Restate the following statements with appropriate conditions, so that they become true. (i) If , then . (ii) If , then . (iii) If , then . (iv) The diagonals of a quadrilateral bisect each other.

We need to add conditions to make the given statements true. Step 1 — Analyze statement (i) Let's look at the first statement. It says: If , then . This statement is not always true. Let's take an example. If and . Here, is , which is true. But is , which is false. We need a condition to make it true. If is a positive number, then means . Since , it must be true that . > Step 2 — Analyze statement (ii) Let's look at the second statement. It says: If , then . This statement is not always true. Let's take an example. If and . Here, is , which is true. But is , which is false. We know that means . This can be written as . So, or . This means or . To make always true, we must exclude (unless ). If and are both positive numbers, then is not possible. > Step 3 — Analyze statement (iii) Let's look at the third statement. It says: If , then . This statement is not always true. Let's expand the left side of the equation. So the given equation becomes: We can subtract from both sides. This means either or (or both). The original statement says "then ". This is not true if and is any other number. Let's take an example. If and . Here, is true. But is , which is false. To make the statement true, we need to ensure must be . If and we know is not , then must be . > Step 4 — Analyze statement (iv) Let's look at the fourth statement. It says: The diagonals of a quadrilateral bisect each other. This statement is not always true. A quadrilateral is any four-sided shape. For example, in a kite, the diagonals do not bisect each other. In a general irregular quadrilateral, the diagonals do not bisect each other. This property is true for specific types of quadrilaterals. For example, in a parallelogram, the diagonals always bisect each other. Rectangles, rhombuses, and squares are all types of parallelograms. > Answer > (i) If and , then . > (ii) If and are positive numbers, then . > (iii) If and , then . > (iv) The diagonals of a parallelogram bisect each other.

Appendix 1: Proofs in Mathematics | A1.1

Question 4

Restate the following statements with appropriate conditions, so that they become true.

(i) If $a^2 > b^2$ , then $a > b$ .

(ii) If $x^2 = y^2$ , then $x = y$ .

(iii) If $(x + y)^2 = x^2 + y^2$ , then $x = 0$ .

(iv) The diagonals of a quadrilateral bisect each other.

Solution

We need to add conditions to make the given statements true.

Step 1 — Analyze statement (i)

Let's look at the first statement. It says: If $a^2 > b^2$ , then $a > b$ .

This statement is not always true.

Let's take an example. If $a = \mathbf{-3}$ and $b = \mathbf{2}$ . $a^2 = (-3)^2 = 9$ $b^2 = (2)^2 = 4$ Here, $a^2 > b^2$ is $\mathbf{9 > 4}$ , which is true. But $a > b$ is $\mathbf{-3 > 2}$ , which is false.

We need a condition to make it true. If $a$ is a positive number, then $a^2 > b^2$ means $a > |b|$ . Since $a > |b|$ , it must be true that $a > b$ .

$\boxed{\text{If } a^2 > b^2 \text{ and } a > 0 \text{, then } a > b.}$

Step 2 — Analyze statement (ii)

Let's look at the second statement. It says: If $x^2 = y^2$ , then $x = y$ .

This statement is not always true.

Let's take an example. If $x = \mathbf{2}$ and $y = \mathbf{-2}$ . $x^2 = (2)^2 = 4$ $y^2 = (-2)^2 = 4$ Here, $x^2 = y^2$ is $\mathbf{4 = 4}$ , which is true. But $x = y$ is $\mathbf{2 = -2}$ , which is false.

We know that $x^2 = y^2$ means $x^2 - y^2 = 0$ . This can be written as $(x - y)(x + y) = 0$ . So, $x - y = 0$ or $x + y = 0$ . This means $x = y$ or $x = -y$ .

To make $x = y$ always true, we must exclude $x = -y$ (unless $x=y=0$ ). If $x$ and $y$ are both positive numbers, then $x = -y$ is not possible.

$\boxed{\text{If } x^2 = y^2 \text{ and } x, y \text{ are positive numbers, then } x = y.}$

Step 3 — Analyze statement (iii)

Let's look at the third statement. It says: If $(x + y)^2 = x^2 + y^2$ , then $x = 0$ .

This statement is not always true.

Let's expand the left side of the equation. $(x + y)^2 = x^2 + 2xy + y^2$ So the given equation becomes: $x^2 + 2xy + y^2 = x^2 + y^2$ We can subtract $x^2 + y^2$ from both sides. $2xy = 0$ This means either $x = \mathbf{0}$ or $y = \mathbf{0}$ (or both).

The original statement says "then $x = 0$ ". This is not true if $y = \mathbf{0}$ and $x$ is any other number. Let's take an example. If $x = \mathbf{5}$ and $y = \mathbf{0}$ . $(5 + 0)^2 = 5^2 = 25$ $x^2 + y^2 = 5^2 + 0^2 = 25$ Here, $(x + y)^2 = x^2 + y^2$ is true. But $x = 0$ is $\mathbf{5 = 0}$ , which is false.

To make the statement true, we need to ensure $x$ must be $0$ . If $2xy = 0$ and we know $y$ is not $0$ , then $x$ must be $0$ .

$\boxed{\text{If } (x + y)^2 = x^2 + y^2 \text{ and } y \ne 0 \text{, then } x = 0.}$

Step 4 — Analyze statement (iv)

Let's look at the fourth statement. It says: The diagonals of a quadrilateral bisect each other.

This statement is not always true.

A quadrilateral is any four-sided shape. For example, in a kite, the diagonals do not bisect each other. In a general irregular quadrilateral, the diagonals do not bisect each other.

This property is true for specific types of quadrilaterals. For example, in a parallelogram, the diagonals always bisect each other. Rectangles, rhombuses, and squares are all types of parallelograms.

$\boxed{\text{The diagonals of a parallelogram bisect each other.}}$

Answer

(i) If $a^2 > b^2$ and $a > 0$ , then $a > b$ . (ii) If $x^2 = y^2$ and $x, y$ are positive numbers, then $x = y$ . (iii) If $(x + y)^2 = x^2 + y^2$ and $y \ne 0$ , then $x = 0$ . (iv) The diagonals of a parallelogram bisect each other.

Question 4

Answer

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