Rational and Irrational Numbers

Q1: Multiple Choice Type

a. Since \( 90 = 2 \times 3 \times 3 \times 5 \), \( \frac{23}{90} \) is not a terminating decimal

Step 1: Check denominator factors \[ 90 = 2 \times 3^2 \times 5 \]Step 2: Condition for terminating decimal
Denominator must have only factors 2 and/or 5
Since factor 3 is present ⇒ non-terminating
Answer: i. True

b. \(\sqrt{27}\) is irrational and \(\sqrt3\) is also irrational, then which of the following is rational:

Step 1: Simplify \[ \sqrt{27} = 3\sqrt{3} \]Step 2: Check options \[ \sqrt{27} – \sqrt{3} = 2\sqrt{3} \quad (\text{irrational}) \\ \sqrt{27} \times \sqrt{3} = \sqrt{81} = 9 \quad (\text{rational}) \]Answer: iii. \( \sqrt{27} \times \sqrt{3} \)

c. If \(x=\sqrt6-\sqrt5\), then \(x-\frac{1}{x}\) is equal to:

Step 1: Find reciprocal \[ \frac{1}{x} = \frac{1}{\sqrt{6}-\sqrt{5}} \times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}} = \sqrt{6}+\sqrt{5} \]Step 2: Subtract \[ x – \frac{1}{x} = (\sqrt{6}-\sqrt{5}) – (\sqrt{6}+\sqrt{5}) = -2\sqrt{5} \]Answer: iv. \( -2\sqrt{5} \)

d. \(\frac{2+\sqrt3}{2-\sqrt3}-\frac{2-\sqrt3}{2+\sqrt3}\) is equal to:

Step 1: Take LCM \[ = \frac{(2+\sqrt{3})^2 – (2-\sqrt{3})^2}{(2+\sqrt{3})(2-\sqrt{3})} \]Step 2: Denominator \[ 4 – 3 = 1 \]Step 3: Numerator \[ (2+\sqrt{3})^2 = 7 + 4\sqrt{3} \\ (2-\sqrt{3})^2 = 7 – 4\sqrt{3} \\ = 8\sqrt{3} \]Answer: iv. \( 8\sqrt{3} \)

e. Statement 1: If \(a=3\sqrt3\) and \(b=\frac{5}{\sqrt{12}}\), then \(a\timesb\) is irrational. Statement 2: \(a\timesb=3\sqrt3\times\frac{5}
{\sqrt{12}}=\frac{15\times\sqrt3}{2\sqrt3}=\frac{15}{2}\)

Step 1: Compute product \[ a \times b = 3\sqrt{3} \times \frac{5}{\sqrt{12}} \\ = \frac{15\sqrt{3}}{2\sqrt{3}} = \frac{15}{2} \]Step 2: Conclusion
Product is rational
Statement 1 is false, Statement 2 is true
Answer: iv. Statement 1 is false and Statement 2 is true

f. Statement 1: If \(x=\sqrt5+2\), then \(x-\frac{1}{x}=4\).
Statement 2: \(\frac{1}{x}=\frac{1}{\sqrt5+2}\times\frac{\sqrt5-2}{\sqrt5-2}=\sqrt5-2\).

Step 1: Find reciprocal \[ \frac{1}{x} = \sqrt{5} – 2 \]Step 2: Subtract \[ x – \frac{1}{x} = (\sqrt{5}+2) – (\sqrt{5}-2) = 4 \]Both statements are true
Answer: i. Both the statements are true

g. Assertion (A): \(x+\frac{1}{x}=4\) and \(\frac{1}{x}=2+\sqrt3\), then \(x=2-\sqrt3\).
Reason (R): \(x+\frac{1}{x}=4\Rightarrowx+2+\sqrt3=4\ \Rightarrowx=2-\sqrt3\).

Step 1: Given \[ x + \frac{1}{x} = 4,\quad \frac{1}{x} = 2 + \sqrt{3} \]Step 2: Substitute \[ x + (2+\sqrt{3}) = 4 \\ x = 4 – (2+\sqrt{3}) = 2 – \sqrt{3} \]Step 3: Check Assertion
Assertion is true
Step 4: Check Reason
Reason shows correct steps leading to result
Answer: iii. Both A and R are true and R is reason for A

h. Assertion (A): \(\sqrt{22},\sqrt{23},\ \sqrt{24},\ \sqrt{25},\ \sqrt{26}\) and \(\sqrt{27}\) are irrational numbers between \(\sqrt{21}\) and \(\sqrt{28}\).
Reason (R): \(\sqrt{25}\) is not an irrational number as \(\sqrt{25}=5\); which is a rational number.

Step 1: Check Assertion \[ \sqrt{25} = 5 \quad (\text{rational}) \]So not all given numbers are irrational ⇒ Assertion is false
Step 2: Check Reason
Reason correctly states that \( \sqrt{25} = 5 \) is rational
Answer: ii. A is false, R is true

Q2: Simplify \( \frac{\sqrt{x^2+y^2}-y}{x-\sqrt{x^2-y^2}} \div \frac{\sqrt{x^2-y^2}+x}{\sqrt{x^2+y^2}+y} \)

Step 1: Convert division into multiplication \[ = \frac{\sqrt{x^2+y^2}-y}{x-\sqrt{x^2-y^2}} \times \frac{\sqrt{x^2+y^2}+y}{\sqrt{x^2-y^2}+x} \]Step 2: Rearrange \[ = \frac{(\sqrt{x^2+y^2}-y)(\sqrt{x^2+y^2}+y)}{(x-\sqrt{x^2-y^2})(\sqrt{x^2-y^2}+x)} \]Step 3: Apply identity \[ (a-b)(a+b) = a^2 – b^2 \]Step 4: Simplify numerator \[ (\sqrt{x^2+y^2})^2 – y^2 = (x^2+y^2) – y^2 = x^2 \]Step 5: Simplify denominator \[ x^2 – (\sqrt{x^2-y^2})^2 = x^2 – (x^2 – y^2) = y^2 \]Step 6: Final result \[ = \frac{x^2}{y^2} = \left(\frac{x}{y}\right)^2 \]Answer: \( \frac{x^2}{y^2} \)

Q3: Evaluate, correct to one place of decimal, the expression \(\frac{5}{\sqrt{20}-\sqrt{10}}\), if \(\sqrt5=2.2\) and \(\sqrt{10}=3.2\).

Step 1: Simplify surd \[ \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \]Step 2: Substitute \[ = \frac{5}{2\sqrt{5} – \sqrt{10}} \]Step 3: Put values \[ = \frac{5}{2(2.2) – 3.2} \\ = \frac{5}{4.4 – 3.2} = \frac{5}{1.2} \]Step 4: Calculate \[ = 4.166\ldots \]Rounded to one decimal place: \[ = 4.2 \]Answer: \( 4.2 \)

Q4: If \( x=\sqrt{3}-\sqrt{2} \), find the value of:

i. \( x+\frac{1}{x} \)

Step 1: Find reciprocal \[ \frac{1}{x} = \frac{1}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ = \sqrt{3} + \sqrt{2} \]Step 2: Add \[ x + \frac{1}{x} = (\sqrt{3}-\sqrt{2}) + (\sqrt{3}+\sqrt{2}) = 2\sqrt{3} \]Answer: \( 2\sqrt{3} \)

ii. \( x^2+\frac{1}{x^2} \)

Step 1: Use identity \[ x^2 + \frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 – 2 \\ = (2\sqrt{3})^2 – 2 = 12 – 2 = 10 \]Answer: \( 10 \)

iii. \( x^3+\frac{1}{x^3} \)

Step 1: Use identity \[ x^3 + \frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 – 3\left(x+\frac{1}{x}\right) \\ = (2\sqrt{3})^3 – 3(2\sqrt{3}) \\ = 8 \cdot 3\sqrt{3} – 6\sqrt{3} = 24\sqrt{3} – 6\sqrt{3} \\ = 18\sqrt{3} \]Answer: \( 18\sqrt{3} \)

iv. \( x^3+\frac{1}{x^3}-3\left(x^2+\frac{1}{x^2}\right)+x+\frac{1}{x} \)

Step 1: Substitute values \[ x^3+\frac{1}{x^3} = 18\sqrt{3},\quad x^2+\frac{1}{x^2} = 10,\quad x+\frac{1}{x} = 2\sqrt{3} \]Step 2: Simplify \[ = 18\sqrt{3} – 3(10) + 2\sqrt{3} \\ = 18\sqrt{3} – 30 + 2\sqrt{3} \\ = 20\sqrt{3} – 30 \]Answer: \( 20\sqrt{3} – 30 \)

Q5: State True or False

i. Negative of an irrational number is irrational

Step 1: Let an irrational number be \( x \)
Step 2: Consider its negative \(- x\)
Step 3: Property
If \( x \) is irrational, then \( -x \) is also irrational
Answer: True

ii. The product of a non-zero rational number and an irrational number is a rational number

Step 1: Let rational number be \( r \neq 0 \) and irrational number be \( x \)
Step 2: Product = \( r \times x\)
Step 3: Property
Product of non-zero rational and irrational number is always irrational
Answer: False

Q6: Draw a line segment of length \(\sqrt{3}\) cm.

Answer: Point \(P\) on the number line represents \(\sqrt{3}\) units (\(\approx 1.73\) cm).

Q7: Draw a line segment of length \(\sqrt{8}\) cm.

Answer: Point \(P\) on the number line represents \(\sqrt{8}\) cm.

Q8: Show that:

i. \( x^3+\frac{1}{x^3}=52 \), if \( x=2+\sqrt{3} \)

Step 1: Find reciprocal \[ \frac{1}{x} = 2 – \sqrt{3} \]Step 2: Find sum \[ x + \frac{1}{x} = (2+\sqrt{3}) + (2-\sqrt{3}) = 4 \]Step 3: Use identity \[ x^3 + \frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 – 3\left(x+\frac{1}{x}\right) \\ = 4^3 – 3(4) = 64 – 12 = 52 \]Answer: \( 52 \)

ii. \( x^2+\frac{1}{x^2}=34 \), if \( x=3+2\sqrt{2} \)

Step 1: Find reciprocal \[ \frac{1}{x} = 3 – 2\sqrt{2} \]Step 2: Find sum \[ x + \frac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2}) = 6 \]Step 3: Use identity \[ x^2 + \frac{1}{x^2} = \left(x+\frac{1}{x}\right)^2 – 2 \\ = 6^2 – 2 = 36 – 2 = 34 \]Answer: \( 34 \)

iii. \( \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}+\frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}}=11 \)

Step 1: Rationalize first term \[ \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}} \times \frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}} \\ = \frac{(3\sqrt{2}-2\sqrt{3})^2}{18 – 12} = \frac{18 + 12 – 12\sqrt{6}}{6} \\ = \frac{30 – 12\sqrt{6}}{6} = 5 – 2\sqrt{6} \]Step 2: Rationalize second term \[ \frac{2\sqrt{3}}{\sqrt{3}-\sqrt{2}} \times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}} \\ = \frac{2\sqrt{3}(\sqrt{3}+\sqrt{2})}{3 – 2} \\ = 2(3 + \sqrt{6}) = 6 + 2\sqrt{6} \]Step 3: Add \[ (5 – 2\sqrt{6}) + (6 + 2\sqrt{6}) = 11 \]Answer: \( 11 \)

Q9: Show that \( x \) is irrational, if:

i. \( x^2 = 6 \)

Step 1: Take square root \[ x = \sqrt{6} \]Step 2: Property
Since 6 is not a perfect square, \( \sqrt{6} \) is irrational
Answer: \( x \) is irrational

ii. \( x^2 = 0.009 \)

Step 1: Convert into fraction \[ 0.009 = \frac{9}{1000} \]Step 2: Take square root \[ x = \sqrt{\frac{9}{1000}} = \frac{3}{\sqrt{1000}} \\ = \frac{3}{10\sqrt{10}} \]Step 3: Property
Since \( \sqrt{10} \) is irrational ⇒ \( x \) is irrational
Answer: \( x \) is irrational

iii. \( x^2 = 27 \)

Step 1: Take square root \[ x = \sqrt{27} = 3\sqrt{3} \]Step 2: Property
Since \( \sqrt{3} \) is irrational ⇒ \( 3\sqrt{3} \) is irrational
Answer: \( x \) is irrational

Q10: Show that \( x \) is rational, if:

i. \( x^2 = 16 \)

Step 1: Take square root \[ x = \sqrt{16} \\ x = \pm 4 \]Step 2: Conclusion
Since \( \pm 4 \) are integers ⇒ rational numbers
Answer: \( x \) is rational

ii. \( x^2 = 0.0004 \)

Step 1: Convert to fraction \[ 0.0004 = \frac{4}{10000} \]Step 2: Take square root \[ x = \sqrt{\frac{4}{10000}} = \frac{2}{100} = \frac{1}{50} \\ x = \pm \frac{1}{50} \]Step 3: Conclusion
Since it is a ratio of integers ⇒ rational
Answer: \( x \) is rational

iii. \( x^2 = 1\frac{7}{9} \)

Step 1: Convert mixed fraction \[ 1\frac{7}{9} = \frac{16}{9} \]Step 2: Take square root \[ x = \sqrt{\frac{16}{9}} = \frac{4}{3} \\ x = \pm \frac{4}{3} \]Step 3: Conclusion
Since it is a ratio of integers ⇒ rational
Answer: \( x \) is rational

Q11: Find the volume of: \( \frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{2025}+\sqrt{2026}} \)

Step 1: General term \[ \frac{1}{\sqrt{n}+\sqrt{n+1}} \]Step 2: Rationalize \[ \frac{1}{\sqrt{n}+\sqrt{n+1}} \times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}} \\ = \frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n} = \sqrt{n+1}-\sqrt{n} \]Step 3: Apply to series \[ (\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (\sqrt{4}-\sqrt{3}) + \cdots + (\sqrt{2026}-\sqrt{2025}) \]Step 4: Telescoping cancellation \[ = -1 + \sqrt{2026} \\ = \sqrt{2026} – 1 \]Answer: \( \sqrt{2026} – 1 \)