1. (Finite) Mathematical Induction In our first lesson on sequences and series, you were told that How can we be certain that this will be true for all counting numbers n ? Let us call this proposition, “P.”

2. (Finite) Mathematical Induction This proposition P is logically equivalent to asserting all of the following, together: … and continuing for all the counting numbers.

3. (Finite) Mathematical Induction Taken by itself, each of these individual propositions is simple to verify: simply compute each side of the equation and check that they are equal. But a proof can’t go on forever. We must find another way to prove P.

4. (Finite) Mathematical Induction The situation could be like this: P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 If there are infinitely many of these, and you need to knock over every one of them, you will never be finished.

5. P6P6 P5P5 P4P4 P3P3 P2P2 (Finite) Mathematical Induction However, if they were arranged like this: P1P1 Even if there are infinitely many of them, one push (to knock over P 1 ) will knock over all of them.

6. (Finite) Mathematical Induction Method for knocking over infinitely many dominos: Method for proving infinitely many propositions: 1.Make sure that each domino, P k, will knock over the next one, P k+1. 1.Prove that each proposition, P k, implies the next one, P k+1. 2.Knock over the first domino, P 1. 2.Verify the first proposition, P 1. P k+1 PkPk “If P k is true, then P k+1 must also be true.

7. (Finite) Mathematical Induction Remark 1: If you include the “k” along with the “k+1” on the left-hand side, you will be glad you did. Remark 2: When you simplify, be flexible. Your goal is to show that two expressions are equivalent.

8. (Finite) Mathematical Induction The two parts of a finite mathematical induction proof: 1)Verify P 1 EASY! 2)Prove: if P k is true, then P k+1 must also be true. P k+1 PkPk Verifying P 1 is usually so easy that it’s hard to say much about it. However, there are some cases where P 1 is actually the difficult part.

9. (Finite) Mathematical Induction The two parts of a finite mathematical induction proof: 1)Verify P 1 EASY! 2)Prove: if P k is true, then P k+1 must also be true. P k+1 PkPk What we must do: Assume that P k is true, and then prove that P k+1 must be true. What can we use? We can use any algebraic techniques, but we can also use P k. How do we use P k ? Usually by substitution: any occurrence of the left-hand side can be replaced with the right-hand side.

10. (Finite) Mathematical Induction The two parts of a finite mathematical induction proof: 1)Verify P 1 EASY! 2)Prove: if P k is true, then P k+1 must also be true. P k+1 PkPk Where do we start? Examine P k+1 and try to find where the left-hand side P k of occurs in it. Replace this occurrence of the left-hand side of P k with the right-hand side.

11. (Finite) Mathematical Induction The two parts of a finite mathematical induction proof: 1)Verify P 1 EASY! 2)Prove: if P k is true, then P k+1 must also be true. Now it should be possible to prove that the two sides are in fact equivalent. Tip: Since one side (the right-hand side) is a single fraction, try making the other side be a single fraction. Tip: Since the denominators are the same, all we need do is prove that the numerators are equivalent.

12. (Finite) Mathematical Induction The two parts of a finite mathematical induction proof: 1)Verify P 1 EASY! 2)Prove: if P k is true, then P k+1 must also be true. Simplify both numerators. Hopefully they will simplify to the same expression. Left-hand side:Right-hand side:

13. (Finite) Mathematical Induction The two parts of a finite mathematical induction proof: 1)Verify P 1 EASY! 2)Prove: if P k is true, then P k+1 must also be true. What did we just accomplish? We just proved to ourselves that if P k is true, then P k+1 must also be true. Now we are ready to write the proof.

14. (Finite) Mathematical Induction Using mathematical induction, we will prove that When n = 1, and So the proposition is true when n = 1.

15. (Finite) Mathematical Induction Induction hypothesis: the proposition is true when n = k. Now, to prove the proposition when n = k + 1, Also, Thus, if the proposition is true for n = k, it is true for n = k + 1. Therefore the proposition is true for all counting numbers n. QED.