RD Chapter 31- Mathematical Reasoning Ex-31.1 |
RD Chapter 31- Mathematical Reasoning Ex-31.2 |
RD Chapter 31- Mathematical Reasoning Ex-31.3 |
RD Chapter 31- Mathematical Reasoning Ex-31.4 |
RD Chapter 31- Mathematical Reasoning Ex-31.5 |

Check the validity of the following statements:

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

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

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

**Answer
1** :

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

We know that 100 is a multiple of 4 as well as 5. So, the given statement is true.

Hence, the statement is true.

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

We know that 125 is a multiple of 5 and not a multiple of 7. So, the given statement is false.

Hence, the statement is false.

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

We know that 60 is a multiple of 3 as well as 5. So, the given statement is true.

Hence, the statement is true.

Check whether the following statement is true or not:

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

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

**Answer
2** :

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

Let us assume that ‘p’ and ‘q’ be the statements given by

p: x and y are odd integers.

q: x + y is an even integer

the given statement can be written as :

if p, then q.

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

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

x + y = (2m+1) + (2n+1)

x + y = (2m+2n+2)

x + y = 2(m+n+1)

x + y is an integer

q is true.

So, p is true and q is true.

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

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

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

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

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

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

So,

xy = 2(n + 1)

Now,

Let x = 2(k + 1)

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

Now take x = 2(k + 1) and y = 2(m + 1)

xy = 2(k + 1).2(m + 1) = 2.2(k + 1)(m + 1)

So, it is also true.

Hence, the statement is true.

Show that the statement

p : “If x is a real number such that x3 + x = 0, then x is 0” is true by

(i) Direct method

(ii) method of Contrapositive

(iii) method of contradiction

**Answer
3** :

**(i)** DirectMethod:

Let us assume that ‘q’ and ‘r’ be the statements given by

q: x is a real number such that x^{3 }+ x=0.

r: x is 0.

The given statement can be written as:

if q, then r.

Let q be true. Then, x is a real number such that x^{3 }+x = 0

x is a real number such that x(x^{2 }+ 1) = 0

x = 0

r is true

Thus, q is true

Therefore, q is true and r is true.

Hence, p is true.

**(ii)** Methodof Contrapositive:

Let r be false. Then,

R is not true

x ≠ 0, x∈R

x(x^{2}+1)≠0, x∈R

q is not true

Thus, -r = -q

Hence, p : q and r is true

**(iii)** Methodof Contradiction:

If possible, let p be false. Then,

P is not true

-p is true

-p (p => r) is true

q and –r is true

x is a real number such that x^{3}+x = 0and x≠ 0

x =0 and x≠0

This is a contradiction.

Hence, p is true.

Show that the following statement is true by the method of the contrapositive

p: “If x is an integer and x2 is odd, then x is also odd.”

**Answer
4** :

Let us assume that ‘q’ and ‘r’ be the statements given

q: x is an integer and x^{2} is odd.

r: x is an odd integer.

The given statement can be written as:

p: if q, then r.

Let r be false. Then,

x is not an odd integer, then x is an even integer

x = (2n) for some integer n

x^{2} = 4n^{2}

x^{2} is an even integer

Thus, q is False

Therefore, r is false and q is false

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

**Show that thefollowing statement is true****“The integer n is even if and only if n ^{2} iseven”**

**Answer
5** :

Let the statements,

p: Integer n is even

q: If n^{2} is even

Let p be true. Then,

Let n = 2k

Squaring both the sides, we get,

n^{2} = 4k^{2}

n^{2} = 2.2k^{2}

n^{2} is an even number.

So, q is true when p is true.

Hence, the given statement is true.

**By giving a counterexample, show that the following statement is not true.****p: “If all the angles of a triangle are equal, thenthe triangle is an obtuse angled triangle.”**

**Answer
6** :

Let us consider a triangle ABC with all angles equal.

Then, each angle of the triangle is equal to 60.

So, ABC is not an obtuse angle triangle.

Hence, the statement “p: If all the angles of a triangle areequal, then the triangle is an obtuse angled triangle” is False.

**Which of thefollowing statements are true and which are false? In each case give a validreason for saying so****(i) p: Each radius of a circle is a chord of thecircle.****(ii) q: The centre of a circle bisect each chord ofthe circle.****(iii) r: Circle is a particular case of an ellipse.****(iv) s: If x and y are integers such that x > y,then – x < – y.****(v) t: √11 is a rational number.**

**Answer
7** :

**(i) **p: Eachradius of a circle is a chord of the circle.

The Radius of the circle is not it chord.

Hence, this statement is False.

**(ii)** q:The centre of a circle bisect each chord of the circle.

A chord does not have to pass through the center.

Hence, this statement is False.

**(iii)** r:Circle is a particular case of an ellipse.

A circle can be an ellipse in a particular case when the circlehas equal axes.

Hence, this statement is true.

**(iv)** s:If x and y are integers such that x > y, then – x < – y.

For any two integers, if x – y id positive then –(x-y) isnegative.

Hence, this statement is true.

**(v)** t:√11 is a rational number.

Square root of prime numbers is irrational numbers.

Hence, this statement is False.

**Determine whetherthe argument used to check the validity of the following statement is correct:****p: “If x ^{2} is irrational, then x isrational.”**

**Answer
8** :

Argument Used: x^{2} = π^{2} isirrational, therefore x = π is irrational.

p: “If x^{2} is irrational, then x is rational.”

Let us take an irrational number given by x = √k, where k is arational number.

Squaring both sides, we get,

x^{2} = k

x^{2} is a rational number and contradicts ourstatement.

Hence, the given argument is wrong.

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