Number System follows how numbers grew to meet human needs: natural numbers for counting, zero and negative integers from Brahmagupta, rational numbers for parts of a whole, and irrational numbers such as √2 and π that fill the gaps rationals leave. Together they form the real number line. Students learn rational arithmetic, number line placement, density, the proof that √2 is irrational, and how decimal expansions separate rationals from irrationals. The chapter carries 7 marks in Class 9 and is the base for Class 10 Real Numbers.
Key Concepts
1. Natural Numbers and the Need to Count
Numbers began with a practical need: keeping count. Picture a herder in a settlement on the banks of the Saraswati river thousands of years ago. Every morning the cattle go out to graze and every evening they come back. With no number words and no written symbols, the herder still needs to know if a calf has wandered off.
The answer is one-to-one correspondence. For every cow that leaves, the herder drops one pebble into a clay pot. For every cow that returns, one pebble comes out. An empty pot in the evening means the herd is safe. Pebbles left over mean cows are missing. Matching one object to exactly one other object in this way is the idea behind the natural numbers:
N = {1, 2, 3, 4, …}
Evidence written in bone. The earliest physical records of counting are tally marks carved into bone.
- The Lebombo Bone, found in the Lebombo Mountains between South Africa and Swaziland, is about 35,000 years old. It has 29 deliberately carved notches of uniform size. It may have been a lunar phase counter.
- The Ishango bone, found near the headwaters of the Nile in the Democratic Republic of Congo, dates to around 20,000 BCE. It has three columns of notches. One column groups them as 11, 13, 17 and 19, which are exactly the prime numbers between 10 and 20. Another column seems to show doubling (multiplication by 2).
The Indian context. The Vedas gave names to every power of 10 up to 1012, which was called parārdha. In the Lalitavistara (4th century BCE), the Buddha describes names up to 1053, called tallakṣhaṇa. Expressing quantities as powers of 10 in the Ṛigveda set the stage for the place-value decimal system developed in India, the system the whole world uses today. That system also paved the way for zero.
Closure. A set is closed under an operation if doing that operation on any two members always gives a member of the same set.
| Operation on natural numbers | Closed? | Reason |
|---|---|---|
| Addition | Yes | The sum of two natural numbers is a natural number: 5 + 4 = 9 |
| Multiplication | Yes | 6 × 7 = 42, again a natural number |
| Subtraction | No | 4 - 9 = -5 and 5 - 5 = 0, neither is a natural number |
| Division | No | 3 ÷ 4 = 3/4, which is a fraction |
One counter-example is enough to show that a set is not closed. This gap in subtraction led to zero and the negative numbers.
2. Zero (Śhūnya): When Nothing Became a Number
For a very long time the number line started at 1. If you had five apples and gave all five away, there was no number for what you had left. The Babylonians and the Mayans used placeholder symbols to mark an empty column in a number, but they did not treat “nothing” as a number you could add, subtract or multiply.
From philosophy to mathematics. In the Upanishads and in Buddhist literature, śhūnyatā (emptiness) described the still mind sought in meditation. Because Indian thinkers valued emptiness, the idea moved into mathematics through Āryabhaṭa and Brahmagupta.
The symbol. The Bakhśhālī Manuscript, dated to the early centuries CE, uses a bold dot called a bindu to stand for zero.
The rules. A symbol becomes a number only when it has rules. In the Brāhmasphuṭasiddhānta (628 CE), Brahmagupta defined zero as the result of subtracting a number from itself: a - a = 0. He then gave the laws of arithmetic with zero:
| Brahmagupta’s rule | In symbols | Example |
|---|---|---|
| Adding zero leaves a number unchanged | a + 0 = a | 17 + 0 = 17 |
| Subtracting zero leaves a number unchanged | a - 0 = a | 9 - 0 = 9 |
| Any number multiplied by zero gives zero | a × 0 = 0 | 250 × 0 = 0 |
Adding zero to the natural numbers gives the whole numbers {0, 1, 2, 3, …}. Division by zero is never allowed: there is no number that, multiplied by 0, gives a non-zero answer. You will see this rule again in the definition of a rational number.
3. Integers: Fortunes, Debts and the Sign Rules
Since 5 - 5 = 0, Brahmagupta asked what 3 - 5 should be, and found the answer in trade. He described two states:
- Fortunes (Dhana): positive numbers, standing for wealth or assets.
- Debts (Ṛiṇa): negative numbers, standing for money owed.
Moving to the left of zero on the number line, he formally introduced negative numbers. Positive natural numbers, their negatives and zero together form the integers:
Z = {…, -3, -2, -1, 0, 1, 2, 3, …}
The letter Z comes from the German word Zahlen, meaning numbers. On the number line, debts lie to the left of zero and fortunes to the right, with each integer the same distance from the next.
Brahmagupta’s rules for integers (we still use them exactly as he stated them over 1,300 years ago):
| Rule in words | Example | Real-life meaning |
|---|---|---|
| A fortune plus a fortune is a fortune | 5 + 4 = 9 | Earnings add up |
| A debt plus a debt is a debt | (-5) + (-4) = -9 | Owe ₹5, borrow ₹4 more, you owe ₹9 |
| Fortune minus zero is a fortune; debt minus zero is a debt | 7 - 0 = 7; -6 - 0 = -6 | Nothing taken away, nothing changes |
| Product of a debt and a fortune is a debt | (-3) × 4 = -12 | Four debts of ₹3 make a debt of ₹12 |
| Product of two debts is a fortune | (-3) × (-4) = 12 | Removing four debts of ₹3 leaves you ₹12 richer |
Why negative × negative is positive. A negative number is a debt. Multiplying by a negative number means removing that debt. If someone takes away four of your debts of ₹3 each, you are ₹12 better off. So (-3) × (-4) = +12.
Subtracting a negative. 10 - (-5) = 15. If you have ₹10 and someone cancels a debt of ₹5 you owed, your position improves by ₹5. Subtracting a debt is the same as adding a fortune: a - (-b) = a + b.
Division follows the same sign pattern. (-20) ÷ 4 = -5 and (-20) ÷ (-4) = 5. Like signs give a positive answer; unlike signs give a negative answer.
Integers are closed under addition, subtraction and multiplication. They are not closed under division, because 3 ÷ 4 is not an integer. That gap leads to fractions.
4. Rational Numbers and Their Arithmetic
Measuring needs parts of a whole: half a cup of ghee, a field shared among three children. Numbers that stand for parts of a whole are fractions. Every positive fraction has an additive inverse, a negative fraction: -3/4 for 3/4, -19/7 for 19/7. The minus sign may sit in front, on the numerator or on the denominator: -1/5 = (-1)/5 = 1/(-5).
All integers and all fractions, positive and negative, together form the rational numbers, written Q (for quotient).
Definition: A rational number is any number that can be expressed in the form p/q, where p and q are integers and q ≠ 0.
The reason for q ≠ 0. p/q means the number which multiplied by q gives p. If q = 0, that number would have to satisfy (number) × 0 = p. By Brahmagupta’s rule, anything times 0 is 0, so no number works when p ≠ 0, and every number works when p = 0. Either way p/0 has no single value, so it is excluded.
Three important observations from NCERT:
- Every integer is rational. 5 = 5/1 and -10 = -10/1. So Q contains the natural numbers, the whole numbers and the integers.
- A rational number has many forms. -1/3 = -2/6 = -3/9 = -10/30 = -2026/6078. These are equivalent rational numbers. You may divide the numerator and denominator by any common factor: 12/30 = 2/5 (dividing both by 6).
- Standard (lowest) form. When we speak of p/q or mark it on the number line, we assume q ≠ 0 and that p and q have no common factor other than 1, that is, p and q are co-prime. Among 1/2, 2/4, 3/6, 6/12, …, 1013/2026, we choose 1/2 to represent all of them.
Brahmagupta’s laws for rational numbers:
| Law | Rule | Condition |
|---|---|---|
| Equality | a/b = c/d if ad = bc | b, d ≠ 0 |
| Addition | a/b + c/b = (a + c)/b | Write both with the same denominator first |
| Subtraction | a/b - c/b = (a - c)/b | Same denominator first |
| Multiplication | (a/b) × (c/d) = ac/bd | b ≠ 0, d ≠ 0 |
| Division | (a/b) ÷ (c/d) = (a/b) × (d/c) = ad/bc | b ≠ 0, d ≠ 0, c ≠ 0 |
| Commutative | a/b + c/d = c/d + a/b and (a/b) × (c/d) = (c/d) × (a/b) | All rationals |
| Distributive | p(q + r) = pq + pr | p, q, r rational |
Making denominators equal. To add 7/12 and 5/8, find the LCM of 12 and 8, which is 24. Then 7/12 = 14/24 and 5/8 = 15/24, so the sum is 29/24.
Closure of Q. Adding, subtracting or multiplying two rational numbers always gives a rational number. Dividing one rational number by another also gives a rational number, as long as you do not divide by zero. So Q is closed under all four operations, with the single exception of division by zero.
| Set | + | - | × | ÷ (non-zero divisor) |
|---|---|---|---|---|
| Natural numbers N | Closed | Not closed | Closed | Not closed |
| Integers Z | Closed | Closed | Closed | Not closed |
| Rational numbers Q | Closed | Closed | Closed | Closed |
5. Rational Numbers on the Number Line, Absolute Value and Distance
Integers on the line. Choose a point and call it 0, the origin. One unit to the right is 1, two units is 2, and so on. One unit to the left is -1, two units is -2. Consecutive integers are equally spaced.
Rationals lie between integers. 1/2 is exactly halfway between 0 and 1. -3/4 lies between -1 and 0.
Method to mark p/q (q > 0):
- Divide the unit interval (the gap between two consecutive integers) into q equal parts.
- Move p parts from 0: to the right if the number is positive, to the left if it is negative.
Example: 3/4. Split the gap from 0 to 1 into four equal parts; the marks are 1/4, 2/4 = 1/2 and 3/4. Move three parts to the right of 0.
Fractions greater than 1. 9/4 = 2 1/4, so it lies between 2 and 3. Split the gap from 2 to 3 into four equal parts and move one part to the right of 2.
Absolute value. The absolute value of a rational number x, written |x|, is its distance from 0 on the number line.
- |5/3| = 5/3, |-5/3| = 5/3 and |0| = 0.
- The absolute value of a positive number is the number itself; for a negative number it is its positive value.
- So |x| ≥ 0 for every rational x.
Distance between two numbers. For rationals a and b, the distance between them on the number line is |a - b|. The distance between 3 and -4 is |3 - (-4)| = |7| = 7. The order does not matter: |-4 - 3| = |-7| = 7.
6. Density of Rational Numbers
Between the integers 1 and 2 there is no other integer, but there is the rational 3/2. Between 1 and 3/2 there is 5/4. This never stops. Rational numbers are dense: between any two rational numbers, however close, there is always another rational number.
The averaging method. For rationals a and b with a < b, the average (a + b)/2 is rational and lies strictly between them.
- Why it is rational: a + b is rational (closure under addition), and dividing a rational by the non-zero rational 2 gives a rational (closure under division).
- Why it lies between: since a < b, adding a to both sides gives 2a < a + b, so a < (a + b)/2. Adding b to both sides of a < b gives a + b < 2b, so (a + b)/2 < b.
Example: between 1 and 3/2, the average is (1 + 3/2)/2 = (5/2)/2 = 5/4. Repeat the process with 1 and 5/4 to get 9/8, and so on without end. So there are infinitely many rational numbers between any two rational numbers.
The common denominator method (fast when you need many numbers). To find 5 rationals between 2/5 and 3/5, rewrite both with a larger denominator so that the numerators are far apart. Multiply numerator and denominator by 6: 2/5 = 12/30 and 3/5 = 18/30. Now 13/30, 14/30, 15/30, 16/30 and 17/30 all lie between them. To find n numbers this way, the two numerators must differ by at least n + 1.
7. Irrational Numbers: √2, √3 and π
Take a square whose sides are each 1 unit long. By the Baudhāyana-Pythagoras Theorem the diagonal d satisfies 12 + 12 = d2, so d2 = 2 and d = √2. Baudhāyana met such lengths while writing his Śhulbasūtra, a manual for building fire altars, around 800 BCE. The ancient Greeks met the same problem a few centuries later.
Definition: Numbers on the number line that cannot be expressed as a ratio of integers p/q (q ≠ 0) are called irrational numbers. The set is often written I.
Proof that √2 is irrational (proof by contradiction). The first proof is credited to Hippasus, a member of the Pythagorean school (c. 400 BCE, as NCERT dates it). In a proof by contradiction you assume the opposite of what you want to prove and show that the assumption leads to something impossible.
- Assume √2 is rational. Then √2 = p/q, where p and q are integers, q ≠ 0, and p and q are co-prime (the fraction is in lowest form).
- Square both sides: 2 = p2/q2.
- Multiply by q2: 2q2 = p2.
- p is even. p2 equals 2 times an integer, so p2 is even. If the square of an integer is even, the integer itself is even. So p is even; write p = 2k for some integer k.
- Substitute: 2q2 = (2k)2 = 4k2.
- Divide by 2: q2 = 2k2.
- q is even. q2 is 2 times an integer, so q2 is even and q is even.
- Contradiction. p and q are both even, so they share the factor 2. This contradicts step 1, where p and q had no common factor other than 1.
Every step after the assumption is correct, so the assumption must be wrong. ∴ √2 is irrational.
Why an even square means an even number: if p = 2m + 1 were odd, p2 = 4m2 + 4m + 1 would be odd. The same argument with the prime 3 in place of 2 proves √3 irrational, and it extends to √5, √7 and √10. It fails for √4 and √9, which equal 2 and 3.
Constructing √2 on the number line.
- On the number line mark O at 0 and A at 1, so OA = 1 unit. Draw a line perpendicular to OA through A.
- On this perpendicular mark B with AB = 1 unit. Join OB. Triangle OAB is right-angled at A, so OB = √(12 + 12) = √2 units.
- With O as centre and radius OB, draw an arc that cuts the number line at P. Then OP = OB = √2, so P represents √2. Drawing the arc on the other side of O gives -√2.
The square root spiral. Repeat the idea. At B draw a perpendicular to OB and mark C with BC = 1 unit. Then OC = √((√2)2 + 12) = √3. At C draw a perpendicular to OC, mark D with CD = 1, and OD = √4 = 2. Continuing gives hypotenuses √2, √3, √4, √5, √6, …, and in general the n-th triangle has hypotenuse √(n + 1). The shape formed is the square root spiral. Any √n can be transferred to the number line with a compass centred at O.
The story of π. π is the ratio of a circle’s circumference to its diameter. Āryabhaṭa (499 CE) gave the approximation 3927/1250 = 3.1416 and called it āsanna (approximate), hinting that no exact fraction could be found. π was proved irrational by Lambert in 1761. In the 14th century Mādhava of Sangamagrama, founder of the Kerala School of Mathematics, found an exact formula as an infinite sum:
π = 4 × (1 - 1/3 + 1/5 - 1/7 + …)
Adding infinitely many terms means finding the value that the running total gets closer and closer to.
8. Real Numbers and Decimal Expansions
The dense rationals together with the irrationals fill the number line completely. This unbroken line is the set of real numbers R. Every real number is a point on the line, and every point on the line is a real number.
How the sets fit together: N is contained in Z, Z is contained in Q, and Q together with the irrationals I makes up R. The irrationals are separate from Q: no number is both rational and irrational.
Decimal expansion of a rational number. Divide the numerator by the denominator. Exactly one of two things happens:
- It terminates: the remainder becomes 0 and the decimal stops. 3/8 = 0.375.
- It repeats: the remainder never becomes 0, and a block of digits loops forever. 5/11 = 0.454545…, written with a bar over the repeating block 45.
Why a rational decimal must repeat if it does not stop. When dividing by 7, the possible non-zero remainders are only 1, 2, 3, 4, 5 and 6. Among any seven successive remainders, two must be the same, and from then on the division repeats itself. So the repeating block of 1/7 is at most 6 digits long. In fact 1/7 = 0.142857142857…, with block 142857.
Predicting the type without dividing. Write p/q in lowest form and factorise q.
- If the prime factors of q are only 2, only 5, or only 2 and 5, the decimal terminates. The reason: you can multiply numerator and denominator by a suitable number to make the denominator a power of 10. For example, 3/20 = 3/(22 × 5) = (3 × 5)/(22 × 52) = 15/100 = 0.15.
- If q has any other prime factor (3, 7, 11, 13, …), the decimal is non-terminating and repeating. 4/15 has 15 = 3 × 5, so it repeats: 4/15 = 0.2666…
- Number of decimal places. If q = 2m × 5n, the decimal ends after the larger of m and n places. 7/40 has 40 = 23 × 5, so it has 3 decimal places: 7/40 = 175/1000 = 0.175.
Converting decimals to p/q.
| Type | Steps | Example |
|---|---|---|
| Terminating | Write over the matching power of 10 and simplify | 0.35 = 35/100 = 7/20 |
| Pure repeating (block starts right after the point) | Let x = the decimal. Multiply by 10n, n = number of repeating digits. Subtract x. Solve. | x = 0.666…; 10x = 6.666…; 9x = 6; x = 2/3 |
| General repeating (some digits before the block) | Let x = the decimal. Multiply by 10m, m = number of non-repeating digits. Multiply that by 10n, n = number of repeating digits. Subtract the two shifted equations. Solve. | x = 0.1666…; 10x = 1.666…; 100x = 16.666…; 90x = 15; x = 1/6 |
Cyclic numbers. The block of 1/7 is 142857. Multiply it by 1 to 6:
142857 × 1 = 142857, × 2 = 285714, × 3 = 428571, × 4 = 571428, × 5 = 714285, × 6 = 857142.
The same six digits appear each time, only rotated in a circle. A number with this property is a cyclic number. It reflects the fact that 2/7, 3/7, …, 6/7 have repeating blocks that are rotations of the block of 1/7.
Decimal expansion of an irrational number. It never ends and never repeats. √2 = 1.4142135623730950488… and π = 3.1415926535897932384…. A decimal such as 1.01001000100001…, where the number of zeros keeps growing, has a clear pattern but no fixed block that repeats, so it is irrational.
0.999… = 1. Let x = 0.999…. Then 10x = 9.999…. Subtracting, 9x = 9, so x = 1. Any terminating decimal also has a form ending in repeating 9s: 2.47 = 2.46999…. So a decimal representation need not be unique.
Formula and Theorem Sheet
| Result | Statement or formula | When to use |
|---|---|---|
| Brahmagupta’s zero rules | a + 0 = a; a - 0 = a; a × 0 = 0; a - a = 0 | Arithmetic with zero |
| Sign rules | (+)(+) = +; (-)(-) = +; (+)(-) = -; a - (-b) = a + b | Integer and rational arithmetic |
| Rational number | p/q, p, q ∈ Z, q ≠ 0 | Classifying numbers |
| Equality test | a/b = c/d exactly when ad = bc | Proving two rationals equal |
| Multiplication | (a/b)(c/d) = ac/bd | b, d ≠ 0 |
| Division | (a/b) ÷ (c/d) = ad/bc | b, c, d ≠ 0 |
| Density | a < (a + b)/2 < b for a < b | Rational between two rationals |
| Irrationality of √2, √3 | Proof by contradiction using co-prime p, q | 3-mark and 5-mark proofs |
| Baudhāyana-Pythagoras | Hypotenuse2 = base2 + perpendicular2 | Constructing √n, square root spiral |
| Terminating test | p/q (lowest form) terminates exactly when q = 2m5n | Predicting decimal type |
| Pure repeating | For x = 0.(block of n digits): (10n - 1)x = the block | 0.abab… = ab/99 |
| General repeating | (10m+n - 10m)x = difference of shifted numbers | 0.1666… = 15/90 |
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Important Definitions
| Term | Definition |
|---|---|
| One-to-one correspondence | Matching each object of one collection with exactly one object of another, as in one pebble for each cow |
| Natural numbers (N) | The counting numbers {1, 2, 3, 4, …} |
| Zero (śhūnya) | The number obtained by subtracting a number from itself, a - a = 0 (Brahmagupta, 628 CE) |
| Integers (Z) | Positive natural numbers, their negatives and zero: {…, -2, -1, 0, 1, 2, …}; Z from German Zahlen |
| Rational number (Q) | A number expressible as p/q where p and q are integers and q ≠ 0 |
| Absolute value |x| | The distance of x from 0 on the number line; always ≥ 0 |
| Density of rationals | Between any two rational numbers there is another rational number, hence infinitely many |
| Irrational number | A number on the number line that cannot be written as p/q with p, q integers and q ≠ 0; its decimal never ends and never repeats |
| Real numbers (R) | All rational and irrational numbers together; they fill the number line with no gaps |
Solved Examples (NCERT-Based)
Example 1: Proving two rationals equal (Exercise Set 3.3 type)
Show that (i) 2/3 = 4/6 and (ii) -3/5 = -6/10.
Solution: Use a/b = c/d exactly when ad = bc.
(i) 2 × 6 = 12 and 3 × 4 = 12. The cross products match, so 2/3 = 4/6.
(ii) (-3) × 10 = -30 and 5 × (-6) = -30. Equal, so -3/5 = -6/10.
Example 2: Adding rationals with different denominators
Find (i) 7/12 + 5/8 and (ii) -4/7 + 3/14.
Solution: (i) LCM of 12 and 8 is 24. 7/12 = 14/24 and 5/8 = 15/24. Sum = 29/24 (lowest form, since 29 is prime and does not divide 24).
(ii) LCM of 7 and 14 is 14. -4/7 = -8/14. Sum = (-8 + 3)/14 = -5/14.
Example 3: Product and quotient
Find (i) 7/11 × 5/8 and (ii) -4/7 ÷ 5/14.
Solution: (i) (7 × 5)/(11 × 8) = 35/88.
(ii) Multiply by the reciprocal: -4/7 × 14/5 = (-4 × 14)/(7 × 5) = -56/35. Divide both by 7: -8/5. Cancelling 14 with 7 before multiplying (14/7 = 2) gives -4 × 2/5 = -8/5 directly.
Example 4: Verifying the distributive law
Show that (1/2 + 3/4) × 8/3 = 1/2 × 8/3 + 3/4 × 8/3.
Solution: LHS: 1/2 + 3/4 = 2/4 + 3/4 = 5/4. Then 5/4 × 8/3 = 40/12 = 10/3.
RHS: 1/2 × 8/3 = 8/6 = 4/3. 3/4 × 8/3 = 24/12 = 2. Sum = 4/3 + 2 = 4/3 + 6/3 = 10/3.
LHS = RHS = 10/3. The distributive law holds.
Example 5: Finding an unknown rational (End-of-Chapter type)
If x/3 + x/5 = 16/15, find the rational number x.
Solution: LCM of 3 and 5 is 15. x/3 + x/5 = 5x/15 + 3x/15 = 8x/15.
So 8x/15 = 16/15. Multiply both sides by 15: 8x = 16, so x = 2.
Example 6: The tailor and the silk (Exercise Set 3.4)
A tailor has 15 3/4 metres of silk. One kurta needs 2 1/4 metres. How many kurtas can he make?
Solution: Convert to improper fractions. 15 3/4 = (60 + 3)/4 = 63/4. 2 1/4 = 9/4.
Number of kurtas = 63/4 ÷ 9/4 = 63/4 × 4/9 = 63/9 = 7 kurtas, with no silk left over.
Example 7: Rational numbers between two rationals
(i) Find three rational numbers between -1/2 and 1/4. (ii) Find five rational numbers between 2/5 and 3/5.
Solution: (i) Averaging: (-1/2 + 1/4)/2 = (-1/4)/2 = -1/8. Average of -1/2 and -1/8: (-4/8 - 1/8)/2 = -5/16. Average of -1/8 and 1/4: (-1/8 + 2/8)/2 = 1/16. So -5/16, -1/8 and 1/16 lie between -1/2 and 1/4.
(ii) Multiply numerator and denominator by 6: 2/5 = 12/30 and 3/5 = 18/30. The numbers 13/30, 14/30, 15/30, 16/30, 17/30 lie between them. In lowest form these are 13/30, 7/15, 1/2, 8/15 and 17/30.
Example 8: Absolute value and distance
Find the distance between (i) -4 and 3, (ii) -3/4 and 5/6 on the number line.
Solution: (i) |3 - (-4)| = |7| = 7 units.
(ii) |-3/4 - 5/6| = |-9/12 - 10/12| = |-19/12| = 19/12 units. Distances are never negative, so the absolute value removes the minus sign.
Example 9: Predicting terminating or repeating (Exercise Set 3.5)
Without long division, decide which of 7/20, 4/15 and 13/250 terminate. Then check.
Solution: All three are in lowest form.
- 20 = 22 × 5. Only 2 and 5, so 7/20 terminates. Check: 7/20 = 35/100 = 0.35.
- 15 = 3 × 5. The factor 3 means 4/15 is non-terminating repeating. Check: 4/15 = 0.2666…
- 250 = 2 × 53. Only 2 and 5, so 13/250 terminates. Check: multiply by 4/4: 52/1000 = 0.052.
Example 10: Converting a general repeating decimal
Express 0.23535… (the block 35 repeats, the 2 does not) in the form p/q.
Solution: Let x = 0.23535…. One non-repeating digit, so multiply by 10: 10x = 2.3535… (equation 1).
Two repeating digits, so multiply equation 1 by 100: 1000x = 235.3535… (equation 2).
Subtract equation 1 from equation 2: 1000x - 10x = 235.3535… - 2.3535…, so 990x = 233.
x = 233/990. 233 is prime and does not divide 990, so this is the lowest form.
Example 11: Classifying numbers (Exercise Set 3.5)
Classify as rational or irrational, and give p/q for the rational ones: (i) √81 (ii) √12 (iii) 0.33333… (iv) 0.123451234512345… (v) 1.01001000100001… (vi) 23.560185612239874790120
Solution:
- √81 = 9 = 9/1. Rational.
- √12 = √(4 × 3) = 2√3. Since √3 is irrational, 2√3 is irrational (if 2√3 = p/q then √3 = p/2q would be rational). Irrational.
- x = 0.333…; 10x = 3.333…; 9x = 3; x = 1/3. Rational.
- Block 12345 repeats (5 digits). 100000x - x = 12345, so x = 12345/99999 = 4115/33333 (dividing by 3). Rational.
- The number of zeros keeps growing, so no fixed block repeats. Irrational.
- The decimal stops, so it is terminating: 23560185612239874790120/1021. Rational.
Example 12: Proving √3 is irrational
Prove that √3 is irrational.
Solution: Suppose √3 is rational. Then √3 = p/q, where p, q are integers, q ≠ 0, and p, q are co-prime.
Squaring: 3 = p2/q2, so 3q2 = p2. Hence 3 divides p2. Since 3 is prime, 3 divides p. Write p = 3k.
Then 3q2 = 9k2, so q2 = 3k2. Hence 3 divides q2, and so 3 divides q.
Now 3 is a common factor of p and q, which contradicts the fact that p and q are co-prime. The assumption was false. ∴ √3 is irrational.
Competency-Based Questions (with answers)
1. Case-based: The Lothal spice trade
A merchant in the port city of Lothal exchanges bags of spices for copper ingots. He receives 15 ingots for every 2 bags. A second merchant pays in debts and fortunes: he takes a loan of ₹850, makes a profit of ₹1,200 the next day, and a loss of ₹450 the following week.
(a) How many ingots does the first merchant get for 12 bags? 12 bags = 6 groups of 2 bags, so 6 × 15 = 90 ingots.
(b) Write the second merchant’s record as one integer expression and find his final position. (-850) + 1200 + (-450) = 350 - 450 = -100. He is in debt by ₹100.
2. Case-based: Night in Ladakh
The temperature in a high-altitude desert in Ladakh is 4 °C at noon. By midnight it has dropped by 15 °C.
(a) Find the midnight temperature. 4 - 15 = -11 °C.
(b) How far apart are the noon and midnight readings on the number line? |4 - (-11)| = 15 degrees, which matches the drop.
(c) Is the set of natural numbers enough to record these temperatures? Justify. No. -11 is a negative integer. The natural numbers are not closed under subtraction (4 - 15 is outside N), so integers are needed.
3. Source-based: The Ishango bone
One column of the Ishango bone (about 20,000 BCE) groups its notches as 11, 13, 17 and 19. Another column appears to show doubling.
(a) What do 11, 13, 17 and 19 have in common? They are all the prime numbers between 10 and 20; each has exactly two factors, 1 and itself.
(b) Give the next three numbers that fit this pattern. 23, 29 and 31, the next three primes.
4. Assertion-Reason
Assertion (A): The decimal expansion of 13/250 terminates. Reason (R): The denominator of a rational number in lowest form that has only 2 and 5 as prime factors can be turned into a power of 10.
Answer: (a) Both A and R are true, and R is the correct explanation of A. 250 = 2 × 53; multiplying by 22/22 gives 52/1000 = 0.052.
5. Assertion-Reason
Assertion (A): Between any two rational numbers there are infinitely many rational numbers. Reason (R): The average of two rational numbers is a rational number lying strictly between them.
Answer: (a) Both A and R are true, and R is the correct explanation of A. Taking averages again and again produces a new rational each time, without end.
6. Assertion-Reason
Assertion (A): 22/7 is an irrational number. Reason (R): π is irrational, and 22/7 is often used for π.
Answer: (d) A is false, R is true. 22/7 is a ratio of two integers, so it is rational. It is only an approximation of π, which Lambert proved irrational in 1761.
7. Data reasoning: Cyclic patterns in 1/13
A student performs long division and gets 1/13 = 0.076923076923…, 2/13 = 0.153846153846…, 3/13 = 0.230769230769… and 4/13 = 0.307692307692….
(a) What is the repeating block of 1/13 and its length? 076923, six digits long.
(b) Which of 2/13, 3/13, 4/13 have blocks that are rotations of 076923? 3/13 (230769) and 4/13 (307692) are rotations. 2/13 has block 153846, which uses different digits, so it belongs to a second cycle.
(c) Why can the block of any fraction with denominator 13 be at most 12 digits long? The only possible non-zero remainders when dividing by 13 are 1 to 12. Among any 13 successive remainders, two must be equal, and the division then loops. So the block has at most 12 digits.
8. Source-based: Mādhava’s series
Mādhava of Sangamagrama wrote π = 4 × (1 - 1/3 + 1/5 - 1/7 + …). Āryabhaṭa earlier gave π ≈ 3927/1250.
(a) Find 4 × (1 - 1/3 + 1/5). 1 - 1/3 + 1/5 = 15/15 - 5/15 + 3/15 = 13/15. Times 4 gives 52/15 ≈ 3.467.
(b) Write 3927/1250 as a decimal and say whether it is exact. 1250 = 2 × 54, so it terminates: 3927/1250 = 31416/10000 = 3.1416. It is an approximation (āsanna); π itself is irrational.
(c) Why did Mādhava need infinitely many terms? Every finite sum of fractions is a rational number, and π is irrational. No finite sum can equal π, so an infinite sum is required.
Important Questions for Exams
1-Mark Questions
- Is zero a rational number? Yes. 0 = 0/1, with 0 and 1 integers and 1 ≠ 0.
- Find |-7/9|. 7/9.
- Write the additive inverse of -19/7. 19/7, because -19/7 + 19/7 = 0.
- Which of these is irrational: √16, 0.1212…, √7, 22/7? √7. (√16 = 4; 0.1212… = 12/99 = 4/33; 22/7 is a ratio of integers.)
- Who defined zero as a - a and in which work? Brahmagupta, in the Brāhmasphuṭasiddhānta (628 CE).
2-Mark Questions
- Without division, state whether 18/125 has a terminating decimal. If so, find the number of decimal places. 18/125 is in lowest form and 125 = 53, so it terminates. Multiply by 23/23: 144/1000 = 0.144, which has 3 decimal places.
- Express 0.4545… in the form p/q. x = 0.4545…; 100x = 45.4545…; 99x = 45; x = 45/99 = 5/11.
- Find 6 rational numbers between 3 and 4. Write 3 = 21/7 and 4 = 28/7. Then 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 all lie between them.
- Let a and b be non-zero rationals with a + 1/b = 0. Without using values, decide whether ab is positive or negative. a = -1/b, so ab = (-1/b) × b = -1. ab is negative.
3-Mark Questions
- Express 2.35777… (only 7 repeating) in the form p/q. Let x = 2.35777…. Two non-repeating decimal digits: 100x = 235.777…. One repeating digit: 1000x = 2357.777…. Subtract: 900x = 2122, so x = 2122/900 = 1061/450.
- Explain, with steps, how to locate √2 on the number line. Mark OA = 1 unit on the number line with O at 0. Draw AB ⊥ OA with AB = 1 unit. By Baudhāyana-Pythagoras, OB = √(1 + 1) = √2. With centre O and radius OB, draw an arc to cut the number line at P. OP = √2, so P represents √2.
- A rational number in lowest form has denominator 23 × 5. How many decimal places does its expansion have? Explain. The denominator is 40. Multiply numerator and denominator by 52 = 25: denominator becomes 1000 = 103. The numerator p is co-prime to 40, so p is odd and not a multiple of 5; 25p ends in 5, so the last digit is non-zero. The decimal has exactly 3 decimal places.
5-Mark Questions
- Prove that √5 is irrational. Hence explain why the same method fails for √9. Assume √5 = p/q with p, q co-prime integers, q ≠ 0. Squaring: 5q2 = p2, so 5 divides p2 and, since 5 is prime, 5 divides p. Write p = 5k: 5q2 = 25k2, so q2 = 5k2, and 5 divides q. Then 5 is a common factor of p and q, contradicting co-primality. Hence √5 is irrational. For √9 the equation 9q2 = p2 has the solution p = 3, q = 1 with p, q co-prime, so no contradiction appears: √9 = 3 is rational.
- Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that x = y = z = 0. Square the first equation: (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) = 0. Since xy + yz + zx = 0, we get x2 + y2 + z2 = 0. Each square is ≥ 0, and three non-negative numbers add to 0 only if each is 0. So x2 = y2 = z2 = 0, which gives x = y = z = 0.
Common Mistakes and Examiner Tips
- Dropping the condition q ≠ 0. Writing “a rational number is p/q” without “p, q integers and q ≠ 0” loses the definition mark. Always write all three parts.
- Calling 22/7 or 3.14 equal to π. Both are rational approximations. π is irrational. If a question asks you to classify 22/7, the answer is rational.
- Skipping “lowest form” in the irrationality proof. The contradiction depends on p and q being co-prime. Say so in step 1, or the final step has nothing to contradict.
- Copying the √2 wording into the √3 proof. For √3 or √5, write “3 divides p2, and 3 is prime, so 3 divides p”. The prime fact is what makes the step valid.
- Testing the denominator before reducing. 6/15 looks like it repeats (15 has the factor 3), but 6/15 = 2/5 = 0.4 terminates. Reduce to lowest form first, then factorise.
- Using the wrong multiplier for repeating decimals. Count repeating digits: one digit repeating uses 10, two use 100, three use 1000. For a general repeating decimal, first shift past the non-repeating digits, then shift by one full block.
- Adding fractions by adding tops and bottoms. 1/2 + 1/3 is 5/6, never 2/5. Make the denominators equal first.
- Forgetting to flip the divisor. a/b ÷ c/d = a/b × d/c. Flip the second fraction only, and remember c must be non-zero.
- Negative absolute values or distances. |x| and |a - b| are never negative. If your distance comes out as -7, you forgot the modulus.
- Treating a patterned decimal as rational. 1.01001000100001… has a pattern but no fixed repeating block, so it is irrational. Rational means a block that repeats unchanged.
Quick Revision Points
- N = {1, 2, 3, …}; Brahmagupta (628 CE) made zero a number: a - a = 0, a × 0 = 0
- Integers Z (from Zahlen): fortunes (dhana) positive, debts (ṛiṇa) negative
- Debt × fortune = debt; debt × debt = fortune; a - (-b) = a + b
- N is closed under + and ×; Z under +, -, ×; Q under +, -, ×, ÷ (divisor non-zero)
- Rational: p/q, p, q integers, q ≠ 0; every integer n = n/1 is rational
- Lowest form: p, q co-prime; a/b = c/d exactly when ad = bc
- Mark p/q: split the unit into q parts, move p parts right (positive) or left (negative)
- |x| = distance from 0, always ≥ 0; distance between a and b = |a - b|
- Density: (a + b)/2 lies between a and b; infinitely many rationals between any two
- Irrational: not expressible as p/q; decimal never ends, never repeats. Examples √2, √3, √5, π
- √2 irrational: Hippasus, proof by contradiction; p and q both come out even
- √2 on the line: OA = 1, AB = 1 perpendicular, OB = √2, arc from O; square root spiral gives √(n + 1)
- Āryabhaṭa: π ≈ 3927/1250 = 3.1416; Lambert (1761) proved π irrational; Mādhava: π = 4(1 - 1/3 + 1/5 - …)
- Real numbers R = rationals and irrationals together; they fill the number line
- Rational decimals terminate or repeat; terminate exactly when the lowest-form denominator is 2m5n
- Pure repeating: multiply by 10n and subtract. General repeating: shift by 10m, then by 10n, subtract
- 0.999… = 1; every terminating decimal has a second form ending in repeating 9s
Weightage and Exam Pattern
Class 9 has no board exam. The chapter is assessed in school exams: periodic tests, the half-yearly exam and the annual exam. In the CBSE Class IX Mathematics course structure for 2026-27, Unit I, Number System, carries 7 marks out of the 80-mark theory paper and is allotted 12 teaching periods. The remaining 20 marks are internal assessment (pen-paper tests and multiple assessment 10, portfolio 5, lab practical 5).
| Unit | Marks (out of 80) |
|---|---|
| I. Number System | 07 |
| II. Algebra | 20 |
| III. Coordinate Geometry | 04 |
| IV. Geometry | 25 |
| V. Mensuration | 14 |
| VI. Statistics and Probability | 10 |
The CBSE question paper design weights questions by typology: about 54% remembering and understanding (definitions, classifying numbers, the terminating test), 24% applying (rational arithmetic, rationals between two numbers, decimal conversion) and 22% analysing, evaluating and creating (irrationality proofs, reasoning about closure, density or decimal places).
The course structure lists these key concepts for the unit: introduction to rational numbers, their representation on the number line, density of rational numbers and its proof, finding rational numbers between two rational numbers, decimal representation of rational numbers, introduction to irrational numbers, the proof of irrationality of √2 and √3, and the square root spiral. This chapter is also the foundation for Class 10 Real Numbers, where the irrationality of √2, √3 and √5 is proved again using prime factorisation in the board year.
Class 9 Maths · Chapter 3 – swipe through all 10 cards to understand the whole chapter.
Natural numbers
Counting began with one-to-one correspondence: one pebble in a pot for each cow that left to graze.
The Ishango bone groups notches as 11, 13, 17, 19, the primes between 10 and 20.
- Lebombo Bone: about 35,000 years old, 29 notches
- Closed under addition and multiplication
- Not closed under subtraction: 4 − 9 = −5
Zero becomes a number
Brahmagupta, in the Brāhmasphuṭasiddhānta (628 CE), defined zero as a number subtracted from itself.
The Bakhśhālī Manuscript used a dot (bindu) as the symbol for zero.
- a + 0 = a
- a − 0 = a
- a × 0 = 0; division by 0 is not defined
Fortunes and debts
Brahmagupta called positive numbers fortunes (dhana) and negative numbers debts (ṛiṇa).
Z comes from the German word Zahlen, meaning numbers.
- Debt + debt = debt: (−5) + (−4) = −9
- Debt × fortune = debt: (−3) × 4 = −12
- Subtracting a debt adds a fortune: 10 − (−5) = 15
Rational numbers
Any number that can be written as p/q with p and q integers and q ≠ 0.
Lowest form means p and q are co-prime: 12/30 = 2/5.
- Every integer is rational: 5 = 5/1
- a/b ÷ c/d = ad/bc with b, c, d ≠ 0
- Closed under +, −, × and ÷ by non-zero numbers
Placing p/q and absolute value
Split the unit interval into q equal parts and move p parts from 0, right for positive, left for negative.
9/4 = 2¼, so split the gap from 2 to 3 into four parts and move one part.
- |x| is the distance of x from 0
- |x| ≥ 0 for every x; |−5/3| = 5/3
- Distance between −4 and 3 is 7
Always another rational
Between any two rational numbers there is another rational, so there are infinitely many.
For many at once, use a common denominator: 2/5 = 12/30, 3/5 = 18/30.
- Between 1 and 3/2 lies 5/4
- 13/30 to 17/30 lie between 2/5 and 3/5
- Density still leaves gaps that irrationals fill
√2 is irrational
The diagonal of a unit square has length √2, which Hippasus proved cannot be p/q.
Proof by contradiction: assume p/q in lowest form, reach a common factor 2.
- The same proof works for √3, √5, √7, √10
- Construct √2: OA = 1, AB = 1 perpendicular, arc from O
- Square root spiral: nth hypotenuse = √(n + 1)
π and Mādhava’s series
π, the ratio of circumference to diameter, is irrational, proved by Lambert in 1761.
22/7 and 3.14 are rational approximations of π.
- Āryabhaṭa (499 CE): π ≈ 3927/1250 = 3.1416
- Mādhava (14th century) founded the Kerala School
- No finite sum of fractions equals π
Terminating or repeating
Rational decimals either stop or repeat a block forever; irrational decimals do neither.
Reduce to lowest form before testing: 6/15 = 2/5 = 0.4.
- 3/20 = 15/100 = 0.15 terminates
- 5/11 = 0.4545… repeats
- √2 = 1.41421356… never ends, never repeats
Repeating decimal to p/q
Multiply to shift the repeating block, subtract, and solve for x.
142857, the block of 1/7, is a cyclic number: its multiples by 1 to 6 rotate its digits.
- 0.666… = 6/9 = 2/3
- 0.1666…: 100x − 10x = 15, x = 1/6
- 0.999… = 1, since 9x = 9
📝 Practice Number System - 10 questions
Concept-check questions · with answers & explanations
Start →Close ✕
Why D: A placeholder only marks that a column is empty; it has no arithmetic rules attached. Computing 7 × 0 = 0 treats zero as a genuine number that can be multiplied, which is exactly the leap Brahmagupta made in 628 CE with his rules a + 0 = a, a – 0 = a and a × 0 = 0.
Why not A: Telling 205 from 25 is precisely what a placeholder already does: it marks the empty tens column.
Why not B: 5 + 4 = 9 is addition of natural numbers, which needs no zero at all.
Why not C: Tally counting of 29 days, as on the Lebombo bone, needs only natural numbers and no zero at all.
Remember: A placeholder marks an empty column; a number obeys rules. Brahmagupta gave zero its rules.
Why A: Step 1 (translate to integers): loan = -850, profit = +1200, loss = -450, so standing = (-850) + 1200 + (-450). Step 2 (fortune larger than debt gives a fortune): -850 + 1200 = +350. Step 3 (debt plus debt, or fortune minus a larger amount): 350 + (-450) = -100. The trader ends Rs 100 in debt.
Why not B: +100 has the right size but the wrong sign; the final loss of 450 exceeds the 350 he had, so he owes money.
Why not C: -800 comes from swapping the signs of the loan and the profit: +850 – 1200 – 450 = -800.
Why not D: 2500 adds all three amounts as fortunes, ignoring that the loan and the loss are debts.
Remember: Write every debt with a minus sign first, then add: -850 + 1200 – 450 = -100.
Why A: Square the first condition: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx). The left side is 0² = 0 and the bracket on the right is 0, so x² + y² + z² = 0. A square of a rational number is never negative, so three squares can add to zero only if each square is zero. Hence x = y = z = 0.
Why not B: If x = y = z = 1 then x + y + z = 3, not 0, so the first condition already fails.
Why not C: If only one were zero, say z = 0, then x = -y and xy = -y², which is 0 only if y = 0 too, contradicting exactly one zero.
Why not D: Three negative numbers cannot add to zero, so the first condition rules this out.
Remember: A sum of squares is zero only when every term is zero.
Why C: Convert to a common denominator 8: -1/2 = -4/8 and 1/4 = 2/8. A number lies between them if its eighths lie between -4 and 2. -1/8 does, since -4 < -1 < 2. In decimals, -0.5 < -0.125 < 0.25.
Why not A: -3/4 = -6/8 is to the left of -4/8, so it is less than -1/2 and outside the interval.
Why not B: 1/2 = 4/8 is greater than 2/8, so it lies to the right of 1/4.
Why not D: 3/8 is greater than 2/8, so it is also beyond 1/4.
Remember: Put everything over one denominator, then compare the numerators.
Why A: Step 1 (common denominator 15): 1 – 1/3 + 1/5 = 15/15 – 5/15 + 3/15 = 13/15. Step 2 (multiply by 4): 4 × 13/15 = 52/15, which is about 3.47. Each further term of the alternating series brings the sum closer to the true value 3.14159…
Why not B: 4 is 4 × 1, the first term alone, before subtracting 1/3 or adding 1/5.
Why not C: 8/3 is 4 × (1 – 1/3) = 4 × 2/3, the sum of only the first two terms.
Why not D: 3.14 is the true value of π to two places, which the series only approaches; no finite number of terms gives it exactly, and three terms are far from it.
Remember: The series alternates: each term pulls the sum back and forth, closing in on π.
Why B: Step 1 (one non-repeating digit, so multiply by 10 to shift it before the point): let x = 0.1666…, then 10x = 1.666… Step 2 (one repeating digit, so multiply by 10 again to move one full cycle): 100x = 16.666… Step 3 (subtract the two shifted equations): 100x – 10x = 16.666… – 1.666… gives 90x = 15, so x = 15/90 = 1/6.
Why not A: 16/99 = 0.161616…, which would be right if the block 16 repeated; here only the 6 repeats.
Why not C: 16/100 = 0.16 exactly, ignoring the repeating 6s.
Why not D: 15/99 uses the correct numerator 15 but the wrong denominator; subtracting 100x – 10x leaves 90x, not 99x.
Remember: Shift past the non-repeating digits first (10 to the m), then past one cycle (10 to the n), then subtract.
Why C: The Bakhśhālī Manuscript already used a dot for zero, but a symbol is only a mark until it has rules. Brahmagupta, in the Brāhmasphuṭasiddhānta of 628 CE, defined zero as a – a and stated the laws for adding, subtracting and multiplying with it, making zero a fully operational number.
Why not A: Āryabhaṭa (499 CE) is known in this chapter for his approximation 3927/1250 for pi; the concept of zeroness passed through his work but the rules came from Brahmagupta.
Why not B: Mādhava (14th century) discovered the infinite series for pi, far later.
Why not D: Baudhāyana (about 800 BCE) wrote the Śhulbasūtra and met irrational lengths like root 2.
Remember: Brahmagupta, 628 CE: a – a = 0 and the three rules of zero.
Why C: A negative number is a debt, and multiplying by a negative number means removing. If someone takes away (the minus) four of your debts each worth Rs 3 (each is -3), you no longer owe Rs 12, so you are Rs 12 richer: (-3) × (-4) = +12. Removing debt is the same as gaining money.
Why not A: Taking on four debts of Rs 3 is (-3) × 4 = -12, a debt times a fortune, which gives a debt.
Why not B: Three fortunes of Rs 4 is 3 × 4 = 12, a product of two positives; it gives the right number but explains a different rule.
Why not D: Paying Rs 3 four times is a loss of Rs 12, that is (-3) × 4 = -12, again a debt.
Remember: Removing debts (minus times minus) makes you richer (plus).
Why B: From a + 1/b = 0 we get a = -1/b. Multiply both sides by b: ab = -1/b × b = -1. So ab equals -1 for every allowed pair, and -1 is negative. Whatever the sign of a, the sign of b is opposite (a = -1/b), so their product is negative.
Why not A: ab = -1, which is negative; a positive product would need a and b of the same sign, but a and 1/b are negatives of each other.
Why not C: ab = 0 would need a = 0 or b = 0, and both are given as non-zero; also 1/b would be undefined for b = 0.
Why not D: The sign of ab does not depend on a at all, since ab is always exactly -1.
Remember: a = -1/b means ab = -1: a number and the negative of its reciprocal multiply to -1.
Why D: No matter how close two rational numbers are, their average is another rational number between them, and repeating the process never ends. Between 3.1415 and 3.1416 lie 3.14151, 3.141515, 3.1415155 and so on without limit. Every terminating decimal is rational, so there are infinitely many rationals in the gap.
Why not A: Decimals with four places look consecutive, but adding a fifth or sixth place produces numbers between them.
Why not B: 3.14155 is one such number, but 3.14152 and 3.141551 are others; picking one does not exhaust the interval.
Why not C: Nine five-place decimals fit, but six-place, seven-place and longer decimals fit too, endlessly.
Remember: Rationals are dense: between any two there is always another, so always infinitely many.
Chapter Navigation
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Frequently Asked Questions
A rational number is any number that can be written as p/q, where p and q are integers and q ≠ 0. Every integer is rational because it can be written over 1, for example 5 = 5/1. A rational number has many equivalent forms, such as 2/5 = 12/30, and we usually write it in lowest form, where p and q are co-prime.
p/q means the number which, multiplied by q, gives p. If q were 0, that number times 0 would have to equal p. Anything multiplied by 0 is 0, so no number works when p is non-zero, and every number works when p is 0. Division by zero therefore has no single answer and is excluded from the definition.
Use proof by contradiction. Assume √2 = p/q with p and q co-prime integers. Squaring gives 2q² = p², so p² is even and p is even. Put p = 2k to get q² = 2k², so q is even too. Now p and q share the factor 2, which contradicts their being co-prime. So √2 cannot be rational.
Write the fraction in lowest form and factorise the denominator. If its only prime factors are 2 and 5, the decimal terminates, because the denominator can be turned into a power of 10. If any other prime appears, the decimal repeats forever. For example 7/20 = 0.35 terminates, while 4/15 = 0.2666… repeats because 15 has the factor 3.
Let x = 0.1666…. One digit does not repeat, so multiply by 10 to get 10x = 1.666…. One digit repeats, so multiply again by 10 to get 100x = 16.666…. Subtracting, 90x = 15, so x = 15/90 = 1/6. The rule is to shift past the non-repeating digits first, then shift by one full repeating block and subtract.
Between any two rational numbers there is always another rational number, so there are infinitely many between any two. The average (a + b)/2 is rational and lies strictly between a and b. For example, between 1 and 3/2 lies 5/4, and between 1 and 5/4 lies 9/8. Even so, the rationals leave gaps that irrationals such as √2 fill.
No. 22/7 is a ratio of two integers, so it is rational, while π is irrational, as Lambert proved in 1761. 22/7 and 3.14 are only approximations. Āryabhaṭa gave the closer value 3927/1250 = 3.1416 and called it āsanna, an approximation. Mādhava later found the exact infinite series π = 4(1 − 1/3 + 1/5 − 1/7 + …).
In the CBSE Class IX Mathematics course structure for 2026-27, Unit I Number System carries 7 marks out of 80 in the theory paper. Common questions include classifying numbers, finding rationals between two numbers, converting repeating decimals to p/q, predicting terminating decimals, and the irrationality proof of √2 or √3, which returns in Class 10 boards.
