- By
**nan** - Follow User

- 75 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Project Part 2' - nan

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Project Part 2

Aaeron Jhonson-Whyte • Akuang Saechao • Allen Saeturn

Section 4.6, Problem 28

- Prove the statement in two ways: (A) by contraposition and (B) by contradiction.

For all integers m and n, if mn is even then m is even or n is even.

Section 4.6, Problem 28: Solution (A)

A) Contraposition:

We have to take the contrapositive of the statement by using the definition of contraposition,

For all integers m and n, if m is odd and n is odd then mn is odd.

Section 4.6, Problem 28: Solution (A)

- Let m and n be odd by definition, as long as 3k and 2k are elements of the integer set.
- m = 2(3k)+1
- n = 2(2k)+1
- mn = (2(3k)+1)(2(2k)+1), substitution
- mn = (6k+1)(4k+1), distribution
- mn = 24k^2+10k+1, simplification
- mn = 2(12k^2+5k)+1,distribution,
- mn is odd by definition, as long as 12k^2+5k is an element of the integer set

Hence, original statement is true by contraposition.

Section 4.6, Problem 28: Solution (B)

B) Contradiction

Make the statement false by method of contradiction,

For some integers m and n, mn is even, and, m is odd and n is odd.

If we do exactly as part (a) then we can see that the contradiction will never prove true because two odd numbers multiplied will be odd. Hence original statement is true by contradiction.

Section 9.3, Problem 18

- Just as the difference rule gives rise to a formula for the probability of the complement of an event, so the addition and inclusion/exclusion rules give rise to formulas for the probability of the union of mutually disjoint events and for a general union of (not necessarily mutually exclusive) events.
- a. Prove that for mutually disjoint events A and B, P(A ∪ B) = P(A) + P(B).
- b. Prove that for any events A and B. P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Section 9.3, Problem 18: Solution (A & B)

- A) We know that the union of A and B are all the elements combined, and if they do not have any thing in common then it will just be both sets added together. Because we know that the elements will only be counted once.
- B) We know that the union of A and B are all the elements combined, but if they do have any in common then it will be the difference of the elements in common, which is the intersection of the two sets. That will cover for the elements that will be counted twice.

Section 9.5, Problem 22

- Each symbol in the Braille code is represented by a rectangular arrangement of six dots, each of which may be raised or flat against a smooth background… Given that at least one of the six dots must be raised, how many symbols can be represented in the Braille code?

Section 9.5, Problem 22: Solution

- The answer can be found by adding the total number of arrangements with the total number of arrangements with one dot raised… two dots raised…three dots raised… and so on until six dots are raised.

6 + 15 + 20 + 15 + 6 + 1 = 63symbols can be represented in the Braille code.

Section 10.3, Problem 7

- Suppose that for all positive integers i, all the entries in the ith row and ith column of the adjacency matrix of a graph are 0. What can you conclude about the graph?
- Ex:

Section 10.3, Problem 7: Solution

- An adjacency matrix in which the entries ith row and column are 0 denotes a graph in which the vertex Vi has no edges connecting it to itself or any other vertex of the graph. In the case of a directed graph (di-graph), the results are the same.

Section 10.3, Problem 19(B)

Let G be the graph with vertices V1, V2, and V3 and with A as its adjacency matrix. Use the answers to part (a) to find the number of walks of length 2 from V1 to V3 and the number of walks of length 3 from V1 to V3. Don’t draw G to solve this problem.

Section 10.3, Problem 19(B)

To find the walks of length n from Vi to Vj,you have to look at the ijth element of An. If A is the adjacency matrix of graph G, the ijth element of Anis equal to the number of walks of length n connecting vertices Vi and Vj.

Therefore, the number of walks of length 2 from V1toV3is 3, and the number of walks of length 3 from V1 toV3 is 15.

Section 10.3, Problem 19(C)

Examine the calculations you performed in answering part (a) to find 5 walks of length 2 from V3toV3. Then draw G and find the walks by visual inspection.

Section 10.3, Problem 19(C)

- Walks of length 2 from V3 to V3:
- V3 E3 V1 E3 V3
- V3 E3 V1 E4 V3
- V3 E4 V1 E4 V3
- V3 E4 V1 E3 V3
- V3 E2 V2 E2 V3

Section 10.3, Problem 19(C)

- Walks of length 2 from V3 to V3:
- V3 E3 V1 E3 V3
- V3 E3 V1 E4 V3
- V3 E4 V1 E4 V3
- V3 E4 V1 E3 V3
- V3 E2 V2 E2 V3

Blaise Pascal (1623 – 1662)

Blaise Pascal was a French mathematician, physicist, inventor, writer and Christian philosopher. He was a child prodigy who was educated by his father, a tax collector in Rouen, France. Pascal's earliest work was in the natural and applied sciences where he made important contributions to the study of fluids, and clarified the concepts of pressure and vacuum by generalizing the work of Evangelista Torricelli. Pascal also wrote in defense of the scientific method. While he was a teenager, he worked on calculation machines, being on of the two inventors of the mechanical calculator. Pascal continued to influence mathematics throughout his life. He created the Traité du triangle arithmétique ("Treatise on the Arithmetical Triangle") of 1653 described a convenient presentation for binomial coefficients, renamed as Pascal's triangle. Pascal also contributed philosophies to mathematics, work to physics, religion, and literature. He lived from 19 June 1623 – 19 August 1662.

Method of Proof by Contradiction:

1. Suppose the statement to be proved is false. That is, suppose that the negation of the statement is true.

2. Show that this supposition leads logically to a contradiction.

3. Conclude that the statement to be proved is true.

Definition of Contraposition:

The contrapositive of a conditional statement of the form “If p then q” is: If ∼q then ∼p.

Definition of Difference:

The difference of B minus A (or relative complement of A in B), denoted B− A,is the set of all elements that are in B and not A.

Definitions UsedDefinition of Odd and Even:

An integer n is even if, and only if, n equals twice some integer. An integer n is odd

if, and only if, n equals twice some integer plus 1.

Symbolically, if n is an integer, then

n is even ⇔ ∃ an integer k such that n=2k.

n is odd ⇔ ∃ an integer k such that n=2k+1

Definition of Union:

The union of A and B, denoted A∪B, is the set of all elements that are in at least one of A or B.

Definition of Intersection:

The intersection of A and B, denoted A ∩ B, is the set of all elements that are common to both A and B.

Download Presentation

Connecting to Server..