## EXERCISE 14.5

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#### Question 1:

Show that the statement

*p*: “If *x* is a real number such that *x*^{3} + 4*x *= 0, then *x* is 0” is true by

(i) direct method

(ii) method of contradiction

(iii) method of contrapositive

#### Answer:

*p*: “If *x* is a real number such that *x*^{3} + 4*x* = 0, then *x* is 0”.

Let *q*: *x* is a real number such that *x*^{3} + 4*x* = 0

*r*: *x* is 0.

(i) To show that statement *p* is true, we assume that *q* is true and then show that* r *is true.

Therefore, let statement *q* be true.

∴ *x*^{3} + 4*x* = 0

*x *(*x*^{2} + 4) = 0

⇒ *x* = 0 or *x*^{2 }+ 4 = 0

However, since *x *is real, it is 0.

Thus, statement *r* is true.

Therefore, the given statement is true.

(ii) To show statement *p* to be true by contradiction, we assume that *p* is not true.

Let *x* be a real number such that *x*^{3} + 4*x* = 0 and let *x* is not 0.

Therefore, *x*^{3} + 4*x* = 0

*x* (*x*^{2} + 4) = 0

*x* = 0 or *x*^{2} + 4 = 0

*x* = 0 or *x*^{2} = – 4

However, *x* is real. Therefore, *x *= 0, which is a contradiction since we have assumed that *x* is not 0.

Thus, the given statement *p* is true.

(iii) To prove statement *p* to be true by contrapositive method, we assume that *r* is false and prove that *q* must be false.

Here, *r* is false implies that it is required to consider the negation of statement *r*. This obtains the following statement.

∼*r*: *x* is not 0.

It can be seen that (*x*^{2} + 4) will always be positive.

*x* ≠ 0 implies that the product of any positive real number with *x* is not zero.

Let us consider the product of *x* with (*x*^{2} + 4).

∴ *x *(*x*^{2} + 4) ≠ 0

⇒ *x*^{3} + 4*x* ≠ 0

This shows that statement *q* is not true.

Thus, it has been proved that

∼*r* ⇒ ∼*q*

Therefore, the given statement *p* is true.

#### Question 2:

Show that the statement “For any real numbers *a* and *b*, *a*^{2} = *b*^{2} implies that *a* = *b*” is not true by giving a counter-example.

#### Answer:

The given statement can be written in the form of “if-then” as follows.

If *a* and *b* are real numbers such that *a*^{2} = *b*^{2}, then *a* = *b*.

Let *p*: *a* and *b* are real numbers such that *a*^{2} = *b*^{2}.

*q*: *a* = *b*

The given statement has to be proved false. For this purpose, it has to be proved that if *p*, then ∼*q*. To show this, two real numbers, *a *and *b*, with *a*^{2} = *b*^{2} are required such that *a* ≠ *b*.

Let *a* = 1 and *b* = –1

*a*^{2} = (1)^{2} = 1 and *b*^{2} = (– 1)^{2} = 1

∴ *a*^{2} = *b*^{2}

However, *a* ≠ *b*

Thus, it can be concluded that the given statement is false.

#### Question 3:

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

*p*: *If x is an integer and x*^{2}* is even, then x is also even.*

#### Answer:

*p*: If *x* is an integer and *x*^{2} is even, then *x* is also even.

Let *q*: *x* is an integer and *x*^{2} is even.

*r*: *x* is even.

To prove that* p *is true by contrapositive method, we assume that *r* is false, and prove that *q* is also false.

Let *x* is not even.

To prove that *q* is false, it has to be proved that *x* is not an integer or *x*^{2} is not even.

*x* is not even implies that *x*^{2} is also not even.

Therefore, statement *q* is false.

Thus, the given statement *p* is true.

#### Question 4:

By giving a counter example, show that the following statements are not true.

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

(ii) *q*: The equation *x*^{2} – 1 = 0 does not have a root lying between 0 and 2.

#### Answer:

(i) The given statement is of the form “if *q* then *r*”.

*q*: All the angles of a triangle are equal.

*r*: The triangle is an obtuse-angled triangle.

The given statement *p* has to be proved false. For this purpose, it has to be proved that if *q*, then ∼*r*.

To show this, angles of a triangle are required such that none of them is an obtuse angle.

It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles are equal, then each of them is of measure 60°, which is not an obtuse angle.

In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle.

Thus, it can be concluded that the given statement *p* is false.

(ii) The given statement is as follows.

*q*: The equation *x*^{2} – 1 = 0 does not have a root lying between 0 and 2.

This statement has to be proved false. To show this, a counter example is required.

Consider *x*^{2} – 1 = 0

*x*^{2} = 1

*x* = ± 1

One root of the equation *x*^{2} – 1 = 0, i.e. the root *x* = 1, lies between 0 and 2.

Thus, the given statement is false.

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#### Question 5:

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

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

(ii) *q*: The centre of a circle bisects each chord of the 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*: is a rational number.

#### Answer:

(i) The given statement *p* is false.

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

(ii) The given statement *q* is false.

If the chord is not the diameter of the circle, then the centre will not bisect that chord.

In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.

(iii) The equation of an ellipse is,

If we put *a* = *b* = 1, then we obtain

*x*^{2} + *y*^{2} = 1, which is an equation of a circle

Therefore, circle is a particular case of an ellipse.

Thus, statement *r* is true.

(iv) *x *> *y*

⇒ –*x* < –*y* (By a rule of inequality)

Thus, the given statement *s* is true.

(v) 11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore, is an irrational number.

Thus, the given statement *t* is false.