^2+(k+1)^3 \;=\;\frac{k^2(k+1)^2}{4} + (k+1)^3

    Simplify the right side:

    . . \frac{k^2(k+1)^2}{4} + (k+1)^3 \;=\(k+1)^2\left[\frac{k^2}{4} + (k+1)\right] \;=\;(k+1)^2\left[\frac{k^2+4k+4}{4}\right]

    . . =\;(k+1)^2\,\frac{(k+2)^2}{4} \;=\;\frac{(k+1)^2(k+2)^2}{4}


    We have proved [2]: . 1^3+2^3+3^3 + \cdots (k+1)^3 \;=\;\frac{(k+1)^2(k+2)^2}{4}

    The inductive proof is complete.
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