Squares, Roots, Reciprocals

1. Square of a Number: \[ a^2 = a \times a \]

Multiplying a real number by itself. For any positive or negative real number \(a\), \(a^2 \ge 0\).

2. Principal Square Root: \[ \sqrt{a} = b \quad \text{such that} \quad b^2 = a \quad (a \ge 0,\; b \ge 0) \]

The non-negative value that, when squared, equals \(a\). Note that \(\sqrt{a^2} = |a|\).

3. Multiplicative Inverse (Reciprocal): \[ \text{Reciprocal of } a = \frac{1}{a} = a^{-1} \quad (a \neq 0) \]

The value that produces a product of \(1\) when multiplied by \(a\): \[ a \times \frac{1}{a} = 1 \]

4. Fractional Powers & Operations: \[ \left(\frac{a}{b}\right)^2 = \frac{a^2}{b^2}, \qquad \sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}, \qquad \left(\frac{a}{b}\right)^{-1} = \frac{b}{a} \quad (a, b \neq 0) \]

EASY (Level 1)

Problem:

Evaluate \(13^2\) and find the principal square root \(\sqrt{169}\).

Step-by-Step Solution:

1. Calculate the square: Multiply \(13\) by itself. \[ 13^2 = 13 \times 13 = 169 \]
2. Find the square root: Identify the non-negative number whose square is \(169\). \[ \sqrt{169} = 13 \]

Answer: \(169\) and \(13\)

MEDIUM (Level 2)

Problem:

Find the reciprocal of \(0.08\), and express the result as an exact whole or mixed fraction.

Step-by-Step Solution:

1. Convert decimal to a common fraction: \[ 0.08 = \frac{8}{100} = \frac{2}{25} \]
2. Apply the reciprocal rule: Invert the fraction \(\frac{2}{25}\). \[ \text{Reciprocal} = \frac{1}{2/25} = \frac{25}{2} = 12.5 \quad \left(\text{or } 12\frac{1}{2}\right) \]
3. Verification: Multiply original by its reciprocal: \(0.08 \times 12.5 = 1\).

Answer: \(12.5\) (or \(\frac{25}{2}\))

HARD (Level 3)

Problem:

Evaluate \(\sqrt{0.0064} + \left(\frac{5}{2}\right)^{-2}\) without using a mathematical table or calculator.

Step-by-Step Solution:

1. Evaluate the square root term: Convert \(0.0064\) into fraction form. \[ \sqrt{0.0064} = \sqrt{\frac{64}{10000}} = \frac{\sqrt{64}}{\sqrt{10000}} = \frac{8}{100} = 0.08 \]
2. Evaluate the negative power (reciprocal and square): \[ \left(\frac{5}{2}\right)^{-2} = \left(\frac{2}{5}\right)^2 = \frac{4}{25} = \frac{16}{100} = 0.16 \]
3. Sum the two parts: \[ 0.08 + 0.16 = 0.24 = \frac{24}{100} = \frac{6}{25} \]

Answer: \(0.24\) (or \(\frac{6}{25}\))

Misconception 1: Confusing reciprocal with negation (or subtracting from 1)

Wrong Thought: "The reciprocal of \(5\) is \(-5\)" or "The reciprocal of \(5\) is \(1 - 5 = -4\)."

Why it happens: Learners confuse the additive inverse (which gives sum \(0\)) with the multiplicative inverse (which gives product \(1\)).

Correct Principle: The reciprocal is \(1 \div a = \frac{1}{a}\). Always test: \(a \times \frac{1}{a} = 1\). Thus, the reciprocal of \(5\) is \(\frac{1}{5} = 0.2\).

Misconception 2: Claiming \(\sqrt{a^2 + b^2} = a + b\)

Wrong Thought: "\(\sqrt{9 + 16} = \sqrt{9} + \sqrt{16} = 3 + 4 = 7\)."

Why it happens: Distributing the radical across addition just like multiplication.

Correct Principle: Square roots do NOT distribute over addition or subtraction: \[ \sqrt{9 + 16} = \sqrt{25} = 5 \neq 7 \]

Misconception 3: Believing squaring always makes a number larger

Wrong Thought: "\((0.5)^2\) must be bigger than \(0.5\)."

Why it happens: Experience with integers (e.g., \(3^2 = 9 > 3\)) creates false overgeneralisation.

Correct Principle: For any number strictly between \(0\) and \(1\), \(x^2 < x\). Example: \((0.5)^2 = 0.25 < 0.5\).