CBSE Class 10 Mathematics Real Numbers Notes Set 10

Kinds of Numbers

Counting numbers 1, 2, 3, … are the natural numbers; adding 0 gives the whole numbers; adding the negatives of the natural numbers gives the integers. A rational number is any number that can be written as \( \frac{p}{q} \) with p, q integers and q ≠ 0. A number that cannot be written this way is irrational. Together they make up the real numbers, and every real number sits on the number line.
A prime number has exactly two factors, 1 and itself (2, 3, 5, 7, 11, …); 1 is not prime. A natural number greater than 1 that is not prime is composite. The modulus of a real number a is |a| = a when a ≥ 0 and |a| = −a when a < 0.

Properties of Irrational Numbers

1. The sum or difference of a rational and an irrational number is irrational.
2. The product or quotient of a non-zero rational and an irrational number is irrational.
3. The sum, difference, product or quotient of two irrational numbers may be rational or irrational: \( \sqrt{2}\times\sqrt{2} = 2 \) is rational but \( \sqrt{2}\times\sqrt{3} = \sqrt{6} \) is not.
4. If a and b are positive rationals and ab is not a perfect square, then \( \sqrt{ab} \) is an irrational number lying between a and b, and \( \frac{a+b}{2} \) is a rational number lying between them.

Proving a Number is Irrational

Start from the key fact: if a prime p divides \( a^2 \), then p divides a. This follows from the uniqueness of prime factorisation, because the prime factors of \( a^2 \) are exactly those of a.
To show \( \sqrt{2} \) is irrational, assume \( \sqrt{2} = \frac{a}{b} \) with a and b coprime. Then \( 2b^2 = a^2 \), so 2 divides a, say a = 2c. Then \( b^2 = 2c^2 \), so 2 divides b as well. Now a and b share the factor 2, contradicting coprimality. Hence \( \sqrt{2} \) is irrational. The same steps work for \( \sqrt{3} \), \( \sqrt{5} \) and the root of any prime.
To show an expression such as \( 7-\sqrt{3} \), \( \frac{\sqrt{5}}{3} \) or \( 2\sqrt{7} \) is irrational, assume it is rational and rearrange. Each time you isolate the root of a prime on one side with a rational expression on the other, which contradicts the irrationality of that root.

Decimal Expansions of Real Numbers

When a fraction \( \frac{p}{q} \) is divided out, either the remainder becomes zero (a terminating decimal) or it never does. If it never does, some remainder must repeat, so the digits repeat in a block (a non-terminating repeating decimal). Both cases are rational numbers. A decimal that neither terminates nor repeats is irrational.
Terminating test: write \( \frac{p}{q} \) in lowest terms. If \( q = 2^n 5^m \) for whole numbers n and m, the decimal terminates, and it has max(n, m) decimal places. If q has any other prime factor, the decimal is non-terminating and repeating. Example: \( \frac{32}{125} = \frac{2^5}{5^3} = \frac{2^8}{10^3} = 0.256 \).
Converting a repeating decimal to a fraction: let x be the number, multiply by a suitable power of 10 so that the repeating parts line up, and subtract. For \( 0.\overline{3} \): 10x − x = 3 so \( x = \frac{1}{3} \).

Euclid's Division Lemma and Algorithm

Euclid's division lemma: for positive integers a and b there are unique whole numbers q and r with a = bq + r and 0 ≤ r < b.
Euclid's algorithm for HCF(a, b) with a > b: divide a by b to get remainder r. If r = 0, then b is the HCF. Otherwise replace (a, b) by (b, r) and repeat. The last non-zero remainder is the HCF, because HCF(a, b) = HCF(b, r) at every step.
Applications: taking b = 2 shows every integer is 2q or 2q + 1; taking b = 4 shows an odd integer is 4q + 1 or 4q + 3; taking b = 3 shows the square of an integer is 3m or 3m + 1, and its cube is 9m, 9m + 1 or 9m + 8.

Fundamental Theorem of Arithmetic

Every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors. For example \( 12600 = 2^3\times 3^2\times 5^2\times 7 \).
A factor tree (repeated division by primes) finds the prime factors. If a number ends with the digit 0, it is divisible by both 2 and 5, so \( 12^n \) can never end in 0 because \( 12^n = 2^{2n}\times 3^n \) has no factor 5. A number of the form 7 × 11 × 13 + 13 is composite because it equals 13 × (7 × 11 + 1).

HCF and LCM by Prime Factorisation

HCF is the product of the common prime factors, each taken with its smallest power. LCM is the product of all prime factors, each taken with its greatest power.
For any two positive integers a and b: HCF(a, b) × LCM(a, b) = a × b. Hence LCM = ab ÷ HCF, HCF = ab ÷ LCM, and one number = (HCF × LCM) ÷ the other. This rule holds for two numbers only, not three.
Word problems: the greatest length or number of groups that divides several quantities exactly is an HCF problem; the time at which periodic events coincide again, or the smallest length measurable by several rods, is an LCM problem.

Worked Examples

Illustration 1. Find the HCF of 196 and 38220 using Euclid's algorithm.
Answer:
38220 = 196 × 195 + 0.
The remainder is 0 at the first step, so the HCF is 196.

Illustration 2. Find the LCM and HCF of 26 and 91 by prime factorisation.
Answer:
\( 26 = 2\times 13 \) and \( 91 = 7\times 13 \).
LCM = 2 × 7 × 13 = 182; HCF = 13.
Check: 182 × 13 = 2366 = 26 × 91.

Illustration 3. Find the HCF and LCM of 17 and 25.
Answer:
Both are co-prime: \( 17 = 17 \) and \( 25 = 5^2 \).
HCF = 1; LCM = 17 × 25 = 425.

Illustration 4. Given LCM(150, 100) = 300, find HCF(150, 100).
Answer:
HCF = (150 × 100) ÷ 300 = 50.

Illustration 5. Prove that \( 7-\sqrt{3} \) is irrational.
Answer:
Suppose \( 7-\sqrt{3} = \frac{p}{q} \) with integers p, q and q ≠ 0.
Then \( \sqrt{3} = \frac{7q-p}{q} \), a rational number.
This contradicts the irrationality of \( \sqrt{3} \), so \( 7-\sqrt{3} \) is irrational.