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CmSc 175 Discrete mathematics
Homework 08 due 03/31
1. Let N be the set of natural numbers. Define a relation R on N x N :
R = { ((a, b), (c, d)) | ad = bc}
Prove that R is an equivalence relation on N x N
a) Reflexifity: we have to show that  (a,b)  N x N, ((a,b),(a,b))  R
Let (a,b)  N x N
ab = ba (Basic algebra)
by the def. of R, ((a,b),(a,b))  R
Therefore R is reflexive
b) Symmetry
We have to show that
 (a,b), (c,d)  N x N if ((a,b),(c,d))  R then ((c,d),(a,b))  R
Note: In order to show that ((c,d),(a,b))  R we need to show that cb = da
Let ((a,b),(c,d))  R
By the def of R,
ad = bc
da = cb (basic algebra)
cb = da (basic algebra)
Therefore, by the def of R, ((c,d),(a,b))  R
Therefore R is symmetric
c) Transitivity
We have to show that
 (a,b), (c,d) , (e,f)  N x N
if ((a,b),(c,d))  R and ((c,d), (e,f))  R then ((a,b),(e,f))  R
Note: In order to show that ((a,b),(e,f))  R we need to show that af = be
Let ((a,b),(c,d))  R and ((c,d), (e,f))  R
By the def. of R we have:
ad = bc
cf = de
Multiply the two sides:
adcf = bcde
Divide both sides by dc:
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af = be Therefore, by the def. of R , ((a,b),(e,f))  R
Therefore R is transitive
Therefore R is a relation of equivalence
2. Let R1 and R2 be any two equivalence relations on the same set A. Examine the
properties of R1  R2 and determine if it is a relation of equivalence or not.
(prove /disprove each property)
R1  R2 is not a relation of equivalence because it is not necessarily transitive.
Here is an example of two relations of equivalence R1 and R2 on the set A = {a,b,c}
R1 = {(a,a),(b,b),(c,c), (a,b), (b,a)}
R2 = {(a,a),(b,b),(c,c), (b,c), (c,b)}
R1  R2 = {(a,a),(b,b),(c,c), (a,b), (b,a), (b,c), (c,b)}, not transitive because the pair (a,c)
is not in the union.
3. For each sentence below do the following:
a.
b.
c.
d.
Choose appropriate predicates
Write the quantified predicate expression
Negate the expression
Translate the negated expression in English
1. No polynomial function is continuous
x, polynomial(x)  ~continuous(x)
x, polynomial(x)  continuous(x)
Some polynomial functions are continuous
2. Some polynomial functions are not continuous
x, polynomial(x)  ~continuous(x)
x, polynomial(x)  continuous(x)
All polynomial functions are continuous
4. Prove the statement: For all integers n, if (n+1)(n-1) is odd then n is even
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Hint: prove the equivalent contrapositive statement using direct proof. This is
called proof by contraposition.
The contrapositive statement is:
If n is odd, then (n+1)(n-1) is even.
Let n be odd, n = 2k + 1. Then n+1 = 2k + 2 = 2(k+1), n – 1 = 2k
(n+1)(n-1) = 2(k+1).2k , even number
The statement “If n is odd, then (n+1)(n-1) is even.” Is equivalent to the statement
“if (n+1)(n-1) is odd then n is even”
Therefore, For all integers n, if (n+1)(n-1) is odd then n is even
5. Let S(n) = 2 + 6 + 18 + …. + 2*3(n-1)
Prove the statement P(n): S(n) = 3n – 1, n ≥1
a. Inductive base
Show that P(1): S(1) = 31 – 1 = 2 is true
By the definition of S(n), S(1) = 2, therefore P(1) is true
b. Inductive step
Show that P(k)  P(k+1) is true.
P(k) : S(k) = 3k – 1, P(k+1) : S(k+1) = 3k+1 – 1
Assume that P(k) is true, i.e. S(k) = 3k – 1
S(k+1) = S(k) + 2*3(k+1-1) = 3k – 1 + 2* 3 k = 3* 3 k – 1 = 3k+1 – 1
Therefore P(k+1) is true,
Therefore P(k)  P(k+1) is true
Therefore by the principle of math induction, S(n) = 3n – 1, n ≥1
6. Given a deck of 52 cards, what is the probability to have a 5-card hand with
exactly 2 spades? Explain your answer.
Let E be the number of hands with exactly 2 spades, and S be the total number of
5-card hands.
E = 13C2 * 39C3 = (13!*39!)/ (11!*2!*36!*3!) = 12*13*37*38*39 / 12 =
13*37*38*39
S = 52C5 = 52! / 47!*5!
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Thus the probability is (13*37*38*39 * 47!*5! ) / 52! =
(13*37*38*39 * 5! ) / (48*49*50*51*52) =
(37*19*13 ) / (4*49*10*17) = 0.27
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