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Transcript of Angie Rangel Jose De Jesus Melendez Carlos Aguilar Drake Jain DISCRETE MATHEMATICS MATH 170 PROJECT...
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Angie Rangel
Jose De Jesus Melendez
Carlos Aguilar
Drake Jain
DISCRETE MATHEMATIC
SMATH 170
PROJECT PART II
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Section 4.6: # 22,
Section 9.2: #33
Section 9.5: #10
Section 10.2: #9, 18
PROJECT PART II
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Section4.6
INDIRECT ARGUMENT:
CONTRADICTION AND CONTRAPOSITION
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Indirect proof
It is a statement that is either true or false but not both
Reduces assumption by reasoning to a contradiction
WHAT IS ARGUMENT BY CONTRADICTION?
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1. Suppose that the statement that has to be proved is false.
2. Logically, show that it leads to a contradiction
3. Conclude that the statement that has to be proved is true
METHOD OF PROOF BY CONTRADICTION
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Prove: For all integers x and y, x²8y+2
So we must suppose that what we went to prove is false.Suppose there are integer x and y such that x²8y+2
Then x² X is also even, so x=2k for some integer k.Then x²=4x²We have 4x²= 2x²=2x² is even and is odd, so they can’t be equal.
Thus we have a contradiction, so there must not be any integers x and y such that x²8y+2
EXAMPLE OF CONTRADICTION
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Form of indirect argument
Logical equivalence between statement & its contrapositive
WHAT IS ARGUMENT BY CONTRAPOSITION?
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Take contrapositive of statement
Prove contrapositive by direct proof
Conclude that original statement is true
STEPS TO PROVE BY CONTRAPOSITION
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∀x in D, if P(x) Q(x).Take contrapositive:
∀x in D, if ~Q(x) ~P(x).
EXAMPLE OF CONTRAPOSITION
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Consider the statement: “For all real numbers r, if r² is irrational then r is irrational.”
a) Write what you would suppose and what you would need to show to prove this statement by contradiction.
b) Write what you would suppose and what you would need to show to prove this statement by contraposition.
QUESTION 22
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a) “For all real numbers r, if r² is irrational then r is irrational.”
Proof by contradiction: Suppose not. That is, suppose that there is a real number r such that r² is irrational and r is rational. Show that this supposition leads logically to a contradiction.
Contradiction:
Let r²=
So, if r²=, then r= (
Statement is false. Original statement is true done by contradiction.
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To prove by contradiction, we must assume that what we went to prove is false.
So we suppose that there exist a real number that if then
So when we let Let r²=√2, r= ( was not rational.
Hence, we have a contradiction.
SUMMARY
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b) “For all real numbers r, if r² is irrational then r is irrational.”
Proof by contraposition: Suppose that r is a real number such that r is irrational. Show that r² is not irrational.
Contraposition:
Let r= a/b, a= 2 and b= 3
So, if r= 2/3 then r²= 4/9
Since r is rational, r² is rational. Original statement is true done by contraposition.
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Section 9.2
MULTIPLICATION RULE
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Pierre-Simon LaplaceMathematician and astronomerHis work led to the development of mathematical astronomy and statisticsThere was no early biography of LaplaceStarring mathematical physicist between Newton and Maxwell
THE MAN BEHIND THE COUNTING
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If the operation consists of k steps and- The first step can be performed in ways- The second step can be performed in ways
.
.
.- The kth step can be performed in ways
Then the entire operation can be performed in - - - ways.
- Discrete Mathematics (Susanna S. Epp)
MULTIPLICATION RULE
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Example: How many 4-digit PIN (Personal Identifi cation Number) numbers are there to create?
1. Look for how many spaces you have
____ ____ ____ ____
In this case, you have 4.
2. Look at how many digits you can choose from.
0 1 2 3 4 5 6 7 8 9
Overall, you have 10 digits to choose from (including the
endpoints). So each space gets to have a choice of the
10
digits.
HOW TO DO IT
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3. Plug it in._10_ _10_ _10_ _10_
4. Multiply the spaces.
_10_ x _10_ x _10_ x _10_
5. There your number of possible PIN numbers
= 10000
6. DONE!
HOW TO DO IT (CONT’D)
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Section 9.2 #33- Six people attend the theater together and sit in a
row with exactly six seats.a. How many ways can they be seated together
in the row?b. Suppose one of the six is a doctor who must
sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat?
c. Suppose six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row?
TRY IT
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a. How many ways can they be seated together in the row?1. How many spaces?
- 62. How many to choose from?
- 63. Plug it in.
_6_ _5_ _4_ _3_ _2_ _1_The reason why each space decrease because as
eachseat is taken, the person who sat down is pulled
from thesituation.
4. Multiply.6 x 5 x 4 x 3 x 2 x 1
5. Answer: 720 ways (or you could leave it as “6 x 5 x 4 x 3 x 2 x 1”)
SOLUTION
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b. Suppose one of the six is a doctor who must sit on the aisle in case she is paged. How many ways can the people be seated together in the row with the doctor in an aisle seat?1. How many spaces?
- 5, because the doctor already occupies a seat2. How many to choose from?
- 53. Plug it in.
_1_ _5_ _4_ _3_ _2_ _1_4. Multiply.
1 x 5 x 4 x 3 x 2 x 15. Answer: 120 ways
SOLUTION
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c. Suppose six people consist of three married couples and each couple wants to sit together with the husband on the left. How many ways can the six be seated together in the row?1. How many spaces?
- 3, because a couple can occupy two seats the number
is reduced to half.2. How many to choose from?
- 3, because a couple counts as 1 person now3. Plug it in.
_3_ _2_ _1_4. Multiply.
3 x 2 x 15. Answer: 6 ways
SOLUTION
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Section 9.5
COUNTING SUBSETS OF A SET:
COMBINATIONS
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INFORMATION TO KNOW:
Formula: C (n, k ) =
- , means “n chose k” = number of subset of size k that can be chosen from n elements.
- n = total number of elements with-in a set
- k = total that are chosen from the set
- Where n and k are both nonnegative integers with
k ≤ n.
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When C(n, 0) :
By definition 0! =1. Therefore, if k is zero and n is any nonnegative integer, then is the number of elements with-in the set of n.
C(n, 0) =
= = 1
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CSUMB has created an indoor-soccer league where teams of 6 (or 6-combination) members must be formed. If there are 31 people who want to play how many teams can be formed?
n = 31 people who want to play
k = 6 people per team
C(31, 6) =
= =
=
= 736,281 diff erent teams can be formed.
EXAMPLE:
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Two new drugs are to be tested using
a group of 60 laboratory mice, each tagged with a
number for identification purposes. Drug A is to be
given to 22 mice, drug B is to be given to another
22 mice, and the remaining 16 mice are to be used
as controls. How many ways can the assignment of
treatments to mice be made? (A single assignment
involves specifying the treatment for each mouse—
whether drug A, drug B, or no drug.)
QUESTION 10
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Solution: Total Mice = 60 = n
Drug A = 22 mice = k (60¦22) Drug B = 22 mice = k (38¦22)
Control = 16 = k (16¦16)
Drug A + Drug B + Control
+ + = + + = + +
=
+ +
= 5,848,876,094,823,595 + 22,239,974,430 + 1
= 5,848,898,334,807,026
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SUMMARY
This question can be solved as an inclusion/exclusion problem. That
is because in the problem is states “A single assignment involves
specifying the treatment for each mouse”, therefore no mouse can be
given more than one drug type. Thus leading it to be and
inclusion/exclusion problem, in which this case it is an exclusion
problem where the number of elements is reduced for each subset
(here the subsets are Drug A, Drug B, and Control).
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Therefore, you start with 60 mice total (n), and 22 mice (k) are to be chosen to be in the first subset (Drug A). So, Drug A= C(60, 22) Since the first combination has taken away 22 mice from the entire set (n), the second subset (Drug B) only has 38 mice (n) in which 22 mice (k) can be chosen. So, Drug B= C(38, 22) Finally for the last subset, since both first subset (Drug A) and the second subset (Drug B) have used up 44 mice out of the 60 total. The last subset (Control) is left with 16 mice (n) in which they can chose 16 (k) to have in the Control group. So, Control= C(16, 16) therefore there is only 1 combination for this subset.
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Section 10.2
TRIALS, PATHS, CIRCUITS
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Walk:An edge can be repeated, as well as vertices, and also the graph does not have not be connected. Trial:Edges cannot be repeated, but vertices can be repeated.Path:Edges, and vertices cannot be repeated, and so one cannot end at the same point they started with. Closed Walk:Edges and vertices can be repeated and you can end at the same point you started with. Circuit:Edges cannot be repeated, but vertices can, also you can end at the same point you started with.
INFORMATION TO KNOW:
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Simple Circuit:Edges cannot be repeated, and only the fi rst and last vertex can be repeated, therefore you can start and end at the same vertex.Euler’s Circuit: A Euler Circuit can be made when a graph is connected (all vertices are connected to one another by an edge) and the degree of every vertex of the graph has a positive even degree. Hamiltonian Circuit:Is an simple circuit that contains every vertex in the graph, in which every vertex appears only once excluding the fi rst and last vertex which are resulting to be the same.
INFORMATION TO KNOW:
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Does the following graph have a Euler’s circuit?
EXAMPLE
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Yes it does because the graph is connected and
also each vertex has a degree of positive even
integer.
Therefore, the Euler circuit is
SOLUTION
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A) G is a connected graph with five vertices of degrees2,2,3,3, and 4.
B) B) G is a connected graph with five vertices of degrees 2,2,4,4, and 6.
C) C) G is a graph with five vertices of degrees 2,2,4,4, and 6.
9. DOES THE GRAPH HAVE AN EULER CIRCUIT? JUSTIFY YOUR
ANSWERS.
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A) No this graph G isn’t a Euler circuit,
however it is a Euler path.
B) Yes by Euler’s theorem 10.2.3, the
connected graph G is a Euler circuit.
C) Not necessary, because we do not know if
the graph is connected or not.
SOLUTION:
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In Euler’s theorem 10.2.3 it states hat a connected graph G
has a Euler circuit if all the vertices have a even positive
degree. with this being said the graph mush be connected
as well with all even degrees. also it is possible to have a
Euler path if there is two degrees of odd integers from
Euler’s theorem 10.2.4 is states that a connected graph G
can be an Euler path if and only if there are exactly two
vertices that are odd positive integers.
SUMMARY:
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A) This is not a Euler circuit, but a Euler path.10.2.4 Euler’s theorem
states that a connected graph G contains a Euler path if and only if
there are exactly two vertices with odd degrees. Graph G with
degrees of 2,2,3,3,and 4 contains exactly two degrees of odd
integers.
B) Yes by Euler’s theorem this is a circuit because all the degrees are
even integers and the graph is connected. The theorem states that
the graph must be connected and contain vertices with all even
degrees. the statement states that the graph G is connected and has
five vertices of all even integers 2,2,4,4, and 6.
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C) The reason why the answer to the statement is not
necessarily is because the Euler theorem 10.2.3 states that the
graph is connected. and the statement does not state that the
graph is connected but it does state that there are five vertices
with degrees of 2,2,4,4, and 6 even though the theorem states
“if and only if the vertices are all even degrees” it still states
that the graph G is connected. This statement has nothing
about the graph being connected or not so with that being said
we can say not necessarily. There is not enough information
provided with it besides that all vertices are even.
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QUESTION 18
Is it possible to take a walk around the city whose
map is shown below, starting and ending at the same
point and crossing each bridge exactly once? If so,
how can this be done?
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SOLUTION
The way that this can be
answered is by the following
circuit:
B →D→C→A→D→E→A→B
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SUMMARY:
This question is looking for a Euler’s circuit that is
because which can be applied by Theorem 10.2.3,
there are five edges who’s each degree is of even
amount. Therefore, since with-in an Euler’s circuit
every vertex has a degree of even amount. Which
by the map vertex B(2), D(4), C(2), A(4), D(4), E(2).
Which means a Euler’s circuit can be made.
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•Therefore, you start with B and the use the first bridge to D.
So, B→D
•Since D has an even amount of bridges connecting to it (degree amount) you can connect to either A, C or E.
So, D→C
•Since, C also has an even amount of bridges connecting to it (degree amount) you can connect to A, you cannot go back to D because the problem states that you cannot repeat the bridge.
So, C→A
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•Then, From A since you have to get to E, you can either go to D then E or just E.
So, A→D
•Finally, from D all that is left is to take the bridge to E and then to A, and one more bridge up to B.
So, D→E→A→B
•Therefore, the answer is:B →D→C→A→D→E→A→B
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Epp, Susanna S. Discrete Mathematics With Applications 4 th Edition. Boston, MA: Brooks/Cole Publishing Company, 2004. Print.
Rowlinson, J. S. "Laplace: The Man." Notes and Records of the Royal Society 60.2 (2006): 221-223.
BIBLIOGRAPHY