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CmSc180 Discrete Mathematics
Homework 07 due 02/29 - Solution
1. Prove by mathematical induction
1 + q + q2 + q3 + … + qn = (q(n+1) – 1) / (q – 1), n = 0, 1, 2, 3, …
where q is a rational number not equal to 1
Let S(n) = 1 + q + q2 + q3 + … + qn
Let P(n) be the statement S(n) = (q(n+1) – 1) / (q – 1)
We have to show that P(n) is true for all non-negative integers
1. Base step
We will show that P(0): S(0) = (q(0+1) – 1) / (q – 1) is true
S(0) = (q(0+1) – 1) / (q – 1) = (q -1) / (q – 1) = 1
By the definition of the sum, S(0) = 1
Therefore P(0) is true
2. Inductive step
We will show that if P(k) is true, then P(k+1) is true
P(k) is the statement S(k) = (q(k+1) – 1) / (q – 1)
P(k+1) is the statement S(k+1) = (q(k+2) – 1) / (q – 1)
S(k+1) = S(k) + qk+1 = (q(k+1) – 1) / (q – 1) + qk+1 =
= (q(k+1) – 1) / (q – 1) + qk+1 (q – 1)/(q – 1) =
= ((q(k+1) – 1) + qk+1 (q – 1)) /(q – 1) =
=((q(k+1) – 1) + qk+2 – q(k+1))) /(q – 1) =
= (qk+2 – 1) /(q – 1)
Therefore P(k+1) is true.
By the principle of mathematical induction P(n) is true for all nonnegative n
2. Prove that for all integers n, n2 – n is even
Hint: consider two cases: n being even and n being odd
Case 1:
Let n be even, i.e. n = 2k
n2 – n = (2k) 2 – 2k = 2k (2k – 1)
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This is a multiple of 2, therefore it is even
Case 2:
Let n be odd, i.e. n = 2k + 1
n2 – n = (2k + 1) 2 – (2k + 1) = (2k + 1)( 2k + 1 – 1) = 2k (2k+1)
This is a multiple of 2, therefore it is even
3. Let the universal set be the set R of all real numbers and let
A = {x  R | -8  x  1}, B = { x  R | 3 < x < 10}.
Find each of the following:
AB
= {x  R | -8  x  1 V 3 < x < 10}
AB
=
~A
= {x  R | x < -8 V 1 < x }
~B
= { x  R | x  3 V 10  x }.
~A  ~B
= {x  R | x < -8 V 1 < x  3 V 10  x }
~A  ~B
= {x | x  R }
~(A  B)
= {x  R | x < -8 V 1 < x  3 V 10  x }
~(A  B)
= {x | x  R }
4. Find the power set of A = { a, {a}}
P(A) = {, {a}, {{a}}, {a, {a}} }
5. Let S = {a, b, c} and for each integer i = 0, 1, 2, 3, let Si be the set of all subsets of
S that have i elements. Write down the sets S0, S1, S2, S3, and the power set P (S)
Is { S0, S1, S2, S3} a partition of P (S)?
S0 = {}
S1 = {{a}, {b}, {c}}
S2 = {{a, b}, {a, c}, {b, c}}
S3 = {{a, b, c}}
Yes, { S0, S1, S2, S3}is a partition of P (S)
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6. Find the first four terms of each of the following sequences defined by an explicit
formula:
a. an = n/(n+1)2, n = 1, 2, 3, ….
a1 = 1/4, a2 = 2/9, a3 = 3/16, a4 = 4/25
b. an = (-1)n+1 n/(n+1), n = 1, 2, 3, . ..
a1 = 1/2, a2 = -2/3, a3 = 3/4, a4 = -4/5
c. an = (-2)n-1 /n, n = 1, 2, 3, . ..
a1 = 1, a2 = -1, a3 = 4/3, a4 = -2
7. Find the explicit formula for each of the following sequences:
a. 1 - 1/2, 1/2 - 1/3, 1/3 - 1/4, 1/4 - 1/5, ….
an = 1/n – 1/(n+1) , n = 1, 2, 3, . .
b. 1/4, 2/9, 3/16, 4/25, 5/36, 6/49, …
an = n/(n+1) 2
n = 1, 2, 3, . .
c. 1/2, -2/3, 3/4, -4/5, 5/6, -6/7, ….
an = (-1) (n+1) n/(n+1)
n = 1, 2, 3, . .
Note: indicate the starting value of n for each formula
8. Let A = { 3, 4, 5}, B = {4, 5, 6}. Let S be the "divides" relation from A to B, i.e.
S= {(x,y)| x  A, y  B and x divides y}
For example, (3,6) is in S, while (3,4) is not in S.
Write down all elements of S and S-1.
Find S S-1 and S  S-1 (write their elements)
S = {(3,6), (4,4), (5,5)}
S-1. = {(6,3), (4,4), (5,5)}
S S-1 = {(3,6), (6,3), (4,4), (5,5)}
S  S-1 = { (4,4), (5,5)}
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9. Let A be the set of all straight lines on the plane. A binary relation R is defined on
A as follows:
R = {(a,b)| a is perpendicular to b}
Determine the properties of R:
reflexive, irreflexive or neither,
symmetric, anti-symmetric or neither,
transitive or not transitive.
Irreflexive, symmetric, not transitive
10. Let A be the set of all straight lines on the plane. A binary relation R is defined on
A as follows:
R = {(a,b)| a is parallel to b}
(assume that each line is parallel to itself)
Determine the properties of R:
reflexive, irreflexive or neither,
symmetric, anti-symmetric or neither,
transitive or not transitive.
Reflexive, symmetric, transitive
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