Rational and Irrational Numbers

Q1: Classify the rational and irrational numbers from the following:

Step 1: Recall:
Rational numbers → Can be written in \( \frac{p}{q} \) form OR terminating/repeating decimal
Irrational numbers → Non-terminating, non-repeating decimals

i. \(5\)

Step: \( 5 = \frac{5}{1} \) → Rational
Answer: Rational

ii. \( \frac{9}{14} \)

Step: Already in \( \frac{p}{q} \) form → Rational
Answer: Rational

iii. \( \sqrt{3} \)

Step: \( \sqrt{3} \) is non-terminating, non-repeating → Irrational
Answer: Irrational

iv. \( \pi \)

Step: \( \pi \) is non-terminating, non-repeating → Irrational
Answer: Irrational

v. \( 3.1416 \)

Step: Terminating decimal → Rational
Answer: Rational

vi. \( \sqrt{4} \)

Step: \( \sqrt{4} = 2 \) → Rational
Answer: Rational

vii. \( -\sqrt{5} \)

Step: \( \sqrt{5} \) is irrational → negative also irrational
Answer: Irrational

viii. \( \sqrt[3]{8} \)

Step: \( \sqrt[3]{8} = 2 \) → Rational
Answer: Rational

ix. \( \sqrt[3]{3} \)

Step: \( \sqrt[3]{3} \) is non-terminating → Irrational
Answer: Irrational

x. \( 2\sqrt{6} \)

Step: \( \sqrt{6} \) is irrational → multiplied by 2 still irrational
Answer: Irrational

xi. \( 0.\overline{36} \)

Step: Repeating decimal → Rational
Answer: Rational

xii. \( 0.202202220\ldots \)

Step: Non-repeating decimal → Irrational
Answer: Irrational

xiii. \( \frac{2}{\sqrt{3}} \)

Step: \( \sqrt{3} \) is irrational → fraction becomes irrational
Answer: Irrational

xiv. \( \frac{22}{7} \)

Step: In \( \frac{p}{q} \) form → Rational
Answer: Rational

Q2: Separate the rationals and irrationals from among the following numbers:

Step 1: Recall:
Rational numbers → Can be written in \( \frac{p}{q} \) form OR terminating/repeating decimal
Irrational numbers → Non-terminating, non-repeating decimals

i. \( -8 \)

Step: \( -8 = \frac{-8}{1} \) → Rational
Answer: Rational

ii. \( \sqrt{25} \)

Step: \( \sqrt{25} = 5 \) → Rational
Answer: Rational

iii. \( \frac{-3}{5} \)

Step: In \( \frac{p}{q} \) form → Rational
Answer: Rational

iv. \( \sqrt{8} \)

Step: \( \sqrt{8} = 2\sqrt{2} \) → Irrational
Answer: Irrational

v. \( 0 \)

Step: \( 0 = \frac{0}{1} \) → Rational
Answer: Rational

vi. \( \pi \)

Step: Non-terminating, non-repeating → Irrational
Answer: Irrational

vii. \( \sqrt[3]{5} \)

Step: Not a perfect cube → Irrational
Answer: Irrational

viii. \( 2.\overline{4} \)

Step: Repeating decimal → Rational
Answer: Rational

ix. \( -\sqrt{3} \)

Step: \( \sqrt{3} \) is irrational → negative also irrational
Answer: Irrational

Step 2: Final Classification:
Rational Numbers:
\( -8,\ \sqrt{25},\ \frac{-3}{5},\ 0,\ 2.\overline{4} \)
Irrational Numbers:
\( \sqrt{8},\ \pi,\ \sqrt[3]{5},\ -\sqrt{3} \)
Answer: Rational → \( -8,\ \sqrt{25},\ \frac{-3}{5},\ 0,\ 2.\overline{4} \)
Irrational → \( \sqrt{8},\ \pi,\ \sqrt[3]{5},\ -\sqrt{3} \)

Q3: Represent each of the following on the real line.

Result: Point P represents the value.

Q4: Write down the values of:

i. \( (2\sqrt{3})^2 \)

Step 1: Use \( (ab)^2 = a^2 b^2 \)
\( (2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 \)
\( = 4 \times 3 = 12 \)
Answer: \( 12 \)

ii. \( \left(\frac{3}{2}\sqrt{2}\right)^2 \)

Step 1:
\( = \left(\frac{3}{2}\right)^2 \times (\sqrt{2})^2 \)
\( = \frac{9}{4} \times 2 = \frac{18}{4} = \frac{9}{2} \)
Answer: \( \frac{9}{2} \)

iii. \( (5 + \sqrt{3})^2 \)

Step 1: Use \( (a+b)^2 = a^2 + b^2 + 2ab \)
\( = 5^2 + (\sqrt{3})^2 + 2 \times 5 \times \sqrt{3} \)
\( = 25 + 3 + 10\sqrt{3} = 28 + 10\sqrt{3} \)
Answer: \( 28 + 10\sqrt{3} \)

iv. \( (\sqrt{6} – 3)^2 \)

Step 1: Use \( (a-b)^2 = a^2 + b^2 – 2ab \)
\( = (\sqrt{6})^2 + 3^2 – 2 \times \sqrt{6} \times 3 \)
\( = 6 + 9 – 6\sqrt{6} = 15 – 6\sqrt{6} \)
Answer: \( 15 – 6\sqrt{6} \)

v. \( (3 + 2\sqrt{5})^2 \)

Step 1:
\( = 3^2 + (2\sqrt{5})^2 + 2 \times 3 \times 2\sqrt{5} \)
\( = 9 + 20 + 12\sqrt{5} = 29 + 12\sqrt{5} \)
Answer: \( 29 + 12\sqrt{5} \)

vi. \( (\sqrt{5} + \sqrt{6})^2 \)

Step 1:
\( = (\sqrt{5})^2 + (\sqrt{6})^2 + 2\sqrt{5}\sqrt{6} \)
\( = 5 + 6 + 2\sqrt{30} = 11 + 2\sqrt{30} \)
Answer: \( 11 + 2\sqrt{30} \)

vii. \( \left(\frac{3}{2\sqrt{2}}\right)^2 \)

Step 1:
\( = \frac{9}{4 \times 2} = \frac{9}{8} \)
Answer: \( \frac{9}{8} \)

viii. \( (5 – 6\sqrt{3})^2 \)

Step 1:
\( = 5^2 + (6\sqrt{3})^2 – 2 \times 5 \times 6\sqrt{3} \)
\( = 25 + 108 – 60\sqrt{3} = 133 – 60\sqrt{3} \)
Answer: \( 133 – 60\sqrt{3} \)

Q5: State, giving reason, whether the given number is rational or irrational:

Step 1: Recall:
Rational + Irrational = Irrational
Non-zero Rational × Irrational = Irrational
\( (a+\sqrt{b})(a-\sqrt{b}) = a^2 – b \) (Rational)

i. \( (3+\sqrt{5}) \)

Step: 3 is rational and \( \sqrt{5} \) is irrational
Rational + Irrational = Irrational
Answer: Irrational

ii. \( (-1+\sqrt{3}) \)

Step: -1 is rational and \( \sqrt{3} \) is irrational
Rational + Irrational = Irrational
Answer: Irrational

iii. \( 5\sqrt{6} \)

Step: 5 is rational and \( \sqrt{6} \) is irrational
Rational × Irrational = Irrational
Answer: Irrational

iv. \( -\sqrt{7} \)

Step: \( \sqrt{7} \) is irrational → negative also irrational
Answer: Irrational

v. \( \frac{\sqrt{6}}{4} \)

Step: \( \sqrt{6} \) is irrational and 4 is rational
Irrational ÷ Rational = Irrational
Answer: Irrational

vi. \( \frac{3}{\sqrt{2}} \)

Step: 3 is rational and \( \sqrt{2} \) is irrational
Rational ÷ Irrational = Irrational
Answer: Irrational

vii. \( (3+\sqrt{3})(3-\sqrt{3}) \)

Step 1: Use identity:
\( (a+b)(a-b) = a^2 – b^2 \)
Step 2:
\( = 3^2 – (\sqrt{3})^2 = 9 – 3 = 6 \)
Step 3: 6 is rational
Answer: Rational

Q6: Show that each of the following is irrational:

Step 1: Recall:
Rational + Irrational = Irrational
Non-zero Rational × Irrational = Irrational

i. \( (2+\sqrt{5})^2 \)

Step 1: Use identity \( (a+b)^2 = a^2 + b^2 + 2ab \)
Step 2:
\( = 2^2 + (\sqrt{5})^2 + 2 \times 2 \times \sqrt{5} \)
Step 3:
\( = 4 + 5 + 4\sqrt{5} = 9 + 4\sqrt{5} \)
Step 4: \( 4\sqrt{5} \) is irrational
Rational + Irrational = Irrational
Answer: Irrational

ii. \( (3-\sqrt{3})^2 \)

Step 1: Use identity \( (a-b)^2 = a^2 + b^2 – 2ab \)
Step 2:
\( = 3^2 + (\sqrt{3})^2 – 2 \times 3 \times \sqrt{3} \)
Step 3:
\( = 9 + 3 – 6\sqrt{3} = 12 – 6\sqrt{3} \)
Step 4: \( 6\sqrt{3} \) is irrational
Hence whole expression is irrational
Answer: Irrational

iii. \( (\sqrt{5}+\sqrt{3})^2 \)

Step 1:
\( = (\sqrt{5})^2 + (\sqrt{3})^2 + 2\sqrt{5}\sqrt{3} \)
Step 2:
\( = 5 + 3 + 2\sqrt{15} = 8 + 2\sqrt{15} \)
Step 3: \( 2\sqrt{15} \) is irrational
So whole expression is irrational
Answer: Irrational

iv. \( \frac{6}{\sqrt{3}} \)

Step 1: Rationalize the denominator:
\( \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} \)
Step 2:
\( = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \)
Step 3: \( \sqrt{3} \) is irrational
So \( 2\sqrt{3} \) is irrational
Answer: Irrational

Q7: Prove that \( \sqrt{5} \) is an irrational number.

Step 1: Assume the contrary (proof by contradiction):
Let \( \sqrt{5} \) be rational
Step 2: Then it can be written in the form:
\( \sqrt{5} = \frac{p}{q} \), where \( p, q \) are integers and \( \gcd(p,q)=1 \)
Step 3: Squaring both sides:
\( 5 = \frac{p^2}{q^2} \)
Step 4: Multiply both sides by \( q^2 \):
\( 5q^2 = p^2 \)
Step 5: This shows \( p^2 \) is divisible by 5
⇒ \( p \) is divisible by 5
Step 6: Let \( p = 5k \)
Step 7: Substitute in equation:
\( 5q^2 = (5k)^2 = 25k^2 \)
\( q^2 = 5k^2 \)
Step 8: This shows \( q \) is also divisible by 5
Step 9: Thus, both \( p \) and \( q \) are divisible by 5
⇒ This contradicts \( \gcd(p,q)=1 \)
Step 10: Hence, our assumption is wrong
Answer: \( \sqrt{5} \) is irrational.

Q8: Write down the examples of 4 distinct irrational numbers.

Step 1: Recall:
Irrational numbers are non-terminating and non-repeating decimals
Square roots of non-perfect squares are irrational
Step 2: Examples of irrational numbers:
\( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \), \( \sqrt{7} \)
Step 3: These are distinct and cannot be written in \( \frac{p}{q} \) form
Answer: \( \sqrt{2},\ \sqrt{3},\ \sqrt{5},\ \sqrt{7} \)

Q9: Prove that \( (\sqrt{3}+\sqrt{7}) \) is irrational.

Step 1: Assume the contrary:
Let \( \sqrt{3}+\sqrt{7} \) be rational
Step 2: Then
\( \sqrt{3}+\sqrt{7} = r \), where \( r \) is rational
Step 3: Rearranging:
\( \sqrt{7} = r – \sqrt{3} \)
Step 4: Square both sides:
\( 7 = (r – \sqrt{3})^2 \)
Step 5: Expand RHS:
\( 7 = r^2 + 3 – 2r\sqrt{3} \)
Step 6: Rearranging:
\( 7 – r^2 – 3 = -2r\sqrt{3} \)
\( 4 – r^2 = -2r\sqrt{3} \)
\( \sqrt{3} = \frac{r^2 – 4}{2r} \)
Step 7: RHS is rational (since \( r \) is rational)
⇒ \( \sqrt{3} \) is rational
Step 8: But we know \( \sqrt{3} \) is irrational
⇒ Contradiction
Step 9: Hence, assumption is wrong
Answer: \( (\sqrt{3}+\sqrt{7}) \) is irrational.

Q10: Prove that \( (\sqrt{2}+\sqrt{3}) \) is irrational.

Step 1: Assume the contrary:
Let \( \sqrt{2}+\sqrt{3} \) be rational
Step 2: Then
\( \sqrt{2}+\sqrt{3} = r \), where \( r \) is rational
Step 3: Rearranging:
\( \sqrt{2} = r – \sqrt{3} \)
Step 4: Square both sides:
\( 2 = (r – \sqrt{3})^2 \)
Step 5: Expand RHS:
\( 2 = r^2 + 3 – 2r\sqrt{3} \)
Step 6: Rearranging:
\( 2 – r^2 – 3 = -2r\sqrt{3} \)
\( -1 – r^2 = -2r\sqrt{3} \)
\( \sqrt{3} = \frac{r^2 + 1}{2r} \)
Step 7: RHS is rational
⇒ \( \sqrt{3} \) is rational
Step 8: But \( \sqrt{3} \) is irrational
⇒ Contradiction
Step 9: Hence, assumption is wrong
Answer: \( (\sqrt{2}+\sqrt{3}) \) is irrational.

Q11: Write two irrational numbers between \( \sqrt{14} \) and \( \sqrt{19} \).

Step 1: Write approximate values:
\( \sqrt{14} \approx 3.74 \)
\( \sqrt{19} \approx 4.35 \)
Step 2: Find numbers between 3.74 and 4.35
Step 3: Choose square roots of non-perfect squares between 14 and 19:
\( \sqrt{15},\ \sqrt{17} \)
Step 4: These are irrational and lie between given numbers
Answer: \( \sqrt{15},\ \sqrt{17} \)

Q12: Write three irrational numbers between \( \sqrt{2} \) and \( \sqrt{7} \).

Step 1: Compare values:
\( \sqrt{2} \approx 1.41 \)
\( \sqrt{7} \approx 2.64 \)
Step 2: Choose numbers between 2 and 7 whose square roots are irrational
Step 3: Take non-perfect squares between 2 and 7:
\( 3, 5, 6 \)
Step 4: Their square roots lie between \( \sqrt{2} \) and \( \sqrt{7} \)
Answer: \( \sqrt{3},\ \sqrt{5},\ \sqrt{6} \)

Q13: State in each case, whether true or false:

i. The sum of two rationals is a rational.

Step 1: Let \(a = \frac{p}{q}\) and \(b = \frac{r}{s}\) be two rational numbers.
Step 2: Their sum is \(a + b = \frac{ps + rq}{qs}\). Since \(ps+rq\) and \(qs\) are integers, the result is a rational number.
Answer: True

ii. The sum of two irrationals is an irrational.

Step 1: Consider two irrational numbers, for example, \(\sqrt{2}\) and \(-\sqrt{2}\).
Step 2: Their sum is \(\sqrt{2} + (-\sqrt{2}) = 0\). Since 0 is a rational number, the sum is not always irrational.
Answer: False

iii. The product of two rationals is a rational.

Step 1: Let \(a = \frac{p}{q}\) and \(b = \frac{r}{s}\) be two rational numbers.
Step 2: Their product is \(a \times b = \frac{pr}{qs}\). Since \(pr\) and \(qs\) are integers, the product is always rational.
Answer: True

iv. The product of two irrationals is an irrational.

Step 1: Consider two irrational numbers, for example, \(\sqrt{2}\) and \(\sqrt{2}\).
Step 2: Their product is \(\sqrt{2} \times \sqrt{2} = 2\). Since 2 is a rational number, the product is not always irrational.
Answer: False

v. The sum of a rational and an irrational is an irrational.

Step 1: Let \(r\) be rational and \(x\) be irrational. Assume their sum \(s = r + x\) is rational.
Step 2: Then \(x = s – r\). Since the difference of two rationals is rational, this implies \(x\) is rational, which contradicts our assumption.
Answer: True

vi. The product of a rational and an irrational is a rational.

Step 1: Consider the rational number 2 and the irrational number \(\sqrt{3}\).
Step 2: Their product is \(2\sqrt{3}\), which is irrational. (Note: The product is only rational if the rational number is 0).
Answer: False

Q14: What are rational numbers? Give ten examples.

Step 1: Definition:
A rational number is any number which can be expressed in the form \( \frac{p}{q} \),
where \( p \) and \( q \) are integers and \( q \ne 0 \)
Step 2: Explanation:
Rational numbers include integers, fractions, and terminating or repeating decimals
Step 3: Examples:
\( 1,\ -2,\ \frac{3}{4},\ \frac{-5}{7},\ 0,\ \frac{9}{2},\ -\frac{11}{3},\ 2.5,\ 0.\overline{6},\ \frac{10}{1} \)
Answer: Rational numbers are numbers of the form \( \frac{p}{q} \), where \( q \ne 0 \).
Examples: \( 1,\ -2,\ \frac{3}{4},\ \frac{-5}{7},\ 0,\ \frac{9}{2},\ -\frac{11}{3},\ 2.5,\ 0.\overline{6},\ \frac{10}{1} \)

Q15: What are irrational numbers? Give ten examples.

Step 1: Definition:
Irrational numbers are numbers which cannot be expressed in the form \( \frac{p}{q} \),
where \( p \) and \( q \) are integers and \( q \ne 0 \)
Step 2: Explanation:
Their decimal expansion is non-terminating and non-repeating
Step 3: Examples:
\( \sqrt{2},\ \sqrt{3},\ \sqrt{5},\ \sqrt{7},\ \sqrt{11},\ \pi,\ e,\ \sqrt[3]{2},\ \sqrt[3]{3},\ 2\sqrt{6} \)
Answer: Irrational numbers are numbers which cannot be written in the form \( \frac{p}{q} \).
Examples: \( \sqrt{2},\ \sqrt{3},\ \sqrt{5},\ \sqrt{7},\ \sqrt{11},\ \pi,\ e,\ \sqrt[3]{2},\ \sqrt[3]{3},\ 2\sqrt{6} \)