Question 1: Check the validity of the following statements:

Answer:

(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.

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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. }

Answer: 

(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.

Answer:

(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

(iii) Method of Contradiction:

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

This is a contradiction.

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 " }

Answer:

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”

Answer:

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”

Answer:

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:

Answer:

(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.

Answer:

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.