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Category > Math Posted 20 Apr 2017 My Price 8.00

State in English the inverse of the contrapositive

I have solved homework set, but have problems with numbers 9, 10, 12. 

Please check all solutions for correctness and provide correct answers for qs 9,10, 12

 

 

Homework Set
Chapter 7 to 13 1. State in English the inverse of the contrapositive of “If it is raining, then my lawn is wet.”
If my lawn is not wet, then it is not raining.
2. Let S be a set, R be a binary relation on S, and x an element of S. Translate the following into a
logical expression with the same meaning:
the negation of the statement “For all x in S, xRx.”
There exists x in the set S, such that x is not related to itself under R.
*Is that right? And why?
3. Let the function f from the positive integers to the positive integers be defined by f(x) = x*x (where
the asterisk * denotes ordinary integer multiplication). Explain why this function f is or is not an onto
function.
A function f from A to B is called onto if for all b in B there is an a in A such that f (a) = b and all
elements in B are used.
F(x) is not onto since, there is no real numbers that will map to 7. If x*x = -7, then x =±
is not an integer √ 7 , which 4. Let sets A and B be defined as follows: A = {a1} and B = {b1, b2}. List as separate sets of ordered
pairs all the one-to-one functions from A to B, or explain why there are no such functions.
(a1, b1) or (a1,b2) 5. Let the set A be defined as A = {a, b, c, d}, and let the relations R and S on the set A be defined as
R = {(d, a), (a, b), (b, c), (b, d) }, and S = {(a, a), (b, d), (d, c)}.
Explain, using the definition of composition of relations on a set (see the second paragraph in Section
5.4 of the zyBook) why the ordered pair (b, d) is or is not an element of the composition of relations R
and S (denoted S o R).
S o R = { (d,a), (a,d) , (b,c)}. Pair (b,d) is not in the element of composition of relations R and S. *Definition: The composition of relations R and S on set A is another relation on A, denoted S ο R.
The pair (a, c) ∈ S ο R if and only if there is a b ∈ A such that (a, b) ∈ R and (b, c) ∈ S.
**How to prove using this definition above
6. What is the value of the variable count after all the loops in the following pseudocode execute?
count:=0
For i= 1 to 2
For j=1 to 2
count:=2i[(j*count)+j]
End-for
End-for
i = 1 j=1
count = 2((1*0)+1)=2
i = I j=2
count =2(2*2)+2) = 12
i= 2 j=1
count =4((1*12)+1) = 52
i = 2 j=2
count=4((2*52)+2) = 424
final count = 424. 8. Explain why a multiplicative inverse mod 7 of 13 does or does not exist. If one does exist, give a
value for it and use appropriate calculations to show that it is a multiplicative inverse of 13 mod 7.
A*B mod C = 1, then A is the multiplicative inverse we are looking for.
A*7 mod 13 = 1; we need to find A
1 * 7 mod 13 = 7;
2* 7 mod 13 = 1; therefore answer is 2. 9. A husband and wife and their two children line up for a photo. How many ways are there for these
four people to line up so that the husband and wife are *not* next to each other? Be sure to show
your work.
3! = Ways all of them can seated
4! = Ways husband and wife can be seated
**3!*4! = 144 Ways? – this is not right, help this to solve 10. Suppose you randomly draw two cards from a standard deck without replacement. What is the
probability that neither card is the ace of spades?
Ways to draw 2 cards = 52*51
Ways to draw queens = 51*50;
51∗50 50 25
= =
Final possibility =
52∗51 52 26 11. Let V = {a, b, c, d, e} be a vertex set and E = { {a,c}, {b,d}, {c,d}, {d,e}, {e,a}} be the edge set
corresponding to V. Explain why the pair (V, E) is or is not a tree.
Pair (V,E) is not a tree because it creates cycle. 12. Let vertex sets V1 and V2 be defined by V1= {1, 2, 3} and V2 = {a, b, c}. Let E1 = { { 1, 2}, {2, 3} }, and
let E2 = { {a, b}, {b, c} } be the edge sets corresponding to the vertex sets V1 and V2, respectively.
Write, as a set of ordered pairs, a function f that is a bijection from V1 to V2, satisfying the following
condition: if x and y are elements in V1 such that {x,y} is in E1, then f(x) and f(y) are elements in V2
such that {f(x),f(y)} is in E2, and show that your function f satisfies this condition.
Note: you do not need to show that your function f is a bijection (though it must be, or you won’t get
any credit), but you DO need to show that it satisfies the condition “if x and y are elements in V1
such that {x,y} is in E1, then f(x) and f(y) are elements in V2 such that {f(x),f(y)} is in E2.” (a,1) (b,2) (c,3)
*how to show that it satisfies the condition

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Status NEW Posted 20 Apr 2017 08:04 AM My Price 8.00

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