Class 9Maths · Number systemsFull chapter

Number Systems

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

Rational and Irrational Numbers

Quick answer A rational number can always be written as p/q (q ≠ 0); an irrational number cannot, and its decimal expansion never terminates or repeats.

A number is called a rational number if it can be written in the form p/q, where p and q are integers and q ≠ 0. Examples: 3/4, −5, 0 and 22/7 are all rational numbers. Every integer is rational because any integer n can be written as n/1.

A number that cannot be written in the form p/q (p, q integers, q ≠ 0) is called an irrational number. Examples: √2, √3, √5 and π are irrational.

Every rational number has a decimal expansion that either terminates (ends), such as 1/8 = 0.125, or is non-terminating recurring (a block of digits repeats forever), such as 1/3 = 0.333... Every irrational number has a decimal expansion that is non-terminating and non-recurring — it goes on forever with no repeating pattern, such as √2 = 1.41421356...

Worked Example: Insert a rational number between 1/4 and 1/2.

  • Write both numbers with the same denominator: 1/4 = 2/8 and 1/2 = 4/8.
  • Any fraction between 2/8 and 4/8 works, for instance 3/8.
  • So 3/8 is a rational number between 1/4 and 1/2 (infinitely many such numbers exist).
Rational number form p/q, where p, q are integers and q ≠ 0
Terminating decimal 1/8 = 0.125 When p/q is written in lowest terms, the decimal terminates exactly when q's only prime factors are 2 and/or 5
Non-terminating recurring decimal 1/3 = 0.333... Rational — repeating block '3'
Non-terminating non-recurring decimal √2 = 1.41421356... Irrational — no repeating pattern
Remember
  • Rational numbers can be written as p/q (q ≠ 0); irrational numbers cannot.
  • Terminating or non-terminating recurring decimals mean the number is rational; non-terminating non-recurring decimals mean it is irrational.
  • Between any two rational numbers there are infinitely many rational numbers (and infinitely many irrational numbers too).
  • √p is irrational whenever p is a positive integer that is not a perfect square.
  • Rational numbers and irrational numbers together make up the real numbers.

Real Numbers and Their Representation on the Number Line

Quick answer Every real number corresponds to a unique point on the number line, and it can be located precisely using geometric construction and successive magnification.

Rational numbers and irrational numbers together are called real numbers. Every real number, rational or irrational, corresponds to exactly one point on the number line, and every point on the number line corresponds to exactly one real number.

Locating √2 on the number line: Mark point O at 0 and point A at 1. At A, draw AB perpendicular to the number line with AB = 1 unit, and join OB. By the Pythagoras theorem, OB² = OA² + AB² = 1² + 1² = 2, so OB = √2. With O as centre and OB as radius, draw an arc cutting the number line at point P. Then OP = OB = √2, so P represents √2.

This idea extends to locate √x for any positive real number x:

  1. Mark a point A on the number line and mark B such that AB = x units.
  2. From B, mark a point C such that BC = 1 unit.
  3. Find the midpoint O of AC and draw a semicircle with centre O and radius OC.
  4. Draw a line through B perpendicular to AC, meeting the semicircle at D. Then BD = √x.
  5. With B as centre and BD as radius, draw an arc cutting the number line at E. The point E represents √x.

Successive magnification is the process of visualising a decimal number on the number line by repeatedly zooming into smaller and smaller intervals, as if viewed through a magnifying glass. For example, to locate 5.37: first zoom into the portion between 5 and 6; divide it into 10 equal parts and zoom into the part between 5.3 and 5.4; divide that part into 10 equal parts again to pinpoint 5.37. Repeating this process any number of times lets us locate any real number, including irrational numbers, as precisely as we like.

Pythagoras for √2 OB² = OA² + AB² = 1² + 1² = 2 ⇒ OB = √2
General square-root construction BD = √x AB = x, BC = 1, O = midpoint of AC, BD ⊥ AC
Remember
  • Real numbers = rational numbers together with irrational numbers; each corresponds to a unique point on the number line.
  • √2 is located using a right triangle with legs 1 and 1, via the Pythagoras theorem.
  • √x, for any positive real x, is located using the construction with AB = x, BC = 1, and BD ⊥ AC.
  • Successive magnification zooms into smaller and smaller subdivisions of the number line to pinpoint a decimal number.
  • This process can locate any real number, rational or irrational, to as many decimal places as needed.

Operations on Real Numbers

Quick answer Combining rational and irrational numbers follows fixed rules, and five key identities help simplify and expand expressions containing square roots.

When rational and irrational numbers are combined by addition, subtraction, multiplication and division, the following results hold:

  • Rational ± Irrational = Irrational (e.g., 3 + √2 is irrational).
  • Non-zero Rational × Irrational = Irrational, and non-zero Rational ÷ Irrational = Irrational (e.g., 5√3 is irrational).
  • Irrational ± Irrational or Irrational × Irrational can be either rational or irrational depending on the numbers involved. For example, √2 + (−√2) = 0 (rational), but √2 + √3 is irrational; also √5 × √5 = 5 (rational), but √5 × √3 = √15 (irrational).

For positive real numbers a and b, the following identities are very useful for simplifying expressions with square roots:

  1. √a × √b = √(ab)
  2. √a ÷ √b = √(a/b)
  3. (√a + √b)(√a − √b) = a − b
  4. (a + √b)(a − √b) = a² − b
  5. (√a + √b)² = a + 2√(ab) + b

Worked Example: Simplify (5 + √7)(5 − √7).

  • Using identity (a + √b)(a − √b) = a² − b with a = 5, b = 7:
  • (5 + √7)(5 − √7) = 5² − 7 = 25 − 7 = 18.

Worked Example: Simplify (√3 + √7)².

  • Using (√a + √b)² = a + 2√(ab) + b with a = 3, b = 7:
  • (√3 + √7)² = 3 + 2√21 + 7 = 10 + 2√21.
Identity I √a × √b = √(ab)
Identity II √a ÷ √b = √(a/b)
Identity III (√a + √b)(√a − √b) = a − b
Identity IV (a + √b)(a − √b) = a² − b
Identity V (√a + √b)² = a + 2√(ab) + b
Remember
  • Rational + Irrational is always irrational; non-zero Rational × Irrational is always irrational.
  • Irrational ± Irrational and Irrational × Irrational can be rational or irrational depending on the numbers.
  • Five key identities let us multiply and expand expressions containing square roots quickly.
  • These identities hold for positive real numbers a and b.

Laws of Exponents for Real Numbers

Quick answer For a > 0, a to the power p/q means the positive q-th root of a to the power p; the familiar laws of exponents extend to these rational powers.

For a real number a > 0 and positive integers p, q, the expression ap/q is defined as the positive q-th root of ap: ap/q = (ap)1/q. In particular, a1/n is the positive n-th root of a, so a1/2 = √a and a1/3 is the cube root of a.

The familiar laws of exponents for integers extend to rational exponents, for a, b > 0:

  • am × an = am+n
  • am ÷ an = am−n
  • (am)n = amn
  • am × bm = (ab)m
  • a0 = 1

Worked Example: Simplify 22/3 × 21/3.

  • Using am × an = am+n: 22/3 × 21/3 = 21 = 2.

Worked Example: Evaluate 64−2/3.

  • 641/3 = 4, since 4³ = 64.
  • So 642/3 = (641/3)² = 4² = 16.
  • Therefore 64−2/3 = 1/642/3 = 1/16.
Definition of rational exponent a^(p/q) = (a^p)^(1/q), a > 0
Product law a^m × a^n = a^(m+n)
Quotient law a^m ÷ a^n = a^(m−n)
Power of a power (a^m)^n = a^(mn)
Negative exponent a^(−m) = 1/a^m
Remember
  • a^(1/n) means the positive n-th root of a, for a > 0.
  • All five laws of exponents for integers continue to hold for rational exponents.
  • A negative exponent means take the reciprocal: a^(−m) = 1/a^m.
  • These laws apply directly when the base a (and b) is a positive real number.

Rationalisation of the Denominator

Quick answer Rationalising means multiplying numerator and denominator by a suitable factor to remove surds from the denominator.

Rationalising the denominator means converting a fraction whose denominator contains a surd (an irrational square root) into an equivalent fraction with a rational denominator, by multiplying the numerator and denominator by a suitable expression called the rationalising factor.

  • The rationalising factor of √a is √a, since √a × √a = a.
  • The rationalising factor of a + √b is a − √b (its conjugate), since (a + √b)(a − √b) = a² − b.
  • The rationalising factor of √a + √b is √a − √b, since (√a + √b)(√a − √b) = a − b.

Worked Example: Rationalise the denominator of 1/(3 + √2).

  • Multiply numerator and denominator by the conjugate (3 − √2):
  • 1/(3 + √2) × (3 − √2)/(3 − √2) = (3 − √2)/(3² − (√2)²) = (3 − √2)/(9 − 2) = (3 − √2)/7.

Worked Example: Rationalise the denominator of 1/(√7 − √5).

  • Multiply numerator and denominator by (√7 + √5):
  • 1/(√7 − √5) × (√7 + √5)/(√7 + √5) = (√7 + √5)/((√7)² − (√5)²) = (√7 + √5)/(7 − 5) = (√7 + √5)/2.
Rationalising √a 1/√a × √a/√a = √a/a
Rationalising a + √b 1/(a+√b) × (a−√b)/(a−√b) = (a−√b)/(a²−b)
Rationalising √a + √b 1/(√a+√b) × (√a−√b)/(√a−√b) = (√a−√b)/(a−b)
Remember
  • Rationalising removes surds from the denominator, making the number easier to compare, add or approximate.
  • To rationalise 1/√a, multiply top and bottom by √a.
  • To rationalise 1/(a ± √b), multiply top and bottom by the conjugate (a ∓ √b).
  • To rationalise 1/(√a ± √b), multiply top and bottom by (√a ∓ √b).
  • The final simplified answer is usually written in the form p + q√r.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

p/q, where p, q are integers and q ≠ 0
Rational number form
1/8 = 0.125
Terminating decimal
1/3 = 0.333...
Non-terminating recurring decimal
√2 = 1.41421356...
Non-terminating non-recurring decimal
OB² = OA² + AB² = 1² + 1² = 2 ⇒ OB = √2
Pythagoras for √2
BD = √x
General square-root construction
√a × √b = √(ab)
Identity I
√a ÷ √b = √(a/b)
Identity II
(√a + √b)(√a − √b) = a − b
Identity III
(a + √b)(a − √b) = a² − b
Identity IV
(√a + √b)² = a + 2√(ab) + b
Identity V
a^(p/q) = (a^p)^(1/q), a > 0
Definition of rational exponent
a^m × a^n = a^(m+n)
Product law
a^m ÷ a^n = a^(m−n)
Quotient law
(a^m)^n = a^(mn)
Power of a power
a^(−m) = 1/a^m
Negative exponent
1/√a × √a/√a = √a/a
Rationalising √a
1/(a+√b) × (a−√b)/(a−√b) = (a−√b)/(a²−b)
Rationalising a + √b
1/(√a+√b) × (√a−√b)/(√a−√b) = (√a−√b)/(a−b)
Rationalising √a + √b

Test yourself

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0 correct · 0/12 answered
Q1 Rational & Irrational Numbers easy

Which of the following is an irrational number?

Q2 Decimal Expansions easy

What type of decimal expansion does 3/8 have?

Q3 Operations on Real Numbers medium

Simplify √8 × √2.

Q4 Laws of Exponents easy

Simplify: 5^(1/2) × 5^(1/2).

Q5 Rationalisation medium

Rationalise the denominator of 1/√5.

Q6 Number Line Representation medium

In the general construction for √x, points A, B, C lie on a line with AB = x and BC = 1; a perpendicular at B meets a semicircle (on AC as diameter) at D. What does BD equal?

Q7 Laws of Exponents medium

Evaluate 64^(2/3).

Q8 Density of Rationals medium

Which statement about rational numbers is true?

Q9 Rationalisation easy

What is the rationalising factor of √7 − √3?

Q10 Operations on Real Numbers easy

Simplify: (√3 + √2)(√3 − √2).

Q11 Rational & Irrational Numbers easy

Which of the following statements is true?

Q12 Decimal Expansions hard

What kind of number is 0.101001000100001... (where the number of zeros between successive 1s keeps increasing)?

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?Rational Numbers

Yes, zero is a rational number.

A number is rational if it can be written as p/q, where p and q are integers and q ≠ 0.

  • We can write 0 as 0/1, 0/2, 0/(−7), and so on.
  • Here p = 0 (an integer) and q can be any non-zero integer.
  • Since this satisfies the definition, 0 is a rational number.
2 Find five rational numbers between 3/5 and 4/5.Rational Numbers

Step 1: Write 3/5 and 4/5 with a common denominator large enough to fit 5 numbers in between it. Multiply numerator and denominator of both by 6:

3/5 = (3×6)/(5×6) = 18/30, and 4/5 = (4×6)/(5×6) = 24/30.

Step 2: Any five fractions with denominator 30 lying between 18/30 and 24/30 are the required numbers:

19/30, 20/30, 21/30, 22/30, 23/30.

So, five rational numbers between 3/5 and 4/5 are 19/30, 20/30, 21/30, 22/30 and 23/30.

3 Locate √5 on the number line.Number Line Representation

Step 1: Draw a number line and mark O at 0 and A at 2, so OA = 2 units.

Step 2: At A, draw AB perpendicular to the number line such that AB = 1 unit.

Step 3: Join OB. By the Pythagoras theorem:

OB² = OA² + AB² = 2² + 1² = 4 + 1 = 5, so OB = √5.

Step 4: With O as centre and radius OB, draw an arc cutting the number line at a point P (on the positive side).

Since OP = OB = √5, the point P represents √5 on the number line.

4 Simplify: (√5 + √2)².Operations on Real Numbers

Using the identity (√a + √b)² = a + 2√(ab) + b, with a = 5 and b = 2:

(√5 + √2)² = (√5)² + 2×√5×√2 + (√2)²

= 5 + 2√10 + 2

= 7 + 2√10.

5 Rationalise the denominator of 1/(7 + 3√2).Rationalisation

Multiply the numerator and denominator by the conjugate of the denominator, (7 − 3√2):

1/(7 + 3√2) = [1 × (7 − 3√2)] / [(7 + 3√2)(7 − 3√2)]

The denominator becomes 7² − (3√2)² = 49 − 9×2 = 49 − 18 = 31.

So, 1/(7 + 3√2) = (7 − 3√2)/31.

6 Simplify (2 + √3)/(2 − √3) by rationalising the denominator.Rationalisation

Multiply numerator and denominator by the conjugate (2 + √3):

(2 + √3)/(2 − √3) × (2 + √3)/(2 + √3) = (2 + √3)² / [(2)² − (√3)²]

Denominator: 4 − 3 = 1.

Numerator: (2 + √3)² = 4 + 4√3 + 3 = 7 + 4√3.

So, (2 + √3)/(2 − √3) = (7 + 4√3)/1 = 7 + 4√3.

Previous-year board questions 4

Q1 Find the value of x, if 2^x = 1/32. CBSE 2023 1 mark

We know that 32 = 2⁵, so 1/32 = 2−5.

So 2x = 2−5.

Since the bases are equal, the exponents must be equal: x = −5.

Q2 Rationalise the denominator of 1/(√7 − √6). CBSE 2022 2 marks

Multiply the numerator and denominator by the conjugate (√7 + √6):

1/(√7 − √6) × (√7 + √6)/(√7 + √6) = (√7 + √6) / [(√7)² − (√6)²]

= (√7 + √6) / (7 − 6) = (√7 + √6)/1 = √7 + √6.

Q3 If x = 3 + 2√2, find the value of √x + 1/√x. CBSE 2019 3 marks

Step 1: Recognise that 3 + 2√2 is a perfect square of a surd expression.

(1 + √2)² = 1² + 2×1×√2 + (√2)² = 1 + 2√2 + 2 = 3 + 2√2.

So x = (1 + √2)², which means √x = 1 + √2.

Step 2: Find 1/√x by rationalising:

1/√x = 1/(1 + √2) × (√2 − 1)/(√2 − 1) = (√2 − 1)/[(√2)² − 1²] = (√2 − 1)/(2 − 1) = √2 − 1.

Step 3: Add the two results:

√x + 1/√x = (1 + √2) + (√2 − 1) = 2√2.

Q4 Simplify by rationalising the denominators: (5 + 3√2)/(5 − 3√2) + (5 − 3√2)/(5 + 3√2). CBSE 2020 5 marks

Let a = 5 + 3√2 and b = 5 − 3√2. We need a/b + b/a = (a² + b²)/(ab).

Step 1: Find ab.

ab = (5 + 3√2)(5 − 3√2) = 5² − (3√2)² = 25 − 18 = 7.

Step 2: Find a + b and use it to get a² + b².

a + b = (5 + 3√2) + (5 − 3√2) = 10.

a² + b² = (a + b)² − 2ab = 10² − 2×7 = 100 − 14 = 86.

Step 3: Combine.

a/b + b/a = (a² + b²)/(ab) = 86/7.

So, the simplified value is 86/7.

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