Discrete Mathematics Lecture 1 - PowerPoint - PowerPoint by YSUoKDY

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									       Discrete Mathematics
        with Applications


                  Chapter 3

Elementary Number Theory and Methods of Proof




                              전 북 대 학 교   전 자 정 보 공 학 부
 Elementary Number Theory and Methods of Proof


3.1 Direct Proof and Counterexample : Introduction

3.2 Direct Proof and Counterexample : Rational Numbers

3.3 Direct Proof and Counterexample : Divisibility

3.4 Direct Proof and Counterexample : Division into Cases

3.6 Indirect Argument : Countradicion and Contraposition



                                    전 북 대 학 교    전 자 정 보 공 학 부
3.1 Direct Proof and Counterexample : Introduction (1/6)

 Term’s definition
    Definition
          용어의 뜻을 명확하게 정한 것
       Theorem
          참인 명제 – 증명된 명제 중에서 기본이 되는 명제.
          다른 명제를 증명할 때 이용되는 명제
       Proof
          이미 알고 있는 사실을 바탕으로 주어진 명제가 참임을 밝히는 것
       Corollary
          증명된 정리나 명제로부터 쉽게 증명이 되는 명제
       Lemma
          다른 정리를 증명하는데 쓰이는 명제
       Axiom or Postulate
          자명한 사실, 즉 증명을 하지 않아도 되는 명확한 명제

                               전 북 대 학 교   전 자 정 보 공 학 부
3.1 Direct Proof and Counterexample : Introduction (2/6)
                                                      A
  이등변삼각형을 생각하면
    Definition
           두 변의 길이가 같은 삼각형( AB = AC )
        Theorem
           이등변 삼각형의 밑각은 같다( ∠B = ∠C )
        Proof
           밑변의 중점을 M이라고 하면,
                                             B              C
            △ ABM ≡ △ ACM임으로 ∠B = ∠C                 M
        Corollary
           BM = CM, BC = 2 × BM = 2 × CM
        Lemma
        Axiom or Postulate
           직선은 임의의 한 점에서 임의의 다른 한 점으로부터 그린다.


                                     전 북 대 학 교   전 자 정 보 공 학 부
3.1 Direct Proof and Counterexample : Introduction (3/6)

  Definition of Even and Odd number( 1/2 )
     Even Number

      An integer n is even if, and only if, n equals twice some integer.

      If n is an integer, then
      n is Even ⇔ k  Z, n = 2ㆍ k

        Odd Number

      An integer n is odd if, and only if, n equals twice some integer plus 1.

      If n is an integer, then
      n is Odd ⇔ k  Z, n = 2ㆍ k + 1



                                             전 북 대 학 교        전 자 정 보 공 학 부
3.1 Direct Proof and Counterexample : Introduction (4/6)

 Example 3.1.1 Even and Odd Integers
  Q
       ①   Is 0 even?
       ②   Is -301 odd?
       ③   If a and b are integers, is 6a2b even?
       ④   If a and b are integers, is 10a + 8b + 1 odd?
       ⑤   Is every integer either even or odd?

  A
       ①   Yes, 0 = 2ㆍ0
       ②   Yes, -301 = 2ㆍ( -151 ) + 1
       ③   Yes, 6a2b = 2ㆍ( 3a2b )
       ④   Yes, 2ㆍ( 5a + 4b ) + 1
       ⑤   The answer is Yes, although the proof is not obvious.




                                            전 북 대 학 교      전 자 정 보 공 학 부
3.1 Direct Proof and Counterexample : Introduction (5/6)

  Definition of Prime and Composite number( 2/2 )
     Prime Number

      An integer n is prime if, and only if, n > 1 and
      for all positive integer r and s, if n = rㆍs, then r = 1 or s = 1.

      n is Prime ⇔ ∀r, s  Z+, n > 1, n = r ㆍs → r = 1 ∨ s = 1.

        Composite Number

     An integer n is composite if, and only if, n > 1 and n = rㆍs
     for some positive integer r and s with r ≠ 1 and s ≠ 1.

     n is Composite ⇔ ∃r, s  Z+, n > 1, n = r ㆍs → r ≠ 1∧ s ≠ 1.



                                               전 북 대 학 교        전 자 정 보 공 학 부
3.1 Direct Proof and Counterexample : Introduction (6/6)

   Example 3.1.2 Prime and Composite numbers
 Q
     ①   Is 1 prime?
     ②   Is it true that every integer greater than 1 is either prime or composite?
     ③   Write the first six prime numbers.
     ④   Write the first six composite numbers.


 A   ① No, A Prime number is required to be greater than 1.
     ② Yes, For any integer greater than 1,
       the two definitions are negations of each other.
     ③ Prime Truth set = { 2, 3, 5, 7, 11, 13, … }
     ④ Composite Truth set = { 4, 6, 8, 9, 10, 12, … }

 ※   ~( Prime Truth set )     ⇔ Composite Truth set
     ~( Composite Truth set ) ⇔ Prime Truth set

                                              전 북 대 학 교        전 자 정 보 공 학 부
            Proving Existential Statements
 Constructive Proofs of Existence
    x  D such that Q(x)
     ↔ Q(x) is true for at least one x in D.

 Example 3.1.3 Constructive Proofs of Existence.
      a.   두 소수의 합이 짝수가 되는 n이 존재함을 증명
              n=10 이라 하면 10 = 5 + 5 = 3 + 7.
              3, 5, 7은 모두 소수


      b.   r, s가 정수라 가정. 22r+18s=2k를 만족하는 k가 존재함을 증명
             k = 11r + 9s 라 하면 2k = 2 ( 11r + 9s ).
             k는 정수 곱의 합이므로 정수임



                                             전 북 대 학 교   전 자 정 보 공 학 부
Disproving Universal Statements by Counterexample (1/2)


   To Disprove a statement means to show that it is False.
   Disproving a statement of the form ∀x  D, P(x) → Q(x)

      ~(∀x  D, P(x) → Q(x) ) ☞ showing that this statement is True.
      ≡ x  D, P(x) ∧ ~Q(x) ☞ find a value of x in D.


   Disproof by Counterexample


       To Disprove a statement of the form “∀x  D, P(x) → Q(x).”,
       find a value of x in D for which P(x) is true and Q(x) is false.
       Such an x is a called a counterexample.


                                              전 북 대 학 교        전 자 정 보 공 학 부
Disproving Universal Statements by Counterexample (2/2)



   Example 3.1.4 Disproof by Counterexample
      Disprove the following statement by finding a counterexample


      Statement : a, b  R, a2 = b2 → a = b
      Counterexample : a = 1, b =-1.
                        a2 = 12 = 1, b2 = (-1)2 = 1,
                        so a2=b2. But a ≠ b since 1 ≠-1.




                                            전 북 대 학 교      전 자 정 보 공 학 부
    Proving Universal Statements (1/3)

 How to Prove : x  D, P(x) → Q(x)
    Example 3.1.5 The Method of Exhaustion

     x  Z, x가 짝수이고 4 ≤ x ≤ 30 이면 x은 두 소수의 합으로 나타냄


      4=2+2       6=3+3       8=3+5       10 = 5 + 5     12 = 5 + 7
     14 =11 + 3 16 = 5 + 11 18 = 7 + 11 20 = 7 + 13 22 = 5 + 17
     24 = 5 + 19 26 = 7 + 19 28 = 11 + 17 30 = 11 + 19




                                         전 북 대 학 교         전 자 정 보 공 학 부
     Proving Universal Statements (2/3)
 Method of Generalizing from the Generic Particular

    To Show that every element of a domain satisfies a certain property,
    Suppose x is a “Particular but Arbitrarily chosen element of the domain”,
    and show that x satisfies the property.
    This is called the method of Direct Proof.

 Method of Direct Proof

    ① Express the statement in the form: x  D, P(x) → Q(x)
    ② Start the proof by supposing x is a particular but arbitrarily chosen
      Element of D for which the P(x) is true.
    ③ Show that Q(x) is true by using definitions,
      previously established results, and the rules for logical inference.


                                            전 북 대 학 교       전 자 정 보 공 학 부
    Proving Universal Statements (3/3)
 Prove that if the sum of any two even integers is even.
     Theorem 3.1.1
     The Sum of any two Even is Even.
     ① Formal Restatement
       m, n  Z, if m and n are even then m + n is Even.

     ② Start the proof by supposing that x  D, P(x)
       m = 2r, n = 2s ( r, s  Z ) ☜ by Even Def.

     ③ Show that the Conclusion
       m + n = 2r + 2s = 2 ( r + s )
       k=r+s (kZ)
       m + n = 2k.
       ∴ m + n is Even.
                                        전 북 대 학 교      전 자 정 보 공 학 부
      Directions for Writing Proofs of Universal
                      Statements
   Standard for writing the final versions of Proofs
    ① Copy the statement of the theorem to be proved on your paper.
    ② Clearly mark the beginning of your proof with the word Proof.
    ③ Make your proof self-contained.
    ④ Write your proof in complete sentences.
    ⑤ Give a reason for each assertion you make in your proof.
    ⑥ Include the “little words” that make the logic of your arguments clear.


   Some general directions for proof writing
    ① Write the theorem to be proved.
    ② Clearly mark the beginning of your proof with the word proof.
    ③ Make your proof self-contained.
    ④ Write proofs in complete English sentences


                                           전 북 대 학 교      전 자 정 보 공 학 부
        Common Mistakes ( 1/4 )
   Some of the most common mistakes people make when writing
    Mathematical Proofs.

    ①   Arguing from examples.
    ②   Using the same letter to mean two different things.
    ③   Jumping to a conclusion.
    ④   Begging the question.
    ⑤   Misuse of the word if.




                                       전 북 대 학 교       전 자 정 보 공 학 부
          Common Mistakes ( 2/4 )
   Arguing from examples.

     This is true because if m = 14 and n = 6, which are both even,
     Then m + n = 20, which is also even.

☞ It is not sufficient to show that the conclusion “m + n is even.” is
    true for m = 14 and n = 6.

   Using the same letter to mean two different things.

     Suppose m and n are odd integers. Then by definition of odd,
     m = 2k + 1 and n = 2k + 1 for some integer k.

☞ Using the same symbol, k, in the expressions for both m and n
   implies that m = 2 k + 1 = n.
☞ This is inconsistent with the supposition that m and n are
   arbitrarily chosen odd integers.
                                         전 북 대 학 교        전 자 정 보 공 학 부
         Common Mistakes ( 3/4 )
   Jumping to a conclusion.

    Suppose m and n are any even integers. By definition of even,
    m = 2r and n = 2s for some integer r and s.
    Then m + n = 2r + 2s. So m + n is even.

    ☞ Crucial calculation 2r + 2s = 2 ( r + s ) is missing.

   Begging the question.( Variation of jumping to a conclusion )

     Suppose m and n are any odd integers.
     When any odd integers are multiplied, their product is odd.
     Hence mㆍn is odd.

☞ First state what it means for the conclusion to be true and
  Later just assumes it to be true.

                                        전 북 대 학 교       전 자 정 보 공 학 부
           Common Mistakes ( 4/4 )
     Misuse of the word if.

Suppose p is a prime number
If p is prime,
Then p cannot be written as a product of two smaller positive integers.




Suppose p is a prime number
Because p is prime,
p cannot be written as a product of two smaller positive integers.




                                          전 북 대 학 교       전 자 정 보 공 학 부
    Showing That an Existential Statement Is False

 To Prove an Existential statement is False, you must Prove a Universal
  statement( Its Negation ) is True.
 Example 3.1.9 Disproving an Existential Statement
    Show that the following statement is False.

                There is a positive integer n such that n2+3n+2 is prime.
                n  Z+, n2+3n+2 is prime.
      Equivalent.
                There is a positive integer n such that n2+3n+2 is not prime.
                n  Z+, n2+3n+2 is not prime.

                Proof :
                Suppose n  Z+.
                n2 + 3n + 2 = ( n + 1 ) ( n + 2 ).
                n + 1 and n + 2  Z+, n + 1 > 1 and n + 2 > 1
                ∴ n2+3n+2 is not prime. ☜ by Prime Def.
                                           전 북 대 학 교        전 자 정 보 공 학 부
   3.2 Direct Proof and Counterexample : Rational
                   Numbers ( 1/2 )
 Definition

          A real number r is rational if, and only if, it can be expressed as
          a quotient of two integers with a nonzero denominator.
          A real number that is not rational is irrational.
          More formally, if r is a real number, then
          r is rational ⇔  integers a and b such that r = a / b and b ≠ 0




                                            전 북 대 학 교        전 자 정 보 공 학 부
   3.2 Direct Proof and Counterexample : Rational
                   Numbers ( 2/2)
 Example 3.2.1 Determining Whether Numbers are Rational or Irrational.
         a.   10 / 3 ?
         b.   –( 5 / 39 ) ?
         c.   0.281 ?
         d.   7?
         e.   0?
         f.   2 / 0 Rational? ☞ No, Not a number.
         g.   2 / 0 Irrational? ☞ No, Not a number.
         h.   0.12121212… ?
         i.   If m, n  Z, m ≠ 0 and n ≠ 0, ( m + n ) / mn ?

        Theorem 3.2.1
        Every integer is a rational number.




                                               전 북 대 학 교       전 자 정 보 공 학 부
  Proving Properties of Rational Numbers

Theorem 3.2.2
The sum of any two rational numbers is rational.
Proof:
Suppose r, s  Q.
r = a / b, s = c / d (단, a, b, c, d는 정수이고 b≠0, d≠0 )
           a c ad  bc
    rs     
           b d   bd
    p = ad + bc, q = bd 라 하면, p, q는 정수이다. q ≠0.
    ∴ r + s는 유리수 정의에 의해 유리수임.




                                     전 북 대 학 교     전 자 정 보 공 학 부
      Deriving New Mathematics from Old
 Properties of Even and Odd integers.
      ①    E + E = E, E × E = E, E - E = E
      ②    O + O = E, O -O = E
      ③    O×O=O
      ④    E×O=E
      ⑤    E+O=O
      ⑥    O -E = O
      ⑦    E -O = O

 Example 3.2.4 The Double of a Rational Number

       Corollary 3.2.3
       The Double of a Rational number is Rational.



                                          전 북 대 학 교   전 자 정 보 공 학 부
3.3 Direct Proof and Counterexample : Divisibility (1/3)

   Divisibility : 분할성, 가분성[나누어 떨어짐]

        If n and d are integer, then
        n is divisible by d if, and only if, n = dㆍk for some integer k.

        Alternatively, we say that
             n is a multiple of d, or                              n
             d is a factor of n, or                           k
                                                                   d
             d is a divisor of n, or
             d divides n.
                                                              nd k

        “d | n” is read “d divides n.”

        Symbolically, if n and d are integers and d ≠ 0.
        d | n ⇔ k  Z, n = dㆍk


                                                 전 북 대 학 교             전 자 정 보 공 학 부
3.3 Direct Proof and Counterexample : Divisibility (2/3)
 Example 3.3.1 Divisibility
        a.   Is a 21 divisible by 3?            d.   Is 32 a multiple of -16?
        b.   Does 5 divide 40?                  e.   Is 6 a factor of 54?
        c.   Does 7 | 42?                       f.   Is 7 a factor of -7?

 Example 3.3.2 Divisors of Zero
        If k is any integer, does k divide 0?
        Yes, ∵ 0 = kㆍ0

 Example The Positive Divisors of a Positive Number
      Suppose a and b are positive integers and a | b. Is a ≤ b?
      Yes, suppose that a | b = k for some integer k.
            b = aㆍk, k ≥ 1
            ∴ a ≤ aㆍk = b

                                                전 북 대 학 교        전 자 정 보 공 학 부
3.3 Direct Proof and Counterexample : Divisibility (3/3)
 Example 3.3.4 Divisors of 1
      Which integers divide 1?
      The only divisors of 1 are 1 and -1.

 Example 3.3.5 Divisibility of Algebraic Expressions
      If a and b are integers, is 3a + 3b divisible by 3?
      Yes, ∵ 3( a + b )

 Example 3.3.6 Checking nonDivisibility
         Does 4 | 15 ?
         No, ∵ 15 / 4 = 3.75, which is not an integer.

 Example 3.3.7 Prime Numbers and Divisibility
      ∀n  Z+, n > 1,
      The only positive integer divisors of Prime are 1 and itself.
                                             전 북 대 학 교      전 자 정 보 공 학 부
     Proving Properties of Divisibility( 1/2 )
 Example 3.3.8 Transitivity of Divisibility (Theorem 3.3.1)
 Theorem 3.3.1
 For all integers a, b, and c, if a divides b and b divides c, then a divides c.
 ∀a, b, c  Z, if a | b and b | c, then a | c.
 Proof:
     Suppose a, b, and c are integers such that a | b and b | c.
     By definition of divisibility
        b = aㆍr c = bㆍs ( r, s  Z )

     By substitution
         c =bㆍ s = ( a ㆍ r ) s = a ( r ㆍs ) ☜ by basic algebra.
         k = rㆍs ( k  Z )
         c=aㆍk
         ∴a|c
                                              전 북 대 학 교        전 자 정 보 공 학 부
Proving Properties of Divisibility( 2/2 )

Theorem 3.3.2
Any integer n > 1 is divisible by a Prime number.
Proof:
    n > 1 정수라 가정.
    n이 소수가 아니면 n = r0s0 (r0, s0 : 정수)
          1 < r0 < n, 1 < s0 < n. ☜ By def. Divisibility that r0| n
    r0가 소수이면 r0는 n을 나누는 소수임.
    만일 소수가 아니면 r0 = r1s1 ( r1, s1 : 정수 )
          1 < r1 < r0, 1 < s1 < r0. ☜ r1| r0, r1| n.
    …
    r0, r1, r2, …, rk
    ( k ≥ 0, 1 < rk < rk – 1 < … < r2 < r1 < r0 < n, ri | n, i = 0, 1, 2, … ,k )
    rk가 소수이어야 함. 그럼 rk는 n을 나누는 소수임.
                                               전 북 대 학 교         전 자 정 보 공 학 부
         Counterexamples and Divisibility
 Example 3.39 Checking a Proposed Divisibility Property


        Statement : for all integers a, b, if a | b and b | a then a = b ?

        Counterexample :
        Let a = 2, b = -2. then
        a|b=        2 | (-2)
        b | a = (-2) | 2,     but a ≠ b ( 2 ≠ -2)

        Therefore, the proposed divisibility property is False.




                                               전 북 대 학 교        전 자 정 보 공 학 부
     The Unique Factorization Theorem
  Theorem 3.3.3 Unique Factorization Theorem for the Integers

  Given any integer n > 1,
  there exist a positive integer k, distinct prime numbers p1, p2, p3, …, pk,
  and positive integer e1, e2, e3, …, ek such that n = p1e1p2e2p3e3…pkek,
  and any other expression of n as a product of prime numbers is identical to
  this except, perhaps, for the order in which the factors are written.

 Example 3.3.19 Writing Integers in Standard Factored Form
      Write 3,300 in standard factored form
      3,300 = 23ㆍ31ㆍ51ㆍ111




                                              전 북 대 학 교     전 자 정 보 공 학 부
3.4 Direct Proof and Counterexample : Division into
    Cases and the Quotient-Remainder Theorem
   Theorem 3.4.1
        Theorem 3.4.1
                                                           2   Quotient
        Given any integer n and positive integer d,
                                                        4 11
        there exist unique integers q and r such that      8
                                                           3   Remainder
        n = dㆍq + r and 0 ≤ r < d.


   Example 3.4.1 The Quotient-Remainder Theorem




                                            전 북 대 학 교   전 자 정 보 공 학 부
                         Div and Mod
 Definition
       Given a nonnegative integer n and a positive integer d,
       n div d = the integer quotient obtained when n is divided by d.
       n mod d = the integer remainder obtained when n is divided by d.

       Symbolically, if n and d are positive integers, then
       n div d = q and n mod d = r ⇔ n = dㆍq + r
       where q and r are integers and 0 ≤ r < d.


 Example 3.4.2 Div and Mod
       Compute 32 Div 9 and 32 Mod 9.




                                           전 북 대 학 교          전 자 정 보 공 학 부
    3.5 Direct Proof and Counterexample V :
                Floor and Ceiling
 Definition
        Given any real number x, the floor of x,
        denoted x  , is defined as follows :
         x  = that unique integer n such that n ≤ x < n + 1

        Symbolically, if x is a real number and n is an integer, then
        x  = n ⇔ n ≤ x < n + 1.
 Definition
        Given any real number x, the ceiling of x,
        denoted x  , is defined as follows :
         x  = that unique integer n such that n -1 < x ≤ n

        Symbolically, if x is a real number and n is an integer, then
         
        =xn ⇔ n - 1 < x ≤ n.

                                             전 북 대 학 교          전 자 정 보 공 학 부
         3.6 Indirect Argument : Contradiction and
                       Contraposition
 Indirect Proof
    Based on the Logical equivalence between
           a statement and negation of its contradiction.☜ Argument by contradiction.
           a statement and its contraposition. ☜ Argument by contraposition.


 Method of Proof by contradiction
  ① Suppose the statement to be proved is False.
  ② Show that this supposition leads logically to a contradiction.
  ③ Conclude that the statement to be proved is True.




                                                전 북 대 학 교         전 자 정 보 공 학 부
3.6 Indirect Argument : Contradiction and
              Contraposition

Theorem 3.6.1
There is no greatest integer.
Proof by Contradiction.
가장 큰 정수 N이 존재한다고 가정.

    N ≥ n (모든 정수 n). M=N + 1이라 하자.
    M은 정수이고 M > N이 된다.
    그래서 M은 N보다 큰 정수이다.
    N이 가장 큰 정수이어야 하는데 그렇지 않다. (모순)




                                전 북 대 학 교   전 자 정 보 공 학 부
3.6 Indirect Argument : Contradiction and
              Contraposition

Theorem 3.6.2
There is no integer that is both even and odd.
Proof by Contradiction
짝수와 홀수인 정수 n이 존재한다고 가정.

    n = 2a (a는 정수), ☜ 짝수 정의에 의해
    n = 2b +1 (b는 정수) ☜ 홀수 정의에 의해
    2a = 2b + 1
    2( a – b ) = 1
    (a–b)=1/2
    a, b가 정수이기 때문에, a – b의 차도 정수여야 함.
    그러나 1 / 2 는 정수가 아님. (모순)



                                     전 북 대 학 교   전 자 정 보 공 학 부
    3.6 Indirect Argument : Contradiction and
                  Contraposition

Theorem 3.6.3
The sum of any rational number and any irrational number is irrational.
Proof by Contradiction
    유리수 r과 무리수 s의 합인 r + s가 유리수 라고 가정

    유리수 정의에 의해
     r = a / b, r + s = c / d (a, b, c, d : 정수, b≠0, d≠0)
    a/b+s=c/d
    s=c/d–a/b=(bc–ad)/bd
    ( b c – a d )와 b d는 모두 정수이고, d b ≠ 0.
    s는 두 정수 ( b c – a d )와 b d의 몫임.
    유리수 정의에 의해 s는 유리수임.(모순)



                                          전 북 대 학 교         전 자 정 보 공 학 부
              Argument by Contraposition
   Proof by Contraposition
    ① Prepare the statement in the form: x  D, P(x) → Q(x)
    ② Rewrite this statement in the form: x  D, ~Q(x) → ~P(x)
    ③ Prove the contrapositive by a direct proof
        a. Suppose x is a element of D such that Q(x) is false.
        b. Show that P(x) is false.
   Example 3.6.4 If the Square of an Integer is even, Then the Integer is Even

       Statement :       ∀n ∈ Z, if n2 is Even then n is even.
       Contraposition : ∀n ∈ Z, if n is not Even then n2 is not even.
       Proof by Contraposition
       n이 홀수라 가정. n = 2k + 1 ( k : 정수 ).
            n2 = ( 2k + 1 )2 = 4k2 + 4k + 1 = 2 ( 2k2 + 2k ) + 1.
            그러나 2k2 + 2k 는 정수.
            n2 = 2 · ( 정수 ) + 1, n2은 홀수임.
                                               전 북 대 학 교        전 자 정 보 공 학 부
 Relation between Proof by Contradiction and Proof
                by Contraposition
 Proposition 3.6.4
     For all integers n, if n2 is even then n is even.
     Proof by Contradiction
     정수 n2은 짝수이고, n은 짝수가 아닌 정수 n이 존재한다고 가정.
     d = 2인 Quotient-Remainder Theorem에 의해 정수는 짝수이거나 홀수임.
     n이 홀수라 가정. n = 2k + 1 (k: 정수).
          n2 = ( 2k + 1 )2 = 4k2 + 4k + 1 = 2( 2k2 + 2k ) + 1.
          그러나 2k2 + 2k는 정수.
          n2 = 2 · ( 정수 ) + 1, n2은 홀수임.
          그럼으로, n2은 짝수이거나 홀수임.
          Theorem 3.6.2를 보면 어떤 정수도 동시에 짝수와 홀수일 수
               없다.




                                   전 북 대 학 교    전 자 정 보 공 학 부

								
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