Abstract Mathematics Lecture 14

Abstract Mathematics

Lecture 14

Mathematical Induction

Outline for Proof by Strong Induction

Proposition

The statements S1, S2, S3, S4, . . . are all true.

Proof.

(Strong induction).

(1) Prove the first statement S1. (Or the first several Sn, if needed.)

Mathematical Induction

Example Proposition

If n ∈ N, then 12|(n4− n2_).

Proof.

We will prove this with strong induction.

First note that the statement is true for the first six positive integers:

For n = 1, 12 divides 14_{− 1}2_{= 0.} For n = 2, 12 divides 24_{− 2}2_{= 12.} For n = 3, 12 divides 34_{− 3}2_{= 72.} For n = 4, 12 divides 44_{− 4}2_{= 240.} For n = 5, 12 divides 54− 52_{= 600.} For n = 6, 12 divides 64− 62_{= 1260.}

Mathematical Induction

Cont.

For k ≥ 6, assume 12|(m4− m2_{) for 1 ≤ m ≤ k. (i.e., S}

1, . . . , Sk are

true.)

We must show Sk+1 is true, that is 12| (k + 1)4− (k + 1)2
. Now,

Sk−5 being true means 12| (k − 5)4− (k − 5)2
. To simplify, put

k − 5 = ` so 12|(`4− `2<sub>), meaning `</sub>4_{− `}2 _{= 12a for a ∈ Z, and}

Mathematical Induction

Then, (k + 1)4− (k + 1)2 _{= (` + 6)</sub>4<sub>− (` + 6)}2 = `4+ 24`3+ 216`2+ 864` + 1296 −(`2<sub>+ 12` + 36)</sub> = (`4− `2_{) + 24`</sub>3<sub>+ 216`}2<sub>+ 852` + 1260</sub> = 12a + 24`3+ 216`2+ 852` + 1260 = 12(a + 2`3+ 18`2+ 71` + 105). Because (a + 2`3+ 18`2+ 71` + 105) ∈ Z, we get 12| (k + 1)4− (k + 1)2
_.

Mathematical Induction

Try to prove the following statement using strong induction:

Proposition