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Mathematical Induction
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
        LHS  25
             32
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
        LHS  25                       RHS  52
             32                            25
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
        LHS  25                       RHS  52
             32                            25
                        LHS  RHS
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
        LHS  25                       RHS  52
             32                            25
                      LHS  RHS
                 Hence the result is true for n = 5
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
        LHS  25                       RHS  52
             32                            25
                      LHS  RHS
                 Hence the result is true for n = 5

Step 2: Assume the result is true for n = k, where k is a positive
        integer > 4
        i.e. 2k  k 2
Mathematical Induction
e.g.v  Prove 2n  n 2 for n  4
Step 1: Prove the result is true for n = 5
        LHS  25                            RHS  52
             32                                 25
                      LHS  RHS
                 Hence the result is true for n = 5

Step 2: Assume the result is true for n = k, where k is a positive
        integer > 4
        i.e. 2k  k 2
Step 3: Prove the result is true for n = k + 1
                         k 1
                                 k  1
                                        2
        i.e. Prove : 2
Proof:
Proof:
         2 k 1
Proof:
         2 k 1  2 2k
Proof:
         2 k 1  2 2k
                 2k 2
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k     k  4
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k         k  4
                 k 2  2k  2k
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k         k  4
                 k 2  2k  2k
                  k 2  2k  8
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k         k  4
                 k 2  2k  2k
                  k 2  2k  8    k  4
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k         k  4
                 k 2  2k  2k
                  k 2  2k  8    k  4
                  k 2  2k  1
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k         k  4
                 k 2  2k  2k
                  k 2  2k  8    k  4
                  k 2  2k  1
                  k  1
                          2
Proof:
         2 k 1  2 2k
                 2k 2
                 k2  k2
                 k2  k k
                 k 2  4k           k  4
                 k 2  2k  2k
                  k 2  2k  8      k  4
                  k 2  2k  1
                  k  1
                          2


                  2  k  1
                     k 1       2
Proof:
      2 k 1  2 2k
              2k 2
              k2  k2
              k2  k k
              k 2  4k            k  4
              k 2  2k  2k
               k 2  2k  8       k  4
               k 2  2k  1
               k  1
                       2


               2  k  1
                  k 1       2


  Hence the result is true for n = k + 1 if it is also true for n = k
Proof:
       2 k 1  2 2k
               2k 2
               k2  k2
               k2  k k
               k 2  4k            k  4
               k 2  2k  2k
                k 2  2k  8       k  4
                k 2  2k  1
                k  1
                        2


                2  k  1
                   k 1       2


   Hence the result is true for n = k + 1 if it is also true for n = k

Step 4: Since the result is true for n = 5, then the result is true for
        all positive integral values of n > 4 by induction .
Proof:
       2 k 1  2 2k
               2k 2
               k2  k2
               k2  k k
               k 2  4k            k  4             Exercise 6N;
               k 2  2k  2k                            6 abc, 8a, 15
                k 2  2k  8       k  4
                k 2  2k  1
                k  1
                        2


                2  k  1
                   k 1       2


   Hence the result is true for n = k + 1 if it is also true for n = k

Step 4: Since the result is true for n = 5, then the result is true for
        all positive integral values of n > 4 by induction .

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11 x1 t14 10 mathematical induction 3 (2012)

  • 2. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4
  • 3. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5
  • 4. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25  32
  • 5. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25
  • 6. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS
  • 7. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS Hence the result is true for n = 5
  • 8. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS Hence the result is true for n = 5 Step 2: Assume the result is true for n = k, where k is a positive integer > 4 i.e. 2k  k 2
  • 9. Mathematical Induction e.g.v  Prove 2n  n 2 for n  4 Step 1: Prove the result is true for n = 5 LHS  25 RHS  52  32  25  LHS  RHS Hence the result is true for n = 5 Step 2: Assume the result is true for n = k, where k is a positive integer > 4 i.e. 2k  k 2 Step 3: Prove the result is true for n = k + 1 k 1  k  1 2 i.e. Prove : 2
  • 11. Proof: 2 k 1
  • 12. Proof: 2 k 1  2 2k
  • 13. Proof: 2 k 1  2 2k  2k 2
  • 14. Proof: 2 k 1  2 2k  2k 2  k2  k2
  • 15. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k
  • 16. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k
  • 17. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4
  • 18. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k
  • 19. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8
  • 20. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4
  • 21. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1
  • 22. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2
  • 23. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2
  • 24. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2 Hence the result is true for n = k + 1 if it is also true for n = k
  • 25. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4  k 2  2k  2k  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2 Hence the result is true for n = k + 1 if it is also true for n = k Step 4: Since the result is true for n = 5, then the result is true for all positive integral values of n > 4 by induction .
  • 26. Proof: 2 k 1  2 2k  2k 2  k2  k2  k2  k k  k 2  4k  k  4 Exercise 6N;  k 2  2k  2k 6 abc, 8a, 15  k 2  2k  8  k  4  k 2  2k  1  k  1 2  2  k  1 k 1 2 Hence the result is true for n = k + 1 if it is also true for n = k Step 4: Since the result is true for n = 5, then the result is true for all positive integral values of n > 4 by induction .