The sum of two even integers must always be divisible by

Question 1 Exercise 3.2

Answer:

SOLUTION:

(i) The sum of any two odd numbers is an even number.

example: 1+3=4, 3+5=8

(ii) The sum of any two even numbers is an even number.

example: 4+4=8, 2+2=4

Video transcript

hello guys welcome to lido homework today we're doing question number one which is what is the sum of any two first one odd numbers so let's take example of any two odd numbers so three three is odd number and five so five is also n number so now we'll add both of them so three plus five which will give us 8 both are odd but the answer is 8 which is even so the sum of any two odd numbers is even okay let's move on to the second part what is the sum of any two even numbers so now let's take any two even numbers so let's take 8 and let's take 4 so sum means addition so 8 plus 4 is 12 12 is also even therefore sum of any two even numbers is also even so answer to both of these questions is even thank you guys for watching the video if you have doubts please let me in the comments below i'll get back to you as soon as possible thank you very much

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1. For propositions P and Q below, state whether or not P º Q. Write a truth table to justify your answer.

P: p Ú q ® r
Q: (p ® r) Ù (q ® r)

ANSWER: P º Q. Truth table is as follows.

p q r p Ú q ® r (p ® r) Ù (q ® r)

T T T T T

T T F F F

T F T T T

T F F F F

F T T T T

F T F F F

F F T T T

F F F T T

2. Determine the truth value of the following proposition. If it is true, prove it using direct proof. If it is false, state the negation explicitly and give a counterexample.

Proposition: The sum of any three consecutive integers is divisible by 3.

ANSWER: True.

Proof: Let three consecutive integers a, b, c be (n+1), (n+2) and (n+3) respectively. Then,

a + b + c = (n+1) + (n+2) + (n+3) = 3n + 6
= 3(n+2)

Since n+2 is an integer, 3(n+2) is divisible by 3.

3. Determine the truth value of the following proposition. If it is true, prove it using direct proof. If it is false, state the negation explicitly and give a counterexample.

Proposition: "a"b such that a, b are real numbers, sqrt(a+b) = sqrt(a) + sqrt(b).

ANSWER: False.

Negation of the proposition is $a$b such that a, b are real numbers, sqrt(a+b) != sqrt(a) + sqrt(b). Examples of such a, b are a =1 and b = 2.

4. Prove the following proposition by using proof by contradiction.

Proposition: If a product of two positive real numbers is greater than 100, then at least one of the numbers is greater than 10.

ANSWER: Proof by contradiction.

The proposition states that, "a"b. such that a, b are real numbers, if a * b > 100, then a > 10 or b > 10. We show this by contradiction.

[Negation of the proposition] Suppose not, that is, $a$b. such that a, b are real numbers, a * b > 100 and a <= 10 and b <= 10 (i.e., (~(p ® (q Ú r)) º p Ù ~q Ù ~r) . Then,

a * b <= 10 * 10 = 100.

But from assumption, a * b > 100. Therefore, contradiction. Therefore, the original proposition is true.

5. (Section 1.6, #4; p. 46) Using induction, verify the following equation is true for every positive integer n.

12 + 22 + 32 + ... + n2 = [n(n+1)(2n+1)] / 6

ANSWER: Proof by induction

Basic Step (n = 1):

LHS = 12 = 1
RHS = [1*2*3]/6 = 6/6 = 1. So LHS = RHS ... (A)

Inductive Step:

Assume 12 + 22 + 32 + ... + n2 = [n(n+1)(2n+1)] / 6 for n >= 1.
Show 12 + 22 + 32 + ... + n2 + (n+1)2 = [(n+1)(n+2)(2n+3)] / 6

LHS = 12 + 22 + 32 + ... + n2 + (n+1)2
           n(n+1)(2n+1)
        = ------------- + (n+1)2   ... by inductive hypothesis                    6            (n+1)[n(2n+1)+6(n+1)]         = ------------------------                         6
           (n+1)(2n2 + n + 6n + 6)
        = ------------------------                         6
           (n+1)(2n2 + 7n + 6)
        = --------------------                         6             (n+1)(2n + 3)(n + 2)         = ---------------------- = RHS     ... (B)
                        6

By (A) and (B), the theorem is true.

6. (Section 1.6, #17; p. 47) Using induction, verify the following inequality.

(1 + x)n >= 1 + nx, for x >= -1 and n = 1,2,...

ANSWER: Proof by induction

Basic Step (n = 1):

LHS = 1+x
RHS = 1+x   So, LHS >= RHS ... (A)

Inductive Step:

Assume (1 + x)n >= 1 + nx, for n >= 1.
Show (1 + x)n+1 >= 1 + (n+1)x

LHS = (1 + x)n+1 = (1 + x)(1 + x)n
                            >= (1 + x) (1 + nx)       by inductive hypothesis
                            = 1 + (n + 1)x + nx2
                            >= 1 + (n + 1)x             since nx2 >= 0                             = RHS        So, LHS >= RHS ... (B)

By (A) and (B), the theorem is true.

7. (Section 1.6, #22; p. 47) By experimenting with small values of n, guess a formula for the given sum:

1 1 1 --- + --- + .. + ----- 1*2 2*3 n(n+1)
then use induction to verify your formula.

ANSWER:

First we claim that
1 1 1 n --- + --- + .. + ----- = --- 1*2 2*3 n(n+1) n+1
for n = 1, 2, ...

Proof: by induction.

Basic Step (n = 1):

LHS = 1/2
RHS = 1/2 So, LHS = RHS ... (A)

Inductive Step:

Assume
1 1 1 n --- + --- + .. + ----- = --- 1*2 2*3 n(n+1) n+1
Show
1 1 1 1 n --- + --- + .. + ----- + ---------- = --- 1*2 2*3 n(n+1) (n+1)(n+2) n+1
LHS
1 1 1 1 --- + --- + .. + ----- + ---------- 1*2 2*3 n(n+1) (n+1)(n+2) n 1 = --- + ---------- n+1 (n+1)(n+2) n(n+2) + 1 = ---------- (n+1)(n+2) n^2 + 2n + 1 = ------------ (n+1)(n+2) (n+1)^2 = ---------- (n+1)(n+2) n+1 = --- n+2
= RHS So, LHS = RHS ... (B)

By (A) and (B), the formula is true.

8. Write an algorithm that counts the number of even numbers in a sequence s = {s1, s2, .., sn}, where si (1 <= i <= n) is an integer.

ANSWER:

procedure count_even(s, n) 1. count := 0 2. i := 1 3. while (i <= n) 4. begin 5. if (si mod 2 = 0) then 6. count := count + 1 7. i := i + 1 8. end 9. return count end count_even

9. Use the Euclidean algorithm to find the greatest common divisor of 527 and 62. Show the values of variables a, b and r in each iteration.

ANSWER: The gcd of 527 and 62 is 31, as shown in the following table.