University of Caloocan City
College of Liberal Arts & Sciences
                        Mathematics Department
               Learning Module in Number Theory (LM1)
Topic            PRELIMINARIES
Sub-Topic        Real Number System, Properties of Integers, Summation and
                 Product Notation.
Duration         3 hours
Introduction     A real number is a number that can be found on the number line. It consists of
                 numbers from negative infinity to positive infinity. Below is a diagram of a real
                 number system.
                                                          Real Numbers (ℝ)
                                   Rational Numbers (ℚ)                     Irrational Numbers (ℚ′ )
                  Non-Integers (Fraction) (ℤ′ )             Integers (ℤ )
                                    Negative Integers (ℤ− )             Whole Numbers (𝕨)
                                                         Zero (𝕆)                 Positive Integers or
                                                                                  Natural Numbers (ℤ+ )
                 The set of integers ℤ = { . . . . −2, −1, 0, 1, 2, 3, . . . . } has the following laws
                 1. Closure Laws
                    For any integers 𝑎 𝑎𝑛𝑑 𝑏, 𝑡ℎ𝑒𝑛
                    𝑎+𝑏 ∈ ℤ,          𝑎−𝑏 ∈ ℤ,          𝑎𝑛𝑑       𝑎∙𝑏 ∈ ℤ
                  However, ℤ is NOT closed under division.
                 2. Commutative Laws
                    For any 𝑎, 𝑏 ∈ ℤ , 𝑡ℎ𝑒𝑛
                    𝑎 + 𝑏 = 𝑏 + 𝑎,   𝑎𝑛𝑑           𝑎∙𝑏 = 𝑏∙𝑎
                 3. Associative Laws
                    For any 𝑎, 𝑏, 𝑐 ∈ ℤ
                    𝑎 + (𝑏 + 𝑐) = (𝑎 + 𝑏) + 𝑐            𝑎𝑛𝑑
                    𝑎 ∙ (𝑏 ∙ 𝑐) = (𝑎 ∙ 𝑏) ∙ 𝑐
                 4. Distributive Laws
                   For any 𝑎, 𝑏, 𝑐 ∈ ℤ
                      (𝑎 + 𝑏)𝑐 = 𝑎𝑐 + 𝑏𝑐          𝑅𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒
                       𝑎(𝑏 + 𝑐) = 𝑎𝑏 + 𝑎𝑐         𝐿𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑣𝑒
                 5. Identity Laws
                    𝑎+0=𝑎             𝑓𝑜𝑟 𝑎𝑛𝑦 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑎
                    𝑎∙1=𝑎            𝑓𝑜𝑟 𝑎𝑛𝑦 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑎
            6. Inverse Laws
                                                                1
              For any 𝑎 ∈ ℤ , there exists inverses – 𝑎 𝑎𝑛𝑑         𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡
                                                                𝑎
                                              1
                𝑎 + (−𝑎) = 0       𝑎𝑛𝑑     𝑎 (𝑎 ) = 1 , 𝑎 ≠ 0
            7. The integers 𝑎 𝑎𝑛𝑑 𝑎 + 1 are consecutive integers and there exists NO
            integer x such that for any integer 𝑎
                            𝑎 <𝑥 <𝑎+1
            8. 𝐼𝑓 𝑎 ∙ 𝑏 = 0, 𝑡ℎ𝑒𝑛 𝑒𝑖𝑡ℎ𝑒𝑟 𝑎 = 0 𝑜𝑟 𝑏 = 0 𝑜𝑟 𝑎 = 𝑏 = 0
            9. 𝐹𝑜𝑟 𝑎𝑛𝑦 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑎
                 𝑎 <𝑎+1        𝑎𝑛𝑑       𝑎−1<𝑎
            10. 𝑎 > 0 𝑚𝑒𝑎𝑛𝑠 𝒂 𝑖𝑠 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑎𝑛𝑑
                 𝑏 > 0 𝑚𝑒𝑎𝑛𝑠 𝒃 𝑖𝑠 𝑎 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
            11. 𝑇ℎ𝑒 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑛 𝑒𝑣𝑒𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑖𝑠 2𝑘 ; 𝑎𝑛𝑑 𝑡ℎ𝑎𝑡 𝑓𝑜𝑟 𝑜𝑑𝑑
                𝑖𝑠 2𝑘 + 1 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑘.
Summation   The summation notation also known as Sigma Notation is a simple form of
Notations   shorthand used to give a concise expression for a sum of the values of a
            variable.
            The above symbol is equivalent to
              ∑𝑛𝑖=1 𝑥𝑖 = 𝑥1 + 𝑥2 + 𝑥3 + ∙ ∙ ∙ ∙ + 𝑥𝑛
            This x indicates that you will find the sum of all the values of 𝑥 starting from 𝑥1
            up to 𝑥𝑛 .
            Properties of Summation
             𝐼𝑓 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑡ℎ𝑒𝑛
            1. ∑𝑛𝑖=1 𝑘 = 𝑘𝑛
               Example:
                 ∑61=1 5 = 5(6) = 30
            2. ∑𝑛𝑖=1 𝑘𝑥𝑖 = 𝑘 ∑𝑛𝑖=1 𝑥𝑖
               Example:
                 ∑4𝑖=1 7𝑥𝑖 = 7 ∑4𝑖=1 𝑥𝑖 = 7(𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 )
            3. ∑𝑛𝑖=1(𝑥𝑖 ± 𝑦𝑖 ) = ∑𝑛𝑖=1 𝑥𝑖 ± ∑𝑛𝑖=1 𝑦𝑖
               Example
                  ∑3𝑖=1(𝑥𝑖 + 𝑦𝑖 ) = ∑3𝑖=1 𝑥𝑖 ± ∑3𝑖=1 𝑦𝑖
                                   = (𝑥1 + 𝑥2 + 𝑥3 ) + (𝑦1 + 𝑦2 + 𝑦3 )
            NOTE: In examples 2 and 3, to evaluate for specific value of the summation the
            values of 𝑥1 , 𝑥2 , 𝑥3 𝑎𝑛𝑑 𝑠𝑜 𝑜𝑛 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑔𝑖𝑣𝑒𝑛 𝑠𝑜 𝑎𝑠 𝑡ℎ𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑦1 , 𝑦2 ….
                   Example 1:
                   𝐼𝑓 𝑥1 = 5 , 𝑥2 = 9, 𝑥3 = −6 𝑎𝑛𝑑 𝑦1 = 7, 𝑦2 = 10 , 𝑦3 = 0
                   Evaluate the following
                      1. ∑3𝑖=1 𝑥𝑖
                      2. ∑3𝑖=1 6𝑦𝑖
                      3. ∑3𝑖=2(𝑥𝑖 + 𝑦𝑖 )
                   Solution.
                       1. ∑3𝑖=1 𝑥𝑖 = 𝑥1 + 𝑥2 + 𝑥3 = 5 + 9 + (−6) = 8
                       2. ∑3𝑖=1 6𝑦𝑖 = 6 ∑3𝑖=1 𝑦𝑖 = 6(𝑦1 + 𝑦2 + 𝑦3 )
                                                 = 6(7 + 10 + 0) = 6(17)
                                                  = 102
                       3. 𝑖=2(𝑥𝑖 + 𝑦𝑖 ) = 𝑖=2 𝑥𝑖 + ∑3𝑖=2 𝑦𝑖
                          ∑  3
                                            ∑ 3
                                            = (𝑥2 + 𝑥3 ) + (𝑦2 + 𝑦3 )
                                            = (9 + (−6)) + (10 + 0)
                                            = (3) + (10)
                                            = 13
                   Example 2:
                      Find the value of ∑5𝑖=1(−1)𝑖 (4𝑖 − 3)
                   Solution.
                    ∑5𝑖=1(−1)𝑖 (4𝑖 − 3) = (−1)1 [4(1) − 3] + (−1)2 [4(2) − 3] +
                          (−1)3 [4(3) − 3] + (−1)4 [4(4) − 3] + (−1)5 [4(5) − 3]
                        = (−1)[1] + (1)[5] + (−1)[9] + (1)[13] + (−1)[17]
                        = −9
Product Notation
                   The product notation also known as Pi Notation is used to indicate repeated
                   multiplication. It is similar to Sigma symbol except that succeeding terms are
                   multiplied instead of added.
                     ∏𝑛𝑖=1 𝑖 = 1 ∙ 2 ∙ 3 ∙ ∙ ∙ 𝑛 = 𝑛!
                   Properties of Product
                   𝐼𝑓 𝑘 𝑖𝑠 𝑎𝑛𝑦 𝑟𝑒𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟
                   1. ∏𝑛𝑖=1 𝑘 = 𝑘 𝑛
                   2. ∏𝑛𝑖=1 𝑥𝑖𝑘 = (∏𝑛𝑖=1 𝑥𝑖 )𝑘
                   3. ∏𝑛𝑖=1 𝑘𝑥𝑖 = 𝑘 ∏𝑛𝑖=1 𝑥𝑖
                   4. ∏𝑛𝑖=1 𝑥𝑖 𝑦𝑖 = (∏𝑛𝑖=1 𝑥𝑖 ) (∏𝑛𝑖=1 𝑦𝑖 )
                   Example 3:
                     Evaluate the following:
                      1. ∏4𝑖=1(−1)𝑖+1
                      2. ∏3𝑖=1(3𝑖 + 1)𝑖
                             Solution.
                             1.   ∏4𝑖=1(−1)𝑖+1 = [(−1)1+1 ] [(−1)2+1 ] [(−1)3+1 ] [(−1)4+1 ]
                                            = [(−1)2 ] [(−1)3 ] + [(−1)4 ][(−1)5 ]
                                            = [1][−1][1 ] [−1 ]
                                            = 1
                             2. ∏3𝑖=1(3𝑖 + 1)𝑖 = [3(1) + 1]1 [3(2) + 1]2 [3(3) + 1]3
                                               = [4]1 [7]2 [10]3
                                               = (4)(49)(1000)
                                               = 196 000
Prepared and Submitted by:
TEODULFO T. UCHI
    Faculty