Step-by-step NCERT solutions for Real Numbers (Chapter 1, CBSE Class 10 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Real Numbers textbook chapter.
All 10 questions in Real Numbers are solved in the PDF. Here's what's inside, exercise by exercise:
3 questions from Real Numbers, solved step by step.
Fundamental Theorem of Arithmetic — HCF & LCM
Using the relation: LCM × HCF = Product of the two numbers
LCM × 9 = 306 × 657
306 × 657 = 306 × 600 + 306 × 57 = 183600 + 17442 = 201042
LCM = 201042 ÷ 9 = 22338
Answer: LCM (306, 657) = 22338
Irrational Numbers — proofs
Assume, for contradiction, that 3 + 2√5 is rational.
Then 3 + 2√5 = r for some rational number r.
Rearranging:
2√5 = r - 3
√5 = (r - 3) / 2
Now, r is rational (by assumption) and 3 is rational, so r - 3 is rational.
Also, 2 is a non-zero rational, so (r - 3)/2 is rational.
This would make √5 rational.
But we proved in Q1 that √5 is irrational. This is a contradiction.
Therefore, our assumption is false, and 3 + 2√5 is irrational.
Answer: 3 + 2√5 is irrational (proved by contradiction, reducing to irrationality of √5).
Fundamental Theorem of Arithmetic — HCF & LCM
A number ends with digit 0 if and only if 10 is a factor of it, i.e., it has both 2 and 5 as prime factors.
Prime factorisation of 6:
6 = 2 × 3
⇒ 6ⁿ = 2ⁿ × 3ⁿ
The prime factorisation of 6ⁿ contains only the primes 2 and 3 — it does not contain 5.
By the Fundamental Theorem of Arithmetic, the prime factorisation of every positive integer is unique.
Since 5 never appears as a prime factor of 6ⁿ, 5 can never divide 6ⁿ.
Therefore 6ⁿ cannot end with the digit 0 for any natural number n.
Answer: 6ⁿ cannot end with the digit 0 for any natural number n, because 5 is not a prime factor of 6ⁿ = 2ⁿ × 3ⁿ.
All 10 questions in Real Numbers are solved step by step in the Solutions PDF above.
Tested in 83 questions across 23 Class 10 board papers (2020–2026). Open a set to see exactly how this chapter is asked:
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