CALCU-1. Parts (a) and (b) of this question are un

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Solution by Steps

step 1

We need to prove that

A \cap B \cap C \subset A \subset A \cup B \cup C

step 2

By definition,

A \cap B \cap C

is the set of all elements that are in

A

,

B

, and

C

. Therefore,

A \cap B \cap C \subset A

step 3

Since

A

is a subset of

A \cup B \cup C

(because

A \cup B \cup C

contains all elements of

A

,

B

, and

C

), we have

A \subset A \cup B \cup C

step 4

Combining these, we get

A \cap B \cap C \subset A \subset A \cup B \cup C

step 5

Similarly,

A \cap B \cap C \subset B \subset A \cup B \cup C

and

A \cap B \cap C \subset C \subset A \cup B \cup C

can be shown using the same logic

Part (b) # Part (i)

step 1

We need to show that

|A_1 \cup A_2| \leq |A_1| + |A_2|

step 2

By the principle of inclusion-exclusion,

|A_1 \cup A_2| = |A_1| + |A_2| - |A_1 \cap A_2|

step 3

Since

|A_1 \cap A_2| \geq 0

, it follows that

|A_1 \cup A_2| \leq |A_1| + |A_2|

# Part (ii)

step 1

We need to show by induction that

|A_1 \cup A_2 \cup \cdots \cup A_k| \leq \sum_{i=1}^k |A_i|

step 2

Base Case: For

k=1

,

|A_1| = |A_1|

, which is true

step 3

Inductive Step: Assume the statement is true for

k=n

, i.e.,

|A_1 \cup A_2 \cup \cdots \cup A_n| \leq \sum_{i=1}^n |A_i|

step 4

We need to show it is true for

k=n+1

. Consider

|A_1 \cup A_2 \cup \cdots \cup A_n \cup A_{n+1}|

step 5

By the principle of inclusion-exclusion,

|A_1 \cup A_2 \cup \cdots \cup A_n \cup A_{n+1}| = |(A_1 \cup A_2 \cup \cdots \cup A_n) \cup A_{n+1}|

step 6

Using part (i),

|(A_1 \cup A_2 \cup \cdots \cup A_n) \cup A_{n+1}| \leq |A_1 \cup A_2 \cup \cdots \cup A_n| + |A_{n+1}|

step 7

By the inductive hypothesis,

|A_1 \cup A_2 \cup \cdots \cup A_n| \leq \sum_{i=1}^n |A_i|

step 8

Therefore,

|A_1 \cup A_2 \cup \cdots \cup A_n \cup A_{n+1}| \leq \sum_{i=1}^n |A_i| + |A_{n+1}| = \sum_{i=1}^{n+1} |A_i|

step 9

By induction, the statement is true for all positive integers

k

Answer

The required proofs are provided for both parts (a) and (b).

Key Concept

Set Theory and Inclusion-Exclusion Principle

Explanation

The problem involves proving subset relations and using the principle of inclusion-exclusion to show inequalities involving the cardinalities of unions of sets.

Solution by Steps

step 1

Let's denote the statements as follows: -

p

: I play footy. -

q

: I study. -

r

: I pass discrete mathematics. The given statements can be translated into logical expressions: 1. If I play footy, I cannot study:

p \rightarrow \neg q

2. Either I play footy, or I pass discrete mathematics:

p \lor r

3. I passed discrete mathematics:

r

4. Therefore, I studied:

q

step 2

We need to construct a truth table to determine the validity of the argument. The truth table will include all possible truth values for

p

,

q

, and

r

, and we will evaluate the expressions

p \rightarrow \neg q

,

p \lor r

, and

q

.

\begin{array}{|c|c|c|c|c|c|} \hline p &amp; q &amp; r &amp; p \rightarrow \neg q &amp; p \lor r &amp; q \\ \hline T &amp; T &amp; T &amp; F &amp; T &amp; T \\ T &amp; T &amp; F &amp; F &amp; T &amp; T \\ T &amp; F &amp; T &amp; T &amp; T &amp; F \\ T &amp; F &amp; F &amp; T &amp; T &amp; F \\ F &amp; T &amp; T &amp; T &amp; T &amp; T \\ F &amp; T &amp; F &amp; T &amp; F &amp; T \\ F &amp; F &amp; T &amp; T &amp; T &amp; F \\ F &amp; F &amp; F &amp; T &amp; F &amp; F \\ \hline \end{array}

step 3

Now, we need to check if the premises lead to the conclusion. We combine the premises using logical AND and check if they imply the conclusion

q

.

(p \rightarrow \neg q) \land (p \lor r) \land r \rightarrow q

Evaluating this for each row in the truth table:

\begin{array}{|c|c|c|c|c|c|c|} \hline p &amp; q &amp; r &amp; (p \rightarrow \neg q) &amp; (p \lor r) &amp; r &amp; (p \rightarrow \neg q) \land (p \lor r) \land r \rightarrow q \\ \hline T &amp; T &amp; T &amp; F &amp; T &amp; T &amp; T \\ T &amp; T &amp; F &amp; F &amp; T &amp; F &amp; T \\ T &amp; F &amp; T &amp; T &amp; T &amp; T &amp; F \\ T &amp; F &amp; F &amp; T &amp; T &amp; F &amp; T \\ F &amp; T &amp; T &amp; T &amp; T &amp; T &amp; T \\ F &amp; T &amp; F &amp; T &amp; F &amp; F &amp; T \\ F &amp; F &amp; T &amp; T &amp; T &amp; T &amp; F \\ F &amp; F &amp; F &amp; T &amp; F &amp; F &amp; T \\ \hline \end{array}

step 4

From the truth table, we see that the expression

(p \rightarrow \neg q) \land (p \lor r) \land r \rightarrow q

is not always true. Specifically, it is false when

p = T

,

q = F

, and

r = T

. Therefore, the argument is not valid

Answer

The argument is not valid.

Key Concept

Truth Table Analysis

Explanation

By constructing a truth table and evaluating the logical expressions, we determined that the given argument does not hold true in all cases, thus it is not valid.

Solution by Steps

step 1

We start by proving that

100 \equiv 0 \pmod{4}

step 2

Since

100 = 4 \times 25

, it is clear that

100

is divisible by

4

. Therefore,

100 \equiv 0 \pmod{4}

step 3

Next, we need to show that

10^k \times 100 \equiv 0 \pmod{4}

for any positive integer

k

step 4

Since

100 \equiv 0 \pmod{4}

, multiplying both sides by

10^k

gives

10^k \times 100 \equiv 10^k \times 0 \equiv 0 \pmod{4}

step 5

Now, we explain why if the number formed by the last 2 digits of

N

is divisible by

4

, then

a_1 \times 10 + a_0 \equiv 0 \pmod{4}

step 6

The number formed by the last 2 digits of

N

is

a_1 \times 10 + a_0

. If this number is divisible by

4

, then

a_1 \times 10 + a_0 \equiv 0 \pmod{4}

step 7

Finally, we prove that

N

is divisible by

4

if and only if the number formed by the last 2 digits of

N

is divisible by

4

step 8

We express

N

as

N = a_k \times 10^k + a_{k-1} \times 10^{k-1} + \cdots + a_2 \times 10^2 + a_1 \times 10 + a_0

step 9

Since

10^k \times 100 \equiv 0 \pmod{4}

for any

k \geq 2

, all terms except

a_1 \times 10 + a_0

are congruent to

0 \pmod{4}

step 10

Therefore,

N \equiv a_1 \times 10 + a_0 \pmod{4}

. If

a_1 \times 10 + a_0 \equiv 0 \pmod{4}

, then

N \equiv 0 \pmod{4}

, proving that

N

is divisible by

4

Answer

N

is divisible by

4

if and only if the number formed by the last 2 digits of

N

is divisible by

4

.

Key Concept

Divisibility by 4

Explanation

A number is divisible by 4 if and only if the number formed by its last two digits is divisible by 4. This is because the higher powers of 10 are all divisible by 4, leaving only the last two digits to determine divisibility.

Solution by Steps

step 1

To find the Theta notation for the function

f(n) = n^{100} + 5n e^n + 3e^n + \log(n)

, we need to identify the term that grows the fastest as

n

approaches infinity

step 2

Among the terms

n^{100}

,

5n e^n

,

3e^n

, and

\log(n)

, the term

5n e^n

grows the fastest because the exponential function

e^n

dominates polynomial and logarithmic growth

step 3

Therefore,

f(n) = \Theta(n e^n)

step 4

For the function

g(n) = (4n - 1)^2 (n - 3)

, we first expand the expression

step 5

Expanding

(4n - 1)^2

gives

16n^2 - 8n + 1

step 6

Multiplying this by

(n - 3)

, we get

g(n) = 16n^3 - 48n^2 - 8n^2 + 24n + n - 3

step 7

Simplifying, we get

g(n) = 16n^3 - 56n^2 + 25n - 3

step 8

The term

16n^3

dominates as

n

approaches infinity

step 9

Therefore,

g(n) = \Theta(n^3)

step 10

For the function

h(n) = \frac{n^3 \log(n) + \log(n) + 100 e^{-n}}{n^2 + 2n + 1}

, we first simplify the denominator

step 11

The denominator

n^2 + 2n + 1

can be approximated as

n^2

for large

n

step 12

The numerator

n^3 \log(n) + \log(n) + 100 e^{-n}

is dominated by

n^3 \log(n)

for large

n

step 13

Therefore,

h(n) \approx \frac{n^3 \log(n)}{n^2} = n \log(n)

step 14

Hence,

h(n) = \Theta(n \log(n))

Answer

f(n) = \Theta(n e^n)

,

g(n) = \Theta(n^3)

,

h(n) = \Theta(n \log(n))

Key Concept

Theta notation

Explanation

Theta notation describes the asymptotic behavior of functions, focusing on the term that grows the fastest as

n

approaches infinity.

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Solution by Steps

step 1

To show that

2^n = \sum_{k=0}^{n} \binom{n}{k}

using a counting argument, we start by considering the number of ways to form a sequence of

n

binary digits (each digit can be either 0 or 1)

step 2

Each of the

n

positions in the sequence can be either 0 or 1, giving us

2

choices per position. Therefore, the total number of sequences is

2^n

step 3

Now, consider the same problem from a different perspective. We can count the number of sequences by considering the number of 1s in the sequence

step 4

For each

k

from

0

to

n

, there are

\binom{n}{k}

ways to choose

k

positions out of

n

to place the 1s, with the remaining positions being 0s

step 5

Summing over all possible values of

k

(from

0

to

n

), we get the total number of sequences as

\sum_{k=0}^{n} \binom{n}{k}

step 6

Since both methods count the same set of sequences, we have

2^n = \sum_{k=0}^{n} \binom{n}{k}

Answer

2^n = \sum_{k=0}^{n} \binom{n}{k}

Key Concept

Counting binary sequences

Explanation

We used two different counting methods to show that the number of binary sequences of length

n

is equal to both

2^n

and

\sum_{k=0}^{n} \binom{n}{k}

.

Solution by Steps

step 1

We need to find the number of integer solutions to the equation

x_1 + x_2 + x_3 = 30

where

x_i \geq 0

for all

i

. This is a classic "stars and bars" problem

step 2

The formula for the number of non-negative integer solutions to the equation

x_1 + x_2 + \cdots + x_k = n

is given by

\binom{n+k-1}{k-1}

step 3

Here,

n = 30

and

k = 3

. So, we need to calculate

\binom{30+3-1}{3-1} = \binom{32}{2}

step 4

Calculate

\binom{32}{2} = \frac{32 \cdot 31}{2 \cdot 1} = 496

(a) Answer

496

Part (b)

step 1

We need to find the number of integer solutions to the equation

x_1 + x_2 + x_3 = 30

where x_i > 0 for all

i

step 2

We can transform this problem by setting

x_i' = x_i - 1

for all

i

. This ensures

x_i' \geq 0

step 3

The transformed equation becomes

x_1' + x_2' + x_3' = 30 - 3 = 27

step 4

Using the "stars and bars" method again, we need to calculate

\binom{27+3-1}{3-1} = \binom{29}{2}

step 5

Calculate

\binom{29}{2} = \frac{29 \cdot 28}{2 \cdot 1} = 406

(b) Answer

406

Part (c)

step 1

We need to find the number of integer solutions to the equation

x_1 + x_2 + x_3 = 30

where

x_1 \geq 0

,

x_2 \geq 1

, and

x_3 \geq 2

step 2

We can transform this problem by setting

x_2' = x_2 - 1

and

x_3' = x_3 - 2

. This ensures

x_2' \geq 0

and

x_3' \geq 0

step 3

The transformed equation becomes

x_1 + x_2' + x_3' = 30 - 1 - 2 = 27

step 4

Using the "stars and bars" method again, we need to calculate

\binom{27+3-1}{3-1} = \binom{29}{2}

step 5

Calculate

\binom{29}{2} = \frac{29 \cdot 28}{2 \cdot 1} = 406

(c) Answer

406

Key Concept

Stars and Bars Method

Explanation

The stars and bars method is a combinatorial technique used to determine the number of ways to distribute indistinguishable objects (stars) into distinguishable bins (bars).

Solution by Steps

step 1

To solve the recurrence relation

a_{n+2} - 6a_{n+1} + 9a_n = 2

with initial conditions

a_0 = 1

and

a_1 = 3

, we first find a constant solution to the non-homogeneous part

step 2

Assume a constant solution

a_n = C

. Substituting into the non-homogeneous equation, we get

C - 6C + 9C = 2

, which simplifies to

4C = 2

. Solving for

C

, we find

C = \frac{1}{2}

step 3

Next, we solve the homogeneous part

a_{n+2} - 6a_{n+1} + 9a_n = 0

. The characteristic equation is

r^2 - 6r + 9 = 0

, which factors to

(r - 3)^2 = 0

. Thus,

r = 3

is a repeated root

step 4

The general solution to the homogeneous equation is

a_n = c_1 3^n + c_2 n 3^n

step 5

Combining the homogeneous and particular solutions, the general solution to the non-homogeneous equation is

a_n = c_1 3^n + c_2 n 3^n + \frac{1}{2}

step 6

Using the initial conditions

a_0 = 1

and

a_1 = 3

, we solve for

c_1

and

c_2

. Substituting

n = 0

into the general solution, we get

1 = c_1 + \frac{1}{2}

, so

c_1 = \frac{1}{2}

step 7

Substituting

n = 1

into the general solution, we get

3 = \frac{1}{2} \cdot 3 + c_2 \cdot 1 \cdot 3 + \frac{1}{2}

, which simplifies to

3 = \frac{3}{2} + 3c_2 + \frac{1}{2}

. Solving for

c_2

, we find

c_2 = \frac{1}{2}

step 8

Therefore, the particular solution to the recurrence relation is

a_n = \frac{1}{2} 3^n + \frac{1}{2} n 3^n + \frac{1}{2}

Answer

a_n = \frac{1}{2} 3^n + \frac{1}{2} n 3^n + \frac{1}{2}

Key Concept

Recurrence Relation Solution

Explanation

The solution involves finding a particular solution to the non-homogeneous part, solving the homogeneous part, and then using initial conditions to find the constants.

Solution by Steps

step 1

To determine if

K_{2,1}

is planar, we first draw the graph.

K_{2,1}

consists of 3 vertices: two vertices in one set

\{a, b\}

and one vertex in the other set

\{1\}

. The edges are

a-1

and

b-1

step 2

Euler's formula for planar graphs is

V - E + F = 2

, where

V

is the number of vertices,

E

is the number of edges, and

F

is the number of faces. For

K_{2,1}

, we have

V = 3

,

E = 2

, and

F = 1

step 3

Substituting into Euler's formula:

3 - 2 + 1 = 2

. Therefore, Euler's formula holds for

K_{2,1}

step 4

To determine if

K_{2,2}

is planar, we draw the graph.

K_{2,2}

consists of 4 vertices: two vertices in one set

\{a, b\}

and two vertices in the other set

\{1, 2\}

. The edges are

a-1

,

a-2

,

b-1

, and

b-2

step 5

Euler's formula for planar graphs is

V - E + F = 2

. For

K_{2,2}

, we have

V = 4

,

E = 4

, and

F = 2

step 6

Substituting into Euler's formula:

4 - 4 + 2 = 2

. Therefore, Euler's formula holds for

K_{2,2}

step 7

To prove by induction that

K_{2,m}

is planar, we start with the base case

m=1

, which we have already shown to be planar

step 8

Assume

K_{2,k}

is planar for some

k \geq 1

. We need to show that

K_{2,k+1}

is also planar

step 9

K_{2,k+1}

can be obtained by adding one more vertex to the set of

k

vertices and connecting it to both vertices in the set

\{a, b\}

. Since

K_{2,k}

is planar by the induction hypothesis, adding one more vertex and two edges does not violate planarity

step 10

Therefore, by induction,

K_{2,m}

is planar for all positive integers

m

step 11

To draw

K_{2,m}

in a planar way, place the two vertices

\{a, b\}

on opposite sides of a circle and distribute the

m

vertices evenly around the circle, connecting each to both

a

and

b

step 12

For an Euler tour in

K_{2,m}

, we need all vertices to have even degrees. If

m

is odd, each vertex in

\{1, 2, \cdots, m\}

has degree 2 (even), but

a

and

b

each have degree

m

(odd). Therefore, there is no Euler tour if

m

is odd

step 13

If

m

is even, each vertex in

\{1, 2, \cdots, m\}

has degree 2 (even), and

a

and

b

each have degree

m

(even). Therefore, there is an Euler tour if

m

is even

Answer

K_{2,1}

and

K_{2,2}

are planar, Euler's formula holds for both.

K_{2,m}

is planar by induction. No Euler tour if

m

is odd; Euler tour exists if

m

is even.

Key Concept

Planarity and Euler's formula for bipartite graphs

Explanation

The key concept is understanding how to apply Euler's formula to determine the planarity of bipartite graphs and using induction to generalize the result. Additionally, the existence of Euler tours depends on the degree of vertices.

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