5.3 mathematical induction ii

5.3 Mathematical Induction II

5.3 Mathematical Induction II 1 / 10

Proof by Induction

Proposition 5.3.2: For all integers n ≥ 3, 2n + 1 < 2n.

Proof:

Let P(n) be the sentence, “2n + 1 < 2n.”

We need to show that P(n) is true for all integers n ≥ 3. That’s infinitelymany statements! How can we prove infinitely many statements?

The same way we did it last time: PROOF BY INDUCTION!


Proof by Induction


Proof: Let P(n) be the sentence,

“2n + 1 < 2n.”




Proof by Induction


Proof: Let P(n) be the sentence, “2n + 1 < 2n.”




Proof by Induction



Base Case: P(3) is true because

2 · 3 + 1 < 23.

Inductive Step: Let k be an integer at least equal to three./Let k ≥ 3 bean integer. (Pick one of these two sentences.) Suppose that 2k + 1 < 2k .We will show that 2(k + 1) + 1 < 2k+1.

2(k + 1) + 1 = 2k + 3 = (2k + 1) + 2.

By the induction hypothesis, (2k + 1) + 2 < (2k) + 2. Since k ≥ 3,

(2k) + 2 < 2k + 2k = 2 · 2k = 2k+1.

Hence 2(k + 1) + 1 < 2k+1.

Conclusion: The result follows by induction. �


Proof by Induction



Base Case: P(3) is true because 2 · 3 + 1 < 23.


2(k + 1) + 1 = 2k + 3 = (2k + 1) + 2.


(2k) + 2 < 2k + 2k = 2 · 2k = 2k+1.

Hence 2(k + 1) + 1 < 2k+1.



Proof by Induction




Inductive Step: Let k be an integer at least equal to three./Let k ≥ 3 bean integer. (Pick one of these two sentences.) Suppose that

2k + 1 < 2k .We will show that 2(k + 1) + 1 < 2k+1.

2(k + 1) + 1 = 2k + 3 = (2k + 1) + 2.


(2k) + 2 < 2k + 2k = 2 · 2k = 2k+1.

Hence 2(k + 1) + 1 < 2k+1.



Proof by Induction




Inductive Step: Let k be an integer at least equal to three./Let k ≥ 3 bean integer. (Pick one of these two sentences.) Suppose that 2k + 1 < 2k .We will show that

2(k + 1) + 1 < 2k+1.

2(k + 1) + 1 = 2k + 3 = (2k + 1) + 2.


(2k) + 2 < 2k + 2k = 2 · 2k = 2k+1.

Hence 2(k + 1) + 1 < 2k+1.



Proof by Induction





2(k + 1) + 1 = 2k + 3 = (2k + 1) + 2.


(2k) + 2 < 2k + 2k = 2 · 2k = 2k+1.

Hence 2(k + 1) + 1 < 2k+1.



Proof by Induction





2(k + 1) + 1 = 2k + 3 = (2k + 1) + 2.


(2k) + 2 < 2k + 2k = 2 · 2k = 2k+1.

Hence 2(k + 1) + 1 < 2k+1.

Conclusion: The result follows by induction. �5.3 Mathematical Induction II 3 / 10

Proof by Induction

Example 5.3.3

Define a sequence a1, a2, . . . as follows: a1 = 2 and ak = 5ak−1 for allintegers k ≥ 2.

1 Write the first four terms of the sequence.

2 Prove that the following claim is true. For each integer n ≥ 1,an = 2 · 5n−1.

5.3. Problem. 11

Claim: 32n − 1 is divisible by 8 for each integer n ≥ 0.

1 Write the three smallest terms in the set.

2 Prove that the claim is true.


A Problem with Trominoes

Example by Solomon Golomb (when he was an student at Harvard)

When can you use trominoes (pictured above) to completely cover acheckerboard?

Formally: For what values of n can a 2nx2n checkerboard be completelycovered by L-shaped trominoes?

None, since 3 does not divide 2nx2n. What if one square from thecheckerboard is removed at random?

Formally: For what values of n can a 2nx2n checkerboard with somesquare removed be completely covered by L-shaped trominoes?

n = 1? n = 2? . . .



Theorem: If n is a positive integer, then a 2nx2n checkerboard with asquare removed can be completely covered by L-shaped trominoes.

Proof. Let P(n) be the sentence, “If any square is removed from a 2nx2n

checkerboard, then it can be completely covered by L-shaped trominoes.”

Show that P(1) is true.

Then P(1) is true because any 2x2 checkerboard with a square removed isan L-shaped tromino.

Show that, for all integers k ≥ 1, if P(k) is true then P(k + 1) is true.

Consider a 2k+1x2k+1 checkerboard with a square removed.







Then P(1) is true because

any 2x2 checkerboard with a square removed isan L-shaped tromino.









Then P(1) is true because any 2x2 checkerboard with a square removed isan L-shaped tromino.






This checkerboard can be viewed as being built by some 2kx2k

checkerboards

as shown below. One will have a missing square.

2k

2k+1

2k

2k+1

Since P(k) is true, the smallercheckerboard that has a missingsquare can be completely coveredby L-shaped trominoes. Add an L-shaped tromino where the other threesmaller checkerboards meet.





checkerboards as shown below. One will have a missing square.

2k

2k+1

2k

2k+1







2k

2k+1

2k

2k+1

Since P(k) is true, the smallercheckerboard that has a missingsquare

can be completely coveredby L-shaped trominoes. Add an L-shaped tromino where the other threesmaller checkerboards meet.






2k

2k+1

2k

2k+1

Since P(k) is true, the smallercheckerboard that has a missingsquare can be completely coveredby L-shaped trominoes.

Add an L-shaped tromino where the other threesmaller checkerboards meet.






2k

2k+1

2k

2k+1







2k

2k+1

2k

2k+1

Since P(k) is true, the rest of eachof the smaller checkerboards can becompletely covered by L-shaped tro-minoes!

Therefore P(k + 1) is true.The result follows by induction! �






2k

2k+1

2k

2k+1

Since P(k) is true, the rest of eachof the smaller checkerboards can becompletely covered by L-shaped tro-minoes! Therefore P(k + 1) is true.

The result follows by induction! �






2k

2k+1

2k

2k+1

Since P(k) is true, the rest of eachof the smaller checkerboards can becompletely covered by L-shaped tro-minoes! Therefore P(k + 1) is true.The result follows by induction! �




2k

2k+1

2k

2k+1

Let P(n) be the sentence, “If any square is removed froma 2nx2n checkerboard, then it can be completely covered byL-shaped trominoes.” Then P(1) is true because any 2x2checkerboard with a square removed is an L-shaped tro-mino. Assume P(k) is true for a positive integer k . Con-sider a 2k+1x2k+1 checkerboard with a square removed. Thischeckerboard can be viewed as being built by some 2kx2k

checkerboards as shown below. One will have a missingsquare. Since P(k) is true, the smaller checkerboard thathas a missing square can be completely covered by L-shapedtrominoes. Add an L-shaped tromino where the other threesmaller checkerboards meet. Since P(k) is true, the rest ofeach of the smaller checkerboards can be completely coveredby L-shaped trominoes. Therefore P(k + 1) is true, and theresult follows by induction. �


5.3 mathematical induction ii

Documents