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Chapter 4

Chapter 4. Sequences and Mathematical Induction. 4.2. Mathematical Induction. Mathematical Induction. A recent method for proving mathematical arguments. De Morgan is credited with its discovery and and name. The validity of proof by mathematical induction is taken as a axiom.

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Chapter 4

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  1. Chapter 4 Sequences and Mathematical Induction

  2. 4.2 Mathematical Induction

  3. Mathematical Induction • A recent method for proving mathematical arguments. • De Morgan is credited with its discovery and and name. • The validity of proof by mathematical induction is taken as a axiom. • an axiom or postulate is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. (Wikipedia)

  4. Principle of Mathematical Induction • Let P(n) be a property that is defined for integers n, and let a be a fixed integer. Suppose the following two statements are true: • P(a) is true. • For all integers k ≥ a, if P(k) is true then P(k+1) is true. • Then the statement, for all integers n ≥ a, P(a) is true

  5. Example • For all integers n ≥ 8, n cents can be obtained using 3 cents and 5 cents. • or, for all integers n ≥ 8, P(n) is true, where P(n) is the sentence “n cents can be obtained using 3 cents and 5 cents.”

  6. Example • This can be proven by exhaustion if we can continue to fill in the table up to $1.00. • The table shows how to obtain k cents using 3 and 5 coins. We must show how to obtain (k+1) cents. • Two cases: • k: 3 + 5, k+1: ? • replace 5 with 3 + 3 • k: 3 + 3 + 3, k+1: ? • replace 3+3+3 with 5+5

  7. Method of Proof Mathematical Induction • Statement: “For all integers n≥a, a property P(n) is true.” • (basis step) Show that the property is true for n = a. • (inductive step) Show that for all integers k≥a, if the property is true for n=k then it is true for n=k+1. • (inductive hypothesis) suppose that the property is true for n=k, where k is any particular but arbitrarily chosen integer with k≥a. • then, show that the property is true for n = k+1

  8. Example Coins Revisited • Proposition 4.2.1: Let P(n) be the property “n cents can be obtained using 3 and 5 cent coins.” Then P(n) is true for all integers n≥8. • Proof: • Show that the property is true for n=8: • The property is true b/c 8=3+5.

  9. Example cont. • Show that for all integers k≥8, if the property true for n=k, then property true for n=k+1 • (inductive hypothesis) Suppose k cents can be obtained using 3 and 5 cent coins for k≥8. • Must show (k+1) cents can be obtained from 3 & 5 coin. • Case (3,5 coin): k+1 can be obtained by replacing the 5 coin with two 3 cent coins. This increments the value by 1 (3+3=6) replaces the 5 cent coin. • Case (3,3,3 coin): k+1 can be obtained by replacing the three 3 coins with two 5 coins. k=b+3+3+3=b+9 and k+1=b+9+1=b+5+5=b+10

  10. Example Formula • Prove with mathematical induction • Identify P(n) • Basis step

  11. Example Formula cont. • Inductive step • assume P(k) is true, k>=1 • show that P(k+1) is true by subing k+1 for n • show that left side 1+2+…k+1 = right side (k+1)(k+2)/2 • 1+2+…+k+1 = (1+2+…+k) + k+1 • sub from inductive hypothesis:

  12. Theorem 4.2.2 • Sum of the First n Integers • For all integers n≥1,

  13. Example • Sum of the First n Integers • Find 2+4+6+…+500 • Get in form of Theorem 4.2.2 (1+2+…+n) • factor out 2: 2(1+2+3+…+250) • sum = 2( n(n+1)/2 ), n = 250 • sum = 2( 250(250+1)/2 ) = 62,750 • Find 5+6+7+8+…+50 • add first 4 terms 1+2+3+4 to problem then subtract back out after computation with 4.2.2 • 1+2+3+4+5+6+7+…+50 – (1+2+3+4) • (50 (50+1)/ 2) – 10 • =1,265

  14. Sum of Geometric Sequence • Prove that , for all integers n≥0 and all real numbers r except 1. • P(n): • Basis: • Inductive: (n=k) • Inductive Hypothesis: (n=k+1)

  15. Theorem 4.2.3 • Sum of Geometric Sequences • For any real number r except 1, and any integer n≥0,

  16. Examples • Sum of Geometric Sequences • Find 1 + 3 + 32 + … + 3m-2 • Sequence is in geometric series, apply 4.2.3 directly • Find 32 + 33 +… + 3m • rearrange into proper geometric sequence by factoring out 32 from sequence • 32(1 + 3 + 32 + … + 3m-2) =

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