Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 1 Mathematical Reasoning Ex 1(b) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 1 Mathematical Reasoning Exercise 1(b)

Question 1.

Check the validity of the following statements.

(i) p: 100 is a multiple of 5 and 4.

(ii) q: 125 is multiple of 5 and 7.

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

Solution:

(i) Here the connective is ‘and’

Step -1: 100 is a multiple of 5 (True)

Step -2: 100 is a multiple of 4 (True)

∴ 100 is a multiple of 5 and 4 is true, i.e. the statement ‘p’ is valid.

(ii) Here the connective is ‘and’

Step – 1: 125 is a multiple of 5 (True)

Step – 2: 125 is a multiple of 7 (false)

∴ The statement ‘q’ is (false), i.e. the statement ‘q’ is not a valid statement

(iii) Here the connective is ‘or’

60 is a multiple of 3 (True)

60 is a multiple of 5 (True)

Thus the component statements are both true.

∴ The statement ‘r’: 60 is a multiple of 3 or is true, i.e. ‘r’ is a valid statement.

Question 2.

Check the validity of the statements given below by the method given against each.

(i) “The sum of an irrational number and a rational number is irrational’’ (by contradiction method).

(ii) “If n is a real number with n > 3, then n^{2} > 9 (by the method of contradiction).

(iii) ‘‘If x, y are integers such that xy is odd then both x and y are odd” (by the method of contrapositive)

(iv) “If x is an integer and x^{2 }is even then x is also even” (method of contrapositive).

Solution:

(i) Let the given statement is false.

i.e. the sum of an irrational number and a rational number is rational.

⇒ An irrational number + a rational number = a rational number

⇒ An irrational number = A rational number + rational number = A rational number

Which is absurd.

We arrive at a contradiction.

This is due to our false assumption. Thus, the given statement is true.

(ii) Let the given statement is false.

i.e. for a natural number n > 3 n^{2} > 9

⇒ n^{2} ≤ 3 ⇒ n ≤ 3

(.’. n is a natural number which contradicts the fact that n > 3)

This contradiction is due to our false assumption.

Thus for any natural number

n > 3, n^{2} > 9.

(iii) Let p: x and y are integers such that xy is odd. and q: both x and y are odd. We shall prove the validity of p → q by using the method of contrapositive.

Now ~ q: It is not true that both x and y are odd, i.e. at least one of x (or y) is even.

Let ~ q is true.

Let x is even and x = 2n for some integer n.

⇒ xy = 2ny for some integer n.

⇒ xy is even (not odd)

⇒ ~ p is true.

Thus by the method of contrapositive, we proved ~ q ⇒ ~ p, i.e. p → q is true.

(iv) Let p: x is an integer and x^{2} is even

q : x is even.

Let ~ q is true

⇒ x is not an even integer

⇒ x is an odd integer

Let x = 2n + 1 for some integer ‘n ’

⇒ x^{2} = (2n + 1)^{2} = 4n^{2} + 4n + 1

= 2 (2n^{2} + 2n) + 1

which is odd.

⇒ p is false ⇒ ~ p is true

Thus ~ q ⇒ ~ p

∴ Thus the statement is true.

i.e x is even ⇒x is also even

Question 3.

By giving counter-examples, show that the following statements are not true:

(i) If measures of all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.

(ii) For every real number x and y, x^{2} = y^{2} ⇒ x = y.

(iii) The equation x^{2} – 1 = 0 does not have any root lying between 0 and 2.

Solution:

(i) Let in ΔABC

∠A = ∠B = ∠C = 60°

Clearly, no angle is obtuse.

∴ The triangle is not obtuse-angled.

(ii) for x = – 2, y = 2

x^{2} = y^{2} = 4 but x ≠ y

∴ x^{2} = y^{2} ⇒ x = y

(iii) x = 1 is a root of x^{2} – 1=0

that lies between 0 and 2.

∴ The given statement is false.

Question 4.

Check the validity of “If I do not work, I sleep.

If I am worried, I will not sleep. Therefore, if I am worried, I will work”.

Solution:

Method-1:

Let p: I work

q: I sleep

r: I am worried

“If I do not work, I sleep” can be written as ~ p → q

“If I am worried, I will not sleep” can be written as r → ~ q

‘If I am worried, I will work’’ can be written as r → p

The given statement can be written as [(~ p → q) ∧ (r → ~ q)] → ( r → p)

Let us draw the truth table

As [(~ p → q) ∧ (r → q)] → (r → p)

is a tautology the given statement is logically valid.

Method – 2:

Let p: I do work

q: I sleep

r: I am worried

“If I do not work, I sleep ” can be represented as ~ p → q

“If I am worried, I will not sleep” can be written as r → ~ q

Thus we can write the given statement as

[(~ p → q) ∧ (r → q)] → (r → p)

But a statement and its contrapositive are equivalent.

Thus this can be written as

[(r → ~ q) ∧ (~ q → p)] → (r → p)

By the principle of syllogism, the given statement is logically valid.

Question 5.

Let a and b be integers. By the law of contrapositive prove that if ab is even then either a is even or b is even:

Solution:

Let p: For integers ‘a and b’ ab is even q: a is even or b is even

we shall check the validity of p → q by the method of contrapositive Let ~ q is true.

⇒ Both a and b are odd

Let a = 2m + 1, b = 2n + 1

for some integers m and n

⇒ ab = (2m + 1) (2n + 1)

= 4mn + 2 (m + n ) + 1

= 2 [2mn + m + n) + 1

Which is odd

⇒ p is false

⇒ ~ p is true

Thus the method of contrapositive p → q is true.

i.e. The given statement is logically valid.