Class 10 Maths NCERT Chapter 1 Real Numbers | Exercise 1.1 to 1.5 Solutions (2025 CBSE)

Lecture: Real Numbers — Easy Language (Class 10, Chapter 1)

This post explains the chapter in simple words so every student can follow. After the short lecture, you will find complete solutions to Exercises 1.1–1.5. Each answer is hidden — click the Show/Hide Answer button for a solution.

What are Real Numbers? (Simple)

Think of numbers you use on the number line: whole numbers, fractions, terminating decimals, repeating decimals and also irrational numbers like \(\sqrt{2}\). All these together are called Real Numbers.

Key ideas — short and easy

• Euclid's division algorithm: If \(a>b>0\), then \(a=bq+r\) with \(0\le r
• Prime factorisation: Every number >1 can be written as product of primes (unique). Use this to find HCF and LCM easily.
• Terminating vs recurring decimals: A fraction in lowest terms \(\dfrac{p}{q}\) has a terminating decimal iff denominator \(q\) has only 2 and 5 as prime factors (i.e., \(q=2^{m}5^{n}\)). Otherwise it repeats.
• Irrational numbers: Numbers that cannot be written as \(\dfrac{p}{q}\) (example \(\sqrt{2}\)). We prove them by contradiction.

Tip for weak students: Always work step-by-step. For Euclid's algorithm write one division per line. For prime factorisation divide by small primes repeatedly (2,3,5,7...). For decimal type decide by factoring denominator.

Exercise 1.1 — Euclid's Division Algorithm

Q1. Use Euclid's algorithm to find HCF of: (i) 135 and 225, (ii) 867 and 255, (iii) 38220 and 196.

(i) 225 = 135×1 + 90
    135 = 90 ×1 + 45
     90 = 45 ×2 + 0
HCF = 45

(ii) 867 = 255×3 + 102
     255 = 102×2 + 51
     102 = 51×2 + 0
HCF = 51

(iii) 38220 = 196×195 + 0
HCF = 196  (since 196 divides 38220 exactly)
          

Q2. Show that any integer of the form 6q + 5 is coprime to 6.

Numbers of the form 6q+5 leave remainder 5 when divided by 6. The divisors of 6 are 1,2,3,6. Such a number is odd, so not divisible by 2. Also 6q+5 ≡ 5 (mod 3), so not divisible by 3. Hence only common factor with 6 is 1 ⇒ coprime.

Exercise 1.2 — Irrational Numbers (Proofs)

Q1. Prove that √2 is irrational.

Assume √2 = p/q in lowest terms (gcd(p,q)=1). Then 2q² = p², so p² is even ⇒ p is even. Put p = 2k. Then 2q² = 4k² ⇒ q² = 2k² ⇒ q is even. Both even contradict gcd(p,q)=1. So √2 is irrational.

Q2. Prove √3, √5 are irrational (short).

Use same method as √2. If √3 = p/q (lowest terms) ⇒ 3q² = p² ⇒ p divisible by 3 ⇒ q divisible by 3 ⇒ contradiction. Similarly for 5.

Q3. Prove √2 + √3 is irrational.

Suppose r = √2 + √3 is rational. Then √3 = r - √2. Square both sides: 3 = r² + 2 - 2r√2 ⇒ 2r√2 = r² - 1 ⇒ √2 = (r² - 1)/(2r) which is rational. But √2 is irrational ⇒ contradiction. So √2 + √3 is irrational.

Q4. Show 3 + 2√5 is irrational.

Assume 3 + 2√5 is rational. Then 2√5 = (rational) - 3 is rational ⇒ √5 rational. But √5 is irrational ⇒ contradiction. Hence 3 + 2√5 is irrational.

Exercise 1.3 — Fundamental Theorem of Arithmetic

Q1. Prime factorise: (a) 140, (b) 156, (c) 38220

(a) 140 = 2×70 = 2×2×35 = 2^2 × 5 × 7
(b) 156 = 2×78 = 2×2×39 = 2^2 × 3 × 13
(c) 38220 = divide by 2: 19110
           = 2×19110 = 2^2×9555
           9555 ÷ 3 = 3185  ⇒ 3×3185
           3185 ÷ 5 = 637   ⇒ 5×637
           637 ÷ 7 = 91     ⇒ 7×91
           91 = 7×13
So 38220 = 2^2 × 3 × 5 × 7^2 × 13
          

Q2. Find HCF and LCM of 900 and 210 using prime powers.

900 = 2^2 × 3^2 × 5^2
210 = 2 × 3 × 5 × 7
HCF = 2^1 × 3^1 × 5^1 = 30
LCM = 2^2 × 3^2 × 5^2 × 7 = 6300
Check: 900×210 = HCF×LCM = 30×6300 = 189000
          

Q3. Prove: If p is prime and p | a² then p | a.

Given p divides a² = a×a. By Euclid's lemma (prime divides product) p must divide a or a. So p divides a.

Exercise 1.4 — Decimal Expansions (Terminating or Recurring)

Q1. Without dividing say whether decimal terminates or recurs: (a) 13/3125, (b) 17/8, (c) 7/15, (d) 21/(2^4·5^3), (e) 11/210

(a) 3125 = 5^5 ⇒ only 5s ⇒ terminating.
(b) 8 = 2^3 ⇒ terminating.
(c) 15 = 3×5 ⇒ contains 3 ⇒ recurring.
(d) Denominator is 2^4×5^3 ⇒ terminating.
(e) 210 = 2×3×5×7 contains primes other than 2,5 ⇒ recurring.

Q2. Convert 13/3125 to decimal (show method).

3125 = 5^5. To make denominator 10^5 multiply numerator & denominator by 2^5 = 32.
13/3125 = (13×32)/(3125×32) = 416/100000 = 0.00416
          

Q3. Why a rational number has either terminating or recurring decimal?

When dividing p by q, remainders are < q. If denominator has only 2 and 5, after finite steps you get remainder 0 ⇒ terminate. Otherwise remainders must repeat (pigeonhole) ⇒ repeating decimal.

Exercise 1.5 — Applications and Theorems

Q1. Prove there are infinitely many primes.

Assume finite primes p1,...,pn. Consider N = p1 p2 ... pn + 1. N leaves remainder 1 when divided by any pi, so none of pi divides N. But N>1 so it has a prime divisor not in the list—contradiction. So primes are infinite.

Q2. Prove: For integers a,b, ab = HCF(a,b) × LCM(a,b).

Write prime factorisation: a = ∏ p_i^{α_i}, b = ∏ p_i^{β_i}. HCF = ∏ p_i^{min(α_i,β_i)}, LCM = ∏ p_i^{max(α_i,β_i)}. Multiply them: exponent becomes α_i+β_i = exponents of a×b. So product equals a×b.

Q3. Quick practice: HCF(306,657) using Euclid's algorithm.

963? (we use 657 and 963 example from earlier). For 306 and 657:
657 = 306×2 + 45
306 = 45 ×6 + 36
45 = 36 ×1 + 9
36 = 9 ×4 + 0
HCF = 9
          

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