Lesson 1: Real Numbers and the Decimal System

India gave the world two gifts that quietly run all of mathematics: the decimal place-value system and zero. Brahmagupta wrote down the rules for zero in 628 CE, and the numerals you use today travelled from India through Arab scholars to the rest of the world. This lesson is the NCERT Class X opening chapter, and it builds the number system carefully: what the real numbers are, how Euclid's division lemma finds highest common factors, why some fractions terminate as decimals and others repeat forever, and how we *prove* that numbers like √2 and √5 can never be written as fractions.

The family of real numbers

Every real number is either rational or irrational.

• A rational number can be written as p/q where p and q are integers and q ≠ 0. Examples: 7/3, −4, 0.375, 2.161616…
• An irrational number cannot be written in that form. Its decimal expansion goes on forever without repeating. Examples: √2, √5, π, 0.10110111011110…

Together they fill the number line completely — there are no gaps. Two facts from Class IX that we now prove properly:

• The decimal expansion of a rational number either terminates or repeats (recurs).
• A prime number p has an irrational square root: √p is irrational.

Euclid's division lemma and HCF

Euclid's division lemma: for positive integers a and b, there exist unique whole numbers q and r with a = bq + r, 0 ≤ r < b.

Repeatedly applying the lemma (always dividing the previous divisor by the remainder) gives Euclid's division algorithm, which ends at remainder 0; the last divisor is the HCF.

Worked example 1 (basic). Find HCF(135, 225).

225 = 135 × 1 + 90
135 = 90 × 1 + 45
90  = 45 × 2 + 0

The last divisor is 45, so HCF(135, 225) = 45.

Worked example 2 (intermediate). Find HCF(96, 404), then use it to find LCM(96, 404).

404 = 96 × 4 + 20
96  = 20 × 4 + 16
20  = 16 × 1 + 4
16  = 4 × 4 + 0        so HCF = 4

For two positive integers, HCF × LCM = product of the numbers, so LCM = (96 × 404)/4 = 96 × 101 = 9696.

Worked example 3 (NCERT exemplar style). Show that every positive even integer is of the form 2m and every positive odd integer is of the form 2m + 1, and hence show that the square of an odd integer is odd. Apply the lemma with b = 2: a = 2q + r with r = 0 or 1. If r = 0, a = 2q (even); if r = 1, a = 2q + 1 (odd). Now (2m + 1)² = 4m² + 4m + 1 = 2(2m² + 2m) + 1, which is again one more than a multiple of 2, so it is odd. This tiny argument is the template for many divisibility proofs: use the lemma to list the possible forms, then test each one.

Terminating and non-terminating decimals

Write the fraction in lowest terms p/q. Factorise q.

• If q is of the form 2ⁿ × 5ᵐ (only 2s and 5s), the decimal terminates.
• If q has any other prime factor (3, 7, 11, …), the decimal is non-terminating and repeating.

Why? A decimal terminates exactly when the fraction can be rewritten with denominator 10ᵏ, and 10ᵏ = 2ᵏ × 5ᵏ contains no other primes.

So 13/64 terminates (64 = 2⁶ — the expansion is 0.203125, ending after 6 places), while 77/210 = 11/30 has denominator 30 = 2 × 3 × 5, and the factor 3 forces it to repeat: 0.3666…

Conversely, any terminating or repeating decimal is rational. For example, 0.375 = 375/1000 = 3/8, and a repeating decimal such as 0.3636… equals 36/99 = 4/11.

Proving irrationality: the contradiction method

Claim: √5 is irrational. Suppose, for contradiction, that √5 = a/b with a, b coprime integers, b ≠ 0. Then a² = 5b², so 5 divides a², which forces 5 to divide a (a standard lemma: if a prime p divides a², then p divides a). Write a = 5c. Then 25c² = 5b², so b² = 5c², and the same argument shows 5 divides b. But then 5 divides both a and b — contradicting the choice of a/b in lowest terms. The assumption collapses, so √5 is irrational. Exactly the same proof works for √2, √3, √7, √11 — any prime under the root.

It follows that combinations like 5 − √3 and 2 + 3√5 are also irrational: if 5 − √3 were a rational r, then √3 = 5 − r would be rational, which it is not.

Competency application: ropes for a school function

A school is decorating a hall for Independence Day. Three ribbons of lengths 36 m, 48 m and 60 m must be cut into pieces of equal maximum length with nothing left over, so that all pieces can be used interchangeably. The piece length must divide all three lengths, and be the largest such: HCF(36, 48, 60) = 12 m. The number of pieces is 36/12 + 48/12 + 60/12 = 3 + 4 + 5 = 12 pieces. Notice the real-world logic: "equal pieces, nothing wasted, as large as possible" always translates into an HCF; "events happening together again" always translates into an LCM.

Common mistakes

• Stopping Euclid's algorithm too early. You must continue until the remainder is exactly 0; the HCF is the *last divisor used*, not the last remainder you happened to compute.
• Claiming a decimal terminates without reducing the fraction first. 6/15 looks as if it has a factor 3, but 6/15 = 2/5, and 2/5 = 0.4 terminates. Always cancel before factorising the denominator.
• Writing HCF × LCM = product for three numbers. That identity holds for *two* numbers only.
• "Proving" √2 irrational from its decimal. Saying "1.414213… never ends, so it is irrational" is not a proof — the contradiction argument using coprime a and b is.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. The HCF of 96 and 404 is: (a) 2 (b) 4 (c) 8 (d) 20. Answer: (b) 4. 404 = 96 × 4 + 20; 96 = 20 × 4 + 16; 20 = 16 × 1 + 4; 16 = 4 × 4 + 0.

Q2. The decimal expansion of 13/64 terminates after how many decimal places? (a) 4 (b) 5 (c) 6 (d) It never terminates. Answer: (c) 6. 64 = 2⁶; a denominator of 2ⁿ terminates after at most n places, and 13/64 = 0.203125 indeed uses all 6.

Section B — Short answer (3 marks) Q3. Prove that √3 is irrational. Marking scheme (CBSE style): 1 mark for assuming √3 = a/b in lowest terms and squaring to get a² = 3b²; 1 mark for concluding 3 | a, substituting a = 3c and deducing 3 | b; 1 mark for stating the contradiction with "a, b coprime" and concluding √3 is irrational.

Section C — Case-based long answer (5 marks) Q4. Three water tanks have capacities 36 L, 48 L and 60 L. A mechanic wants a single measuring can that can fill each tank exactly (a whole number of cans each time), and wants the largest such can. (i) Find the capacity of the can. (2 marks) (ii) In how many trips does he fill the 48 L tank? (1 mark) (iii) If instead he wants one can that measures all *three together* in the fewest trips, explain whether HCF or LCM reasoning changes. (2 marks) Marking scheme: (i) Required capacity = HCF(36, 48, 60) (1 mark); Euclid/prime-factor steps giving 12 L (1 mark). (ii) 48/12 = 4 trips (1 mark). (iii) The can must divide each capacity, so it is still a common divisor — HCF reasoning; LCM would apply if the can's multiples had to match the capacities, which is not the situation described (2 marks for correct reasoning — award 1 mark for identifying "HCF" and 1 for the justification).

Going deeper: HCF and LCM together

When a problem gives you the HCF and LCM, the product identity becomes a solving tool. Example: two positive integers have HCF 6, LCM 72, and sum 42. Find them. Let the numbers be 6a and 6b with a, b coprime. Then LCM = 6ab = 72 gives ab = 12, and the sum gives 6(a + b) = 42, so a + b = 7. Coprime factor pairs of 12: (1, 12) sums to 13 ✗; (3, 4) sums to 7 ✓. So a = 3, b = 4 and the numbers are 18 and 24. (Check: HCF(18, 24) = 6 ✓, LCM = 72 ✓.) The method — factor out the HCF first — converts an LCM condition into a coprime condition, which is far easier to search.

Going deeper: repeating decimals are rational — constructively

The claim "every repeating decimal is a fraction" can be *shown*, not just asserted. Let x = 0.363636… Then 100x = 36.363636… Subtracting: 99x = 36, so x = 36/99 = 4/11. The multiplier is 10ᵏ where k is the length of the repeating block. Mixed cases work the same way with one shift: for y = 0.41666…, first 10y = 4.1666… and 100y = 41.666…, subtract: 90y = 37.5, y = 37.5/90 = 375/900 = 5/12. (Check: 5/12 = 0.41666… ✓.)

Going deeper: irrational arithmetic — handle with care

Sums and products of irrationals are *usually* irrational, but not always, and board one-markers exploit the exceptions:

• √2 × √8 = √16 = 4, rational! (Two irrationals, rational product.)
• (3 + √2) + (3 − √2) = 6, rational. Conjugate pairs cancel the irrational part.
• √2 + √8 = √2 + 2√2 = 3√2, still irrational.

The safe statements: rational ± irrational is always irrational; a *nonzero* rational × irrational is always irrational. Beyond those, compute — do not guess.

Final drills

Drill 1 — prime factorisation route. Find HCF and LCM of 72 and 120 by prime factorisation. 72 = 2³ × 3² and 120 = 2³ × 3 × 5. HCF takes the *minimum* power of each common prime: 2³ × 3 = 24. LCM takes the *maximum* power of every prime present: 2³ × 3² × 5 = 360. Check: 24 × 360 = 8640 = 72 × 120 ✓. Euclid's algorithm and factorisation must always agree — use one to check the other.

Drill 2 — decimal classification. 7/80: 80 = 2⁴ × 5, only 2s and 5s, so it terminates — indeed 7/80 = 0.0875, ending after max(4, 1) = 4 places. And the fraction form of 0.235: 235/1000 = 47/200 (and 200 = 2³ × 5² confirms it must terminate — it does, at 3 places).

Drill 3 — the fundamental theorem. Every composite number factorises into primes *uniquely* (apart from order). This is why Drill 1 is well-defined at all: 360 has only one prime signature, 2³ × 3² × 5, so "the" HCF and "the" LCM of two numbers are single values, not matters of taste.