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Rational and Irrational Numbers Class 9 ICSE Important Questions with Answers

The 13 questions below are Rational and Irrational Numbers Class 9 ICSE important questions, each with a full step-by-step answer: extra questions beyond the textbook exercises, covering 6 topics of ICSE Class 9 Maths in the question formats used in the exam.

By The Classmate AI Editorial Team

Reviewed by Classmate AI Team · 17 October 2026

Last updated 25 October 2026

13
Questions
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Topics
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Key concepts
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With answers
Quick answer

ICSE Class 9 Rational and Irrational Numbers questions cover telling rational numbers from irrational ones, the place of both among the real numbers, simplifying and comparing surds, rationalising denominators such as 15−3\dfrac{1}{\sqrt5-\sqrt3}1/(√5-√3), finding rational and irrational numbers between two numbers, proving that 2\sqrt2√2, 3\sqrt3√3 and 5\sqrt5√5 are irrational, and showing 2\sqrt2√2 and 3\sqrt3√3 on the number line.

Syllabus reference (2026–27): ICSE 2028 Mathematics syllabus (PDF) · CISCE (Council for the Indian School Certificate Examinations)

About Rational and Irrational Numbers

In this ICSE Class 9 Maths chapter Rational and Irrational Numbers you classify numbers as rational (pq\frac{p}{q}p/q form) or irrational (non-terminating non-recurring decimals), operate on surds, rationalise denominators, and locate irrational numbers such as 2\sqrt2√2 and 3\sqrt3√3 on the number line using geometric constructions.

Rational vs irrational numbersRational and irrational numbers between two numbersSurds: simplifying and comparingRationalising the denominatorProofs of irrationalityRepresenting surds on the number line

Key concepts & formulas

Rational and irrational

A rational number can be written as pq\dfrac{p}{q}p/q with q≠0q\neq0q≠0 and has a terminating or recurring decimal; an irrational number (e.g. 2, π\sqrt2,\ \pi√2, π) has a non-terminating, non-recurring decimal.

Rationalising factor

To rationalise 1a+b\dfrac{1}{a+\sqrt b}1/(a+√b) multiply by the conjugate a−ba−b\dfrac{a-\sqrt b}{a-\sqrt b}(a-√b)/(a-√b), using (a+b)(a−b)=a2−b(a+\sqrt b)(a-\sqrt b)=a^2-b(a+√b)(a-√b)=a²-b.

Laws of surds

a×b=ab\sqrt a\times\sqrt b=\sqrt{ab}√a× √b=√ab, ab=ab\dfrac{\sqrt a}{\sqrt b}=\sqrt{\dfrac{a}{b}}√a/√b=a/b, and (a)2=a(\sqrt a)^2=a(√a)²=a for a,b≥0a,b\ge0a,b≥0.

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Rational and Irrational Numbers Questions and Answers

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MCQs with answers (1 mark)

Q1MCQEasy1 mark

Which of the following is an irrational number?

  1. (a)

    16\sqrt{16}√16

  2. (b)

    0.3‾0.\overline{3}0.3

  3. (c)

    7\sqrt7√7

  4. (d)

    227\dfrac{22}{7}22/7

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Answer: (c) 7\sqrt7√7.

16=4\sqrt{16}=4√16=4 and 227\dfrac{22}{7}22/7 are rational, 0.3‾=130.\overline{3}=\dfrac130.3=1/3 is a recurring (rational) decimal, but 7\sqrt7√7 is non-terminating and non-recurring, hence irrational.

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Q2MCQEasy1 mark

The rationalising factor of 5\sqrt5√5 is:

  1. (a)

    555

  2. (b)

    5\sqrt5√5

  3. (c)

    −5-\sqrt5-√5

  4. (d)

    15\dfrac{1}{\sqrt5}1/√5

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Answer: (b) 5\sqrt5√5.

Multiplying 5×5=5\sqrt5\times\sqrt5=5√5× √5=5, a rational number, so 5\sqrt5√5 is the rationalising factor.

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Q3MCQModerate1 mark

Which of the following is the greatest: 232\sqrt32√3, 323\sqrt23√2, 15\sqrt{15}√15, 17\sqrt{17}√17?

  1. (a)

    232\sqrt32√3

  2. (b)

    323\sqrt23√2

  3. (c)

    15\sqrt{15}√15

  4. (d)

    17\sqrt{17}√17

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Answer: (b) 323\sqrt23√2.

Write each number as a single square root: 23=4×3=122\sqrt3=\sqrt{4\times3}=\sqrt{12}2√3=√4×3=√12 and 32=9×2=183\sqrt2=\sqrt{9\times2}=\sqrt{18}3√2=√9×2=√18. Comparing 12, 18, 15, 17\sqrt{12},\ \sqrt{18},\ \sqrt{15},\ \sqrt{17}√12, √18, √15, √17, the largest number under the root is 181818, so 323\sqrt23√2 is the greatest.

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Q4MCQHOTS1 mark

If x=5+2x=\sqrt5+2x=√5+2, then x+1xx+\dfrac{1}{x}x+1/x equals:

  1. (a)

    252\sqrt52√5

  2. (b)

    444

  3. (c)

    222

  4. (d)

    5\sqrt5√5

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Answer: (a) 252\sqrt52√5.

1x=15+2=5−25−4=5−2\dfrac{1}{x}=\dfrac{1}{\sqrt5+2}=\dfrac{\sqrt5-2}{5-4}=\sqrt5-21/x=1/(√5+2)=(√5-2)/(5-4)=√5-2. So x+1x=(5+2)+(5−2)=25.x+\dfrac1x=(\sqrt5+2)+(\sqrt5-2)=2\sqrt5.x+1/x=(√5+2)+(√5-2)=2√5.

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Assertion–Reason questions (1 mark)

Q5Assertion–ReasonModerate1 mark

Assertion (A): 2+3\sqrt2+\sqrt3√2+√3 is an irrational number.

Reason (R): The sum of two irrational numbers is always irrational.

  1. (a)

    Both A and R are true and R is the correct explanation of A

  2. (b)

    Both A and R are true but R is not the correct explanation of A

  3. (c)

    A is true but R is false

  4. (d)

    A is false but R is true

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Answer: (c) 2+3\sqrt2+\sqrt3√2+√3 is indeed irrational, so A is true. But R is false: the sum of two irrationals need not be irrational, e.g. (2+3)+(2−3)=4(2+\sqrt3)+(2-\sqrt3)=4(2+√3)+(2-√3)=4, which is rational.

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Very short answer questions (2 marks)

Q6Very ShortEasy2 marks

Insert one rational and one irrational number between 13\dfrac131/3 and 12\dfrac121/2.

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As decimals, 13=0.333…\dfrac13=0.333\ldots1/3=0.333… and 12=0.5\dfrac12=0.51/2=0.5.

Rational number: 0.4=250.4=\dfrac{2}{5}0.4=2/5 lies between them.

Irrational number: 0.4040040004…0.4040040004\ldots0.4040040004… (a non-terminating, non-recurring decimal) lies between 0.333…0.333\ldots0.333… and 0.50.50.5, so it is a valid irrational number in the interval.

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Q7Very ShortModerate2 marks

Simplify: 45−320+45\sqrt{45}-3\sqrt{20}+4\sqrt5√45-3√20+4√5.

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Write each surd in terms of 5\sqrt5√5:

45=9×5=35\sqrt{45}=\sqrt{9\times5}=3\sqrt5√45=√9×5=3√5 and 20=4×5=25\sqrt{20}=\sqrt{4\times5}=2\sqrt5√20=√4×5=2√5.

∴ 45−320+45=35−3(25)+45=35−65+45=5.\therefore\ \sqrt{45}-3\sqrt{20}+4\sqrt5=3\sqrt5-3(2\sqrt5)+4\sqrt5=3\sqrt5-6\sqrt5+4\sqrt5=\sqrt5.∴ √45-3√20+4√5=3√5-3(2√5)+4√5=3√5-6√5+4√5=√5.

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Short answer questions (3 marks)

Q8Short AnswerModerate3 marks

Rationalise the denominator and simplify: 3+23−2\dfrac{3+\sqrt2}{3-\sqrt2}(3+√2)/(3-√2).

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Multiply numerator and denominator by the conjugate 3+23+\sqrt23+√2:

3+23−2×3+23+2=(3+2)2(3)2−(2)2.\dfrac{3+\sqrt2}{3-\sqrt2}\times\dfrac{3+\sqrt2}{3+\sqrt2}=\dfrac{(3+\sqrt2)^2}{(3)^2-(\sqrt2)^2}.(3+√2)/(3-√2)× (3+√2)/(3+√2)=((3+√2)²)/((3)²-(√2)²).

Numerator: (3+2)2=9+62+2=11+62(3+\sqrt2)^2=9+6\sqrt2+2=11+6\sqrt2(3+√2)²=9+6√2+2=11+6√2.

Denominator: 9−2=79-2=79-2=7.

∴ 3+23−2=11+627.\therefore\ \dfrac{3+\sqrt2}{3-\sqrt2}=\dfrac{11+6\sqrt2}{7}.∴ (3+√2)/(3-√2)=(11+6√2)/7.

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Q9Short AnswerModerate3 marks

If 7+57−5=a+b35\dfrac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}=a+b\sqrt{35}(√7+√5)/(√7-√5)=a+b√35, where aaa and bbb are rational, find aaa and bbb.

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Multiply the numerator and denominator by the conjugate 7+5\sqrt7+\sqrt5√7+√5:
7+57−5×7+57+5=(7+5)2(7)2−(5)2=7+235+57−5=12+2352=6+35.\dfrac{\sqrt7+\sqrt5}{\sqrt7-\sqrt5}\times\dfrac{\sqrt7+\sqrt5}{\sqrt7+\sqrt5}=\dfrac{(\sqrt7+\sqrt5)^2}{(\sqrt7)^2-(\sqrt5)^2}=\dfrac{7+2\sqrt{35}+5}{7-5}=\dfrac{12+2\sqrt{35}}{2}=6+\sqrt{35}.(√7+√5)/(√7-√5)× (√7+√5)/(√7+√5)=((√7+√5)²)/((√7)²-(√5)²)=(7+2√35+5)/(7-5)=(12+2√35)/2=6+√35.
Comparing 6+356+\sqrt{35}6+√35 with a+b35a+b\sqrt{35}a+b√35: a=6a=6a=6 and b=1b=1b=1.

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Q10Short AnswerHOTS3 marks

Prove that 3\sqrt3√3 is an irrational number. (The same method proves that 2\sqrt2√2 and 5\sqrt5√5 are irrational.)

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Assume, to the contrary, that 3\sqrt3√3 is rational. Then 3=pq\sqrt3=\dfrac{p}{q}√3=p/q where p,qp,qp,q are integers with q≠0q\neq0q≠0 and gcd⁡(p,q)=1\gcd(p,q)=1gcd (p,q)=1 (in lowest terms).

Squaring: 3=p2q2⇒p2=3q23=\dfrac{p^2}{q^2}\Rightarrow p^2=3q^23=p²/q²⇒ p²=3q². So 3∣p23\mid p^23| p², hence 3∣p3\mid p3| p. Let p=3kp=3kp=3k.

Then (3k)2=3q2⇒9k2=3q2⇒q2=3k2(3k)^2=3q^2\Rightarrow 9k^2=3q^2\Rightarrow q^2=3k^2(3k)²=3q²⇒ 9k²=3q²⇒ q²=3k², so 3∣q23\mid q^23| q², hence 3∣q3\mid q3| q.

Thus 333 divides both ppp and qqq, contradicting gcd⁡(p,q)=1\gcd(p,q)=1gcd (p,q)=1. Therefore our assumption is wrong and 3\sqrt3√3 is irrational.

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Long answer questions (5 marks)

Q11Long AnswerHOTS5 marks

If x=5+35−3x=\dfrac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}x=(√5+√3)/(√5-√3) and y=5−35+3y=\dfrac{\sqrt5-\sqrt3}{\sqrt5+\sqrt3}y=(√5-√3)/(√5+√3), find the value of x2+y2+xyx^2+y^2+xyx²+y²+xy.

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First rationalise xxx:

x=5+35−3×5+35+3=(5+3)25−3=5+215+32=8+2152=4+15.x=\dfrac{\sqrt5+\sqrt3}{\sqrt5-\sqrt3}\times\dfrac{\sqrt5+\sqrt3}{\sqrt5+\sqrt3}=\dfrac{(\sqrt5+\sqrt3)^2}{5-3}=\dfrac{5+2\sqrt{15}+3}{2}=\dfrac{8+2\sqrt{15}}{2}=4+\sqrt{15}.x=(√5+√3)/(√5-√3)× (√5+√3)/(√5+√3)=((√5+√3)²)/(5-3)=(5+2√15+3)/2=(8+2√15)/2=4+√15.

Similarly y=4−15y=4-\sqrt{15}y=4-√15.

Now x+y=(4+15)+(4−15)=8x+y=(4+\sqrt{15})+(4-\sqrt{15})=8x+y=(4+√15)+(4-√15)=8 and xy=(4+15)(4−15)=16−15=1xy=(4+\sqrt{15})(4-\sqrt{15})=16-15=1xy=(4+√15)(4-√15)=16-15=1.

Use x2+y2+xy=(x+y)2−xyx^2+y^2+xy=(x+y)^2-xyx²+y²+xy=(x+y)²-xy:

x2+y2+xy=(8)2−1=64−1=63.x^2+y^2+xy=(8)^2-1=64-1=63.x²+y²+xy=(8)²-1=64-1=63.

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Q12Long AnswerModerate5 marks

Represent 3\sqrt{3}√3 on the number line using a geometric construction, describing the steps.

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Construction steps:

  1. Draw a number line and mark OOO at 000 and AAA at 111, so OA=1OA=1OA=1 unit.

  2. At AAA, draw AB⊥OAAB\perp OAAB⊥ OA with AB=1AB=1AB=1 unit. Then OB=12+12=2OB=\sqrt{1^2+1^2}=\sqrt2OB=√(1²+1²)=√2 (by Pythagoras).

  3. At BBB, draw BC⊥OBBC\perp OBBC⊥ OB with BC=1BC=1BC=1 unit. Then OC=(2)2+12=2+1=3OC=\sqrt{(\sqrt2)^2+1^2}=\sqrt{2+1}=\sqrt3OC=√((√2)²+1²)=√(2+1)=√3.

  4. With centre OOO and radius OCOCOC, draw an arc cutting the number line at point PPP. Then OP=3OP=\sqrt3OP=√3, so PPP represents 3\sqrt3√3.

ICSE Class 9 Maths — Rational and Irrational Numbers: Represent \sqrt{3} on the number line using a geometric construction, describing the steps.
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Case study questions (4 marks)

Q13Case-basedModerate4 marks

A carpenter is cutting square tiles. He calculates side lengths that come out as surds and must simplify them.

He records three lengths (in cm): 50\sqrt{50}√50, 72\sqrt{72}√72 and 12\dfrac{1}{\sqrt2}1/√2.

(i) Simplify 50\sqrt{50}√50 in the form a2a\sqrt2a√2.

(ii) Simplify 72\sqrt{72}√72 in the form b2b\sqrt2b√2.

(iii) Rationalise 12\dfrac{1}{\sqrt2}1/√2.

(iv) Find 50+72\sqrt{50}+\sqrt{72}√50+√72 in simplest surd form.

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(i) 50=25×2=52\sqrt{50}=\sqrt{25\times2}=5\sqrt2√50=√25×2=5√2 cm.

(ii) 72=36×2=62\sqrt{72}=\sqrt{36\times2}=6\sqrt2√72=√36×2=6√2 cm.

(iii) 12=12×22=22\dfrac{1}{\sqrt2}=\dfrac{1}{\sqrt2}\times\dfrac{\sqrt2}{\sqrt2}=\dfrac{\sqrt2}{2}1/√2=1/√2× √2/√2=√2/2 cm.

(iv) 50+72=52+62=112\sqrt{50}+\sqrt{72}=5\sqrt2+6\sqrt2=11\sqrt2√50+√72=5√2+6√2=11√2 cm.

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All ICSE Class 9 Maths Chapters

Rational and Irrational NumbersCompound Interest (Without Using Formula)Compound Interest (Using Formula)ExpansionsFactorisationSimultaneous (Linear) EquationsIndices (Exponents)LogarithmsTriangles (Congruency in Triangles)Isosceles TrianglesInequalitiesMid-point and Its Converse (Including Intercept Theorem)Pythagoras TheoremRectilinear Figures (Quadrilaterals)Construction of PolygonsArea TheoremsCircleStatistics (Classification of Data, Tabulation)Mean and MedianArea and Perimeter of Plane FiguresSolids (Surface Area and Volume of 3-D Solids)Trigonometrical RatiosTrigonometrical Ratios of Standard AnglesSolution of Right TrianglesComplementary AnglesCo-ordinate GeometryGraphical Solution (Simultaneous Linear Equations, Graphically)Distance Formula

Frequently asked questions

  • What are the important questions in Rational and Irrational Numbers ICSE Class 9?
    Identifying rational and irrational numbers, simplifying and comparing surds, rationalising denominators, inserting rational and irrational numbers between two numbers, proving 3\sqrt3√3 irrational, and showing 3\sqrt3√3 on the number line. This page on Rational and Irrational Numbers has 13 of them, each with a worked answer.
  • How do you rationalise the denominator of (3 + √2)/(3 − √2)?
    Multiply the numerator and denominator by the conjugate 3+23+\sqrt23+√2. The denominator becomes 9−2=79-2=79-2=7 and the numerator (3+2)2=11+62(3+\sqrt2)^2=11+6\sqrt2(3+√2)²=11+6√2, so the answer is 11+627\dfrac{11+6\sqrt2}{7}(11+6√2)/7. The conjugate works because it turns the denominator into a difference of squares with no surd left.
  • How do you compare surds like 2√3, 3√2, √15 and √17?
    Write each as a single square root: 23=122\sqrt3=\sqrt{12}2√3=√12 and 32=183\sqrt2=\sqrt{18}3√2=√18. Then compare the numbers under the root: 12<15<17<1812<15<17<1812<15<17<18, so 323\sqrt23√2 is the greatest and 232\sqrt32√3 the smallest.
  • How do you simplify √45 − 3√20 + 4√5?
    Take out square factors so every term is a multiple of 5\sqrt5√5: 45=35\sqrt{45}=3\sqrt5√45=3√5 and 320=653\sqrt{20}=6\sqrt53√20=6√5. Then 35−65+45=53\sqrt5-6\sqrt5+4\sqrt5=\sqrt53√5-6√5+4√5=√5. Only like surds can be added or subtracted.
  • What is the first chapter of ICSE Class 9 Maths?
    Rational and Irrational Numbers is Chapter 1, followed by Compound Interest (Without Using Formula) in Chapter 2. Indices and Logarithms (Chapters 7 and 8) also use the surd skills learnt here.
  • How do I find a and b in questions like (√7 + √5)/(√7 − √5) = a + b√35?
    Rationalise, then compare the rational and irrational parts. Multiplying by 7+5\sqrt7+\sqrt5√7+√5 gives 12+2352=6+35\dfrac{12+2\sqrt{35}}{2}=6+\sqrt{35}(12+2√35)/2=6+√35, so a=6a=6a=6 and b=1b=1b=1. This compare-the-parts step is what the 3-mark question tests.
  • Can I download ICSE Rational and Irrational Numbers questions?
    Yes. Enter your email in the free PDF box on this page to open all 13 surd and number-system questions with full working as a printable worksheet.

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