1. Mathematical Induction
Section 11.4
Mathematical Induction
Copyright ©2013, 2009, 2006, 2005 Pearson Education, Inc.

2. Objectives
List the statements of an infinite sequence that is defined by a formula.
Do proofs by mathematical induction.

3. Sequences of Statements
Infinite sequences of statements occur often in mathematics. In an infinite sequence of statements, there is a statement for each natural number. Example: List the first four statements in the sequence that can be obtained from each of the following. a) log n < n b)    + (2n  1) = n2

4. Example
List the first four statements in the sequence that can be obtained from each of the following. a) log n < n b)    + (2n  1) = n2

5. Example (cont)
a) This time, Sn is “log n < n.” S1: log 1 < 1 S2: log 2 < 2 S3: log 3 < 3 S4: log 4 < 4 b) This time, Sn is “    + (2n  1) = n2.” S1: 1 = 12 S2: = 22 S3: = 32 S4: = 42

6. Proving Infinite Sequences of Statements
Mathematical induction can be used to try to prove that all statements in an infinite sequence of statements are true. The statements usually have the form: “For all natural numbers n, Sn”, where Sn is some mathematical sentence.

7. The Principle of Mathematical Induction
We can prove an infinite sequence of statements Sn by showing
the following.
Basis step. S1 is true.
Induction step. For all natural numbers k, Sk Sk+1.
*When you are learning to do proofs by mathematical induction,
it is helpful to first write out Sn , S1, Sk , and Sk + 1. This helps to
identify what is to be assumed and what is to be deduced.

8. Example Prove: For every natural number n,
   + (2n  1) = n2.
Proof: We first list Sn , S1, Sk , and Sk+1.
Sn:    + (2n  1) = n2
S1: = 12
Sk:    + (2k  1) = k2
Sk + 1:    + (2k  1) + [2(k + 1)  1] = (k + 1)2
Basis step. S1, as listed, is true.
Induction step. We let k be any natural number. We assume Sk to be true and try to show that it implies that Sk+1 is true.

9. Example continued Now, Sk is 1 + 3 + 5 +    + (2k  1) = k2.
Starting with the left side of Sk+1 and substituting k2 for
   + (2k  1), we have
   + (2k  1) + [2(k + 1)  1]
= k2 + [2(k + 1)  1] = k2 + 2k + 1 = (k + 1)2.
We have derived Sk+1 from Sk. Thus we have shown that for all natural numbers k, Sk Sk+1. The basis step and the induction step tell us that the proof is complete.

10. Example Prove: For every natural number n, n < 2n.
Proof: We first list Sn , S1, Sk , and Sk+1.
Sn: n < 2n
S1: < 21
Sk: k < 2k
Sk + 1: k + 1 < 2k+1
Basis step. S1, as listed, is true since 21 = 2 and 1 < 2.
Induction step. We let k be any natural number. We assume Sk to be true and try to show that it implies that Sk+1 is true.

11. Example continued Now, k < 2k This is Sk.
2k < 2  2k Multiplying by 2 on both sides
2k < 2k Adding exponents on the right
k + k < 2k Rewriting 2k as k + k
Since k is any natural number, we now that 1 k. Thus,
k + 1  k + k. Adding k on both sides
Putting the results k k + k and k + k < 2k+1 together gives
us k + 1 < 2k+1 . This is Sk+1.
We have derived Sk+1 from Sk. Thus we have shown that for all natural numbers k, Sk Sk+1. The basis step and the induction step tell us that the proof is complete.