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Define logic and explain the logical operators conjunction, disjunction, negation, and bi-implication. Prove De Morgan's law for disjunction using a truth table.

Computer Science

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1a) Logic is the systematic study of valid reasoning and argumentation. It provides principles and methods to analyze and evaluate arguments, ensuring that conclusions follow necessarily from premises.

1b)

  • i) Conjunction (∧\land): A logical operator that combines two propositions, PP and QQ, into a new proposition P∧QP \land Q (read as "P and Q"). The conjunction is true if and only if both PP and QQ are true; otherwise, it is false.
  • ii) Disjunction (∨\lor): A logical operator that combines two propositions, PP and QQ, into a new proposition P∨QP \lor Q (read as "P or Q"). The disjunction is true if at least one of PP or QQ is true (inclusive or); it is false only if both PP and QQ are false.
  • iii) Negation (¬\neg): A logical operator that takes a single proposition, PP, and forms a new proposition ¬P\neg P (read as "not P"). The negation is true if and only if PP is false, and false if and only if PP is true.
  • iv) Bi-implication (or Equivalence) (↔\leftrightarrow): A logical operator that combines two propositions, PP and QQ, into a new proposition P↔QP \leftrightarrow Q (read as "P if and only if Q"). The bi-implication is true if PP and QQ have the same truth value (both true or both false); otherwise, it is false.

1c) To prove ¬(K∨L)=¬K∧¬L\neg(K \lor L) = \neg K \land \neg L, we can use a truth table:

KKLLK∨LK \lor L¬(K∨L)\neg(K \lor L)¬K\neg K¬L\neg L¬K∧¬L\neg K \land \neg L
TTTFFFF
TFTFFTF
FTTFTFF
FFFTTTT

Since the column for ¬(K∨L)\neg(K \lor L) is identical to the column for ¬K∧¬L\neg K \land \neg L, the equivalence is proven.

2a) We will prove by mathematical induction that for all positive integers nn, 1+2+3+⋯+n=n(n+1)21+2+3+\dots+n = \frac{n(n+1)}{2}.

  1. Base Case

    For n=1n=1:
    LHS: 1
    RHS: 1(1+1)2=1×22=1\frac{1(1+1)}{2} = \frac{1 \times 2}{2} = 1
    Since LHS = RHS, the formula holds for n=1n=1.

  2. Inductive Hypothesis

    Assume the formula holds for some positive integer kk.

    1+2+3+⋯+k=k(k+1)21+2+3+\dots+k = \frac{k(k+1)}{2}
  3. Inductive Step

    We need to show that the formula holds for n=k+1n=k+1. That is, we need to show:

    1+2+3+⋯+k+(k+1)=(k+1)((k+1)+1)2=(k+1)(k+2)21+2+3+\dots+k+(k+1) = \frac{(k+1)((k+1)+1)}{2} = \frac{(k+1)(k+2)}{2}

    Start with the LHS:

    1+2+3+⋯+k+(k+1)1+2+3+\dots+k+(k+1)

    Using the inductive hypothesis, substitute the sum up to kk:

    =k(k+1)2+(k+1)= \frac{k(k+1)}{2} + (k+1)

    Find a common denominator:

    =k(k+1)2+2(k+1)2= \frac{k(k+1)}{2} + \frac{2(k+1)}{2}

    Factor out (k+1)(k+1):

    =(k+1)(k+2)2= \frac{(k+1)(k+2)}{2}

    This is the RHS.

  4. Conclusion

    By the principle of mathematical induction, the formula 1+2+3+⋯+n=n(n+1)21+2+3+\dots+n = \frac{n(n+1)}{2} holds for all positive integers nn.

    2b) To expand (a+b)6(a+b)^6 using Pascal's triangle, we first find the coefficients from the 6th row of Pascal's triangle.
    Row 0: 1
    Row 1: 1 1
    Row 2: 1 2 1
    Row 3: 1 3 3 1
    Row 4: 1 4 6 4 1
    Row 5: 1 5 10 10 5 1
    Row 6: 1 6 15 20 15 6 1

    Using these coefficients, the expansion is:

    (a+b)6=1a6b0+6a5b1+15a4b2+20a3b3+15a2b4+6a1b5+1a0b6(a+b)^6 = 1a^6b^0 + 6a^5b^1 + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6a^1b^5 + 1a^0b^6 (a+b)6=a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6(a+b)^6 = \boxed{a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6}

    3a)

    • Permutation: A permutation is an arrangement of objects in a specific order. The order of selection or arrangement matters. For example, if we select two letters from A, B, C, the permutations are AB, BA, AC, CA, BC, CB.
    • Combination: A combination is a selection of objects where the order of selection does not matter. It is simply a group or subset of items. For example, if we select two letters from A, B, C, the combinations are {A, B}, {A, C}, {B, C}.

    3b) To find the number of permutations from the letters of the word STATISTICS, we first count the total number of letters and the frequency of each repeated letter.
    The word is STATISTICS.
    Total number of letters, n=10n = 10.

    Frequencies of repeated letters:

    • S appears 3 times (n1=3n_1 = 3)
    • T appears 3 times (n2=3n_2 = 3)
    • A appears 1 time
    • I appears 2 times (n3=2n_3 = 2)
    • C appears 1 time

    The formula for permutations with repetitions is n!n1!n2!…nk!\frac{n!}{n_1! n_2! \dots n_k!}.

  5. Identify nn and the frequencies of repeated letters.

    n=10n = 10
    nS=3n_S = 3
    nT=3n_T = 3
    nI=2n_I = 2

  6. Apply the permutation formula.

    Number of permutations=10!3!×3!×2!\text{Number of permutations} = \frac{10!}{3! \times 3! \times 2!}
  7. Calculate the factorials.

    10!=3,628,80010! = 3,628,800
    3!=3×2×1=63! = 3 \times 2 \times 1 = 6
    2!=2×1=22! = 2 \times 1 = 2

  8. Substitute and simplify.

    Number of permutations=3,628,8006×6×2\text{Number of permutations} = \frac{3,628,800}{6 \times 6 \times 2} Number of permutations=3,628,80072\text{Number of permutations} = \frac{3,628,800}{72} Number of permutations=50,400\text{Number of permutations} = 50,400

    The number of permutations that can be made from the letters of the word STATISTICS is 50,400\boxed{50,400}.

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