Question 1: Check the validity of the following statements:

(i)  p :100  is a multiple of 4 and 5. $Since, 100 is a multiple of 4 and 5, the statement is true. Hence, it is a valid statement. (ii) q: 125 is a multiple of 5 and7.$

Since, 125 is a multiple of 5 but not a multiple of 7, the statement is not true. Hence, it is not a valid statement.

(iii)  r: 60  is a multiple of 3 or 5. \$

Since, 60 is a multiple of 3 and 5, the statement is true. Hence, it is a valid statement.

$\\$

Question 2: Check whether the following statement are true or not:

$\displaystyle (i) \ p: \text{ If } x \text{ and } y \text{ are odd integers, then } x + y \text{ is an even integer }$

$\displaystyle (ii) \ q: \text{ If } x, y \text{ are integers such that } xy \text{ is even, then at least one of } \\ \\ x \text{ and } y \text{ is an even integer. }$

(i)      $\displaystyle p:$ If $\displaystyle x$ and $\displaystyle y$ are odd integers, then $\displaystyle x + y$ is an even integer.

Let us assume that $\displaystyle 'p'$ and $\displaystyle 'q'$ be the statements given by

$\displaystyle p: x$ and $\displaystyle y$ are odd integers.

$\displaystyle q: x + y$ is an even integer

the given statement can be written as :

if $\displaystyle p,$ then $\displaystyle q.$

Let $\displaystyle p$ be true. Then, $\displaystyle x$ and $\displaystyle y$ are odd integers

$\displaystyle x = 2m+1, y = 2n+1$ for some integers $\displaystyle m, n$

$\displaystyle x + y = (2m+1) + (2n+1)$

$\displaystyle x + y = (2m+2n+2)$

$\displaystyle x + y = 2(m+n+1)$

$\displaystyle x + y$ is an integer

$\displaystyle q$ is true.

So, $\displaystyle p$ is true and $\displaystyle q$ is true.

Hence, “if p, then q “is a true statement.

(ii)     $\displaystyle q:$ if $\displaystyle x, y$ are integer such that $\displaystyle xy$ is even, then at least one of $\displaystyle x$ and $\displaystyle y$ is an even integer.

Let us assume that $\displaystyle p$ and $\displaystyle q$ be the statements given by

$\displaystyle p: x$ and $\displaystyle y$ are integers and $\displaystyle xy$ is an even integer.

$\displaystyle q:$ At least one of $\displaystyle x$ and $\displaystyle y$ is even.

Let $\displaystyle p$ be true, and then $\displaystyle xy$ is an even integer.

So,

$\displaystyle xy = 2(n + 1)$

Now,

Let $\displaystyle x = 2(k + 1)$

Since, $\displaystyle x$ is an even integer, $\displaystyle xy = 2(k + 1). y$ is also an even integer.

Now take $\displaystyle x = 2(k + 1) \text{ and } y = 2(m + 1)$

$\displaystyle xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)$

So, it is also true.

Hence, the statement is true.

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Question 3: Show that the statement: $\displaystyle p : \text{ If } x \text{ is a real number such that } x^3+x = 0, \text{ then } x \text{ is } 0 "$ is true by

(i) direct method      (ii) method of contrapositive      (iii) method of contradiction.

(i) direct method

Let us assume that $\displaystyle 'q'$ and $\displaystyle 'r'$ be the statements given by

$\displaystyle q: x$ is a real number such that $\displaystyle x^3 + x = 0.$

$\displaystyle r: x \text{ is } 0.$

The given statement can be written as:

if $\displaystyle q,$ then $\displaystyle r.$

Let $\displaystyle q$ be true. Then, $\displaystyle x$ is a real number such that $\displaystyle x^3 + x = 0$

$\displaystyle x$ is a real number such that $\displaystyle x(x^2 + 1) = 0$

$\displaystyle x = 0$

$\displaystyle r$ is true

Thus, $\displaystyle q$ is true

Therefore, $\displaystyle q$ is true and $\displaystyle r$ is true.

Hence, $\displaystyle p$ is true.

(ii) Method of Contrapositive:

Let $\displaystyle r$ be false. Then,

$\displaystyle R$ is not true

$\displaystyle x \neq 0, x \in R$

$\displaystyle x(x^2 + 1) \neq 0, x \in R$

$\displaystyle q$ is not true

Thus, $\displaystyle -r = -q$

Hence, $\displaystyle p : q$ and $\displaystyle r$ is true

If possible, let $\displaystyle p$ be false.

Then, $\displaystyle p$ is not true

$\displaystyle -p$ is true

$\displaystyle -p (p \geq r)$ is true

$\displaystyle q$ and $\displaystyle -r$ is true $\displaystyle x$ is a real number such that $\displaystyle x^3 + x = 0 \text{ and } x \neq 0$

$\displaystyle x = 0 \text{ and } x \neq 0$

Hence, $\displaystyle p$ is true.

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Question 4: Show that the following statement is true by the method of contrapositive:

$\displaystyle p: \text{"If } x \text{ is an integer and } x^2 \text{is odd , then } x \text{is also odd " }$

Let us assume that $\displaystyle 'q'$ and $\displaystyle 'r'$ be the statements given

$\displaystyle q: x$ is an integer and $\displaystyle x^2$ is odd.

$\displaystyle r: x$ is an odd integer.

The given statement can be written as:

$\displaystyle p:$ if $\displaystyle q,$ then $\displaystyle r.$

Let $\displaystyle r$ be false.

Then, $\displaystyle x$ is not an odd integer, then $\displaystyle x$ is an even integer

$\displaystyle x = (2n)$ for some integer n

$\displaystyle x^2 = 4n^2$

$\displaystyle x^2$ is an even integer

Thus, $\displaystyle q$ is False

Therefore, $\displaystyle r$ is false and $\displaystyle q$ is false

Hence, $\displaystyle p:$ “ if $\displaystyle q,$ then $\displaystyle r$” is a true statement.

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Question 5: Show that the following statement is true “The integer $\displaystyle n$ is even if and only if $\displaystyle n^2$ is even”

Let the statements,

$\displaystyle p:$ Integer $\displaystyle n$ is even

$\displaystyle q:$ If $\displaystyle n^2$ is even Let $\displaystyle p$ be true. Then,

Let $\displaystyle n = 2k$

Squaring both the sides, we get,

$\displaystyle n^2 = 4k^2$

$\displaystyle n^2 = 2.2k^2$

$\displaystyle n^2$ is an even number.

So, $\displaystyle q$ is true when $\displaystyle p$ is true.

Hence, the given statement is true.

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Question 6: By giving a counter example, show that the following statement is not true.

p: “If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle”

Let $\displaystyle PQR$ be triangle in which $\displaystyle P = 60^{\circ}, Q = 60^{\circ}, R = 60^{\circ},$ then the $\displaystyle \triangle PQR$ is not an obtuse angled triangle.

Therefore given statement is not true.

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Question 7: Which of the following statements are true and which are false? In each case give a valid reason for saying so:

(i) $\displaystyle p :$ Each radius of a circle is a chord of the circle.

The given statement is false.

According the the definition of a chord, it should intersect the circumference of a circle at two distinct points.

(ii) $\displaystyle q:$ The center of a circle bisects each chord of the circle.

The given statement is false.

If a chord is not a diameter of a circle, then the center does not bisect that chord. In other words, the center of a circle only bisects the diameter, which is the chord of the circle.

(iii) $\displaystyle r :$ Circle is a particular case of an ellipse.

The statement is true.

$\displaystyle \text{ Equation of an ellipse: } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

If we put $\displaystyle a = b = 1,$ then we obtain $\displaystyle x^2 + y^2 = 1$ which is the equation of a circle. Therefore, a circle is a particular case of an ellipse.

(iv) $\displaystyle s :$If $\displaystyle x$ and $\displaystyle y$ are integers such that $\displaystyle x > y,$ then $\displaystyle - x <-y.$

The statement is true.

$x > y \ \ \ \Rightarrow -x < -y$  (By the rule of inequality)

(v) $\displaystyle t: \sqrt{11}$ is a rational number.

The given statement is false.

11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore $\sqrt{11}$ is an irrational number.

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Question 8: Determine whether the argument used to check the validity of the following statement is correct:

$\displaystyle p: \text{ "If } x^2 \text{ is irrational, then } x \text{ is rational " }$

The statement is true because the number $\displaystyle x^2 = \pi^2$ is irrational, therefore $\displaystyle x = \pi$ is irrational.

Argument Used: $\displaystyle x^2 = \pi^2$ is irrational, therefore $\displaystyle x = \pi$ is irrational.

$\displaystyle p:$ “If $\displaystyle x^2$ is irrational, then $\displaystyle x$ is rational.”

Let us take an irrational number given by $\displaystyle x = \sqrt{k},$ where k is a rational number.

Squaring both sides, we get,

$\displaystyle x^2 = k$

$\displaystyle x^2$ is a rational number and contradicts our statement.

Hence, the given argument is wrong.