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.

Understand the Question Mathematical statements of the form "If , then " must hold true for all possible values satisfying the hypothesis . If there exists even a single counterexample where is true but is false, the statement is mathematically false. To make such statements true, we identify the counterexamples and impose additional conditions on the variables or shapes involved. Part (i) If , then . Step 1 · Find Counterexample and Add Condition Consider and : Here, () is true, but () is false. For to imply , must be a positive number (). Answer > (i) If and , then . Part (ii) If , then . Step 1 · Factorise and Restrict Signs Consider and : Here, is true, but is false. Algebraically: To ensure , we must rule out by restricting and to be positive numbers (or having the same sign). Answer > (ii) If and are positive numbers, then . Part (iii) If , then . Step 1 · Expand and Set Non-Zero Condition Expanding the equation: If and : Here holds, but . Therefore, for to necessarily hold, we require . Answer > (iii) If and , then . Part (iv) The diagonals of a quadrilateral bisect each other. Step 1 · Identify the Specific Quadrilateral Type In a general quadrilateral (such as a kite or an irregular four-sided figure), diagonals do not necessarily bisect each other. Diagonals bisect each other specifically in a parallelogram (and its special cases: rectangle, rhombus, square). Answer > (iv) The diagonals of a parallelogram bisect each other. Common Mistakes Assuming Absolute Values Imply Ordered Numbers: Forgetting that squaring negatives makes them positive, so only implies , not necessarily . Neglecting the Zero Product Property: yields two possibilities ( or ). To force , we must explicitly state . Overgeneralising Geometric Properties: Assuming properties of special quadrilaterals (like parallelograms) apply to all general quadrilaterals.

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

Understand the Question

Mathematical statements of the form "If $P$ , then $Q$ " must hold true for all possible values satisfying the hypothesis $P$ .
If there exists even a single counterexample where $P$ is true but $Q$ is false, the statement is mathematically false.
To make such statements true, we identify the counterexamples and impose additional conditions on the variables or shapes involved.

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

Step 1 · Find Counterexample and Add Condition

Consider $a = -3$ and $b = 2$ :

\begin{aligned} a^2 &= (-3)^2 = 9 \\ b^2 &= (2)^2 = 4 \end{aligned}

Here, $a^2 > b^2$ ( $9 > 4$ ) is true, but $a > b$ ( $-3 > 2$ ) is false.

For $a^2 > b^2$ to imply $a > b$ , $a$ must be a positive number ( $a > 0$ ).

Answer

(i) If $a^2 > b^2$ and $a > 0$ , then $a > b$ .

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

Step 1 · Factorise and Restrict Signs

Consider $x = 2$ and $y = -2$ :

\begin{aligned} x^2 &= (2)^2 = 4 \\ y^2 &= (-2)^2 = 4 \end{aligned}

Here, $x^2 = y^2$ is true, but $x = y$ is false.

Algebraically:

\begin{aligned} x^2 &= y^2 \\ x^2 - y^2 &= 0 \\ (x - y)(x + y) &= 0 \\ \implies x = y \quad &\text{or} \quad x = -y \end{aligned}

To ensure $x = y$ , we must rule out $x = -y$ by restricting $x$ and $y$ to be positive numbers (or having the same sign).

Answer

(ii) If $x^2 = y^2$ and $x, y$ are positive numbers, then $x = y$ .

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

Step 1 · Expand and Set Non-Zero Condition

Expanding the equation:

\begin{aligned} (x + y)^2 &= x^2 + y^2 \\ x^2 + 2xy + y^2 &= x^2 + y^2 \\ 2xy &= 0 \\ \implies x = 0 \quad &\text{or} \quad y = 0 \end{aligned}

If $y = 0$ and $x = 5$ :

\begin{aligned} (5 + 0)^2 &= 5^2 = 25 \\ x^2 + y^2 &= 5^2 + 0^2 = 25 \end{aligned}

Here $(x+y)^2 = x^2+y^2$ holds, but $x \ne 0$ .

Therefore, for $x = 0$ to necessarily hold, we require $y \ne 0$ .

Answer

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

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

Step 1 · Identify the Specific Quadrilateral Type

In a general quadrilateral (such as a kite or an irregular four-sided figure), diagonals do not necessarily bisect each other.

Diagonals bisect each other specifically in a parallelogram (and its special cases: rectangle, rhombus, square).

Answer

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

Common Mistakes

Assuming Absolute Values Imply Ordered Numbers: Forgetting that squaring negatives makes them positive, so $a^2 > b^2$ only implies $|a| > |b|$ , not necessarily $a > b$ .
Neglecting the Zero Product Property: $2xy = 0$ yields two possibilities ( $x = 0$ or $y = 0$ ). To force $x = 0$ , we must explicitly state $y \ne 0$ .
Overgeneralising Geometric Properties: Assuming properties of special quadrilaterals (like parallelograms) apply to all general quadrilaterals.

Question 4

More questions in A1.1